power electronics, mohan 2nd ed solutions manual
TRANSCRIPT
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Chapter 19 Problem Solutions
19-1. Intrinsic temperature is reached when the intrinsic carrier density ni equals the lowest doping density in the pn juinction structure (the n-side in this problem). Thus
ni(Ti) = Nd = 1014 = 1010 exp ÎÍÈ
˚˙˘
-!q!Eg2k !
ÓÌÏ
˛˝¸1
Ti!-!
1300
Solving for Ti using Eg = 1.1 eV, k = 1.4x10-23 [1/°K] yields
Ti = 262 °C or 535 °K.
19-2. N-side resistivity rn = 1
q!mn!Nd =
1(1.6x10-19)(1500)(1014)
= 43.5 ohm-cm
P-side resistivity rp = 1
q!mp!Na =
1(1.6x10-19)(500)(1018)
= 0.013 ohm-cm
19-3. Material is n-type with Nd = 1013 cm-3 >> ni = 1010 cm-3. Hence use approximateformulas given in Chapter 19.
n = Nd = 1013 cm-3 ; p = n2i
Nd =
1020
1013 = 107 cm-3
19-4. po = n2i [300]Nd
; 2po = n2i [300!+!T]
Nd
2 n2i [300] = n
2i [300 + T] ; 2x1010 = 1010 exp Î
ÍÈ
˚˙˘
-!q!Eg2k !
ÓÌÏ
˛˝¸1
T!-!1
300
Solving for T yields T = q!Eg!300
(q!Eg!-!k!300!ln(2)) = 305.2 °K
DT = 305.2 - 300 = 5.2 °K.
19-5. I1 = Is exp(q!V1k!T ; 10 I1 = Is exp(
q!V1!+!dVk!T ) ; dV =
k!Tq ln(10) = 60 mV
19-6. (a) xn(0) = depletion layer width on n-side at zero bias; xp(0) = depletion layer width on p-side at zero bias.
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Chapter 19 Problem Solutions
19-1. Intrinsic temperature is reached when the intrinsic carrier density ni equals the lowest doping density in the pn juinction structure (the n-side in this problem). Thus
ni(Ti) = Nd = 1014 = 1010 exp ÎÍÈ
˚˙˘
-!q!Eg2k !
ÓÌÏ
˛˝¸1
Ti!-!
1300
Solving for Ti using Eg = 1.1 eV, k = 1.4x10-23 [1/°K] yields
Ti = 262 °C or 535 °K.
19-2. N-side resistivity rn = 1
q!mn!Nd =
1(1.6x10-19)(1500)(1014)
= 43.5 ohm-cm
P-side resistivity rp = 1
q!mp!Na =
1(1.6x10-19)(500)(1018)
= 0.013 ohm-cm
19-3. Material is n-type with Nd = 1013 cm-3 >> ni = 1010 cm-3. Hence use approximateformulas given in Chapter 19.
n = Nd = 1013 cm-3 ; p = n2i
Nd =
1020
1013 = 107 cm-3
19-4. po = n2i [300]Nd
; 2po = n2i [300!+!T]
Nd
2 n2i [300] = n
2i [300 + T] ; 2x1010 = 1010 exp Î
ÍÈ
˚˙˘
-!q!Eg2k !
ÓÌÏ
˛˝¸1
T!-!1
300
Solving for T yields T = q!Eg!300
(q!Eg!-!k!300!ln(2)) = 305.2 °K
DT = 305.2 - 300 = 5.2 °K.
19-5. I1 = Is exp(q!V1k!T ; 10 I1 = Is exp(
q!V1!+!dVk!T ) ; dV =
k!Tq ln(10) = 60 mV
19-6. (a) xn(0) = depletion layer width on n-side at zero bias; xp(0) = depletion layer width on p-side at zero bias.
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Chapter 19 Problem Solutions
19-1. Intrinsic temperature is reached when the intrinsic carrier density ni equals the lowest doping density in the pn juinction structure (the n-side in this problem). Thus
ni(Ti) = Nd = 1014 = 1010 exp ÎÍÈ
˚˙˘
-!q!Eg2k !
ÓÌÏ
˛˝¸1
Ti!-!
1300
Solving for Ti using Eg = 1.1 eV, k = 1.4x10-23 [1/°K] yields
Ti = 262 °C or 535 °K.
19-2. N-side resistivity rn = 1
q!mn!Nd =
1(1.6x10-19)(1500)(1014)
= 43.5 ohm-cm
P-side resistivity rp = 1
q!mp!Na =
1(1.6x10-19)(500)(1018)
= 0.013 ohm-cm
19-3. Material is n-type with Nd = 1013 cm-3 >> ni = 1010 cm-3. Hence use approximateformulas given in Chapter 19.
n = Nd = 1013 cm-3 ; p = n2i
Nd =
1020
1013 = 107 cm-3
19-4. po = n2i [300]Nd
; 2po = n2i [300!+!T]
Nd
2 n2i [300] = n
2i [300 + T] ; 2x1010 = 1010 exp Î
ÍÈ
˚˙˘
-!q!Eg2k !
ÓÌÏ
˛˝¸1
T!-!1
300
Solving for T yields T = q!Eg!300
(q!Eg!-!k!300!ln(2)) = 305.2 °K
DT = 305.2 - 300 = 5.2 °K.
19-5. I1 = Is exp(q!V1k!T ; 10 I1 = Is exp(
q!V1!+!dVk!T ) ; dV =
k!Tq ln(10) = 60 mV
19-6. (a) xn(0) = depletion layer width on n-side at zero bias; xp(0) = depletion layer width on p-side at zero bias.
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Chapter 19 Problem Solutions
19-1. Intrinsic temperature is reached when the intrinsic carrier density ni equals the lowest doping density in the pn juinction structure (the n-side in this problem). Thus
ni(Ti) = Nd = 1014 = 1010 exp ÎÍÈ
˚˙˘
-!q!Eg2k !
ÓÌÏ
˛˝¸1
Ti!-!
1300
Solving for Ti using Eg = 1.1 eV, k = 1.4x10-23 [1/°K] yields
Ti = 262 °C or 535 °K.
19-2. N-side resistivity rn = 1
q!mn!Nd =
1(1.6x10-19)(1500)(1014)
= 43.5 ohm-cm
P-side resistivity rp = 1
q!mp!Na =
1(1.6x10-19)(500)(1018)
= 0.013 ohm-cm
19-3. Material is n-type with Nd = 1013 cm-3 >> ni = 1010 cm-3. Hence use approximateformulas given in Chapter 19.
n = Nd = 1013 cm-3 ; p = n2i
Nd =
1020
1013 = 107 cm-3
19-4. po = n2i [300]Nd
; 2po = n2i [300!+!T]
Nd
2 n2i [300] = n
2i [300 + T] ; 2x1010 = 1010 exp Î
ÍÈ
˚˙˘
-!q!Eg2k !
ÓÌÏ
˛˝¸1
T!-!1
300
Solving for T yields T = q!Eg!300
(q!Eg!-!k!300!ln(2)) = 305.2 °K
DT = 305.2 - 300 = 5.2 °K.
19-5. I1 = Is exp(q!V1k!T ; 10 I1 = Is exp(
q!V1!+!dVk!T ) ; dV =
k!Tq ln(10) = 60 mV
19-6. (a) xn(0) = depletion layer width on n-side at zero bias; xp(0) = depletion layer width on p-side at zero bias.
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Chapter 19 Problem Solutions
19-1. Intrinsic temperature is reached when the intrinsic carrier density ni equals the lowest doping density in the pn juinction structure (the n-side in this problem). Thus
ni(Ti) = Nd = 1014 = 1010 exp ÎÍÈ
˚˙˘
-!q!Eg2k !
ÓÌÏ
˛˝¸1
Ti!-!
1300
Solving for Ti using Eg = 1.1 eV, k = 1.4x10-23 [1/°K] yields
Ti = 262 °C or 535 °K.
19-2. N-side resistivity rn = 1
q!mn!Nd =
1(1.6x10-19)(1500)(1014)
= 43.5 ohm-cm
P-side resistivity rp = 1
q!mp!Na =
1(1.6x10-19)(500)(1018)
= 0.013 ohm-cm
19-3. Material is n-type with Nd = 1013 cm-3 >> ni = 1010 cm-3. Hence use approximateformulas given in Chapter 19.
n = Nd = 1013 cm-3 ; p = n2i
Nd =
1020
1013 = 107 cm-3
19-4. po = n2i [300]Nd
; 2po = n2i [300!+!T]
Nd
2 n2i [300] = n
2i [300 + T] ; 2x1010 = 1010 exp Î
ÍÈ
˚˙˘
-!q!Eg2k !
ÓÌÏ
˛˝¸1
T!-!1
300
Solving for T yields T = q!Eg!300
(q!Eg!-!k!300!ln(2)) = 305.2 °K
DT = 305.2 - 300 = 5.2 °K.
19-5. I1 = Is exp(q!V1k!T ; 10 I1 = Is exp(
q!V1!+!dVk!T ) ; dV =
k!Tq ln(10) = 60 mV
19-6. (a) xn(0) = depletion layer width on n-side at zero bias; xp(0) = depletion layer width on p-side at zero bias.
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Chapter 19 Problem Solutions
19-1. Intrinsic temperature is reached when the intrinsic carrier density ni equals the lowest doping density in the pn juinction structure (the n-side in this problem). Thus
ni(Ti) = Nd = 1014 = 1010 exp ÎÍÈ
˚˙˘
-!q!Eg2k !
ÓÌÏ
˛˝¸1
Ti!-!
1300
Solving for Ti using Eg = 1.1 eV, k = 1.4x10-23 [1/°K] yields
Ti = 262 °C or 535 °K.
19-2. N-side resistivity rn = 1
q!mn!Nd =
1(1.6x10-19)(1500)(1014)
= 43.5 ohm-cm
P-side resistivity rp = 1
q!mp!Na =
1(1.6x10-19)(500)(1018)
= 0.013 ohm-cm
19-3. Material is n-type with Nd = 1013 cm-3 >> ni = 1010 cm-3. Hence use approximateformulas given in Chapter 19.
n = Nd = 1013 cm-3 ; p = n2i
Nd =
1020
1013 = 107 cm-3
19-4. po = n2i [300]Nd
; 2po = n2i [300!+!T]
Nd
2 n2i [300] = n
2i [300 + T] ; 2x1010 = 1010 exp Î
ÍÈ
˚˙˘
-!q!Eg2k !
ÓÌÏ
˛˝¸1
T!-!1
300
Solving for T yields T = q!Eg!300
(q!Eg!-!k!300!ln(2)) = 305.2 °K
DT = 305.2 - 300 = 5.2 °K.
19-5. I1 = Is exp(q!V1k!T ; 10 I1 = Is exp(
q!V1!+!dVk!T ) ; dV =
k!Tq ln(10) = 60 mV
19-6. (a) xn(0) = depletion layer width on n-side at zero bias; xp(0) = depletion layer width on p-side at zero bias.
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xn(0) + xp(0) = Wo = !2!e!fc!(Na!+!Nd)
q!Na!Nd (1)
fc = k!Tq ln
ÎÍÍÈ
˚˙˙˘
!Na!Nd
n2i
= 0.026 ln ÎÍÈ
˚˙˘
!1014!1015
1020 = 0.54 eV
Conservation of charge: q Na xp = q Nd xn (2)
Solving (1) and (2) simultaneously gives using the numerical values given in the problemstatement gives:
Wo = 2.8 microns ; xn(0) = 2.55 microns ; xp(0) = 0.25 microns
(b) Electric field profile triangular-shaped as shown in Fig. 19-9b. Maximum electric at zero bias given by
Emax = 2!fcWo
= (2)!(0.54)(2.8x10-4)
= 3,900 V/cm
(c) From part a) fc = 0.54 eV
(d)C(V)
A = e
Wo 1!+!Vfc
; C(V) = space-charge capacitance at reverse voltage V.
C(0)A =
(11.7)(8.9x10-14)2.8x10-4 = 3.7x10-9 F/cm2
C(50)A =
(11.7)(8.9x10-14)
2.8x10-4 1!+!50
0.54 = 3.8x10-10 F/cm2
(e) I = Is exp(qVkT ) ; exp(
qVkT ) = exp (
0.70.026 ) = 5x1011
Is = q n2i ÎÍ
ÍÈ
˚˙˙˘
!Dnt
Na!t !+!Dpt
Nd!t A
= (1.6x10-19)(1020) ÎÍÍÈ
˚˙˙˘
!(38)(10-6)
(1015)(10-6)!+!
(13)(10-6)(1014)(10-6)
(2) 2 Is = 6.7x10-14 A ;
I = (6.7x10-14 )(5x1011) = 34 mA
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19-7. Resistance R = r!LA ;
LA =
0.020.01 = 2 cm-1
At 25 °C, Nd = 1014 >> ni so r = 1
q!mn!Nd =
1(1.6x10-19)(1500)(1014!)
= 41.7 W-cm
R(25 °C) = (41.7)(2) = 83.4 ohms
At 250 °C (523 °K), ni[523] = 1010 exp ÎÍÈ
˚˙˘
-!(1.6x10-19)!(1.1)(2)(1.4x10-23)
!ÓÌÏ
˛˝¸1
523!-!1
300 =
(1010)(7.6x103) = 7.6x1013 which is an appreciable fraction of Nd = 1014. Thus we should solve Eqs. (19-2) and (19-3) exactly for no and po rather than using equations similiar to Eq. (19-4). Solving Eqs. (19-2) and (19-3) for Nd >> Na yields
no = Nd2 Î
ÍÈ
˚˙˘
1!+! 1!+!4!n
2i
N2d
and po = n2i
no . Putting in numerical values yields
no = 1014
2 ÎÍÍÈ
˚˙˙˘
1!+! 1!+!(4)(7.6x1013)2!
(1014)2 = 1.4x1014 and
po = 5.8x1027
1014 = 5.8x1013
Assuming temperature-independent mobilities (not a valid assumption but no other information is given in text or the problem statement), resistance is
R(250 °C) = r(250!° C)!!L
A ; r(250 °C) ≈ 1
q!mn!no!+!q!mp!po
= 1
(1.6x10-19)(1500)(1.4x1014!)!+!(1.6x10-19)(500)(5.8x1013!) = 26.2 W-cm ;
R(250 °C) ≈ (26.2)(2) = 52.4 ohms
19-8. BVBD = e!(Na!+!Nd)!E
2BD
2!q!Na!Nd =
(11.7)(8.9x1014)(1015!+!1014)(3x105)2
(2)(1.6x10-19)(1015)(1014)
= 3,340 volts
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19-7. Resistance R = r!LA ;
LA =
0.020.01 = 2 cm-1
At 25 °C, Nd = 1014 >> ni so r = 1
q!mn!Nd =
1(1.6x10-19)(1500)(1014!)
= 41.7 W-cm
R(25 °C) = (41.7)(2) = 83.4 ohms
At 250 °C (523 °K), ni[523] = 1010 exp ÎÍÈ
˚˙˘
-!(1.6x10-19)!(1.1)(2)(1.4x10-23)
!ÓÌÏ
˛˝¸1
523!-!1
300 =
(1010)(7.6x103) = 7.6x1013 which is an appreciable fraction of Nd = 1014. Thus we should solve Eqs. (19-2) and (19-3) exactly for no and po rather than using equations similiar to Eq. (19-4). Solving Eqs. (19-2) and (19-3) for Nd >> Na yields
no = Nd2 Î
ÍÈ
˚˙˘
1!+! 1!+!4!n
2i
N2d
and po = n2i
no . Putting in numerical values yields
no = 1014
2 ÎÍÍÈ
˚˙˙˘
1!+! 1!+!(4)(7.6x1013)2!
(1014)2 = 1.4x1014 and
po = 5.8x1027
1014 = 5.8x1013
Assuming temperature-independent mobilities (not a valid assumption but no other information is given in text or the problem statement), resistance is
R(250 °C) = r(250!° C)!!L
A ; r(250 °C) ≈ 1
q!mn!no!+!q!mp!po
= 1
(1.6x10-19)(1500)(1.4x1014!)!+!(1.6x10-19)(500)(5.8x1013!) = 26.2 W-cm ;
R(250 °C) ≈ (26.2)(2) = 52.4 ohms
19-8. BVBD = e!(Na!+!Nd)!E
2BD
2!q!Na!Nd =
(11.7)(8.9x1014)(1015!+!1014)(3x105)2
(2)(1.6x10-19)(1015)(1014)
= 3,340 volts
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19-9. E2max = E
2BD ≈
4!f2c!BVBD
W2o!fc
; Eq. (19-13); or W
2o
!fc = 4!BVBD
E2BD
W2(BVBD) = W
2o!BVBD
!fc ; Eq. (19-11) ; Inserting W
2o
!fc = 4!BVBD
E2BD
and taking
the square root yields W (BVBD) ≈ 2!BVBD
E2BD.
19-10. Lp = Dp!t = (13)(10-6) = 36 microns ; Ln = Dn!t = (39)(10-6) = 62 microns
19-11. Assume a one-sided step junction with Na >> Nd
I1 = q n2i A
Dp!t1Nd!t1
exp(q!Vk!T ) ; I2 = q n
2i A
Dp!t2Nd!t2
exp(q!Vk!T )
I2I1
= 2 = t1t2 ; Thus 4 t2 = t1
19-12. s = q mp p + q mn n ; np = n2i ; Combining yeilds s = q mp
n2in + q mn n
dsdn = 0 = - q mp
n2i
n2 + q mn ; Solving for n yields n = ni mpmn and p = ni
mnmp
p = 1010 1500500 = 1.7x1010 cm-3; n = 1010
5001500 = 6x109 cm-3;
Thus minimum conductivity realized when silicon is slightly p-type.
Inserting p and n back into the equation for conductivity yields smin = 2 q ni mp!mn .
Putting in numerical values smin = (2)(1.6x10-19)(1010) (500)(1500) = 2.8x10-6 mhos-cm
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19-9. E2max = E
2BD ≈
4!f2c!BVBD
W2o!fc
; Eq. (19-13); or W
2o
!fc = 4!BVBD
E2BD
W2(BVBD) = W
2o!BVBD
!fc ; Eq. (19-11) ; Inserting W
2o
!fc = 4!BVBD
E2BD
and taking
the square root yields W (BVBD) ≈ 2!BVBD
E2BD.
19-10. Lp = Dp!t = (13)(10-6) = 36 microns ; Ln = Dn!t = (39)(10-6) = 62 microns
19-11. Assume a one-sided step junction with Na >> Nd
I1 = q n2i A
Dp!t1Nd!t1
exp(q!Vk!T ) ; I2 = q n
2i A
Dp!t2Nd!t2
exp(q!Vk!T )
I2I1
= 2 = t1t2 ; Thus 4 t2 = t1
19-12. s = q mp p + q mn n ; np = n2i ; Combining yeilds s = q mp
n2in + q mn n
dsdn = 0 = - q mp
n2i
n2 + q mn ; Solving for n yields n = ni mpmn and p = ni
mnmp
p = 1010 1500500 = 1.7x1010 cm-3; n = 1010
5001500 = 6x109 cm-3;
Thus minimum conductivity realized when silicon is slightly p-type.
Inserting p and n back into the equation for conductivity yields smin = 2 q ni mp!mn .
Putting in numerical values smin = (2)(1.6x10-19)(1010) (500)(1500) = 2.8x10-6 mhos-cm
![Page 315: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/315.jpg)
19-9. E2max = E
2BD ≈
4!f2c!BVBD
W2o!fc
; Eq. (19-13); or W
2o
!fc = 4!BVBD
E2BD
W2(BVBD) = W
2o!BVBD
!fc ; Eq. (19-11) ; Inserting W
2o
!fc = 4!BVBD
E2BD
and taking
the square root yields W (BVBD) ≈ 2!BVBD
E2BD.
19-10. Lp = Dp!t = (13)(10-6) = 36 microns ; Ln = Dn!t = (39)(10-6) = 62 microns
19-11. Assume a one-sided step junction with Na >> Nd
I1 = q n2i A
Dp!t1Nd!t1
exp(q!Vk!T ) ; I2 = q n
2i A
Dp!t2Nd!t2
exp(q!Vk!T )
I2I1
= 2 = t1t2 ; Thus 4 t2 = t1
19-12. s = q mp p + q mn n ; np = n2i ; Combining yeilds s = q mp
n2in + q mn n
dsdn = 0 = - q mp
n2i
n2 + q mn ; Solving for n yields n = ni mpmn and p = ni
mnmp
p = 1010 1500500 = 1.7x1010 cm-3; n = 1010
5001500 = 6x109 cm-3;
Thus minimum conductivity realized when silicon is slightly p-type.
Inserting p and n back into the equation for conductivity yields smin = 2 q ni mp!mn .
Putting in numerical values smin = (2)(1.6x10-19)(1010) (500)(1500) = 2.8x10-6 mhos-cm
![Page 316: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/316.jpg)
19-9. E2max = E
2BD ≈
4!f2c!BVBD
W2o!fc
; Eq. (19-13); or W
2o
!fc = 4!BVBD
E2BD
W2(BVBD) = W
2o!BVBD
!fc ; Eq. (19-11) ; Inserting W
2o
!fc = 4!BVBD
E2BD
and taking
the square root yields W (BVBD) ≈ 2!BVBD
E2BD.
19-10. Lp = Dp!t = (13)(10-6) = 36 microns ; Ln = Dn!t = (39)(10-6) = 62 microns
19-11. Assume a one-sided step junction with Na >> Nd
I1 = q n2i A
Dp!t1Nd!t1
exp(q!Vk!T ) ; I2 = q n
2i A
Dp!t2Nd!t2
exp(q!Vk!T )
I2I1
= 2 = t1t2 ; Thus 4 t2 = t1
19-12. s = q mp p + q mn n ; np = n2i ; Combining yeilds s = q mp
n2in + q mn n
dsdn = 0 = - q mp
n2i
n2 + q mn ; Solving for n yields n = ni mpmn and p = ni
mnmp
p = 1010 1500500 = 1.7x1010 cm-3; n = 1010
5001500 = 6x109 cm-3;
Thus minimum conductivity realized when silicon is slightly p-type.
Inserting p and n back into the equation for conductivity yields smin = 2 q ni mp!mn .
Putting in numerical values smin = (2)(1.6x10-19)(1010) (500)(1500) = 2.8x10-6 mhos-cm
![Page 317: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/317.jpg)
Chapter 20 Problem Solutions
20-1. Nd = 1.3x1017BVBD! =
1.3x10172500 = 5x1013 cm-3 ; W(2500 V) = (10-5)(2500) = 250
microns
20-2. Drift region length of 50 microns is much less than the 250 microns found in the previousproblem (20-1) for the same drift region doping density. Hence this must be a punch-through structure and Eq. (20-9) applies.
BVBD = (2x105)(5x10-3) - (1.6x10-19)(5x1013)(5x10-3)2
(2)(11.7)(8.9x10-14) = 900 V
20-3. Von = Vj + Vdrift ; Vj = k!Tq ln ÎÍ
È˚˙!
IIs
; For one-sided step junction Is = q!A!n
2i !Lp
!Nd!to ;
Evaluating Is yields (1.6x10-19)(2)(1010)2! (13)(2x10-6)
!(5x1013)(2x10-6) = 1.6x10-9 A
Vd = K1 I + K2 (I)2/3 Eq. (20-16) with I = forward bias current through the diode.
K1 = Wd
q!mo!A!nb =
5x10-3
(1.6x10-19)(900)(2)(1017) = 1.7x10-4
K2 = 3 W
4d
q2!m3o!n
2b!A2!to
= 3 (5x10-3)4
(1.6x10-19)2!(900)3!(1017)2!(2)2!(2x10-6)
= 7.5x10-4
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Chapter 20 Problem Solutions
20-1. Nd = 1.3x1017BVBD! =
1.3x10172500 = 5x1013 cm-3 ; W(2500 V) = (10-5)(2500) = 250
microns
20-2. Drift region length of 50 microns is much less than the 250 microns found in the previousproblem (20-1) for the same drift region doping density. Hence this must be a punch-through structure and Eq. (20-9) applies.
BVBD = (2x105)(5x10-3) - (1.6x10-19)(5x1013)(5x10-3)2
(2)(11.7)(8.9x10-14) = 900 V
20-3. Von = Vj + Vdrift ; Vj = k!Tq ln ÎÍ
È˚˙!
IIs
; For one-sided step junction Is = q!A!n
2i !Lp
!Nd!to ;
Evaluating Is yields (1.6x10-19)(2)(1010)2! (13)(2x10-6)
!(5x1013)(2x10-6) = 1.6x10-9 A
Vd = K1 I + K2 (I)2/3 Eq. (20-16) with I = forward bias current through the diode.
K1 = Wd
q!mo!A!nb =
5x10-3
(1.6x10-19)(900)(2)(1017) = 1.7x10-4
K2 = 3 W
4d
q2!m3o!n
2b!A2!to
= 3 (5x10-3)4
(1.6x10-19)2!(900)3!(1017)2!(2)2!(2x10-6)
= 7.5x10-4
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Chapter 20 Problem Solutions
20-1. Nd = 1.3x1017BVBD! =
1.3x10172500 = 5x1013 cm-3 ; W(2500 V) = (10-5)(2500) = 250
microns
20-2. Drift region length of 50 microns is much less than the 250 microns found in the previousproblem (20-1) for the same drift region doping density. Hence this must be a punch-through structure and Eq. (20-9) applies.
BVBD = (2x105)(5x10-3) - (1.6x10-19)(5x1013)(5x10-3)2
(2)(11.7)(8.9x10-14) = 900 V
20-3. Von = Vj + Vdrift ; Vj = k!Tq ln ÎÍ
È˚˙!
IIs
; For one-sided step junction Is = q!A!n
2i !Lp
!Nd!to ;
Evaluating Is yields (1.6x10-19)(2)(1010)2! (13)(2x10-6)
!(5x1013)(2x10-6) = 1.6x10-9 A
Vd = K1 I + K2 (I)2/3 Eq. (20-16) with I = forward bias current through the diode.
K1 = Wd
q!mo!A!nb =
5x10-3
(1.6x10-19)(900)(2)(1017) = 1.7x10-4
K2 = 3 W
4d
q2!m3o!n
2b!A2!to
= 3 (5x10-3)4
(1.6x10-19)2!(900)3!(1017)2!(2)2!(2x10-6)
= 7.5x10-4
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I Vj Vdrift Von0 A 0 V 0 V 0 V1 0.53 0.001 0.5310 0.59 0.005 0.59100 0.65 0.033 0.681000 0.71 0.25 0.963000 0.74 0.67 1.41 • •
•
•
•
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1 10 100 1000 10000
Von in volts
Forward current in amperes
20-4. a) Von(t) = Rdrift I(t) >> Vj ≈ 1 V ; Rdrift = r LA ; r =
1q!mn!Nd
r = 1
(1.6x10-19)(1500)(5x1013) = 85 ohm-cm ;
LA =
5x10-32 = 2.5x10-3
Rdrift = (85)(2.5x10-3) = 0.21 ohms ; I(t) = 2.5x108 t ; 0 < t < 4 microseconds
Von(t) = (0.21)(2.5x108 t ) = 5.3x107 t Volts ; 0 < t < 4 microseconds
Von(4 ms) = (5.3x107)(4x10-6) = 212 volts
b) Von(t) = Rdrift (t) I(t) ; Rdrift (t) = 0.21[1 - 2.5x107 t]
Von(t) = {0.21[1 - 2.5x107 t]} { 2.5x108 t } = 53[t - 0.25 t2] ; t in microseconds
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I Vj Vdrift Von0 A 0 V 0 V 0 V1 0.53 0.001 0.5310 0.59 0.005 0.59100 0.65 0.033 0.681000 0.71 0.25 0.963000 0.74 0.67 1.41 • •
•
•
•
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1 10 100 1000 10000
Von in volts
Forward current in amperes
20-4. a) Von(t) = Rdrift I(t) >> Vj ≈ 1 V ; Rdrift = r LA ; r =
1q!mn!Nd
r = 1
(1.6x10-19)(1500)(5x1013) = 85 ohm-cm ;
LA =
5x10-32 = 2.5x10-3
Rdrift = (85)(2.5x10-3) = 0.21 ohms ; I(t) = 2.5x108 t ; 0 < t < 4 microseconds
Von(t) = (0.21)(2.5x108 t ) = 5.3x107 t Volts ; 0 < t < 4 microseconds
Von(4 ms) = (5.3x107)(4x10-6) = 212 volts
b) Von(t) = Rdrift (t) I(t) ; Rdrift (t) = 0.21[1 - 2.5x107 t]
Von(t) = {0.21[1 - 2.5x107 t]} { 2.5x108 t } = 53[t - 0.25 t2] ; t in microseconds
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0
50
100
150
200
250
0 0.5 1 1.5 2 2.5 3 3.5 4
Von involts
With carrier injection
No carrier injection
20-5. toff = trr + t3 = trr + IF
diR/dt = trr + 2000
2.5x108 = trr + 8 ms
trr = 2!t!IFdiR/dt ; t = 4x10-12(BVBD)2 = 4x10-12(2000)2 = 16 ms
trr = (2)(1.6x10-5)(2x103)
2.5x108 = 16 ms ; toff = 8 ms + 16 ms = 24 ms
20-6. Assume a non-punch-through structure for the Schottky diode.
Nd = 1.3x1017BVBD
= 1.3x1017
150 = 8.7x1014 cm-3
Wd = 10-5 BVBD = (10-5) (150) = 15 microns
20-7. Vdrift = (100 A) (Rdrift) = 2 V ; Rdrift = 0.02 ohms ; 0.02 = 1
q!mn!Nd
LA
A = 2x10-3
!(1.6x10-19)(1500)(1015)(2x10-2) = 0.42 cm2
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0
50
100
150
200
250
0 0.5 1 1.5 2 2.5 3 3.5 4
Von involts
With carrier injection
No carrier injection
20-5. toff = trr + t3 = trr + IF
diR/dt = trr + 2000
2.5x108 = trr + 8 ms
trr = 2!t!IFdiR/dt ; t = 4x10-12(BVBD)2 = 4x10-12(2000)2 = 16 ms
trr = (2)(1.6x10-5)(2x103)
2.5x108 = 16 ms ; toff = 8 ms + 16 ms = 24 ms
20-6. Assume a non-punch-through structure for the Schottky diode.
Nd = 1.3x1017BVBD
= 1.3x1017
150 = 8.7x1014 cm-3
Wd = 10-5 BVBD = (10-5) (150) = 15 microns
20-7. Vdrift = (100 A) (Rdrift) = 2 V ; Rdrift = 0.02 ohms ; 0.02 = 1
q!mn!Nd
LA
A = 2x10-3
!(1.6x10-19)(1500)(1015)(2x10-2) = 0.42 cm2
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0
50
100
150
200
250
0 0.5 1 1.5 2 2.5 3 3.5 4
Von involts
With carrier injection
No carrier injection
20-5. toff = trr + t3 = trr + IF
diR/dt = trr + 2000
2.5x108 = trr + 8 ms
trr = 2!t!IFdiR/dt ; t = 4x10-12(BVBD)2 = 4x10-12(2000)2 = 16 ms
trr = (2)(1.6x10-5)(2x103)
2.5x108 = 16 ms ; toff = 8 ms + 16 ms = 24 ms
20-6. Assume a non-punch-through structure for the Schottky diode.
Nd = 1.3x1017BVBD
= 1.3x1017
150 = 8.7x1014 cm-3
Wd = 10-5 BVBD = (10-5) (150) = 15 microns
20-7. Vdrift = (100 A) (Rdrift) = 2 V ; Rdrift = 0.02 ohms ; 0.02 = 1
q!mn!Nd
LA
A = 2x10-3
!(1.6x10-19)(1500)(1015)(2x10-2) = 0.42 cm2
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0
50
100
150
200
250
0 0.5 1 1.5 2 2.5 3 3.5 4
Von involts
With carrier injection
No carrier injection
20-5. toff = trr + t3 = trr + IF
diR/dt = trr + 2000
2.5x108 = trr + 8 ms
trr = 2!t!IFdiR/dt ; t = 4x10-12(BVBD)2 = 4x10-12(2000)2 = 16 ms
trr = (2)(1.6x10-5)(2x103)
2.5x108 = 16 ms ; toff = 8 ms + 16 ms = 24 ms
20-6. Assume a non-punch-through structure for the Schottky diode.
Nd = 1.3x1017BVBD
= 1.3x1017
150 = 8.7x1014 cm-3
Wd = 10-5 BVBD = (10-5) (150) = 15 microns
20-7. Vdrift = (100 A) (Rdrift) = 2 V ; Rdrift = 0.02 ohms ; 0.02 = 1
q!mn!Nd
LA
A = 2x10-3
!(1.6x10-19)(1500)(1015)(2x10-2) = 0.42 cm2
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20-8. Use Eq. (20-9) to solve for Nd ; Nd = [EBD Wd - BVBD] ÎÍÍÈ
˚˙˙˘2!e
q!W2d
= [(2x105)(2x10-3) - 300] (2)(11.7)(8.9x10-14)(1.6x10-19)(2x10-3)2
= Nd = 3.4x1014 cm-3
20-9.RA
npt =
Wd(npt)q!mn!Nnpt
; Nnpt = Nd of non-punch-throuth (npt) diode
RA
pt =
Wd(pt)q!mn!Npt
; Npt = Nd of punch-through (pt) diode
Wd(npt) = e!EBD!q!Nnpt
; Derived from Eqs. (19-11), (19-12), and (19-13)
Wd(pt) = e!EBD!q!Npt
ÎÍÍÈ
˚˙˙˘
1!+_ ! 1!-!2!q!Npt!BVBD
e!E2BD
2!q!Npt!BVBD
e!E2BD
= q!Npte!EBD
NnptNnpt
2!BVBD
EBD =
q!Nnpte!EBD
2!BVBD
EBD
NptNnpt
= 1
Wd(npt) Wd(npt) NptNnpt
= x = NptNnpt
; Wd(pt) = Wd(npt) 1x [ ]1!+_ ! 1!-!x
If Npt << Nnpt (x << 1) Wd(pt) ≈ 0.5 Wd(npt) ; Eq. (20-10)
Limit of 1x [ ]1!+_ ! 1!-!x as x approaches is infinite for the plus root and 0.5 for the
minus root. Hence the minus root is the correct choice to use.
RA
pt =
Wd(npt)!1x![ ]1!-!! 1!-!x
q!mn!Npt =
Wd(npt)!1x![ ]1!-!! 1!-!x
q!mn!Npt
Nnpt!Nnpt
= Wd(npt)
!q!mn!Nnpt
1x2 [1 - 1!-!x ] =
RA
npt
1x2 [1 - 1!-!x ]
ddx ÎÍ
È˚˙!
1x2!! 1!-!x! = 0 =
-2x3 [1 - 1!-!x ] +
12!x2! 1!-!x
Solving for x yields x = 89 i.e. Npt =
89 Nnpt ; Wd(pt) = 0.75 Wd(npt)
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20-8. Use Eq. (20-9) to solve for Nd ; Nd = [EBD Wd - BVBD] ÎÍÍÈ
˚˙˙˘2!e
q!W2d
= [(2x105)(2x10-3) - 300] (2)(11.7)(8.9x10-14)(1.6x10-19)(2x10-3)2
= Nd = 3.4x1014 cm-3
20-9.RA
npt =
Wd(npt)q!mn!Nnpt
; Nnpt = Nd of non-punch-throuth (npt) diode
RA
pt =
Wd(pt)q!mn!Npt
; Npt = Nd of punch-through (pt) diode
Wd(npt) = e!EBD!q!Nnpt
; Derived from Eqs. (19-11), (19-12), and (19-13)
Wd(pt) = e!EBD!q!Npt
ÎÍÍÈ
˚˙˙˘
1!+_ ! 1!-!2!q!Npt!BVBD
e!E2BD
2!q!Npt!BVBD
e!E2BD
= q!Npte!EBD
NnptNnpt
2!BVBD
EBD =
q!Nnpte!EBD
2!BVBD
EBD
NptNnpt
= 1
Wd(npt) Wd(npt) NptNnpt
= x = NptNnpt
; Wd(pt) = Wd(npt) 1x [ ]1!+_ ! 1!-!x
If Npt << Nnpt (x << 1) Wd(pt) ≈ 0.5 Wd(npt) ; Eq. (20-10)
Limit of 1x [ ]1!+_ ! 1!-!x as x approaches is infinite for the plus root and 0.5 for the
minus root. Hence the minus root is the correct choice to use.
RA
pt =
Wd(npt)!1x![ ]1!-!! 1!-!x
q!mn!Npt =
Wd(npt)!1x![ ]1!-!! 1!-!x
q!mn!Npt
Nnpt!Nnpt
= Wd(npt)
!q!mn!Nnpt
1x2 [1 - 1!-!x ] =
RA
npt
1x2 [1 - 1!-!x ]
ddx ÎÍ
È˚˙!
1x2!! 1!-!x! = 0 =
-2x3 [1 - 1!-!x ] +
12!x2! 1!-!x
Solving for x yields x = 89 i.e. Npt =
89 Nnpt ; Wd(pt) = 0.75 Wd(npt)
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RA
npt = 0.84
RA
npt
20-10. Cjo = eA
Wd(0) ; Area A determined by on-state voltage and current. Depletion-layer
width Wd(0) set by breakdown voltage. Wd(0) same for both diodes.
Wd(150) = Wd(0) 1!+!1500.7 = (10-5)(150) = 15 mm ;
Wd(0) = 15
(150)/(0.7) ≈ 1 mm
Schottky diode area ; Vdrift = 2 volts = Wd(150)
q!mn!Nd!ASchottky IF
Drift region doping density Nd same for both diodes. Nd = 1.3x1017
150
= 8.7x1014 cm-3
ASchottky = (1.5x10-3)(300)
(2)(1.6x10-19)(1500)(8.7x1014) = 1.07 cm2
PN junction diode area: Vdrift = 2 volts = K1 IF + K2 (IF)2/3 ; Eq. (20-16)
K1 = Wd(150)
q!mo!Apn!nb =
1.5x10-3
(1.6x10-19)(900)(Apn)(1017) =
10-4Apn
K2 = 3 W
4d(150)
q2!m3o!n
2b!A
2pn!to
= 3 (1.5x10-3)4
(1.6x10-19)2!(900)3!(1017)2!(A2pn)!(2x10-6)
K2 = 2.4x10-4 (Apn)-0.67 ; 2 volts = 10-4Apn
(300) + 2.4x10-4 (Apn)-0.67 (300)0.67
Apn = 0.015 + 0.00154 (Apn)0.333 ; Solve by successive approximation ;
Apn ≈ 0.017 cm2
Schottky diode Cjo = (11.7)(8.9x10-14)(1.07)
10-4 ≈ 11 nF = 0.011 mF
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RA
npt = 0.84
RA
npt
20-10. Cjo = eA
Wd(0) ; Area A determined by on-state voltage and current. Depletion-layer
width Wd(0) set by breakdown voltage. Wd(0) same for both diodes.
Wd(150) = Wd(0) 1!+!1500.7 = (10-5)(150) = 15 mm ;
Wd(0) = 15
(150)/(0.7) ≈ 1 mm
Schottky diode area ; Vdrift = 2 volts = Wd(150)
q!mn!Nd!ASchottky IF
Drift region doping density Nd same for both diodes. Nd = 1.3x1017
150
= 8.7x1014 cm-3
ASchottky = (1.5x10-3)(300)
(2)(1.6x10-19)(1500)(8.7x1014) = 1.07 cm2
PN junction diode area: Vdrift = 2 volts = K1 IF + K2 (IF)2/3 ; Eq. (20-16)
K1 = Wd(150)
q!mo!Apn!nb =
1.5x10-3
(1.6x10-19)(900)(Apn)(1017) =
10-4Apn
K2 = 3 W
4d(150)
q2!m3o!n
2b!A
2pn!to
= 3 (1.5x10-3)4
(1.6x10-19)2!(900)3!(1017)2!(A2pn)!(2x10-6)
K2 = 2.4x10-4 (Apn)-0.67 ; 2 volts = 10-4Apn
(300) + 2.4x10-4 (Apn)-0.67 (300)0.67
Apn = 0.015 + 0.00154 (Apn)0.333 ; Solve by successive approximation ;
Apn ≈ 0.017 cm2
Schottky diode Cjo = (11.7)(8.9x10-14)(1.07)
10-4 ≈ 11 nF = 0.011 mF
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PN junction diode Cjo = (11.7)(8.9x10-14)(0.017)
10-4 ≈ 180 pF
20-11.BVcylBVp
= 2 r2 (1 + 1r ) ln (1 +
1r ) - 2r
•
••
••
• • • •
00.10.20.30.40.50.60.70.80.9
1
0.1 1 10
BVcyl / BVp
r = R/(2W n)
20-12.BVcylBVp
= 9501000 = 0.95 ; From graph in problem 20-11, r ≈ 6 =
R2!Wn
R ≈ 12 Wn
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PN junction diode Cjo = (11.7)(8.9x10-14)(0.017)
10-4 ≈ 180 pF
20-11.BVcylBVp
= 2 r2 (1 + 1r ) ln (1 +
1r ) - 2r
•
••
••
• • • •
00.10.20.30.40.50.60.70.80.9
1
0.1 1 10
BVcyl / BVp
r = R/(2W n)
20-12.BVcylBVp
= 9501000 = 0.95 ; From graph in problem 20-11, r ≈ 6 =
R2!Wn
R ≈ 12 Wn
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PN junction diode Cjo = (11.7)(8.9x10-14)(0.017)
10-4 ≈ 180 pF
20-11.BVcylBVp
= 2 r2 (1 + 1r ) ln (1 +
1r ) - 2r
•
••
••
• • • •
00.10.20.30.40.50.60.70.80.9
1
0.1 1 10
BVcyl / BVp
r = R/(2W n)
20-12.BVcylBVp
= 9501000 = 0.95 ; From graph in problem 20-11, r ≈ 6 =
R2!Wn
R ≈ 12 Wn
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Chapter 21 Problem Solutions
21-1. NPN BJT ; BVCEO = BVCBO
b1/4 ; PNP BJT ; BVCEO = BVCBO
b1/6
B
**
** * * * * *
** * * * * * *
J
••
• • • •••••
• • • • •••
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1 10 100
BVCEOBVCBO
PNP
NPN
21-2. The flow of large reverse base currents when emitter current is still flowing in the forward direction will lead to emitter current crowding towards the center of the emitter. This accenuates the possibility of second breakdown. See Fig. 21-8b.
When emitter-open switching is used, there is no emitter current and thus no emitter current crowding and second breakdown is much less likely to occur.
21-3. a) Idealized BJT current and voltage waveforms in step-down converter
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Chapter 21 Problem Solutions
21-1. NPN BJT ; BVCEO = BVCBO
b1/4 ; PNP BJT ; BVCEO = BVCBO
b1/6
B
**
** * * * * *
** * * * * * *
J
••
• • • •••••
• • • • •••
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1 10 100
BVCEOBVCBO
PNP
NPN
21-2. The flow of large reverse base currents when emitter current is still flowing in the forward direction will lead to emitter current crowding towards the center of the emitter. This accenuates the possibility of second breakdown. See Fig. 21-8b.
When emitter-open switching is used, there is no emitter current and thus no emitter current crowding and second breakdown is much less likely to occur.
21-3. a) Idealized BJT current and voltage waveforms in step-down converter
![Page 335: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/335.jpg)
Chapter 21 Problem Solutions
21-1. NPN BJT ; BVCEO = BVCBO
b1/4 ; PNP BJT ; BVCEO = BVCBO
b1/6
B
**
** * * * * *
** * * * * * *
J
••
• • • •••••
• • • • •••
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1 10 100
BVCEOBVCBO
PNP
NPN
21-2. The flow of large reverse base currents when emitter current is still flowing in the forward direction will lead to emitter current crowding towards the center of the emitter. This accenuates the possibility of second breakdown. See Fig. 21-8b.
When emitter-open switching is used, there is no emitter current and thus no emitter current crowding and second breakdown is much less likely to occur.
21-3. a) Idealized BJT current and voltage waveforms in step-down converter
![Page 336: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/336.jpg)
T/2
I oVd
V (t)CEi (t)
C
t d,off t rv t fit d,on t ri t fv
BJT power dissipation Pc = 1T ıÛ
0
T!vCE(t)!iC(t)!dt = {Eon + Esw }fs ; fs =
1T
Eon = VCE,on Io {T2 + td,off - td,on} ≈ (2)(40)
12!fs
= 40fs
Joules
Esw = Eri + Efv + Erv + Efi
Eri = Vd!Io!tri
2 = (100)(40)(2x10-7)
2 = 4x10-4 Joules
Efv = Vd!Io!tfv
2 = (100)(40)(1x10-7)
2 = 2x10-4 Joules
Erv = Vd!Io!trv
2 = (100)(40)(1x10-7)
2 = 2x10-4 Joules
Efi = Vd!Io!tfi
2 = (100)(40)(2x10-7)
2 = 4x10-4 Joules
Esw = 1.2x10-3 Joules ; Pc = [40fs
+ 1.2x10-3] fs = 40 + 1.2x10-3 fs
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40
80
033 kHz
fs
[watts]
Pc
b) Tj = Rqja Pc + Ta ; Pc,max = 150!-!25
1 = 125 watts = 40 + 1.2x10-3 fs,max
fs,max = (125!-!40)1.2x10-3 = 71 kHz
21-4. From problem 21-3, Pc = {Eon + Esw }fs where Esw is proportional to the switching times. If the switching times vary with temperature then Esw can be written as
Esw(Tj) = Esw(25 °C){1 + a[Tj - 25]} with a = 0.4
(125!-!25) = 4x10-3
110 - 50 = Rqja {40 + 1.2x10-3[1+0.004(110-25)] (2.5x104)} Rqja = 0.75 °C/watt ; This represents an upper limit.
21-5. Beta begins to drop when the excessd carrier density in the base becomes comparable to the doping density in the base, i.e. nbase(x) ≈ Na in an NPN BJT.
IC = q!Dn!Na!A
!Wb =
(1.6x10-19)(38)(1016)(1)(3x10-4)
= 200 A
21-6. The diode can carry the larger current. The lateral voltage drops in the BJT base limit the maximum current that the transistor can carry. These lateral voltage drops lead to emitter current crowding. The diode has no such limitation.
21-7. The base doping is not much larger than the collector doping so that the CB depletion layer protrudes a significant amount into the base. This encroachment may stretch across the base and reach the EB depletion layer before the desired blocking voltage is reached. At reach-through xp will equal the difference between the base width WB and the base-side protusion of the base-emitter depletion layer width WEB,depl.The reach-through voltage is approximately estimated as shown below.
Estimate of base-emitter base-side protusion = WEB,depl at zero bias.
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40
80
033 kHz
fs
[watts]
Pc
b) Tj = Rqja Pc + Ta ; Pc,max = 150!-!25
1 = 125 watts = 40 + 1.2x10-3 fs,max
fs,max = (125!-!40)1.2x10-3 = 71 kHz
21-4. From problem 21-3, Pc = {Eon + Esw }fs where Esw is proportional to the switching times. If the switching times vary with temperature then Esw can be written as
Esw(Tj) = Esw(25 °C){1 + a[Tj - 25]} with a = 0.4
(125!-!25) = 4x10-3
110 - 50 = Rqja {40 + 1.2x10-3[1+0.004(110-25)] (2.5x104)} Rqja = 0.75 °C/watt ; This represents an upper limit.
21-5. Beta begins to drop when the excessd carrier density in the base becomes comparable to the doping density in the base, i.e. nbase(x) ≈ Na in an NPN BJT.
IC = q!Dn!Na!A
!Wb =
(1.6x10-19)(38)(1016)(1)(3x10-4)
= 200 A
21-6. The diode can carry the larger current. The lateral voltage drops in the BJT base limit the maximum current that the transistor can carry. These lateral voltage drops lead to emitter current crowding. The diode has no such limitation.
21-7. The base doping is not much larger than the collector doping so that the CB depletion layer protrudes a significant amount into the base. This encroachment may stretch across the base and reach the EB depletion layer before the desired blocking voltage is reached. At reach-through xp will equal the difference between the base width WB and the base-side protusion of the base-emitter depletion layer width WEB,depl.The reach-through voltage is approximately estimated as shown below.
Estimate of base-emitter base-side protusion = WEB,depl at zero bias.
![Page 339: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/339.jpg)
40
80
033 kHz
fs
[watts]
Pc
b) Tj = Rqja Pc + Ta ; Pc,max = 150!-!25
1 = 125 watts = 40 + 1.2x10-3 fs,max
fs,max = (125!-!40)1.2x10-3 = 71 kHz
21-4. From problem 21-3, Pc = {Eon + Esw }fs where Esw is proportional to the switching times. If the switching times vary with temperature then Esw can be written as
Esw(Tj) = Esw(25 °C){1 + a[Tj - 25]} with a = 0.4
(125!-!25) = 4x10-3
110 - 50 = Rqja {40 + 1.2x10-3[1+0.004(110-25)] (2.5x104)} Rqja = 0.75 °C/watt ; This represents an upper limit.
21-5. Beta begins to drop when the excessd carrier density in the base becomes comparable to the doping density in the base, i.e. nbase(x) ≈ Na in an NPN BJT.
IC = q!Dn!Na!A
!Wb =
(1.6x10-19)(38)(1016)(1)(3x10-4)
= 200 A
21-6. The diode can carry the larger current. The lateral voltage drops in the BJT base limit the maximum current that the transistor can carry. These lateral voltage drops lead to emitter current crowding. The diode has no such limitation.
21-7. The base doping is not much larger than the collector doping so that the CB depletion layer protrudes a significant amount into the base. This encroachment may stretch across the base and reach the EB depletion layer before the desired blocking voltage is reached. At reach-through xp will equal the difference between the base width WB and the base-side protusion of the base-emitter depletion layer width WEB,depl.The reach-through voltage is approximately estimated as shown below.
Estimate of base-emitter base-side protusion = WEB,depl at zero bias.
![Page 340: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/340.jpg)
40
80
033 kHz
fs
[watts]
Pc
b) Tj = Rqja Pc + Ta ; Pc,max = 150!-!25
1 = 125 watts = 40 + 1.2x10-3 fs,max
fs,max = (125!-!40)1.2x10-3 = 71 kHz
21-4. From problem 21-3, Pc = {Eon + Esw }fs where Esw is proportional to the switching times. If the switching times vary with temperature then Esw can be written as
Esw(Tj) = Esw(25 °C){1 + a[Tj - 25]} with a = 0.4
(125!-!25) = 4x10-3
110 - 50 = Rqja {40 + 1.2x10-3[1+0.004(110-25)] (2.5x104)} Rqja = 0.75 °C/watt ; This represents an upper limit.
21-5. Beta begins to drop when the excessd carrier density in the base becomes comparable to the doping density in the base, i.e. nbase(x) ≈ Na in an NPN BJT.
IC = q!Dn!Na!A
!Wb =
(1.6x10-19)(38)(1016)(1)(3x10-4)
= 200 A
21-6. The diode can carry the larger current. The lateral voltage drops in the BJT base limit the maximum current that the transistor can carry. These lateral voltage drops lead to emitter current crowding. The diode has no such limitation.
21-7. The base doping is not much larger than the collector doping so that the CB depletion layer protrudes a significant amount into the base. This encroachment may stretch across the base and reach the EB depletion layer before the desired blocking voltage is reached. At reach-through xp will equal the difference between the base width WB and the base-side protusion of the base-emitter depletion layer width WEB,depl.The reach-through voltage is approximately estimated as shown below.
Estimate of base-emitter base-side protusion = WEB,depl at zero bias.
![Page 341: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/341.jpg)
40
80
033 kHz
fs
[watts]
Pc
b) Tj = Rqja Pc + Ta ; Pc,max = 150!-!25
1 = 125 watts = 40 + 1.2x10-3 fs,max
fs,max = (125!-!40)1.2x10-3 = 71 kHz
21-4. From problem 21-3, Pc = {Eon + Esw }fs where Esw is proportional to the switching times. If the switching times vary with temperature then Esw can be written as
Esw(Tj) = Esw(25 °C){1 + a[Tj - 25]} with a = 0.4
(125!-!25) = 4x10-3
110 - 50 = Rqja {40 + 1.2x10-3[1+0.004(110-25)] (2.5x104)} Rqja = 0.75 °C/watt ; This represents an upper limit.
21-5. Beta begins to drop when the excessd carrier density in the base becomes comparable to the doping density in the base, i.e. nbase(x) ≈ Na in an NPN BJT.
IC = q!Dn!Na!A
!Wb =
(1.6x10-19)(38)(1016)(1)(3x10-4)
= 200 A
21-6. The diode can carry the larger current. The lateral voltage drops in the BJT base limit the maximum current that the transistor can carry. These lateral voltage drops lead to emitter current crowding. The diode has no such limitation.
21-7. The base doping is not much larger than the collector doping so that the CB depletion layer protrudes a significant amount into the base. This encroachment may stretch across the base and reach the EB depletion layer before the desired blocking voltage is reached. At reach-through xp will equal the difference between the base width WB and the base-side protusion of the base-emitter depletion layer width WEB,depl.The reach-through voltage is approximately estimated as shown below.
Estimate of base-emitter base-side protusion = WEB,depl at zero bias.
![Page 342: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/342.jpg)
Wo,EB = 2!e!fc!(NaB!+!NdE)
!q!NaB!NdE ; fcE =
k!Tq ln
ÎÍÍÈ
˚˙˙˘
!NaB!NdE
n2i
fcE = 0.26 ln ÎÍÈ
˚˙˘
!1034
1020 = 0.84 V
Wo,EB = (2)(11.7)(8.9x10-14)(0.84)(1019!+!1015)
(1.6x10-19)(1019)(1015) = 0.33 microns
Estimate of collector-base base-side protusion of WCB,depl.
CB depletion layer thickness W(V) = Wo,CB 1!+!Vfc = xp + xn ;
xp = protrusion of CB depletion layer into p-type base region. xp = W(V)
11 using xp Na= xn Nd (charge neutrality).
Wo,CB = 2!e!fcC!(NaB!+!NdC)
!q!NaB!NdC ; fcC =
k!Tq ln
ÎÍÍÈ
˚˙˙˘
!NaB!NdC
n2i
fcC = 0.26 ln ÎÍÈ
˚˙˘
!1029
1020 = 0.54 V
Wo,CB = (2)(11.7)(8.9x10-14)(0.54)(1014!+!1015)
(1.6x10-19)(1014)(1015) = 2.8 microns
{(3 - 0.33)x10-4}(11) = 2.8x10-4 1!+!V
0.54 : Solving for V yields V = 59 volts. Reach-through voltage of 59 V is much less than the expected value of BVBD = 1000 V.
21-8. BVEBO = 10 V = 1.3x1017
!NaB ; NaB = acceptor doping density in base = 1.3x1016 cm-3
BVCBO = b1/4 BVCEO = (5)1/4(1000) = ≈ 1500 Volts ≈ 1.3x1017
!NdC
NdC = collector drift region donor density = 8.7x1013 cn -3
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Wo,EB = 2!e!fc!(NaB!+!NdE)
!q!NaB!NdE ; fcE =
k!Tq ln
ÎÍÍÈ
˚˙˙˘
!NaB!NdE
n2i
fcE = 0.26 ln ÎÍÈ
˚˙˘
!1034
1020 = 0.84 V
Wo,EB = (2)(11.7)(8.9x10-14)(0.84)(1019!+!1015)
(1.6x10-19)(1019)(1015) = 0.33 microns
Estimate of collector-base base-side protusion of WCB,depl.
CB depletion layer thickness W(V) = Wo,CB 1!+!Vfc = xp + xn ;
xp = protrusion of CB depletion layer into p-type base region. xp = W(V)
11 using xp Na= xn Nd (charge neutrality).
Wo,CB = 2!e!fcC!(NaB!+!NdC)
!q!NaB!NdC ; fcC =
k!Tq ln
ÎÍÍÈ
˚˙˙˘
!NaB!NdC
n2i
fcC = 0.26 ln ÎÍÈ
˚˙˘
!1029
1020 = 0.54 V
Wo,CB = (2)(11.7)(8.9x10-14)(0.54)(1014!+!1015)
(1.6x10-19)(1014)(1015) = 2.8 microns
{(3 - 0.33)x10-4}(11) = 2.8x10-4 1!+!V
0.54 : Solving for V yields V = 59 volts. Reach-through voltage of 59 V is much less than the expected value of BVBD = 1000 V.
21-8. BVEBO = 10 V = 1.3x1017
!NaB ; NaB = acceptor doping density in base = 1.3x1016 cm-3
BVCBO = b1/4 BVCEO = (5)1/4(1000) = ≈ 1500 Volts ≈ 1.3x1017
!NdC
NdC = collector drift region donor density = 8.7x1013 cn -3
![Page 344: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/344.jpg)
Set base width Wbase by calculating the protrusion, xp, of the CB depletion layer into the base at VCB = 1500 volts = BVCBO; Wbase = xp(BVCBO)
Approximate width of CB depletion layer at breakdown WCB(BVCBO) ≈ (10-5)(1500) = 150 mm
xp(BVCBO) = xn(BVCBO)!NdC
NaB = xn(BVCBO)
8.7x1013
1.3x1016 = xn(BVCBO) 6.7x10-3
xp(BVCBO) << xn(BVCBO) so xn(BVCBO) ≈ WCB(BVCBO) ≈ 150 mm
Hence xp(BVCBO) = (1.5x10-2 cm)(6.7x10-3) = 1 micron = Wbase
21-9. Beta = 150 = bD bM + bD + bM = 20 bM + 20 + bM
bM = 150!-!20
21 = 6.2
21-10. Must first ascertain the operating states of the two transistors.Two likely choices including (1) driver BJT saturated and main BJT active and (2) driver BJT and main BJT both saturated. Initially assume driver BJT saturated and main BJT active.
10 I B,M
B,MI
0.02 W
0.8 V
0.6 V
0.8 V
-
+
-
+
-
+
C
B
E
10 IB,M + IB,M = 100 AIB,M = IC,D = 9.1A IC,M = 91 AVCE,D = 0.2 +(.02)(9.1) = 0.382 VVCE,M = VCE,D + 0.8 V = 1.18 vBut a saturated main BJT with IC,M = 91 Agoing through 0.02 ohms generatesa voltage drop of 1.8 V which is > 1.18 V.Hence main BJT must be saturated.
![Page 345: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/345.jpg)
Set base width Wbase by calculating the protrusion, xp, of the CB depletion layer into the base at VCB = 1500 volts = BVCBO; Wbase = xp(BVCBO)
Approximate width of CB depletion layer at breakdown WCB(BVCBO) ≈ (10-5)(1500) = 150 mm
xp(BVCBO) = xn(BVCBO)!NdC
NaB = xn(BVCBO)
8.7x1013
1.3x1016 = xn(BVCBO) 6.7x10-3
xp(BVCBO) << xn(BVCBO) so xn(BVCBO) ≈ WCB(BVCBO) ≈ 150 mm
Hence xp(BVCBO) = (1.5x10-2 cm)(6.7x10-3) = 1 micron = Wbase
21-9. Beta = 150 = bD bM + bD + bM = 20 bM + 20 + bM
bM = 150!-!20
21 = 6.2
21-10. Must first ascertain the operating states of the two transistors.Two likely choices including (1) driver BJT saturated and main BJT active and (2) driver BJT and main BJT both saturated. Initially assume driver BJT saturated and main BJT active.
10 I B,M
B,MI
0.02 W
0.8 V
0.6 V
0.8 V
-
+
-
+
-
+
C
B
E
10 IB,M + IB,M = 100 AIB,M = IC,D = 9.1A IC,M = 91 AVCE,D = 0.2 +(.02)(9.1) = 0.382 VVCE,M = VCE,D + 0.8 V = 1.18 vBut a saturated main BJT with IC,M = 91 Agoing through 0.02 ohms generatesa voltage drop of 1.8 V which is > 1.18 V.Hence main BJT must be saturated.
![Page 346: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/346.jpg)
Set base width Wbase by calculating the protrusion, xp, of the CB depletion layer into the base at VCB = 1500 volts = BVCBO; Wbase = xp(BVCBO)
Approximate width of CB depletion layer at breakdown WCB(BVCBO) ≈ (10-5)(1500) = 150 mm
xp(BVCBO) = xn(BVCBO)!NdC
NaB = xn(BVCBO)
8.7x1013
1.3x1016 = xn(BVCBO) 6.7x10-3
xp(BVCBO) << xn(BVCBO) so xn(BVCBO) ≈ WCB(BVCBO) ≈ 150 mm
Hence xp(BVCBO) = (1.5x10-2 cm)(6.7x10-3) = 1 micron = Wbase
21-9. Beta = 150 = bD bM + bD + bM = 20 bM + 20 + bM
bM = 150!-!20
21 = 6.2
21-10. Must first ascertain the operating states of the two transistors.Two likely choices including (1) driver BJT saturated and main BJT active and (2) driver BJT and main BJT both saturated. Initially assume driver BJT saturated and main BJT active.
10 I B,M
B,MI
0.02 W
0.8 V
0.6 V
0.8 V
-
+
-
+
-
+
C
B
E
10 IB,M + IB,M = 100 AIB,M = IC,D = 9.1A IC,M = 91 AVCE,D = 0.2 +(.02)(9.1) = 0.382 VVCE,M = VCE,D + 0.8 V = 1.18 vBut a saturated main BJT with IC,M = 91 Agoing through 0.02 ohms generatesa voltage drop of 1.8 V which is > 1.18 V.Hence main BJT must be saturated.
![Page 347: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/347.jpg)
B,MI
0.02 W
0.8 V
0.6 V
0.8 V
-
+
-
+
-
+
C
B
E
0.02 W
0.6 V-
+I C,M
Neglecting IB,DIC,M + IC,D = 100 A(.02)IC,M -0.6 = (.02) I C,D +0.2IC,D = 30 A ; IC,M = 70 AVCE,M = 0.2 + (70)(.02) = 1.6 VPDarl = VCE,M [IC,M + IC,D]
PDarl = (1.6 V)(100 A) = 160 W
21-11. CEBO = e!AE
WEBO ; CCBO = e!AC
WCBO
WEBO = zero-bias emitter-base depletion layer thicknessWCBO = zero-bias collector-base depletion layer thickness
WEBO = 2!e!fcE!(NaB!+!NdE)
!q!NaB!NdE ; fcE =
k!Tq ln
ÎÍÍÈ
˚˙˙˘
!NaB!NdE
n2i
fcE = 0.026 ln ÎÍÈ
˚˙˘
!(1019)(1016)
1020 = 0.89 V ;
WEBO = (2)(11.7)(8.9x10-14)(0.89)(1019!+!1016)
(1.6x10-19)(1019)(1016) = 0.34 microns
CEBO = (11.7)(8.9x10-14)(0.3)
3.4x10-5 = 9.2 nF
WCBO = 2!e!fcC!(NaB!+!NdC)
!q!NaB!NdC ; fcC =
k!Tq ln
ÎÍÍÈ
˚˙˙˘
!NaB!!NdC
n2i
fcC = 0.026 ln ÎÍÈ
˚˙˘
!(1.3x1016)(8.7x1013)
1020 = 0.6 V ;
![Page 348: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/348.jpg)
B,MI
0.02 W
0.8 V
0.6 V
0.8 V
-
+
-
+
-
+
C
B
E
0.02 W
0.6 V-
+I C,M
Neglecting IB,DIC,M + IC,D = 100 A(.02)IC,M -0.6 = (.02) I C,D +0.2IC,D = 30 A ; IC,M = 70 AVCE,M = 0.2 + (70)(.02) = 1.6 VPDarl = VCE,M [IC,M + IC,D]
PDarl = (1.6 V)(100 A) = 160 W
21-11. CEBO = e!AE
WEBO ; CCBO = e!AC
WCBO
WEBO = zero-bias emitter-base depletion layer thicknessWCBO = zero-bias collector-base depletion layer thickness
WEBO = 2!e!fcE!(NaB!+!NdE)
!q!NaB!NdE ; fcE =
k!Tq ln
ÎÍÍÈ
˚˙˙˘
!NaB!NdE
n2i
fcE = 0.026 ln ÎÍÈ
˚˙˘
!(1019)(1016)
1020 = 0.89 V ;
WEBO = (2)(11.7)(8.9x10-14)(0.89)(1019!+!1016)
(1.6x10-19)(1019)(1016) = 0.34 microns
CEBO = (11.7)(8.9x10-14)(0.3)
3.4x10-5 = 9.2 nF
WCBO = 2!e!fcC!(NaB!+!NdC)
!q!NaB!NdC ; fcC =
k!Tq ln
ÎÍÍÈ
˚˙˙˘
!NaB!!NdC
n2i
fcC = 0.026 ln ÎÍÈ
˚˙˘
!(1.3x1016)(8.7x1013)
1020 = 0.6 V ;
![Page 349: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/349.jpg)
WCBO = (2)(11.7)(8.9x10-14)(0.6)(1.3x1016!+!8.7x1013)
(1.6x10-19)(1.3x1016)(8.7x1013) = 2.1 microns
CCBO = (11.7)(8.9x10-14)(3)
2.1x10-4 = 14.8 nF
21-12. Equivalent circuit for turn-on delay time, td,on, calculation.
+-
10 WCCB
CBEVBEVin
+
--8 V
8 V
Vin
t
VBE(t) = 8 - 16 exp ÎÍÈ
˚˙-t
!t ; t = (10 W)(CBE + CCB)
At t = td,on VBE(td,on) = 0.7 V ; td,on = t ln ÎÍÈ
˚˙16
7.3
The space-charge capacitances are nonlinear functions of the voltages across them. Need to find an average value for each of the two capacitors. During the td,on interval, the voltage VCB changes from 108 V to 100 volts and thus will be considered a constant. Hence CCB will be given by
CCB = CCBO
! 1!+!VCBfcC
= 1.5x10-8
1!+!1000.6
= 1.2 nF
The voltage VBE changes from -8 V to 0.7 volts during the same interval. We must find the average value of CBE.
![Page 350: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/350.jpg)
WCBO = (2)(11.7)(8.9x10-14)(0.6)(1.3x1016!+!8.7x1013)
(1.6x10-19)(1.3x1016)(8.7x1013) = 2.1 microns
CCBO = (11.7)(8.9x10-14)(3)
2.1x10-4 = 14.8 nF
21-12. Equivalent circuit for turn-on delay time, td,on, calculation.
+-
10 WCCB
CBEVBEVin
+
--8 V
8 V
Vin
t
VBE(t) = 8 - 16 exp ÎÍÈ
˚˙-t
!t ; t = (10 W)(CBE + CCB)
At t = td,on VBE(td,on) = 0.7 V ; td,on = t ln ÎÍÈ
˚˙16
7.3
The space-charge capacitances are nonlinear functions of the voltages across them. Need to find an average value for each of the two capacitors. During the td,on interval, the voltage VCB changes from 108 V to 100 volts and thus will be considered a constant. Hence CCB will be given by
CCB = CCBO
! 1!+!VCBfcC
= 1.5x10-8
1!+!1000.6
= 1.2 nF
The voltage VBE changes from -8 V to 0.7 volts during the same interval. We must find the average value of CBE.
![Page 351: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/351.jpg)
CBE =
ıÙÙÛ
8
0
!CEBO
! 1!+!VEBfcE
!dVEB
ıÛ
8
0dVEB
= CEBO!fcE
(-8) ÎÍÈ
˚˙˘
1!+!0
fcE!!-!! 1!+!8
fcE
CBE = (9.2x10-9)(0.89)
8 [ 1!+!8
0.89 - 1] = 2.2 nF
td,on = (10) [2.2x10-9 + 1.2x10-9] ln ÎÍÈ
˚˙16
7.3 = (3.4x10-8)(0.78) ≈ 27 nanoseconds
![Page 352: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/352.jpg)
Chapter 22 Problem Solutions
22-1. The capacitance of the gate-source terminals can be modeled as two capacitors connected electrically in series. Cox is the capacitance of the oxide layer and is a constant independent of vGS. Cdepl(vGS) is the capacitance of the depletion layer which increases in thickness as VGS increases as is shown in Fig. 22-6. Any increase in the depletion layer thickness reduces the value of Cdepl(vGS) and hence the atotal gate-source capacitance Cgs. However once vGS becomes equal to or greater than VGS(th), the depletion layer thickness becomes constant because the formation of the inversion layer jillustrated in Fig. 22-6c shields the depletion layer from further increases in vGS(any additional increase in vGS is dropped across the oxide layer). Thus both components of Cgs are constant for vGS > VGS(th).
22-2. a) Idealized MOSFET waveforms shown below. To dimensions the waveforms, we need the numerical values of the various waveform parameters. The voltage and current amplitude parameters are given in the problem statement as: Vd = 300 V, VGS(th) = 4 V, VGS,Io = 7 V, Io = 10 A. To complete the dimensioning, we must calculate the various switching times.
![Page 353: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/353.jpg)
Chapter 22 Problem Solutions
22-1. The capacitance of the gate-source terminals can be modeled as two capacitors connected electrically in series. Cox is the capacitance of the oxide layer and is a constant independent of vGS. Cdepl(vGS) is the capacitance of the depletion layer which increases in thickness as VGS increases as is shown in Fig. 22-6. Any increase in the depletion layer thickness reduces the value of Cdepl(vGS) and hence the atotal gate-source capacitance Cgs. However once vGS becomes equal to or greater than VGS(th), the depletion layer thickness becomes constant because the formation of the inversion layer jillustrated in Fig. 22-6c shields the depletion layer from further increases in vGS(any additional increase in vGS is dropped across the oxide layer). Thus both components of Cgs are constant for vGS > VGS(th).
22-2. a) Idealized MOSFET waveforms shown below. To dimensions the waveforms, we need the numerical values of the various waveform parameters. The voltage and current amplitude parameters are given in the problem statement as: Vd = 300 V, VGS(th) = 4 V, VGS,Io = 7 V, Io = 10 A. To complete the dimensioning, we must calculate the various switching times.
![Page 354: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/354.jpg)
T/2
I oVd
V (t)CE
i (t)C
t d,off t rv t fit d,on t ri t fv
-VGG
0
VGS(th)
VGS,Io
VGG
V (t)GS
td,on estimate - Use equivalent circuit of Fig. 22-12a.
Governing equation is dvGS
dt + vGS
RG(Cgs!+!Cgd) = VGG
RG(Cgs!+!Cgd) ;
Boundary condition vGS(0) = - VGGSolution is vGS(t) = VGG - 2 VGG e-t/t ; t = RG(Cgs + Cgd) ;
At t = td,on , vGS = VGS(th). Solving for td,on yields
![Page 355: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/355.jpg)
td,on = RG(Cgs + Cgd) ln ÎÍÈ
˚˙˘
!2!VGG
!VGG!-!VGS(th)
td,on = (50) (1.15x10-9) ln ÎÍÈ
˚˙!
(2)(15)(15!-!4) = 58 ns
tri estimate - Use equivalent circuit of Fig. 22-12b.
vGS(t) still given by governing equation given above in td,on estimate. Changing time origin to when vGS = VGS(th) yields;
vGS(t) = VGG + [VGG - VGS(th)] e-t/t . The drain current is given by
iD(t) = Cgd d(Vd!-!vGS)
dt + gm[vGS(t) - VGs(th)] ; gm = 10
7!-!4 = 3.3 mhos
At t = tri, iD = Io. Substituting vGS(t) into iD(t) and solving for tri yields
tri = RG(Cgs + Cgd) ln ÎÍÈ
˚˙˘
!(VGG!-!VGS(th)){gm!+ !
CgdRG(Cgs!+!Cgd)}
gm(VGG!-!VGS(th))!-!Io
tri = (50)(1.15x10-9) ln ÎÍÍÈ
˚˙˙˘
!(15!-!4)(3.3!+!
1.5x10-10
(50)(1.15x10-9))
(3.3)(15!-!4)!-!10 = 21 ns
tfv estimate - Use equivalent circuit of Fig. 22-12c.
vGS approximately constant at VGS,Io = Iogm
+ VGS(th) during this interval.
Governing equation is Cgd dvDS
dt = - ÎÍÈ
˚˙˘
!VGG!-!VGS(th)!-!
Iogm
RG with vDS(0) = Vd.
Solution is given by
vDS(t) = Vd - ÎÍÈ
˚˙˘
VGG!-!VGS(th)!-!Iogm
t
RGCgd ; At t = tfv, vDS = 0.
Solving for tfv yields
![Page 356: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/356.jpg)
tfv = RG!Cgd!Vd
VGG!-!VGS(th)!-!Iogm
= (50)(1.5x10-10) 300
(15!-!4!-!3) = 300 ns
td,off estimate - use equivalent circuit of Fig. 22-12d with the input voltagge VGGreversed.
vGS(t) = - VGG + 2 VGG e-t/t ; At t = td,off, vGS = VGS,Io. Solving for td,off
td,off = RG(Cgs + Cgd) ln
ÎÍÈ
˚˙˘
!2!VGG
VGG!+!VGS(th)!+!Iogm
= (50)(1.15x10-9) ln ÎÍÍÈ
˚˙˙˘
!(2)(15)
103.3!+!4!+!15
= 18 ns
trv estimate - Use equivalent circuit of Fig. 22-12c with the input voltage VGGreversed. vGS approximately constant at VGS,Io as in previous of tfv. Governing equation is
Cgd d{vDS!-!VGS,Io}
dt = VGG!+!VGS,Io
RG with vDS(0) = 0. Solution given by
vDS(t) = VGG!+!VGS,Io
RG!Cgd t . At t = trv , vDS = Vd . Solving for trv yields
trv = Vd!RG!Cgd
VGG!+!VGS,Io =
(300)(50)(1.5x10-10)(15!+!7) = 100 ns
tfi estimate - use equivalent circuit of Fig. 22-12b with the input voltage VGGreversed. Governing equation the same as in previous calculation of tri. At t = 0, vGS(0) = VGS,Io. Solution in this caae is given by
vGS(t) = - VGG + [VGS,Io + VGG] e-t/t ; At t = tfi, vGS = VGS(th). Solving for tfi
tfi = RG(Cgs + Cgd) ln ÎÍÈ
˚˙˘
!VGG!+!VGS,IoVGG+!VGS(th)
= (50)(1.15x10-9)ln ÎÍÈ
˚˙!
15!+!715!+!4 = 9 ns
b) Estimate the power dissipated in the MOSFET in the same manner as was done for theBJT in problem 21-3.Waveforms for the MOSFET are the same as for the BJT except forappropriate re-labeling of the currents and voltages.
![Page 357: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/357.jpg)
Eri = (0.5)(300)(10)(2.1x10-8) = 3x10-5 Joules
Efv = (0.5)(300)(10)(3x10-7) = 4.5x10-4 JoulesEon = Io VDS,on [0.5 T - td,on + td,off] ; VDS,on = Io rDS,on = (10)(0.5) = 5 VT >> td,on and td,offEon = (10)(300)(0.5)(5x10-5) = 1.25x10-3 Joules
Erv = (0.5)(300)(10)(10-7) = 1.5x10-4 Joules
Efi = (0.5)(300)(10)(9x10-9) = 1.5x10-5 Joules
Pc = (1.95x10-3)(2x104) = 39 watts
22-3. Use test conditons to estimate Cgd. Then estimate the switching times in the circuit with the 150 ohm load. Test circuit waveforms are shown below.
![Page 358: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/358.jpg)
Eri = (0.5)(300)(10)(2.1x10-8) = 3x10-5 Joules
Efv = (0.5)(300)(10)(3x10-7) = 4.5x10-4 JoulesEon = Io VDS,on [0.5 T - td,on + td,off] ; VDS,on = Io rDS,on = (10)(0.5) = 5 VT >> td,on and td,offEon = (10)(300)(0.5)(5x10-5) = 1.25x10-3 Joules
Erv = (0.5)(300)(10)(10-7) = 1.5x10-4 Joules
Efi = (0.5)(300)(10)(9x10-9) = 1.5x10-5 Joules
Pc = (1.95x10-3)(2x104) = 39 watts
22-3. Use test conditons to estimate Cgd. Then estimate the switching times in the circuit with the 150 ohm load. Test circuit waveforms are shown below.
![Page 359: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/359.jpg)
V (t)G
VGG
V (t)GS
VGS(th)
VGS,Io
t
t d,ont = t r i fv t = tf i rv
td,offt
VGG
Vd
V (t)DS
I o
i (t)D
t
Equivalent circuit during voltage and current rise and fall intervals:
+- Cgs
Cgd
V (t)G
R D
Vd
g (V - V )m GS GS(th)
RG
Governing equation using Miller capacitance approximation:
![Page 360: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/360.jpg)
dvGSdt +
vGSt =
!VG(t)t ; t = RG [Cgs + Cgd{1 + gmRD}] ;
During tri = tfv interval, VG(t) = VGG. Solution is
vGS(t) = VGG + {VGS(th) - VGG} e-t/t ; At t = tri, VGS = VGS(th) + Vd
gmRD ;
Solving for tri = tfv yields
tri = tfv = t ln
ÎÍÈ
˚˙˘
!VGG!-!VGS(th)
VGG!-!VGS(th)!-!!Vd
!gm!RD
During trv = tfi, vG(t) = 0 and solution is vGS(t) = VGS,Io e-t/t .At t = trv, vGS(t) = VGS(th). Solving for trv yields
trv = t ln ÎÍÈ
˚˙˘
!VGS(th)!+!
Vd!gm!RD
VGS(th) . Invert equation for tri to find Cgd. Result is
Cgd =
ÓÔÌÔÏ
Ô˝Ô
tri
!ln!
ÎÍÈ
˚˙˘
!VGG!-!VGS(th)
VGG!-!VGS(th)!-!!!Vd
!gm!RD
!!-!!RG!Cgs ÓÌÏ
˛˝¸1
RG(1!+!gmRD)
Cgd = ÓÔÌÔÏ
Ô˝Ô3x10-8
!ln!ÎÍÈ
˚˙!
15!-!415!-!4!-!1
!!-!!5x10-9 ÓÌÏ
˛˝¸1
5(1!+!25) = 2.3x10-9 F = 2.3 nF
Solving for switching times in circuit with RD = 150 ohms.
t = [10-9 + 2.3x10-9 {1 + 150}][100] = 35 ms
tri = 3.5x10-5 ln ÎÍÈ
˚˙!
15!-!415!-!4!-!2 = 7 ms ; trv = 3.5x10-5 ln ÎÍ
È˚˙!
4!+!24 = 14 ms
22-4. Waveforms for vDS(t) and iD(t) shown in previous problem. Power dissipation in MOSFET given by
<PMOSFET> = [Eon + Esw] fs ; fs = 1T ; Eon = [ID]2 rDS,on(Tj)
T2 ;
![Page 361: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/361.jpg)
dvGSdt +
vGSt =
!VG(t)t ; t = RG [Cgs + Cgd{1 + gmRD}] ;
During tri = tfv interval, VG(t) = VGG. Solution is
vGS(t) = VGG + {VGS(th) - VGG} e-t/t ; At t = tri, VGS = VGS(th) + Vd
gmRD ;
Solving for tri = tfv yields
tri = tfv = t ln
ÎÍÈ
˚˙˘
!VGG!-!VGS(th)
VGG!-!VGS(th)!-!!Vd
!gm!RD
During trv = tfi, vG(t) = 0 and solution is vGS(t) = VGS,Io e-t/t .At t = trv, vGS(t) = VGS(th). Solving for trv yields
trv = t ln ÎÍÈ
˚˙˘
!VGS(th)!+!
Vd!gm!RD
VGS(th) . Invert equation for tri to find Cgd. Result is
Cgd =
ÓÔÌÔÏ
Ô˝Ô
tri
!ln!
ÎÍÈ
˚˙˘
!VGG!-!VGS(th)
VGG!-!VGS(th)!-!!!Vd
!gm!RD
!!-!!RG!Cgs ÓÌÏ
˛˝¸1
RG(1!+!gmRD)
Cgd = ÓÔÌÔÏ
Ô˝Ô3x10-8
!ln!ÎÍÈ
˚˙!
15!-!415!-!4!-!1
!!-!!5x10-9 ÓÌÏ
˛˝¸1
5(1!+!25) = 2.3x10-9 F = 2.3 nF
Solving for switching times in circuit with RD = 150 ohms.
t = [10-9 + 2.3x10-9 {1 + 150}][100] = 35 ms
tri = 3.5x10-5 ln ÎÍÈ
˚˙!
15!-!415!-!4!-!2 = 7 ms ; trv = 3.5x10-5 ln ÎÍ
È˚˙!
4!+!24 = 14 ms
22-4. Waveforms for vDS(t) and iD(t) shown in previous problem. Power dissipation in MOSFET given by
<PMOSFET> = [Eon + Esw] fs ; fs = 1T ; Eon = [ID]2 rDS,on(Tj)
T2 ;
![Page 362: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/362.jpg)
ID = VdRD
= 300150 = 2 A ; rDS,on(Tj) = 2 ÎÍ
È˚˙1!+!
Tj!-!25150 = 2 ÎÍ
È˚˙0.833!+!
Tj150
Eon = (4)(2) ÎÍÈ
˚˙0.833!+!
Tj150
12fs
= {3.32 + 0.027 Tj} 1fs
Esw = 1T ıÙ
Û
0
tri
Vd!ID(1!-!ttri
)(ttri
)dt + 1T ıÙ
Û
0
tfi
Vd!ID(1!-!ttfi
)(ttfi
)dt = Vd!ID
6 [tri + tfi]
Esw = (300)(2)
6 [7x10-6 + 14x10-6] = 2.1x10-3 joules
<PMOSFET> = {3.32 + 0.027 Tj} 1fs
fs + 2.1x10-3 fs
<PMOSFET> = {3.32 + 0.027 Tj} +[ 2.1x10-3][104] = 24.3 + 0.027 Tj
B B B B B
0
5
10
15
20
25
30
0 10 20 30 40 50 60 70 80 90 100
PMOSFET
Watts
Temperature [ °K]
22-5. Von = on-state voltage of three MOSFETs in parallel = Io reff
![Page 363: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/363.jpg)
ID = VdRD
= 300150 = 2 A ; rDS,on(Tj) = 2 ÎÍ
È˚˙1!+!
Tj!-!25150 = 2 ÎÍ
È˚˙0.833!+!
Tj150
Eon = (4)(2) ÎÍÈ
˚˙0.833!+!
Tj150
12fs
= {3.32 + 0.027 Tj} 1fs
Esw = 1T ıÙ
Û
0
tri
Vd!ID(1!-!ttri
)(ttri
)dt + 1T ıÙ
Û
0
tfi
Vd!ID(1!-!ttfi
)(ttfi
)dt = Vd!ID
6 [tri + tfi]
Esw = (300)(2)
6 [7x10-6 + 14x10-6] = 2.1x10-3 joules
<PMOSFET> = {3.32 + 0.027 Tj} 1fs
fs + 2.1x10-3 fs
<PMOSFET> = {3.32 + 0.027 Tj} +[ 2.1x10-3][104] = 24.3 + 0.027 Tj
B B B B B
0
5
10
15
20
25
30
0 10 20 30 40 50 60 70 80 90 100
PMOSFET
Watts
Temperature [ °K]
22-5. Von = on-state voltage of three MOSFETs in parallel = Io reff
![Page 364: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/364.jpg)
reff = r1!r2!r3
r1!r2!+!r2!r3!+!r3!r1 ; r1 etc. = on-state resistance of MOSFET #1 etc.
r1(Tj) = r1(25 °C) ÎÍÈ
˚˙1!+!0.8!!
Tj!-!25100 ; r1(105 °C) = (1.64) r1(25 °C) etc.
r1(105 °C) = 2.95 W ; r2(105 °C) = 3.28 W ; r3(105 °C) = 3.61 W
reff(105 °C) = (2.95)(3.28)(3.61)
[(2.95)(3.28)!+!(3.28)(3.61)!+!(3.61)(2.95)] = 1.09 ohms
For the ith MOSFET, Pi = Von
2
2!ri =
Io2!reff
2
2!ri ; Assume a 50% duty cycle and ignore
switching losses.
P1 = (5)2(1.09)2
(2)(2.95) = 5 W ; P2 = (5)2(1.09)2
(2)(3.28) = 4.5 W ; P3 = (5)2(1.09)2
(2)(3.61) = 4.1 W
22-6. Hybrid switch would combine the low on-state losses of the BJT and the faster switching of the MOSFET. In order to obtain these advantages, The MOSFET would be turned on before the BJT and turned off after the BJT. The waveforms shown below indicate the relative timing. The switch blocks Vd volts in the off-state and conducts Io amps in the on-state.
![Page 365: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/365.jpg)
reff = r1!r2!r3
r1!r2!+!r2!r3!+!r3!r1 ; r1 etc. = on-state resistance of MOSFET #1 etc.
r1(Tj) = r1(25 °C) ÎÍÈ
˚˙1!+!0.8!!
Tj!-!25100 ; r1(105 °C) = (1.64) r1(25 °C) etc.
r1(105 °C) = 2.95 W ; r2(105 °C) = 3.28 W ; r3(105 °C) = 3.61 W
reff(105 °C) = (2.95)(3.28)(3.61)
[(2.95)(3.28)!+!(3.28)(3.61)!+!(3.61)(2.95)] = 1.09 ohms
For the ith MOSFET, Pi = Von
2
2!ri =
Io2!reff
2
2!ri ; Assume a 50% duty cycle and ignore
switching losses.
P1 = (5)2(1.09)2
(2)(2.95) = 5 W ; P2 = (5)2(1.09)2
(2)(3.28) = 4.5 W ; P3 = (5)2(1.09)2
(2)(3.61) = 4.1 W
22-6. Hybrid switch would combine the low on-state losses of the BJT and the faster switching of the MOSFET. In order to obtain these advantages, The MOSFET would be turned on before the BJT and turned off after the BJT. The waveforms shown below indicate the relative timing. The switch blocks Vd volts in the off-state and conducts Io amps in the on-state.
![Page 366: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/366.jpg)
v = vDS CE
i D
i C
vGS
vBE
VDS,on VCE,on
I or I ; r < 1o
(1 - r)I o
22-7. BVDSS ≈ 1.3x1017
Ndrift = 750 volts ; Ndrift = 1.7x1014 cm-3
Wdrift ≈ (10-5)(750) = 75 microns ;Wd,body = protrusion of drain depletion layer into body region
≈ Wdrift!Ndrift
Nbody =
(75)(1.7x1014)5x1016 ≈ 0.3 microns
Even though body-source junction is shorted, there is a depletion layer associated with it which is contained entirely on the body side of the junction. This must be included in the estimate of the required length of the body region.
Ws,body ≈ 2!e!fc
q!Na,body ; fc =
k!Tq ln
ÎÍÍÈ
˚˙˙˘
!Na!Nd
ni2 ;
![Page 367: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/367.jpg)
v = vDS CE
i D
i C
vGS
vBE
VDS,on VCE,on
I or I ; r < 1o
(1 - r)I o
22-7. BVDSS ≈ 1.3x1017
Ndrift = 750 volts ; Ndrift = 1.7x1014 cm-3
Wdrift ≈ (10-5)(750) = 75 microns ;Wd,body = protrusion of drain depletion layer into body region
≈ Wdrift!Ndrift
Nbody =
(75)(1.7x1014)5x1016 ≈ 0.3 microns
Even though body-source junction is shorted, there is a depletion layer associated with it which is contained entirely on the body side of the junction. This must be included in the estimate of the required length of the body region.
Ws,body ≈ 2!e!fc
q!Na,body ; fc =
k!Tq ln
ÎÍÍÈ
˚˙˙˘
!Na!Nd
ni2 ;
![Page 368: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/368.jpg)
fc = 0.026 ln ÎÍÈ
˚˙˘
!(1019)(5x1016)
(1020) = 0.94
Ws,body ≈ (2)(11.7)(8.9x10-14)(0.94)
(1.6x10-19)(5x1016) ≈ 0.16 microns
In order to avoid reach-through , Wbody > Wd,body + Ws,body = 0.3 + 0.16 = 0.46 microns
22-8. Displacement current = Cgd dvGD
dt ≈ Cgd dvDS
dt ; vDS ≈ vGD>> vGS
BJT will turn on if Rbody Cgd dvDS
dt = 0.7 V
dvDS
dt > 0.7
Rbody!Cgd will turn on the BJT.
22-9. VGS,max = (0.67) EBD tox = (0.67) (5x106) (5x10-6) = 16.7 volts
22-10. a) iD = mn!Cox!N!Wcell!(vGS!-!VGS(th))
2!L
Cox = e
tox =
(11.7)(8.9x10-14)10-5 = 1.04x10-7 F/cm2
N = 2!iD!L
mn!Cox!Wcell!(vGS!-!VGS(th))
N = !(2)(100)(10-4)
(1500)(1.04x10-7)(2x10-3)(15!-!4) ≈ 5,800 cells
b) Icell = 1005800 = 17 milliamps
22-11. Von = 4 volts = Ion Ron = (10 A) Ron ; Ron = 0.4 ohms
Ron = Wdrift
q!mn!Nd!A : Wdrift = 10-5 BVDSS = (10-5)(800) = 80 microns
![Page 369: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/369.jpg)
fc = 0.026 ln ÎÍÈ
˚˙˘
!(1019)(5x1016)
(1020) = 0.94
Ws,body ≈ (2)(11.7)(8.9x10-14)(0.94)
(1.6x10-19)(5x1016) ≈ 0.16 microns
In order to avoid reach-through , Wbody > Wd,body + Ws,body = 0.3 + 0.16 = 0.46 microns
22-8. Displacement current = Cgd dvGD
dt ≈ Cgd dvDS
dt ; vDS ≈ vGD>> vGS
BJT will turn on if Rbody Cgd dvDS
dt = 0.7 V
dvDS
dt > 0.7
Rbody!Cgd will turn on the BJT.
22-9. VGS,max = (0.67) EBD tox = (0.67) (5x106) (5x10-6) = 16.7 volts
22-10. a) iD = mn!Cox!N!Wcell!(vGS!-!VGS(th))
2!L
Cox = e
tox =
(11.7)(8.9x10-14)10-5 = 1.04x10-7 F/cm2
N = 2!iD!L
mn!Cox!Wcell!(vGS!-!VGS(th))
N = !(2)(100)(10-4)
(1500)(1.04x10-7)(2x10-3)(15!-!4) ≈ 5,800 cells
b) Icell = 1005800 = 17 milliamps
22-11. Von = 4 volts = Ion Ron = (10 A) Ron ; Ron = 0.4 ohms
Ron = Wdrift
q!mn!Nd!A : Wdrift = 10-5 BVDSS = (10-5)(800) = 80 microns
![Page 370: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/370.jpg)
fc = 0.026 ln ÎÍÈ
˚˙˘
!(1019)(5x1016)
(1020) = 0.94
Ws,body ≈ (2)(11.7)(8.9x10-14)(0.94)
(1.6x10-19)(5x1016) ≈ 0.16 microns
In order to avoid reach-through , Wbody > Wd,body + Ws,body = 0.3 + 0.16 = 0.46 microns
22-8. Displacement current = Cgd dvGD
dt ≈ Cgd dvDS
dt ; vDS ≈ vGD>> vGS
BJT will turn on if Rbody Cgd dvDS
dt = 0.7 V
dvDS
dt > 0.7
Rbody!Cgd will turn on the BJT.
22-9. VGS,max = (0.67) EBD tox = (0.67) (5x106) (5x10-6) = 16.7 volts
22-10. a) iD = mn!Cox!N!Wcell!(vGS!-!VGS(th))
2!L
Cox = e
tox =
(11.7)(8.9x10-14)10-5 = 1.04x10-7 F/cm2
N = 2!iD!L
mn!Cox!Wcell!(vGS!-!VGS(th))
N = !(2)(100)(10-4)
(1500)(1.04x10-7)(2x10-3)(15!-!4) ≈ 5,800 cells
b) Icell = 1005800 = 17 milliamps
22-11. Von = 4 volts = Ion Ron = (10 A) Ron ; Ron = 0.4 ohms
Ron = Wdrift
q!mn!Nd!A : Wdrift = 10-5 BVDSS = (10-5)(800) = 80 microns
![Page 371: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/371.jpg)
fc = 0.026 ln ÎÍÈ
˚˙˘
!(1019)(5x1016)
(1020) = 0.94
Ws,body ≈ (2)(11.7)(8.9x10-14)(0.94)
(1.6x10-19)(5x1016) ≈ 0.16 microns
In order to avoid reach-through , Wbody > Wd,body + Ws,body = 0.3 + 0.16 = 0.46 microns
22-8. Displacement current = Cgd dvGD
dt ≈ Cgd dvDS
dt ; vDS ≈ vGD>> vGS
BJT will turn on if Rbody Cgd dvDS
dt = 0.7 V
dvDS
dt > 0.7
Rbody!Cgd will turn on the BJT.
22-9. VGS,max = (0.67) EBD tox = (0.67) (5x106) (5x10-6) = 16.7 volts
22-10. a) iD = mn!Cox!N!Wcell!(vGS!-!VGS(th))
2!L
Cox = e
tox =
(11.7)(8.9x10-14)10-5 = 1.04x10-7 F/cm2
N = 2!iD!L
mn!Cox!Wcell!(vGS!-!VGS(th))
N = !(2)(100)(10-4)
(1500)(1.04x10-7)(2x10-3)(15!-!4) ≈ 5,800 cells
b) Icell = 1005800 = 17 milliamps
22-11. Von = 4 volts = Ion Ron = (10 A) Ron ; Ron = 0.4 ohms
Ron = Wdrift
q!mn!Nd!A : Wdrift = 10-5 BVDSS = (10-5)(800) = 80 microns
![Page 372: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/372.jpg)
fc = 0.026 ln ÎÍÈ
˚˙˘
!(1019)(5x1016)
(1020) = 0.94
Ws,body ≈ (2)(11.7)(8.9x10-14)(0.94)
(1.6x10-19)(5x1016) ≈ 0.16 microns
In order to avoid reach-through , Wbody > Wd,body + Ws,body = 0.3 + 0.16 = 0.46 microns
22-8. Displacement current = Cgd dvGD
dt ≈ Cgd dvDS
dt ; vDS ≈ vGD>> vGS
BJT will turn on if Rbody Cgd dvDS
dt = 0.7 V
dvDS
dt > 0.7
Rbody!Cgd will turn on the BJT.
22-9. VGS,max = (0.67) EBD tox = (0.67) (5x106) (5x10-6) = 16.7 volts
22-10. a) iD = mn!Cox!N!Wcell!(vGS!-!VGS(th))
2!L
Cox = e
tox =
(11.7)(8.9x10-14)10-5 = 1.04x10-7 F/cm2
N = 2!iD!L
mn!Cox!Wcell!(vGS!-!VGS(th))
N = !(2)(100)(10-4)
(1500)(1.04x10-7)(2x10-3)(15!-!4) ≈ 5,800 cells
b) Icell = 1005800 = 17 milliamps
22-11. Von = 4 volts = Ion Ron = (10 A) Ron ; Ron = 0.4 ohms
Ron = Wdrift
q!mn!Nd!A : Wdrift = 10-5 BVDSS = (10-5)(800) = 80 microns
![Page 373: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/373.jpg)
Nd = 1.3x1017BVDSS
= 1.3x1017
800 ≈ 1.6x1014 cm-3
A = 8x10-3
(1.6x10-19)(1500)(1.6x1014)(0.4) ≈ 0.5 cm2
10!A0.5!cm2 = 20
Acm2 << the allowable maximum of 200
Acm2 , so estimate is alright.
22-12. Cgs ≈ Cox N Wcell L = (1.04x10-7)(5.8x103)(2x10-3)(10-4) = 121 pF
22-13. Two overstress possibilities, overvoltage across drain-source terminals because of stray inductance and excessive power dissipation. Check for overvoltage first.
VDS(turn-off) = Vd + L didt = 100 + (10-7) ÎÍ
È˚˙!
1005x10-8 = 300 V > BVDSS = 150 V
Check for excessive power dissipation.
Pallowed = Tj,max!-!Ta
Rq,j-a =
150!-!501 = 100 watts ; Pdissipated = [Eon + Esw] fs
Eonfs = Io
2!rDS(on)2 =
(100)2(0.01)2 = 50 watts
Esw = Vd!Io
2 [tri + tfi + trv +tfv] = (100)(100)
2 [(2)(5x10-8) + (2)(2x10-7)]
Esw = 2.5x10-3 joules ; Eswfs = (2.5x10-3)(3x104) = 75 watts
Pdissipated = 50 + 75 = 125 watts > Pallowed = 100 watts
MOSFET overstressed by both overvoltages and excessive power dissipation.
22-14. Gate current which charges/discharges Cgs and Cgd during turn-on and turn-off is approximately constant during the four time intervals. However during the current rise and fall times the voltages VGS and VGD change by only a few tens of volts. However during the voltage rise and fall times VGD changes by approximately Vd which is much larger than a few tens of volts. Thus we have:
![Page 374: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/374.jpg)
Nd = 1.3x1017BVDSS
= 1.3x1017
800 ≈ 1.6x1014 cm-3
A = 8x10-3
(1.6x10-19)(1500)(1.6x1014)(0.4) ≈ 0.5 cm2
10!A0.5!cm2 = 20
Acm2 << the allowable maximum of 200
Acm2 , so estimate is alright.
22-12. Cgs ≈ Cox N Wcell L = (1.04x10-7)(5.8x103)(2x10-3)(10-4) = 121 pF
22-13. Two overstress possibilities, overvoltage across drain-source terminals because of stray inductance and excessive power dissipation. Check for overvoltage first.
VDS(turn-off) = Vd + L didt = 100 + (10-7) ÎÍ
È˚˙!
1005x10-8 = 300 V > BVDSS = 150 V
Check for excessive power dissipation.
Pallowed = Tj,max!-!Ta
Rq,j-a =
150!-!501 = 100 watts ; Pdissipated = [Eon + Esw] fs
Eonfs = Io
2!rDS(on)2 =
(100)2(0.01)2 = 50 watts
Esw = Vd!Io
2 [tri + tfi + trv +tfv] = (100)(100)
2 [(2)(5x10-8) + (2)(2x10-7)]
Esw = 2.5x10-3 joules ; Eswfs = (2.5x10-3)(3x104) = 75 watts
Pdissipated = 50 + 75 = 125 watts > Pallowed = 100 watts
MOSFET overstressed by both overvoltages and excessive power dissipation.
22-14. Gate current which charges/discharges Cgs and Cgd during turn-on and turn-off is approximately constant during the four time intervals. However during the current rise and fall times the voltages VGS and VGD change by only a few tens of volts. However during the voltage rise and fall times VGD changes by approximately Vd which is much larger than a few tens of volts. Thus we have:
![Page 375: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/375.jpg)
Nd = 1.3x1017BVDSS
= 1.3x1017
800 ≈ 1.6x1014 cm-3
A = 8x10-3
(1.6x10-19)(1500)(1.6x1014)(0.4) ≈ 0.5 cm2
10!A0.5!cm2 = 20
Acm2 << the allowable maximum of 200
Acm2 , so estimate is alright.
22-12. Cgs ≈ Cox N Wcell L = (1.04x10-7)(5.8x103)(2x10-3)(10-4) = 121 pF
22-13. Two overstress possibilities, overvoltage across drain-source terminals because of stray inductance and excessive power dissipation. Check for overvoltage first.
VDS(turn-off) = Vd + L didt = 100 + (10-7) ÎÍ
È˚˙!
1005x10-8 = 300 V > BVDSS = 150 V
Check for excessive power dissipation.
Pallowed = Tj,max!-!Ta
Rq,j-a =
150!-!501 = 100 watts ; Pdissipated = [Eon + Esw] fs
Eonfs = Io
2!rDS(on)2 =
(100)2(0.01)2 = 50 watts
Esw = Vd!Io
2 [tri + tfi + trv +tfv] = (100)(100)
2 [(2)(5x10-8) + (2)(2x10-7)]
Esw = 2.5x10-3 joules ; Eswfs = (2.5x10-3)(3x104) = 75 watts
Pdissipated = 50 + 75 = 125 watts > Pallowed = 100 watts
MOSFET overstressed by both overvoltages and excessive power dissipation.
22-14. Gate current which charges/discharges Cgs and Cgd during turn-on and turn-off is approximately constant during the four time intervals. However during the current rise and fall times the voltages VGS and VGD change by only a few tens of volts. However during the voltage rise and fall times VGD changes by approximately Vd which is much larger than a few tens of volts. Thus we have:
![Page 376: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/376.jpg)
Nd = 1.3x1017BVDSS
= 1.3x1017
800 ≈ 1.6x1014 cm-3
A = 8x10-3
(1.6x10-19)(1500)(1.6x1014)(0.4) ≈ 0.5 cm2
10!A0.5!cm2 = 20
Acm2 << the allowable maximum of 200
Acm2 , so estimate is alright.
22-12. Cgs ≈ Cox N Wcell L = (1.04x10-7)(5.8x103)(2x10-3)(10-4) = 121 pF
22-13. Two overstress possibilities, overvoltage across drain-source terminals because of stray inductance and excessive power dissipation. Check for overvoltage first.
VDS(turn-off) = Vd + L didt = 100 + (10-7) ÎÍ
È˚˙!
1005x10-8 = 300 V > BVDSS = 150 V
Check for excessive power dissipation.
Pallowed = Tj,max!-!Ta
Rq,j-a =
150!-!501 = 100 watts ; Pdissipated = [Eon + Esw] fs
Eonfs = Io
2!rDS(on)2 =
(100)2(0.01)2 = 50 watts
Esw = Vd!Io
2 [tri + tfi + trv +tfv] = (100)(100)
2 [(2)(5x10-8) + (2)(2x10-7)]
Esw = 2.5x10-3 joules ; Eswfs = (2.5x10-3)(3x104) = 75 watts
Pdissipated = 50 + 75 = 125 watts > Pallowed = 100 watts
MOSFET overstressed by both overvoltages and excessive power dissipation.
22-14. Gate current which charges/discharges Cgs and Cgd during turn-on and turn-off is approximately constant during the four time intervals. However during the current rise and fall times the voltages VGS and VGD change by only a few tens of volts. However during the voltage rise and fall times VGD changes by approximately Vd which is much larger than a few tens of volts. Thus we have:
![Page 377: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/377.jpg)
Current rise/fall times proportional to [Cgs + Cgd] VGGIG
Voltage rise/fall times proportional to Cgd VdIG
Cgs roughly the same size as Cgd and Vd >> VGG
Hence voltage switching times much greater than current switching times.
![Page 378: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/378.jpg)
Chapter 23 Problem Solutions
23-1. vs(t) = 2 Vs sin(wt) ; iL(t) = vs(t)RL
; a < wt < p
<PSCR> = 12p ıÛ
a
p
[(1)iL(wt)!+!{iL(wt)}2Ron]!d(wt)
<PSCR> = Vs
2!pRL {1 + cos(a)} +
1π Î
ÍÈ
˚˙˘
!VsRL
2Ron [π - a +
sin(2a)2 ]
23-2. 120 °F = 49 °C ; Tj,max = 125 °C ; <PSCR>|max = Tj,max!-!Ta,max
Rqja
<PSCR>|max = 125!-!49
0.1 = 760 Watts
Check <PSCR> at a = 0
<PSCR> = 2202π!(1) [1 + cos(0)] +
1π ÎÍ
È˚˙!
2201
2(2x10-3)[π - 0 +
sin(0)2 ] = 107 watts
<PSCR> = 107 watts less than the allowable 760 watts. Hence trigger angle of zero where maximum load power is delivered is permissible.
Average load power <PL> = 12π ıÛ
a
π{iL(wt)}2RL!d(wt) ; iL(wt) = 2 (220) sin(wt)
<PL> = 12π ıÛ
a
π{ 2!(220)}2sin2(wt)!d(wt) =
(220)26.28 [3.14 - a +
sin(2a)2 ]
For a = 0 <PL> = 24.2 kW
23-3. PSCR(t) = instantaneous power dissipated in the SCR during turn-on.
PSCR(t) = vAK(t) iA(t) = VAK {1 - ttf
} dIdt t during tf
P(t) = power density = PSCR(t)
A(t) = watts per cm2 ; A(t) = conducting area of SCR
![Page 379: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/379.jpg)
Chapter 23 Problem Solutions
23-1. vs(t) = 2 Vs sin(wt) ; iL(t) = vs(t)RL
; a < wt < p
<PSCR> = 12p ıÛ
a
p
[(1)iL(wt)!+!{iL(wt)}2Ron]!d(wt)
<PSCR> = Vs
2!pRL {1 + cos(a)} +
1π Î
ÍÈ
˚˙˘
!VsRL
2Ron [π - a +
sin(2a)2 ]
23-2. 120 °F = 49 °C ; Tj,max = 125 °C ; <PSCR>|max = Tj,max!-!Ta,max
Rqja
<PSCR>|max = 125!-!49
0.1 = 760 Watts
Check <PSCR> at a = 0
<PSCR> = 2202π!(1) [1 + cos(0)] +
1π ÎÍ
È˚˙!
2201
2(2x10-3)[π - 0 +
sin(0)2 ] = 107 watts
<PSCR> = 107 watts less than the allowable 760 watts. Hence trigger angle of zero where maximum load power is delivered is permissible.
Average load power <PL> = 12π ıÛ
a
π{iL(wt)}2RL!d(wt) ; iL(wt) = 2 (220) sin(wt)
<PL> = 12π ıÛ
a
π{ 2!(220)}2sin2(wt)!d(wt) =
(220)26.28 [3.14 - a +
sin(2a)2 ]
For a = 0 <PL> = 24.2 kW
23-3. PSCR(t) = instantaneous power dissipated in the SCR during turn-on.
PSCR(t) = vAK(t) iA(t) = VAK {1 - ttf
} dIdt t during tf
P(t) = power density = PSCR(t)
A(t) = watts per cm2 ; A(t) = conducting area of SCR
![Page 380: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/380.jpg)
Chapter 23 Problem Solutions
23-1. vs(t) = 2 Vs sin(wt) ; iL(t) = vs(t)RL
; a < wt < p
<PSCR> = 12p ıÛ
a
p
[(1)iL(wt)!+!{iL(wt)}2Ron]!d(wt)
<PSCR> = Vs
2!pRL {1 + cos(a)} +
1π Î
ÍÈ
˚˙˘
!VsRL
2Ron [π - a +
sin(2a)2 ]
23-2. 120 °F = 49 °C ; Tj,max = 125 °C ; <PSCR>|max = Tj,max!-!Ta,max
Rqja
<PSCR>|max = 125!-!49
0.1 = 760 Watts
Check <PSCR> at a = 0
<PSCR> = 2202π!(1) [1 + cos(0)] +
1π ÎÍ
È˚˙!
2201
2(2x10-3)[π - 0 +
sin(0)2 ] = 107 watts
<PSCR> = 107 watts less than the allowable 760 watts. Hence trigger angle of zero where maximum load power is delivered is permissible.
Average load power <PL> = 12π ıÛ
a
π{iL(wt)}2RL!d(wt) ; iL(wt) = 2 (220) sin(wt)
<PL> = 12π ıÛ
a
π{ 2!(220)}2sin2(wt)!d(wt) =
(220)26.28 [3.14 - a +
sin(2a)2 ]
For a = 0 <PL> = 24.2 kW
23-3. PSCR(t) = instantaneous power dissipated in the SCR during turn-on.
PSCR(t) = vAK(t) iA(t) = VAK {1 - ttf
} dIdt t during tf
P(t) = power density = PSCR(t)
A(t) = watts per cm2 ; A(t) = conducting area of SCR
![Page 381: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/381.jpg)
A(t) = π [ro + us t]2 - π ro2 = π [2 ro us t + (us t)2]
dTj = 1Cv
ıÛ0
tfP(t)!dt =
1Cv
ıÙÛ
0
tf
VAK!{1!-!!ttf
!}!dIdt !!t!
π![2!ro!us!t!+!(us!t)2]!dt
dTj = 1Cv
VAK
2πrous dIdt
ıÙÙÛ
0
tf
!ÎÍÍÈ
˚˙˙˘
!1!-!!
ttf
1!+!ust2ro
!dt ; Let a = a' = 1, b = -1tf
, and b' = us2ro
Integral becomes ıÙÛ
0
tf
!ÎÍÈ
˚˙!
a!+!bta'!+!b't !dt ; Using integral tables
ıÙÛ
0
tf
!ÎÍÈ
˚˙!
a!+!bta'!+!b't !dt =
btfb' +
[ab'!-!ab][b']2
ln[a' + b' tf] ;
b = - 1
2x10-5 = - 5x104 sec-1 ; b' = 104
(2)(0.5) = 104 sec-1
Evaluating the integral yields
ıÙÛ
0
tf
!ÎÍÈ
˚˙!
a!+!bta'!+!b't !dt =
-1104 + Î
ÍÈ
˚˙˘
10-4!+!!5x104
108 ln[1 + (104)(2x10-5)] = 9.4x10-6 sec
With Cv = 1.75 Joule/(°C-cm3), the expression for dTj becomes
dTj = (103)(!9.4x10-6)
(2π)(1.75)(104)(0.5) dIdt = 125 - 25 = 100 °C ; Solving for
dIdt yields
dIdt =
1001.7x10-7 = 590 A/ms
![Page 382: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/382.jpg)
23-4. Advantages - Shorter N- region length Wd means shorter carrier lifetimes can be accomodated and thus faster switching times.
Disadvantages - Junction J1 is now a P+ - N+ junction with a low breakdown voltage. Since J3 already has a low breakdown voltage, the modified thyristor has no significant reverse blocking capability.
23-5. ton = (4!-!0.5)!cm
us =
3.5104 = 350 microseconds
23-6. Lateral voltage drops caused by base currents cause current density nonuniformities. At large currents, these nonuniformities become severe and the increasing possibility of second breakdown limit the total current that the BJT can safely conduct.
In the thyristor, no significant gate current is needed to keep the thyristor on and there is consequently no lateral current flow and thus lateral voltage drop. The current density is uniform across the entire cross-sectional area of the thyristor and there is much less likelyhood of second breakdown.
23-7. a) Breakover in a thyristor is not due to impact ionization. However in a well-designed thyristor, the value of the breakover voltage is an appreciable fraction of the actual avalanche breakdown voltage. Thus an estimate of the n1 thickness and doping level based avalanche breakdown would be a reasonable first attempt.
Nd = 1.3x1017
2x103 = 6.5x1013 cm-3 ; Wd ≈ (10-5)(2x103) = 200 microns
b) t = qWd
2
kT(mn!+!mp) = (1.6x10-19)(2x10-2)2
(1.4x10-23)(300)(900) = 17 microseconds
Used (mn + mp) = 900 cm2/V-sec which is value appropriate to large excess carrier
densities (approaching 1017 cm-3)
c) Von = I Rdrift ; Ignore I2/3 contribution as it is usually small compared to the linear term.
Rdrift ≈ Wd
q!(mn!+!mp)!nb!A ; 2
2000 = 10-3 = 0.02
(1.6x10-19)(900)(1017)!A ;
Solving for A gives A = 0.02
(1.6x10-19)(900)(1017)(10-3) = 1.4 cm2
![Page 383: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/383.jpg)
23-4. Advantages - Shorter N- region length Wd means shorter carrier lifetimes can be accomodated and thus faster switching times.
Disadvantages - Junction J1 is now a P+ - N+ junction with a low breakdown voltage. Since J3 already has a low breakdown voltage, the modified thyristor has no significant reverse blocking capability.
23-5. ton = (4!-!0.5)!cm
us =
3.5104 = 350 microseconds
23-6. Lateral voltage drops caused by base currents cause current density nonuniformities. At large currents, these nonuniformities become severe and the increasing possibility of second breakdown limit the total current that the BJT can safely conduct.
In the thyristor, no significant gate current is needed to keep the thyristor on and there is consequently no lateral current flow and thus lateral voltage drop. The current density is uniform across the entire cross-sectional area of the thyristor and there is much less likelyhood of second breakdown.
23-7. a) Breakover in a thyristor is not due to impact ionization. However in a well-designed thyristor, the value of the breakover voltage is an appreciable fraction of the actual avalanche breakdown voltage. Thus an estimate of the n1 thickness and doping level based avalanche breakdown would be a reasonable first attempt.
Nd = 1.3x1017
2x103 = 6.5x1013 cm-3 ; Wd ≈ (10-5)(2x103) = 200 microns
b) t = qWd
2
kT(mn!+!mp) = (1.6x10-19)(2x10-2)2
(1.4x10-23)(300)(900) = 17 microseconds
Used (mn + mp) = 900 cm2/V-sec which is value appropriate to large excess carrier
densities (approaching 1017 cm-3)
c) Von = I Rdrift ; Ignore I2/3 contribution as it is usually small compared to the linear term.
Rdrift ≈ Wd
q!(mn!+!mp)!nb!A ; 2
2000 = 10-3 = 0.02
(1.6x10-19)(900)(1017)!A ;
Solving for A gives A = 0.02
(1.6x10-19)(900)(1017)(10-3) = 1.4 cm2
![Page 384: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/384.jpg)
23-4. Advantages - Shorter N- region length Wd means shorter carrier lifetimes can be accomodated and thus faster switching times.
Disadvantages - Junction J1 is now a P+ - N+ junction with a low breakdown voltage. Since J3 already has a low breakdown voltage, the modified thyristor has no significant reverse blocking capability.
23-5. ton = (4!-!0.5)!cm
us =
3.5104 = 350 microseconds
23-6. Lateral voltage drops caused by base currents cause current density nonuniformities. At large currents, these nonuniformities become severe and the increasing possibility of second breakdown limit the total current that the BJT can safely conduct.
In the thyristor, no significant gate current is needed to keep the thyristor on and there is consequently no lateral current flow and thus lateral voltage drop. The current density is uniform across the entire cross-sectional area of the thyristor and there is much less likelyhood of second breakdown.
23-7. a) Breakover in a thyristor is not due to impact ionization. However in a well-designed thyristor, the value of the breakover voltage is an appreciable fraction of the actual avalanche breakdown voltage. Thus an estimate of the n1 thickness and doping level based avalanche breakdown would be a reasonable first attempt.
Nd = 1.3x1017
2x103 = 6.5x1013 cm-3 ; Wd ≈ (10-5)(2x103) = 200 microns
b) t = qWd
2
kT(mn!+!mp) = (1.6x10-19)(2x10-2)2
(1.4x10-23)(300)(900) = 17 microseconds
Used (mn + mp) = 900 cm2/V-sec which is value appropriate to large excess carrier
densities (approaching 1017 cm-3)
c) Von = I Rdrift ; Ignore I2/3 contribution as it is usually small compared to the linear term.
Rdrift ≈ Wd
q!(mn!+!mp)!nb!A ; 2
2000 = 10-3 = 0.02
(1.6x10-19)(900)(1017)!A ;
Solving for A gives A = 0.02
(1.6x10-19)(900)(1017)(10-3) = 1.4 cm2
![Page 385: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/385.jpg)
23-4. Advantages - Shorter N- region length Wd means shorter carrier lifetimes can be accomodated and thus faster switching times.
Disadvantages - Junction J1 is now a P+ - N+ junction with a low breakdown voltage. Since J3 already has a low breakdown voltage, the modified thyristor has no significant reverse blocking capability.
23-5. ton = (4!-!0.5)!cm
us =
3.5104 = 350 microseconds
23-6. Lateral voltage drops caused by base currents cause current density nonuniformities. At large currents, these nonuniformities become severe and the increasing possibility of second breakdown limit the total current that the BJT can safely conduct.
In the thyristor, no significant gate current is needed to keep the thyristor on and there is consequently no lateral current flow and thus lateral voltage drop. The current density is uniform across the entire cross-sectional area of the thyristor and there is much less likelyhood of second breakdown.
23-7. a) Breakover in a thyristor is not due to impact ionization. However in a well-designed thyristor, the value of the breakover voltage is an appreciable fraction of the actual avalanche breakdown voltage. Thus an estimate of the n1 thickness and doping level based avalanche breakdown would be a reasonable first attempt.
Nd = 1.3x1017
2x103 = 6.5x1013 cm-3 ; Wd ≈ (10-5)(2x103) = 200 microns
b) t = qWd
2
kT(mn!+!mp) = (1.6x10-19)(2x10-2)2
(1.4x10-23)(300)(900) = 17 microseconds
Used (mn + mp) = 900 cm2/V-sec which is value appropriate to large excess carrier
densities (approaching 1017 cm-3)
c) Von = I Rdrift ; Ignore I2/3 contribution as it is usually small compared to the linear term.
Rdrift ≈ Wd
q!(mn!+!mp)!nb!A ; 2
2000 = 10-3 = 0.02
(1.6x10-19)(900)(1017)!A ;
Solving for A gives A = 0.02
(1.6x10-19)(900)(1017)(10-3) = 1.4 cm2
![Page 386: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/386.jpg)
Resulting current density is 20001.4 = 1430 A/cm2 which is excessively large. Probably
should use nb ≈ 1016 cm-3 which would give a current density of 140 A/cm2, a more realistic value.
23-8. Cathode area = 3000200 = 15 cm2 = 0.65 Asi ; Asi =
150.65 = 23 cm2
23 cm2 = π Rsi2 ; Rsi =
23π = 2.7 cm
23-9.dvAK
dt |max ≈ IBO
Cj2(0) ; Cj2(0) = zero bias value of junction J2 space charge capacitance
Cj2(0) ≈ e!A
Wdepl(0) ; Wdepl(0) ≈ 2!e!fj2q!Nd
; fj2 = kTq ln[
NaNdni
2 ]
fj2 = 0.026 ln[(1014)(1017)
(1020) ] = 0.66 V ;
Wdepl(0) ≈ (2)(11.7)(8.9x10-14)(0.66)
(1.6x10-19)(1014) = 2.9 microns
Cj2(0) = (11.7)(8.9x10-14)(10)
2.9x10-4 = 36 nF
dvAKdt |max =
0.053.6x10-8 = 1.4x106 V/sec or 1.4 V per microsecond
![Page 387: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/387.jpg)
Resulting current density is 20001.4 = 1430 A/cm2 which is excessively large. Probably
should use nb ≈ 1016 cm-3 which would give a current density of 140 A/cm2, a more realistic value.
23-8. Cathode area = 3000200 = 15 cm2 = 0.65 Asi ; Asi =
150.65 = 23 cm2
23 cm2 = π Rsi2 ; Rsi =
23π = 2.7 cm
23-9.dvAK
dt |max ≈ IBO
Cj2(0) ; Cj2(0) = zero bias value of junction J2 space charge capacitance
Cj2(0) ≈ e!A
Wdepl(0) ; Wdepl(0) ≈ 2!e!fj2q!Nd
; fj2 = kTq ln[
NaNdni
2 ]
fj2 = 0.026 ln[(1014)(1017)
(1020) ] = 0.66 V ;
Wdepl(0) ≈ (2)(11.7)(8.9x10-14)(0.66)
(1.6x10-19)(1014) = 2.9 microns
Cj2(0) = (11.7)(8.9x10-14)(10)
2.9x10-4 = 36 nF
dvAKdt |max =
0.053.6x10-8 = 1.4x106 V/sec or 1.4 V per microsecond
![Page 388: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/388.jpg)
Resulting current density is 20001.4 = 1430 A/cm2 which is excessively large. Probably
should use nb ≈ 1016 cm-3 which would give a current density of 140 A/cm2, a more realistic value.
23-8. Cathode area = 3000200 = 15 cm2 = 0.65 Asi ; Asi =
150.65 = 23 cm2
23 cm2 = π Rsi2 ; Rsi =
23π = 2.7 cm
23-9.dvAK
dt |max ≈ IBO
Cj2(0) ; Cj2(0) = zero bias value of junction J2 space charge capacitance
Cj2(0) ≈ e!A
Wdepl(0) ; Wdepl(0) ≈ 2!e!fj2q!Nd
; fj2 = kTq ln[
NaNdni
2 ]
fj2 = 0.026 ln[(1014)(1017)
(1020) ] = 0.66 V ;
Wdepl(0) ≈ (2)(11.7)(8.9x10-14)(0.66)
(1.6x10-19)(1014) = 2.9 microns
Cj2(0) = (11.7)(8.9x10-14)(10)
2.9x10-4 = 36 nF
dvAKdt |max =
0.053.6x10-8 = 1.4x106 V/sec or 1.4 V per microsecond
![Page 389: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/389.jpg)
Chapter 24 Problem Solutions
24-1. Cross-sectional view of GTO gate-cathode area with reverse gate current flowing.
N2
P2
GateCathode
Center Line
t++ -- vGKvGK
RR
-i G
The negative gate current, -iG, flowing through the P2 layher beneath the N2 cathode layer develops a lateral voltage drop vGK as indicated. Maximum negative vGK = BVJ3
|vGK| = |IG,max| RGK < BVJ3 in order to avoid breakdown or IG,max < BVJ3RGK
RGK = R
2!N = rp2!W4!N!L!t ; N = number of cathode islands in parallel.
IG,max = BVJ3RGK
= IA,maxboff : Solving for IA,max yields
IA,max = 4!N!L!t!boff!BVJ3
rp2!W
24-2. Assume that the current is communtated from the GTO to the turn-off snubber and associated stray inductance linearly as a function of time. That is the inductor current
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Chapter 24 Problem Solutions
24-1. Cross-sectional view of GTO gate-cathode area with reverse gate current flowing.
N2
P2
GateCathode
Center Line
t++ -- vGKvGK
RR
-i G
The negative gate current, -iG, flowing through the P2 layher beneath the N2 cathode layer develops a lateral voltage drop vGK as indicated. Maximum negative vGK = BVJ3
|vGK| = |IG,max| RGK < BVJ3 in order to avoid breakdown or IG,max < BVJ3RGK
RGK = R
2!N = rp2!W4!N!L!t ; N = number of cathode islands in parallel.
IG,max = BVJ3RGK
= IA,maxboff : Solving for IA,max yields
IA,max = 4!N!L!t!boff!BVJ3
rp2!W
24-2. Assume that the current is communtated from the GTO to the turn-off snubber and associated stray inductance linearly as a function of time. That is the inductor current
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iLs = Io ttfi
; Assume that just prior to the end of the current fall time interval, the
voltage across the snubber capacitor has built up to approximately Vd.
vAK,max = 1.5 Vd = Ls diLs
dt + vcap = Ls
Iotfi
+ Vd ; Solving for Ls yields
Ls = Vd!tfi2!Io
24-3. Equivalent circuit during tgq shown below.
-
+VGG-
LG
PNPN J3 forward
biased during tgq
i (t)G
LGdiGdt = - VGG- ; iG(t) =
-!VGG-LG
t ; At t = tgq want iG = - Io
boff ; Solve for LG
LG = boff!VGG-!tgq
Io =
(5)(15)(5x10-6)(500) = 0.75 microhenries
Equivalent circuit during tw2 interval.
![Page 392: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/392.jpg)
iLs = Io ttfi
; Assume that just prior to the end of the current fall time interval, the
voltage across the snubber capacitor has built up to approximately Vd.
vAK,max = 1.5 Vd = Ls diLs
dt + vcap = Ls
Iotfi
+ Vd ; Solving for Ls yields
Ls = Vd!tfi2!Io
24-3. Equivalent circuit during tgq shown below.
-
+VGG-
LG
PNPN J3 forward
biased during tgq
i (t)G
LGdiGdt = - VGG- ; iG(t) =
-!VGG-LG
t ; At t = tgq want iG = - Io
boff ; Solve for LG
LG = boff!VGG-!tgq
Io =
(5)(15)(5x10-6)(500) = 0.75 microhenries
Equivalent circuit during tw2 interval.
![Page 393: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/393.jpg)
-
+
VGG-
LG
i (t)G
+
-BV
J3
LGdiGdt = BVJ3 - VGG- ; iG(0) = -
Ioboff ; iG(t) = -
Ioboff +
!BVJ3!-!VGG-!LG
t
At t = tw2 , iG = 0 ; solving for tw2 yields
tw2 = Io!LG
boff![BVJ3!-!VGG-] =
(500)(7.5x10-7)(5)(25!-!15) = 7.5 microseconds
![Page 394: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/394.jpg)
Chapter 25 Problem Solutions
25-1.Ron(MOS)
A proprotional to 1
mmajority ; mn = 3 mp ; Hence
Ron(p-channel)A = 3
Ron(n-channel)A
Ron(IGBT)A proportional to
1dn(mn!+!mp) ; dn = excess carrier density
dn = dp so p-channel IGBTs have the same Ron as n-channel IGBTs
25-2. Turn-off waveforms of short versus long lifetime IGBTs
i D
long lifetime
short lifetimeI (long)BJT
I (short)BJTt
Long lifetime IGBT -
a. BJT portion of the device has a larger beta and thus the BJT section carries the largest fraction of the IGBT current. Thus IBJT(long) > IBJT(short).
b. Longer lifetime leads to longer BJT turn-off times.
Short lifetime IGBT -
a. BJT beta smaller. MOSFET section of the device carries most of the current.
b. Shorter lifetime means less stored charge in the BJT section and thus faster turn-off.
![Page 395: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/395.jpg)
Chapter 25 Problem Solutions
25-1.Ron(MOS)
A proprotional to 1
mmajority ; mn = 3 mp ; Hence
Ron(p-channel)A = 3
Ron(n-channel)A
Ron(IGBT)A proportional to
1dn(mn!+!mp) ; dn = excess carrier density
dn = dp so p-channel IGBTs have the same Ron as n-channel IGBTs
25-2. Turn-off waveforms of short versus long lifetime IGBTs
i D
long lifetime
short lifetimeI (long)BJT
I (short)BJTt
Long lifetime IGBT -
a. BJT portion of the device has a larger beta and thus the BJT section carries the largest fraction of the IGBT current. Thus IBJT(long) > IBJT(short).
b. Longer lifetime leads to longer BJT turn-off times.
Short lifetime IGBT -
a. BJT beta smaller. MOSFET section of the device carries most of the current.
b. Shorter lifetime means less stored charge in the BJT section and thus faster turn-off.
![Page 396: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/396.jpg)
25-3.
P P+N -
Body region of MOSFET section of IGBT
Collector junction of pnp BJT section of IGBT Base of pnp BJT
Emitter of pnp BJT
VDS1
V > VDS1DS2
Depletion layer Effective base width
Drain of IGBT
Significant encroachment intoa the base of the PNP BJT section by the depletion layer of the blocking junction. The effective base width is thus lowered and the beta increases as vDS increases. This is base width modulation and it results in a lower output resistance ro (steeper slope in the active region of the iD-vDS characteristics).
![Page 397: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/397.jpg)
P P+N -
Body region of MOSFET section of IGBT
Collector junction of pnp BJT section of IGBT Base of pnp BJT
Emitter of pnp BJT
Depletion layer
Drain of IGBT
N+
VDS
Effective base width independent of VDS
Depletion encroaches into the N- layer but the advance is halted at moderate vDS values
by the N+ buffer layer. The PNP base width becomes constant and so the effective resistance ro remains large.
25-4. One dimensional model of n-channel IGBT
P P+N- N+Source
Drain
1017 1014
1019
1019
25 m m
N+
1019
Reverse blocking junction is the P+ - N+ junction because of body-source short.
BVRB ≈ 1.3x1017
1019 < 1 volt. No reverse blocking capability.
Forward breakdown - limited by P - N- junction.
![Page 398: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/398.jpg)
P P+N -
Body region of MOSFET section of IGBT
Collector junction of pnp BJT section of IGBT Base of pnp BJT
Emitter of pnp BJT
Depletion layer
Drain of IGBT
N+
VDS
Effective base width independent of VDS
Depletion encroaches into the N- layer but the advance is halted at moderate vDS values
by the N+ buffer layer. The PNP base width becomes constant and so the effective resistance ro remains large.
25-4. One dimensional model of n-channel IGBT
P P+N- N+Source
Drain
1017 1014
1019
1019
25 m m
N+
1019
Reverse blocking junction is the P+ - N+ junction because of body-source short.
BVRB ≈ 1.3x1017
1019 < 1 volt. No reverse blocking capability.
Forward breakdown - limited by P - N- junction.
![Page 399: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/399.jpg)
BVFB = 1.3x1017
1014 ≈ 1300 volts ; But Wdepl(1300 V) = (10-5)(1300) = 130 microns
130 microns > 25 micron drift region length. Hence forward blocking limited bypunch-through.
BVFB = (2x105)(2.5x10-3) - (1.6x10-19)(1014)(2.5x10-3)2
(2)(11.7)(8.9x10-14) = 453 volts
25-5. IGBT current - Ion,IGBT ; Von(IGBT) = Vj + Ion,IGBT Ron,IGBT
Assume Vj ≈ 0.8 V ; Exact value not critical to an approximate estimate of Ion,IGBT.
Ion,IGBT ≈ 3!-!0.8
Ron,IGBT! ; Ron,IGBT ≈
Wdq!(mn!+!mp)!nb!A
Wd = (10-5)(750) = 75 mm ; Ron,IGBT = 7.5x10-3
(1.6x10-19)(900)(1016)(2) = 2.6x10-3 W
Ion,IGBT ≈ 2.2
2.6x10-3 ≈ 850 amps
MOSFET current - Ion,MOS ; Von(MOS) = Ion,MOS Ron,MOS
Ion,MOS = Von(MOS)Ron,MOS
; Ron,MOS = Wd
q!mn!!Nd!A ; Wd = 75 mm
Nd = 1.3x1017
750 = 1.7x1014 cm-3
Ron,MOS = 7.5x10-3
(1.6x10-19)(1.5x103)(1.7x1014)(2) = 0.09 ohms
Ion,MOS = 3
0.09 = 33 amps
25-6. Von(PT) = Vj,PT + Ion,PT Ron,PT = Vj,NPT + Ion,NPT Ron,NPT
Ion,PTIon,NPT
≈ Ron,NPTRon,PT
since Vj,NPT ≈ Vj,PT
![Page 400: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/400.jpg)
BVFB = 1.3x1017
1014 ≈ 1300 volts ; But Wdepl(1300 V) = (10-5)(1300) = 130 microns
130 microns > 25 micron drift region length. Hence forward blocking limited bypunch-through.
BVFB = (2x105)(2.5x10-3) - (1.6x10-19)(1014)(2.5x10-3)2
(2)(11.7)(8.9x10-14) = 453 volts
25-5. IGBT current - Ion,IGBT ; Von(IGBT) = Vj + Ion,IGBT Ron,IGBT
Assume Vj ≈ 0.8 V ; Exact value not critical to an approximate estimate of Ion,IGBT.
Ion,IGBT ≈ 3!-!0.8
Ron,IGBT! ; Ron,IGBT ≈
Wdq!(mn!+!mp)!nb!A
Wd = (10-5)(750) = 75 mm ; Ron,IGBT = 7.5x10-3
(1.6x10-19)(900)(1016)(2) = 2.6x10-3 W
Ion,IGBT ≈ 2.2
2.6x10-3 ≈ 850 amps
MOSFET current - Ion,MOS ; Von(MOS) = Ion,MOS Ron,MOS
Ion,MOS = Von(MOS)Ron,MOS
; Ron,MOS = Wd
q!mn!!Nd!A ; Wd = 75 mm
Nd = 1.3x1017
750 = 1.7x1014 cm-3
Ron,MOS = 7.5x10-3
(1.6x10-19)(1.5x103)(1.7x1014)(2) = 0.09 ohms
Ion,MOS = 3
0.09 = 33 amps
25-6. Von(PT) = Vj,PT + Ion,PT Ron,PT = Vj,NPT + Ion,NPT Ron,NPT
Ion,PTIon,NPT
≈ Ron,NPTRon,PT
since Vj,NPT ≈ Vj,PT
![Page 401: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/401.jpg)
BVFB = 1.3x1017
1014 ≈ 1300 volts ; But Wdepl(1300 V) = (10-5)(1300) = 130 microns
130 microns > 25 micron drift region length. Hence forward blocking limited bypunch-through.
BVFB = (2x105)(2.5x10-3) - (1.6x10-19)(1014)(2.5x10-3)2
(2)(11.7)(8.9x10-14) = 453 volts
25-5. IGBT current - Ion,IGBT ; Von(IGBT) = Vj + Ion,IGBT Ron,IGBT
Assume Vj ≈ 0.8 V ; Exact value not critical to an approximate estimate of Ion,IGBT.
Ion,IGBT ≈ 3!-!0.8
Ron,IGBT! ; Ron,IGBT ≈
Wdq!(mn!+!mp)!nb!A
Wd = (10-5)(750) = 75 mm ; Ron,IGBT = 7.5x10-3
(1.6x10-19)(900)(1016)(2) = 2.6x10-3 W
Ion,IGBT ≈ 2.2
2.6x10-3 ≈ 850 amps
MOSFET current - Ion,MOS ; Von(MOS) = Ion,MOS Ron,MOS
Ion,MOS = Von(MOS)Ron,MOS
; Ron,MOS = Wd
q!mn!!Nd!A ; Wd = 75 mm
Nd = 1.3x1017
750 = 1.7x1014 cm-3
Ron,MOS = 7.5x10-3
(1.6x10-19)(1.5x103)(1.7x1014)(2) = 0.09 ohms
Ion,MOS = 3
0.09 = 33 amps
25-6. Von(PT) = Vj,PT + Ion,PT Ron,PT = Vj,NPT + Ion,NPT Ron,NPT
Ion,PTIon,NPT
≈ Ron,NPTRon,PT
since Vj,NPT ≈ Vj,PT
![Page 402: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/402.jpg)
Ron,NPTRon,PT
= Wd,NPT!Wd,PT
≈ 2 assuming doping level in PT drift region is much less than
the doping level in the NPT drift region.
Hence Ion,PT
Ion,NPT ≈ 2
25-7. Cv dT = dQ ; dQ = PV dt ; P = power dissipated in IGBT during overcurrent transient.
V = volume in IGBT where power is dissipated.Duration of transient = dt.
P = Iov2 Ron ; Iov =
!V!Cv!dT
dt!Ron ; V ≈ A Wdrift
Ron = Wdrift
q!(mn!+!mp)!nb!A ; Iov = q!(mn!+!mp)!nb!A2!Cv!dT
dt
Iov = (1.6x10-19)(900)(1016)(0.5)2(1.75)(100)
(10-5) ≈ 2.5x103 amps
Estimate is overly optimistic because it ignores any other ohmic losses in the device such as channel resistance or resistance of the heavily doped source and drain diffusions. However IGBTs nominally rated at 100 A have repeatedly been tested at 1000 A for 10 microseconds or less and survived.
25-8. The IGBT has the smaller values of Cgd and Cgs because its effective cross-sectional area is smaller than that of the MOSFET. The IGBT has a smaller area even though its current rating is identical to the MOSFET's rating because the IGBT utilizes conductivity modulation of the drift region to significantly reduce the specific on-state resistance.
25-9. Check dvDS
dt at turn-off ; dvDS
dt = Vd!trv
; trv < 0.75 microseconds
dvDS
dt > Vd!toff
= 700
!3x10-7 = 2330 V/ms > 800 V/ms limit.
Device is overstressed by an overly large dvDS
dt .
![Page 403: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/403.jpg)
Ron,NPTRon,PT
= Wd,NPT!Wd,PT
≈ 2 assuming doping level in PT drift region is much less than
the doping level in the NPT drift region.
Hence Ion,PT
Ion,NPT ≈ 2
25-7. Cv dT = dQ ; dQ = PV dt ; P = power dissipated in IGBT during overcurrent transient.
V = volume in IGBT where power is dissipated.Duration of transient = dt.
P = Iov2 Ron ; Iov =
!V!Cv!dT
dt!Ron ; V ≈ A Wdrift
Ron = Wdrift
q!(mn!+!mp)!nb!A ; Iov = q!(mn!+!mp)!nb!A2!Cv!dT
dt
Iov = (1.6x10-19)(900)(1016)(0.5)2(1.75)(100)
(10-5) ≈ 2.5x103 amps
Estimate is overly optimistic because it ignores any other ohmic losses in the device such as channel resistance or resistance of the heavily doped source and drain diffusions. However IGBTs nominally rated at 100 A have repeatedly been tested at 1000 A for 10 microseconds or less and survived.
25-8. The IGBT has the smaller values of Cgd and Cgs because its effective cross-sectional area is smaller than that of the MOSFET. The IGBT has a smaller area even though its current rating is identical to the MOSFET's rating because the IGBT utilizes conductivity modulation of the drift region to significantly reduce the specific on-state resistance.
25-9. Check dvDS
dt at turn-off ; dvDS
dt = Vd!trv
; trv < 0.75 microseconds
dvDS
dt > Vd!toff
= 700
!3x10-7 = 2330 V/ms > 800 V/ms limit.
Device is overstressed by an overly large dvDS
dt .
![Page 404: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/404.jpg)
Ron,NPTRon,PT
= Wd,NPT!Wd,PT
≈ 2 assuming doping level in PT drift region is much less than
the doping level in the NPT drift region.
Hence Ion,PT
Ion,NPT ≈ 2
25-7. Cv dT = dQ ; dQ = PV dt ; P = power dissipated in IGBT during overcurrent transient.
V = volume in IGBT where power is dissipated.Duration of transient = dt.
P = Iov2 Ron ; Iov =
!V!Cv!dT
dt!Ron ; V ≈ A Wdrift
Ron = Wdrift
q!(mn!+!mp)!nb!A ; Iov = q!(mn!+!mp)!nb!A2!Cv!dT
dt
Iov = (1.6x10-19)(900)(1016)(0.5)2(1.75)(100)
(10-5) ≈ 2.5x103 amps
Estimate is overly optimistic because it ignores any other ohmic losses in the device such as channel resistance or resistance of the heavily doped source and drain diffusions. However IGBTs nominally rated at 100 A have repeatedly been tested at 1000 A for 10 microseconds or less and survived.
25-8. The IGBT has the smaller values of Cgd and Cgs because its effective cross-sectional area is smaller than that of the MOSFET. The IGBT has a smaller area even though its current rating is identical to the MOSFET's rating because the IGBT utilizes conductivity modulation of the drift region to significantly reduce the specific on-state resistance.
25-9. Check dvDS
dt at turn-off ; dvDS
dt = Vd!trv
; trv < 0.75 microseconds
dvDS
dt > Vd!toff
= 700
!3x10-7 = 2330 V/ms > 800 V/ms limit.
Device is overstressed by an overly large dvDS
dt .
![Page 405: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/405.jpg)
Ron,NPTRon,PT
= Wd,NPT!Wd,PT
≈ 2 assuming doping level in PT drift region is much less than
the doping level in the NPT drift region.
Hence Ion,PT
Ion,NPT ≈ 2
25-7. Cv dT = dQ ; dQ = PV dt ; P = power dissipated in IGBT during overcurrent transient.
V = volume in IGBT where power is dissipated.Duration of transient = dt.
P = Iov2 Ron ; Iov =
!V!Cv!dT
dt!Ron ; V ≈ A Wdrift
Ron = Wdrift
q!(mn!+!mp)!nb!A ; Iov = q!(mn!+!mp)!nb!A2!Cv!dT
dt
Iov = (1.6x10-19)(900)(1016)(0.5)2(1.75)(100)
(10-5) ≈ 2.5x103 amps
Estimate is overly optimistic because it ignores any other ohmic losses in the device such as channel resistance or resistance of the heavily doped source and drain diffusions. However IGBTs nominally rated at 100 A have repeatedly been tested at 1000 A for 10 microseconds or less and survived.
25-8. The IGBT has the smaller values of Cgd and Cgs because its effective cross-sectional area is smaller than that of the MOSFET. The IGBT has a smaller area even though its current rating is identical to the MOSFET's rating because the IGBT utilizes conductivity modulation of the drift region to significantly reduce the specific on-state resistance.
25-9. Check dvDS
dt at turn-off ; dvDS
dt = Vd!trv
; trv < 0.75 microseconds
dvDS
dt > Vd!toff
= 700
!3x10-7 = 2330 V/ms > 800 V/ms limit.
Device is overstressed by an overly large dvDS
dt .
![Page 406: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/406.jpg)
Check switching losses Psw = Vd Io !tri!+!trv!+!!tfi!+!tfv
2 fs
Psw = (700)(100) (3x10-7!+!7.5x10-7)
2 (2.5x104) = 875 W
Allowable power loss = Tj,max!-!Ta
Rqja =
150!-!250.5 = 250 watts
Switching losses exceed allowable losses. Module is overstressed by too muchpower dissipation.
![Page 407: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/407.jpg)
Chapter 26 Problem Solutions
26-1. Equivalent circuit for JFET in active region.
+-
CGD
CGS
ro
mvGS
vGSvDS
+
-
+
-
Equivelent circuit for JFET Linearized I-V characteristicsin the blocking state.
CGD
CGSvGS
vDS
+
-
+
-
0 -V GS2GS1-Vi D
VDS1 VDS20
26-2. Drive circuit configuration
+
-
VDD
R L
2R
1R
Vdrive
MOSFET off
VDS = VKG = - VGK = VDD!R2R1!+!R2
Negative enough to insure thatthe FCT is off.
MOSFET on
VDS = VKG = - VGK = 0
and FCT is on.
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Chapter 26 Problem Solutions
26-1. Equivalent circuit for JFET in active region.
+-
CGD
CGS
ro
mvGS
vGSvDS
+
-
+
-
Equivelent circuit for JFET Linearized I-V characteristicsin the blocking state.
CGD
CGSvGS
vDS
+
-
+
-
0 -V GS2GS1-Vi D
VDS1 VDS20
26-2. Drive circuit configuration
+
-
VDD
R L
2R
1R
Vdrive
MOSFET off
VDS = VKG = - VGK = VDD!R2R1!+!R2
Negative enough to insure thatthe FCT is off.
MOSFET on
VDS = VKG = - VGK = 0
and FCT is on.
![Page 409: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/409.jpg)
MOSFET characteirstics:- High current sinking capability- Low Ron ; Low BVDss
26-3. Vdrift = Wd
2
(mn!+!!mp)!t ; BVBD = EBD!Wd
2 ; Vdrift,GaAs = Vdrift,Si
Wd
2(Si)(mn!+!!mp)|Si!tSi
= Wd
2(GaAs)(mn!+!!mp)|GaAs!tGaAs
tGaAtSi =
(mn!+!!mp)|Si(mn!+!!mp)|GaAs
ÎÍÈ
˚˙˘EBD(Si)
EBD(GaAs) 2
(mn + mp)|Si = 2000 cm2/V-sec ; (mn + mp)|GaAs = 9000 cm2/V-sec
EBD(Si) = 300 kV/cm ; EBD(GaAs) = 400 kV/cm
tGaAtSi =
29 ÎÍ
È˚˙3
4 2
= 0.125 ; GaAs has the shorter lifetime.
26-4. EBD = 107 V/cm ; BVBD = EBD tox
tox = 103
107 = 10-4 cm = 1 micron
26-5. IA,max = (105)(1.5x10-2) = 1500 amperes
26-6. P-MCT fabricated in silicon can turn-off three times more current than an identicalN-MCT due to the difference between the mobilities in the n-channel OFF-FET in the P-MCT and p-channel OFF-FET in the N-MCT. Hence
IA,max = (3)(105)(1.5x10-2) = 4500 amperes
26-7. Assume an n-type drift region since mn > mp.
![Page 410: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/410.jpg)
MOSFET characteirstics:- High current sinking capability- Low Ron ; Low BVDss
26-3. Vdrift = Wd
2
(mn!+!!mp)!t ; BVBD = EBD!Wd
2 ; Vdrift,GaAs = Vdrift,Si
Wd
2(Si)(mn!+!!mp)|Si!tSi
= Wd
2(GaAs)(mn!+!!mp)|GaAs!tGaAs
tGaAtSi =
(mn!+!!mp)|Si(mn!+!!mp)|GaAs
ÎÍÈ
˚˙˘EBD(Si)
EBD(GaAs) 2
(mn + mp)|Si = 2000 cm2/V-sec ; (mn + mp)|GaAs = 9000 cm2/V-sec
EBD(Si) = 300 kV/cm ; EBD(GaAs) = 400 kV/cm
tGaAtSi =
29 ÎÍ
È˚˙3
4 2
= 0.125 ; GaAs has the shorter lifetime.
26-4. EBD = 107 V/cm ; BVBD = EBD tox
tox = 103
107 = 10-4 cm = 1 micron
26-5. IA,max = (105)(1.5x10-2) = 1500 amperes
26-6. P-MCT fabricated in silicon can turn-off three times more current than an identicalN-MCT due to the difference between the mobilities in the n-channel OFF-FET in the P-MCT and p-channel OFF-FET in the N-MCT. Hence
IA,max = (3)(105)(1.5x10-2) = 4500 amperes
26-7. Assume an n-type drift region since mn > mp.
![Page 411: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/411.jpg)
MOSFET characteirstics:- High current sinking capability- Low Ron ; Low BVDss
26-3. Vdrift = Wd
2
(mn!+!!mp)!t ; BVBD = EBD!Wd
2 ; Vdrift,GaAs = Vdrift,Si
Wd
2(Si)(mn!+!!mp)|Si!tSi
= Wd
2(GaAs)(mn!+!!mp)|GaAs!tGaAs
tGaAtSi =
(mn!+!!mp)|Si(mn!+!!mp)|GaAs
ÎÍÈ
˚˙˘EBD(Si)
EBD(GaAs) 2
(mn + mp)|Si = 2000 cm2/V-sec ; (mn + mp)|GaAs = 9000 cm2/V-sec
EBD(Si) = 300 kV/cm ; EBD(GaAs) = 400 kV/cm
tGaAtSi =
29 ÎÍ
È˚˙3
4 2
= 0.125 ; GaAs has the shorter lifetime.
26-4. EBD = 107 V/cm ; BVBD = EBD tox
tox = 103
107 = 10-4 cm = 1 micron
26-5. IA,max = (105)(1.5x10-2) = 1500 amperes
26-6. P-MCT fabricated in silicon can turn-off three times more current than an identicalN-MCT due to the difference between the mobilities in the n-channel OFF-FET in the P-MCT and p-channel OFF-FET in the N-MCT. Hence
IA,max = (3)(105)(1.5x10-2) = 4500 amperes
26-7. Assume an n-type drift region since mn > mp.
![Page 412: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/412.jpg)
MOSFET characteirstics:- High current sinking capability- Low Ron ; Low BVDss
26-3. Vdrift = Wd
2
(mn!+!!mp)!t ; BVBD = EBD!Wd
2 ; Vdrift,GaAs = Vdrift,Si
Wd
2(Si)(mn!+!!mp)|Si!tSi
= Wd
2(GaAs)(mn!+!!mp)|GaAs!tGaAs
tGaAtSi =
(mn!+!!mp)|Si(mn!+!!mp)|GaAs
ÎÍÈ
˚˙˘EBD(Si)
EBD(GaAs) 2
(mn + mp)|Si = 2000 cm2/V-sec ; (mn + mp)|GaAs = 9000 cm2/V-sec
EBD(Si) = 300 kV/cm ; EBD(GaAs) = 400 kV/cm
tGaAtSi =
29 ÎÍ
È˚˙3
4 2
= 0.125 ; GaAs has the shorter lifetime.
26-4. EBD = 107 V/cm ; BVBD = EBD tox
tox = 103
107 = 10-4 cm = 1 micron
26-5. IA,max = (105)(1.5x10-2) = 1500 amperes
26-6. P-MCT fabricated in silicon can turn-off three times more current than an identicalN-MCT due to the difference between the mobilities in the n-channel OFF-FET in the P-MCT and p-channel OFF-FET in the N-MCT. Hence
IA,max = (3)(105)(1.5x10-2) = 4500 amperes
26-7. Assume an n-type drift region since mn > mp.
![Page 413: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/413.jpg)
MOSFET characteirstics:- High current sinking capability- Low Ron ; Low BVDss
26-3. Vdrift = Wd
2
(mn!+!!mp)!t ; BVBD = EBD!Wd
2 ; Vdrift,GaAs = Vdrift,Si
Wd
2(Si)(mn!+!!mp)|Si!tSi
= Wd
2(GaAs)(mn!+!!mp)|GaAs!tGaAs
tGaAtSi =
(mn!+!!mp)|Si(mn!+!!mp)|GaAs
ÎÍÈ
˚˙˘EBD(Si)
EBD(GaAs) 2
(mn + mp)|Si = 2000 cm2/V-sec ; (mn + mp)|GaAs = 9000 cm2/V-sec
EBD(Si) = 300 kV/cm ; EBD(GaAs) = 400 kV/cm
tGaAtSi =
29 ÎÍ
È˚˙3
4 2
= 0.125 ; GaAs has the shorter lifetime.
26-4. EBD = 107 V/cm ; BVBD = EBD tox
tox = 103
107 = 10-4 cm = 1 micron
26-5. IA,max = (105)(1.5x10-2) = 1500 amperes
26-6. P-MCT fabricated in silicon can turn-off three times more current than an identicalN-MCT due to the difference between the mobilities in the n-channel OFF-FET in the P-MCT and p-channel OFF-FET in the N-MCT. Hence
IA,max = (3)(105)(1.5x10-2) = 4500 amperes
26-7. Assume an n-type drift region since mn > mp.
![Page 414: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/414.jpg)
MOSFET characteirstics:- High current sinking capability- Low Ron ; Low BVDss
26-3. Vdrift = Wd
2
(mn!+!!mp)!t ; BVBD = EBD!Wd
2 ; Vdrift,GaAs = Vdrift,Si
Wd
2(Si)(mn!+!!mp)|Si!tSi
= Wd
2(GaAs)(mn!+!!mp)|GaAs!tGaAs
tGaAtSi =
(mn!+!!mp)|Si(mn!+!!mp)|GaAs
ÎÍÈ
˚˙˘EBD(Si)
EBD(GaAs) 2
(mn + mp)|Si = 2000 cm2/V-sec ; (mn + mp)|GaAs = 9000 cm2/V-sec
EBD(Si) = 300 kV/cm ; EBD(GaAs) = 400 kV/cm
tGaAtSi =
29 ÎÍ
È˚˙3
4 2
= 0.125 ; GaAs has the shorter lifetime.
26-4. EBD = 107 V/cm ; BVBD = EBD tox
tox = 103
107 = 10-4 cm = 1 micron
26-5. IA,max = (105)(1.5x10-2) = 1500 amperes
26-6. P-MCT fabricated in silicon can turn-off three times more current than an identicalN-MCT due to the difference between the mobilities in the n-channel OFF-FET in the P-MCT and p-channel OFF-FET in the N-MCT. Hence
IA,max = (3)(105)(1.5x10-2) = 4500 amperes
26-7. Assume an n-type drift region since mn > mp.
![Page 415: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/415.jpg)
Rdrift = Wd
q!mn!Nd!A ; Rdrift,sp = Rdrift A = Wd
q!mn!Nd
Using Eq. (20-1) Nd = e!EBD
2
2!q!BVBD ; Using Eq. (20-3) Wd =
2!!BVBDEBD
Substituting into the expression for Rdrift,sp yields
Rdrift,sp = 2!!BVBD
EBD
1q!e
2!q!BVBDe!EBD
2 = !!!4!!(BVBD)2
e!mn!(EBD)3
26-8. Silicon : Rdrift,sp = (4)(500)2
(11.7)(1500)(8.9x10-14)(3x105)3 = 0.024 ohms-cm2
GaAs: Rdrift,sp = (4)(500)2
(12.8)(8500)(8.9x10-14)(4x105)3 = 0.0016 ohms-cm2
6H-SiC: Rdrift,sp = (4)(500)2
(10)(600)(8.9x10-14)(2x106)3 = 2.3x10-4 ohms-cm2
Diamond: Rdrift,sp = (4)(500)2
(5.5)(2200)(8.9x10-14)(107)3 = 9.3x10-7 ohms-cm2
26-9. Diamond is the most suitable material for high temperature operation. It has the largest bandgap (by almost a factor of two) and thus the smallest intrinsic carrier density at anygiven temperature. This statement presumes that the phase change listed for diamond inthe table of material properties exceeds the sublimation temperature of SiC (1800 °C).
26-10. Eq. (20-1): Nd = e!EBD
2
2!q!BVBD
For GaAs: Nd = (12.8)(8.9x10-14)(4x105)2
(2)(1.6x10-19)(BVBD) =
5.7x1017BVBD
For 6H-SiC: Nd = (10)(8.9x10-14)(2x106)2
(2)(1.6x10-19)(BVBD) =
1.1x1019BVBD
![Page 416: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/416.jpg)
Rdrift = Wd
q!mn!Nd!A ; Rdrift,sp = Rdrift A = Wd
q!mn!Nd
Using Eq. (20-1) Nd = e!EBD
2
2!q!BVBD ; Using Eq. (20-3) Wd =
2!!BVBDEBD
Substituting into the expression for Rdrift,sp yields
Rdrift,sp = 2!!BVBD
EBD
1q!e
2!q!BVBDe!EBD
2 = !!!4!!(BVBD)2
e!mn!(EBD)3
26-8. Silicon : Rdrift,sp = (4)(500)2
(11.7)(1500)(8.9x10-14)(3x105)3 = 0.024 ohms-cm2
GaAs: Rdrift,sp = (4)(500)2
(12.8)(8500)(8.9x10-14)(4x105)3 = 0.0016 ohms-cm2
6H-SiC: Rdrift,sp = (4)(500)2
(10)(600)(8.9x10-14)(2x106)3 = 2.3x10-4 ohms-cm2
Diamond: Rdrift,sp = (4)(500)2
(5.5)(2200)(8.9x10-14)(107)3 = 9.3x10-7 ohms-cm2
26-9. Diamond is the most suitable material for high temperature operation. It has the largest bandgap (by almost a factor of two) and thus the smallest intrinsic carrier density at anygiven temperature. This statement presumes that the phase change listed for diamond inthe table of material properties exceeds the sublimation temperature of SiC (1800 °C).
26-10. Eq. (20-1): Nd = e!EBD
2
2!q!BVBD
For GaAs: Nd = (12.8)(8.9x10-14)(4x105)2
(2)(1.6x10-19)(BVBD) =
5.7x1017BVBD
For 6H-SiC: Nd = (10)(8.9x10-14)(2x106)2
(2)(1.6x10-19)(BVBD) =
1.1x1019BVBD
![Page 417: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/417.jpg)
Rdrift = Wd
q!mn!Nd!A ; Rdrift,sp = Rdrift A = Wd
q!mn!Nd
Using Eq. (20-1) Nd = e!EBD
2
2!q!BVBD ; Using Eq. (20-3) Wd =
2!!BVBDEBD
Substituting into the expression for Rdrift,sp yields
Rdrift,sp = 2!!BVBD
EBD
1q!e
2!q!BVBDe!EBD
2 = !!!4!!(BVBD)2
e!mn!(EBD)3
26-8. Silicon : Rdrift,sp = (4)(500)2
(11.7)(1500)(8.9x10-14)(3x105)3 = 0.024 ohms-cm2
GaAs: Rdrift,sp = (4)(500)2
(12.8)(8500)(8.9x10-14)(4x105)3 = 0.0016 ohms-cm2
6H-SiC: Rdrift,sp = (4)(500)2
(10)(600)(8.9x10-14)(2x106)3 = 2.3x10-4 ohms-cm2
Diamond: Rdrift,sp = (4)(500)2
(5.5)(2200)(8.9x10-14)(107)3 = 9.3x10-7 ohms-cm2
26-9. Diamond is the most suitable material for high temperature operation. It has the largest bandgap (by almost a factor of two) and thus the smallest intrinsic carrier density at anygiven temperature. This statement presumes that the phase change listed for diamond inthe table of material properties exceeds the sublimation temperature of SiC (1800 °C).
26-10. Eq. (20-1): Nd = e!EBD
2
2!q!BVBD
For GaAs: Nd = (12.8)(8.9x10-14)(4x105)2
(2)(1.6x10-19)(BVBD) =
5.7x1017BVBD
For 6H-SiC: Nd = (10)(8.9x10-14)(2x106)2
(2)(1.6x10-19)(BVBD) =
1.1x1019BVBD
![Page 418: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/418.jpg)
Rdrift = Wd
q!mn!Nd!A ; Rdrift,sp = Rdrift A = Wd
q!mn!Nd
Using Eq. (20-1) Nd = e!EBD
2
2!q!BVBD ; Using Eq. (20-3) Wd =
2!!BVBDEBD
Substituting into the expression for Rdrift,sp yields
Rdrift,sp = 2!!BVBD
EBD
1q!e
2!q!BVBDe!EBD
2 = !!!4!!(BVBD)2
e!mn!(EBD)3
26-8. Silicon : Rdrift,sp = (4)(500)2
(11.7)(1500)(8.9x10-14)(3x105)3 = 0.024 ohms-cm2
GaAs: Rdrift,sp = (4)(500)2
(12.8)(8500)(8.9x10-14)(4x105)3 = 0.0016 ohms-cm2
6H-SiC: Rdrift,sp = (4)(500)2
(10)(600)(8.9x10-14)(2x106)3 = 2.3x10-4 ohms-cm2
Diamond: Rdrift,sp = (4)(500)2
(5.5)(2200)(8.9x10-14)(107)3 = 9.3x10-7 ohms-cm2
26-9. Diamond is the most suitable material for high temperature operation. It has the largest bandgap (by almost a factor of two) and thus the smallest intrinsic carrier density at anygiven temperature. This statement presumes that the phase change listed for diamond inthe table of material properties exceeds the sublimation temperature of SiC (1800 °C).
26-10. Eq. (20-1): Nd = e!EBD
2
2!q!BVBD
For GaAs: Nd = (12.8)(8.9x10-14)(4x105)2
(2)(1.6x10-19)(BVBD) =
5.7x1017BVBD
For 6H-SiC: Nd = (10)(8.9x10-14)(2x106)2
(2)(1.6x10-19)(BVBD) =
1.1x1019BVBD
![Page 419: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/419.jpg)
For diamond: Nd = (5.5)(8.9x10-14)(107)2
(2)(1.6x10-19)(BVBD) =
1.5x1020BVBD
Eq. (20-3): Wd = 2!!BVBD
EBD
For GaAs: Wd = 2!!BVBD
4x105 = 5x10-6 BVBD [cm]
For 6H-SiC: Wd = 2!!BVBD
2x106 = 10-6 BVBD [cm]
For diamond: Wd = 2!!BVBD
107 = 2x10-7 BVBD [cm]
26-11. Use equations from problem 26-10.
Material Nd Wd
GaAs 2.9x1015 cm-3 10-2 cm
6H-SiC 5.5x1016 cm-3 2x10-3 cm
Diamond 7.5x1017 cm-3 4x10-5 cm
26-12. Tj = Rqjc Pdiode + Tcase : Rqjc = C.• (k)-1
k = thermal conductivity and C = constant
Using silicon diode data: C = (Tj!!-!!Tcase)!k
Pdiode =
(150!-!50)(!1.5)200 = 0.75 cm-1
Rqjc(GaAs) = 0.750.5 = 1.5 °C/W : Rqjc(SiC) =
0.755 = 0.15°C/W
Rqjc(diamond) = 0.7520 = 0.038°C/W
Tj(GaAs) = (1.5)(200) + 50 = 350 °C : Tj(SiC) = (0.15)(200) + 50 = 80 °C
![Page 420: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/420.jpg)
For diamond: Nd = (5.5)(8.9x10-14)(107)2
(2)(1.6x10-19)(BVBD) =
1.5x1020BVBD
Eq. (20-3): Wd = 2!!BVBD
EBD
For GaAs: Wd = 2!!BVBD
4x105 = 5x10-6 BVBD [cm]
For 6H-SiC: Wd = 2!!BVBD
2x106 = 10-6 BVBD [cm]
For diamond: Wd = 2!!BVBD
107 = 2x10-7 BVBD [cm]
26-11. Use equations from problem 26-10.
Material Nd Wd
GaAs 2.9x1015 cm-3 10-2 cm
6H-SiC 5.5x1016 cm-3 2x10-3 cm
Diamond 7.5x1017 cm-3 4x10-5 cm
26-12. Tj = Rqjc Pdiode + Tcase : Rqjc = C.• (k)-1
k = thermal conductivity and C = constant
Using silicon diode data: C = (Tj!!-!!Tcase)!k
Pdiode =
(150!-!50)(!1.5)200 = 0.75 cm-1
Rqjc(GaAs) = 0.750.5 = 1.5 °C/W : Rqjc(SiC) =
0.755 = 0.15°C/W
Rqjc(diamond) = 0.7520 = 0.038°C/W
Tj(GaAs) = (1.5)(200) + 50 = 350 °C : Tj(SiC) = (0.15)(200) + 50 = 80 °C
![Page 421: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/421.jpg)
For diamond: Nd = (5.5)(8.9x10-14)(107)2
(2)(1.6x10-19)(BVBD) =
1.5x1020BVBD
Eq. (20-3): Wd = 2!!BVBD
EBD
For GaAs: Wd = 2!!BVBD
4x105 = 5x10-6 BVBD [cm]
For 6H-SiC: Wd = 2!!BVBD
2x106 = 10-6 BVBD [cm]
For diamond: Wd = 2!!BVBD
107 = 2x10-7 BVBD [cm]
26-11. Use equations from problem 26-10.
Material Nd Wd
GaAs 2.9x1015 cm-3 10-2 cm
6H-SiC 5.5x1016 cm-3 2x10-3 cm
Diamond 7.5x1017 cm-3 4x10-5 cm
26-12. Tj = Rqjc Pdiode + Tcase : Rqjc = C.• (k)-1
k = thermal conductivity and C = constant
Using silicon diode data: C = (Tj!!-!!Tcase)!k
Pdiode =
(150!-!50)(!1.5)200 = 0.75 cm-1
Rqjc(GaAs) = 0.750.5 = 1.5 °C/W : Rqjc(SiC) =
0.755 = 0.15°C/W
Rqjc(diamond) = 0.7520 = 0.038°C/W
Tj(GaAs) = (1.5)(200) + 50 = 350 °C : Tj(SiC) = (0.15)(200) + 50 = 80 °C
![Page 422: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/422.jpg)
Tj(diamond) = (0.038)(200) + 50 = 57.4 °C
26-13. Pinch-off of the channel occurs when the depletion region of the gate-channel (P+N-) junction is equal to W/2 where W is the width of the channel. Occurs at a gate-source voltage of -Vp. The other half of the channel is depleted by the depletion region from the gate-channel junction on the other side of the channel. See Figs. 26-1 and 26-3.
W2 = Wo 1!+!
Vpfc ; fc =
kTq ln
ÎÍÍÈ
˚˙˙˘Na!Nd
ni2 ; Wo =
2!e!fcq!Nd
Solving for Vp yields Vp = fc ÎÍÈ
˚˙W
2!Wo 2 - fc
fc = 0.026 ln ÎÍÈ
˚˙˘(1019)(2x1014)
1010 = 0.8 V
Wo = (2)(11.7)(8.9x10-14)(0.8)
(1.6x10-19)(2x1014) = 2.3 microns
Vp = (0.8) ÎÍÈ
˚˙10
(2)(2.3) 2 - 0.8 = 3.8 - 0.8 = 3 V
26-14. Single cell of the multi-cell JFET shown below. See Fig. 26-1 for a fuller picture of the multi-cell nature of the JFET. The diagram on the left indicates the various contributions to the on-state resistance and the figure on the right shows the various geometrical factors that determine the resistance. Each cell is d centimeters deep in the direction perpendicular to the plane (page) of the diagram. The gate-source voltage is set at zero.
![Page 423: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/423.jpg)
Tj(diamond) = (0.038)(200) + 50 = 57.4 °C
26-13. Pinch-off of the channel occurs when the depletion region of the gate-channel (P+N-) junction is equal to W/2 where W is the width of the channel. Occurs at a gate-source voltage of -Vp. The other half of the channel is depleted by the depletion region from the gate-channel junction on the other side of the channel. See Figs. 26-1 and 26-3.
W2 = Wo 1!+!
Vpfc ; fc =
kTq ln
ÎÍÍÈ
˚˙˙˘Na!Nd
ni2 ; Wo =
2!e!fcq!Nd
Solving for Vp yields Vp = fc ÎÍÈ
˚˙W
2!Wo 2 - fc
fc = 0.026 ln ÎÍÈ
˚˙˘(1019)(2x1014)
1010 = 0.8 V
Wo = (2)(11.7)(8.9x10-14)(0.8)
(1.6x10-19)(2x1014) = 2.3 microns
Vp = (0.8) ÎÍÈ
˚˙10
(2)(2.3) 2 - 0.8 = 3.8 - 0.8 = 3 V
26-14. Single cell of the multi-cell JFET shown below. See Fig. 26-1 for a fuller picture of the multi-cell nature of the JFET. The diagram on the left indicates the various contributions to the on-state resistance and the figure on the right shows the various geometrical factors that determine the resistance. Each cell is d centimeters deep in the direction perpendicular to the plane (page) of the diagram. The gate-source voltage is set at zero.
![Page 424: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/424.jpg)
Tj(diamond) = (0.038)(200) + 50 = 57.4 °C
26-13. Pinch-off of the channel occurs when the depletion region of the gate-channel (P+N-) junction is equal to W/2 where W is the width of the channel. Occurs at a gate-source voltage of -Vp. The other half of the channel is depleted by the depletion region from the gate-channel junction on the other side of the channel. See Figs. 26-1 and 26-3.
W2 = Wo 1!+!
Vpfc ; fc =
kTq ln
ÎÍÍÈ
˚˙˙˘Na!Nd
ni2 ; Wo =
2!e!fcq!Nd
Solving for Vp yields Vp = fc ÎÍÈ
˚˙W
2!Wo 2 - fc
fc = 0.026 ln ÎÍÈ
˚˙˘(1019)(2x1014)
1010 = 0.8 V
Wo = (2)(11.7)(8.9x10-14)(0.8)
(1.6x10-19)(2x1014) = 2.3 microns
Vp = (0.8) ÎÍÈ
˚˙10
(2)(2.3) 2 - 0.8 = 3.8 - 0.8 = 3 V
26-14. Single cell of the multi-cell JFET shown below. See Fig. 26-1 for a fuller picture of the multi-cell nature of the JFET. The diagram on the left indicates the various contributions to the on-state resistance and the figure on the right shows the various geometrical factors that determine the resistance. Each cell is d centimeters deep in the direction perpendicular to the plane (page) of the diagram. The gate-source voltage is set at zero.
![Page 425: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/425.jpg)
Rd
R t
R c
R s
P+P+
drain
source
P+P+ W - 2Wo
W
W + Wg
W + W /2o g
l c
l gsWg
l - W - W /2gd o g
Rs = lgs
q!mn!Nd!d!W = 10-3
(1.6x10-19)(1500)(2x1014)(0.07)(10-3) = 298 ohms
Rc = lc
q!mn!Nd!d!(W!-!2Wo)
= 10-3
(1.6x10-19)(1500)(2x1014)(0.07)(10-3!-!4.6x10-4) = 552 ohms
Rt estimate. Treat the region of thickness Wo + Wg/2 as though it has an average width
given by (W!-!2!Wo)!+!(W!+!Wg)
2 = W + Wg/2 - Wo. Rt now approximately given by
Rt = Wo!+!Wg/2
q!mn!Nd!d!(W!+!Wg/2!-!Wo)
Rt = (10-3!+!5x10-4)
(1.6x10-19)(1500)(2x1014)(0.07)(10-3!+!5x10-4!-!2.3x10-4) = 351 ohms
Rd = (lgd!-!Wo!-Wg/2)
q!mn!Nd!d!(W!+!Wg) =
Rd = (35x10-4!-!2.3x10-4!-!5x10-4)
(1.6x10-19)(1500)(2x1014)(0.07)(10-3!+10-3) = 412 ohms
Total resistance of a single cell is Rcell = Rs + Rc + Rt + Rd
![Page 426: Power Electronics, Mohan 2nd Ed Solutions Manual](https://reader031.vdocument.in/reader031/viewer/2022013121/552c54784a7959e17c8b4707/html5/thumbnails/426.jpg)
Rcell = 298 + 552 + 351 + 412 = 1613 ohms
There are N = 28 cells in parallel so the the net on-state resistance is
Ron = Rcell
N = 161328 = 58 ohms
26-15. As the drain-source voltage increases, the reverse-bias on the gate-drain pn junction increases. The depletion region of the two adjacent P+ regions merge and then grow towards the drain. The drift region of length lgd and doping Nd must contain this depletion region and will determine the breakdown voltage. The short length of the drift region suggests that punch-through will limit the breakdown voltage. Check for this possibility first.
Non-punch-through estimate:
BV = 1.3x1017
2x1014 = 650 V ; Wd > (10-5)(6.5x102) = 65 microns > 35 microns
Hence this is a punch-through structure. Use Eq. (21-21) and EBD = 2x105 V/cm
BV = (2x105)(3.5x10-3) - (1.6x10-19)(2x1014)(3.5x10-3)2
(2)(11.7)(8.9x10-14) = 700 - 189
BV = 511 V
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Rcell = 298 + 552 + 351 + 412 = 1613 ohms
There are N = 28 cells in parallel so the the net on-state resistance is
Ron = Rcell
N = 161328 = 58 ohms
26-15. As the drain-source voltage increases, the reverse-bias on the gate-drain pn junction increases. The depletion region of the two adjacent P+ regions merge and then grow towards the drain. The drift region of length lgd and doping Nd must contain this depletion region and will determine the breakdown voltage. The short length of the drift region suggests that punch-through will limit the breakdown voltage. Check for this possibility first.
Non-punch-through estimate:
BV = 1.3x1017
2x1014 = 650 V ; Wd > (10-5)(6.5x102) = 65 microns > 35 microns
Hence this is a punch-through structure. Use Eq. (21-21) and EBD = 2x105 V/cm
BV = (2x105)(3.5x10-3) - (1.6x10-19)(2x1014)(3.5x10-3)2
(2)(11.7)(8.9x10-14) = 700 - 189
BV = 511 V
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Chapter 27 Problem Solutions
27-1. a. During turn-off of the GTO, Io communtates linearly to Cs.
Cs dvCdt = Io
ttfi
; dvCdt =
dvAKdt =
Io!tCs!tfi
< 5x107 V/s
Maximum dvAK
dt occurs at tfi. Solving for Cs yields
Cs > Io ÎÍÈ
˚˙!
dvAKdt
-1 =
5005x107 = 10 microfarads
Rs chosen on basis of limiting discharge current from Cs to safe level when GTO turns on. ICs,max = IAM - Io - Irr . Assume irr = 0.2 Io. Then
ICs,max = 1000 - 500 -100 = 400 A
Rs = !500!400 ≈ 1.3 ohms
Snubber recovery time = 2.3 Rs Cs = (2.3)(10-5)(1.3) = 30 microseconds.
b. Power dissipated in snubber PRs ≈ fsw!Cs!Vd
2
2
PRs = (0.5)(103)(10-5)(500)2 = 1.25 kW
27-2. L sdiAdt max = Vd ; Ls ≈
5003x108 ≈ 1.7 microhenries.
Voltage across GTO at turn-off = Vd + Io Rs : Assume Io
Rs = 0.2 Vd
Rs = (0.2)(500)
(500) = 0.2 ohms.
27-3. vCs(t) = Vd - Vd cos(wot) + Vd Cbase
Cs sin(wot) = Vd + K sin(wot - f)
vCs,max = Vd + K ; K sin(wot - f) = K sin(wot) cos(f) - Kcos(wot) sin(f)
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Chapter 27 Problem Solutions
27-1. a. During turn-off of the GTO, Io communtates linearly to Cs.
Cs dvCdt = Io
ttfi
; dvCdt =
dvAKdt =
Io!tCs!tfi
< 5x107 V/s
Maximum dvAK
dt occurs at tfi. Solving for Cs yields
Cs > Io ÎÍÈ
˚˙!
dvAKdt
-1 =
5005x107 = 10 microfarads
Rs chosen on basis of limiting discharge current from Cs to safe level when GTO turns on. ICs,max = IAM - Io - Irr . Assume irr = 0.2 Io. Then
ICs,max = 1000 - 500 -100 = 400 A
Rs = !500!400 ≈ 1.3 ohms
Snubber recovery time = 2.3 Rs Cs = (2.3)(10-5)(1.3) = 30 microseconds.
b. Power dissipated in snubber PRs ≈ fsw!Cs!Vd
2
2
PRs = (0.5)(103)(10-5)(500)2 = 1.25 kW
27-2. L sdiAdt max = Vd ; Ls ≈
5003x108 ≈ 1.7 microhenries.
Voltage across GTO at turn-off = Vd + Io Rs : Assume Io
Rs = 0.2 Vd
Rs = (0.2)(500)
(500) = 0.2 ohms.
27-3. vCs(t) = Vd - Vd cos(wot) + Vd Cbase
Cs sin(wot) = Vd + K sin(wot - f)
vCs,max = Vd + K ; K sin(wot - f) = K sin(wot) cos(f) - Kcos(wot) sin(f)
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Chapter 27 Problem Solutions
27-1. a. During turn-off of the GTO, Io communtates linearly to Cs.
Cs dvCdt = Io
ttfi
; dvCdt =
dvAKdt =
Io!tCs!tfi
< 5x107 V/s
Maximum dvAK
dt occurs at tfi. Solving for Cs yields
Cs > Io ÎÍÈ
˚˙!
dvAKdt
-1 =
5005x107 = 10 microfarads
Rs chosen on basis of limiting discharge current from Cs to safe level when GTO turns on. ICs,max = IAM - Io - Irr . Assume irr = 0.2 Io. Then
ICs,max = 1000 - 500 -100 = 400 A
Rs = !500!400 ≈ 1.3 ohms
Snubber recovery time = 2.3 Rs Cs = (2.3)(10-5)(1.3) = 30 microseconds.
b. Power dissipated in snubber PRs ≈ fsw!Cs!Vd
2
2
PRs = (0.5)(103)(10-5)(500)2 = 1.25 kW
27-2. L sdiAdt max = Vd ; Ls ≈
5003x108 ≈ 1.7 microhenries.
Voltage across GTO at turn-off = Vd + Io Rs : Assume Io
Rs = 0.2 Vd
Rs = (0.2)(500)
(500) = 0.2 ohms.
27-3. vCs(t) = Vd - Vd cos(wot) + Vd Cbase
Cs sin(wot) = Vd + K sin(wot - f)
vCs,max = Vd + K ; K sin(wot - f) = K sin(wot) cos(f) - Kcos(wot) sin(f)
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K sin(wot) cos(f) - K cos(wot) sin(f) = Vd Cbase
Cs sin(wot) - Vd cos(wot)
K cos(f) = Vd Cbase
Cs and K sin(f) = Vd ;
[ ]K!cos(f) 2 + [ ]K!sin(f) 2 = K2 = Vd2
CbaseCs
+ Vd2
vCs,max = Vd + K = Vd + Vd 1!+!Cbase
Cs
27-4. a. Equivalent circuit after diode reverse recovery.
L = 10 mH
Rs
Cs
200 V+
-i L
iL(0+) = Irr ; During reverse recovery L diRdt = 200 V
diRdt =
Irrtrr
= 20010-5 = 2x107 A/sec ; Irr = (2x107)(3x10-7) = 6 A
b. vCs,max = 500 V = 200 + 200 1!+!Cbase
Cs
1!+!Cbase
Cs = 1.5 ;
CbaseCs
= 1.5 ≈ 1.25
Cbase = (10-5) ÎÍÈ
˚˙˘62
(200)2 = 9 nF
Cs = 9!nF1.25 ≈ 7 nF
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K sin(wot) cos(f) - K cos(wot) sin(f) = Vd Cbase
Cs sin(wot) - Vd cos(wot)
K cos(f) = Vd Cbase
Cs and K sin(f) = Vd ;
[ ]K!cos(f) 2 + [ ]K!sin(f) 2 = K2 = Vd2
CbaseCs
+ Vd2
vCs,max = Vd + K = Vd + Vd 1!+!Cbase
Cs
27-4. a. Equivalent circuit after diode reverse recovery.
L = 10 mH
Rs
Cs
200 V+
-i L
iL(0+) = Irr ; During reverse recovery L diRdt = 200 V
diRdt =
Irrtrr
= 20010-5 = 2x107 A/sec ; Irr = (2x107)(3x10-7) = 6 A
b. vCs,max = 500 V = 200 + 200 1!+!Cbase
Cs
1!+!Cbase
Cs = 1.5 ;
CbaseCs
= 1.5 ≈ 1.25
Cbase = (10-5) ÎÍÈ
˚˙˘62
(200)2 = 9 nF
Cs = 9!nF1.25 ≈ 7 nF
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27-5. Use the circuit shown in problem 27-4.
Cs = Cbase = (10-5) ÎÍÈ
˚˙˘62
(200)2 = 9 nF
Rs = 1.3 Rbase = (1.3) ÎÍÈ
˚˙200
6 = 43 ohms
vCs,max = (1.5)(200) = 300 V
27-6. P = WR fsw = ÎÍÈ
˚˙˘Ls!Irr
2!+!Cs!Vin2
2 fsw
WR = (0.5)(10-5)(6)2 + (0.5)(9x10-9)(200)2 = 3.6x10-4 Joules
P = WR fsw = (3.6x10-4 )(2x104) = 7.2 watts
27-7. a. BJT waveforms (trv assumed to be zero for Cs = 0)
I o
i Cs
tt f i
t
vCE
Vd
C = 0s
C > 0s
0
0
Power dissipation for Cs = 0 is Pc = Vd!Io!tfi
2 fsw
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27-5. Use the circuit shown in problem 27-4.
Cs = Cbase = (10-5) ÎÍÈ
˚˙˘62
(200)2 = 9 nF
Rs = 1.3 Rbase = (1.3) ÎÍÈ
˚˙200
6 = 43 ohms
vCs,max = (1.5)(200) = 300 V
27-6. P = WR fsw = ÎÍÈ
˚˙˘Ls!Irr
2!+!Cs!Vin2
2 fsw
WR = (0.5)(10-5)(6)2 + (0.5)(9x10-9)(200)2 = 3.6x10-4 Joules
P = WR fsw = (3.6x10-4 )(2x104) = 7.2 watts
27-7. a. BJT waveforms (trv assumed to be zero for Cs = 0)
I o
i Cs
tt f i
t
vCE
Vd
C = 0s
C > 0s
0
0
Power dissipation for Cs = 0 is Pc = Vd!Io!tfi
2 fsw
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27-5. Use the circuit shown in problem 27-4.
Cs = Cbase = (10-5) ÎÍÈ
˚˙˘62
(200)2 = 9 nF
Rs = 1.3 Rbase = (1.3) ÎÍÈ
˚˙200
6 = 43 ohms
vCs,max = (1.5)(200) = 300 V
27-6. P = WR fsw = ÎÍÈ
˚˙˘Ls!Irr
2!+!Cs!Vin2
2 fsw
WR = (0.5)(10-5)(6)2 + (0.5)(9x10-9)(200)2 = 3.6x10-4 Joules
P = WR fsw = (3.6x10-4 )(2x104) = 7.2 watts
27-7. a. BJT waveforms (trv assumed to be zero for Cs = 0)
I o
i Cs
tt f i
t
vCE
Vd
C = 0s
C > 0s
0
0
Power dissipation for Cs = 0 is Pc = Vd!Io!tfi
2 fsw
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Pc = (200)(25)(4x10-7)
2 (2x104) = 20 W
Power dissipation for Cs = Cs1
Pc = Wc fsw ; Wc = ıÙÛ
0
tfi
Io(1!-!ttfi
)!!Io
2!Cs1!t2tfi
!!d!t = Vd!Io!tfi
12
Pc = (200)(25)(4x10-7)(2x104)
12 = 3.3 watts
Factor of six reduction in the turn-off losses.
b. BJT losses increase at turn-on only becaue of energy stored in Cs being dissipatedin the BJT, but also because the time to complete turn-on is extended as shown in
Fig. 27-14a. This extended duration of traversal of the active region also increases the turn-on losses.
During the turn-on interval, the collector-emitter voltage is given by (assuming that the external circuit dominates the transient)
Cs1 dvCE
dt = - diCdt t - Irr ; vCE(t) = Vd -
diCdt
t22!Cs1
- Irr t
Cs1
Seting the expression for vCE(t) equal to zero and solving for the timeDT = t2 - (tri + trr) (see Fig. 27-14a) required for vCE to reach zero yields
DT = - Irr
diC/dt + ÎÍÈ
˚˙˘
!!Irr
diC/dt2!+!
2!Vd!Cs1diC/dt !
Note that DT = 0 if Cs1 = 0 which is consistent with the assumption that the external circuit and not the BJT that dominates the turn-on transient. Extra energy disspated in the BJT at turn-on due to Cs1 is thus
ıÛ
0
DTvCE(t)!iC(t)!dt = Vd Irr DT + Î
ÍÈ
˚˙˘Vd
2 !diCdt !!!-!
[Io!+!Irr]!Irr2!Cs1
DT2
- (Io!+!3!Irr)
2!Cs1 diCdt DT3 - ÎÍ
È˚˙diC
dt 2 DT4
8!Cs1
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The increase in the BJT loss is ÎÍÍÈ
˚˙˙˘
ıÛ
0
DTvCE(t)!iC(t)!dt fs where fs is the switching
frequency. Numerical evaluation of DT gives DT = 0.29 ms (Irr = 10 A and Cs1 = 25 nF).
Evaluation of the loss gives 1.3x10-3 fs = 26.5 watts
27-8. a. Ds shorts out the snubber resistance during the BJT turn-off. Hence Cs1 is directly across the BJT as in problem 27-7a. Thus the loss reduction is the same as in problem 27-7a.
b. Equivalent circuit during BJT turn-off after free-wheeling diode reverse recovery is shown below.
Rs
sCI + tr rdi Cdt
vCE+
-vC
+
-v (0 ) = V+C d
vCE = CsRs dvCdt + vC ;; Cs
dvCdt = - Irr -
diCdt t
Combining equations and solving for vCE(t) yields
vCE(t) = Vd - Irr Rs - ÎÍÈ
˚˙˘
!IrrCs
!+!Rs!diCdt t -
diCdt
t22!Cs
At t = D T, vCE = 0 and turn-on is completed.
D T = - Irr!+!Rs!Cs!!
diCdt
!diCdt !
+ ÎÍÍÈ
˚˙˙˘
Irr!+!Rs!Cs!!diCdt
!diCdt !
2!+!!
2!CsdiCdt
![Vd!-!Irr!Rs]
D T goes to zero when Rs = VdIrr
. hence there is no increase in power dissipation
in the BJT due to the presence of Cs.
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The increase in the BJT loss is ÎÍÍÈ
˚˙˙˘
ıÛ
0
DTvCE(t)!iC(t)!dt fs where fs is the switching
frequency. Numerical evaluation of DT gives DT = 0.29 ms (Irr = 10 A and Cs1 = 25 nF).
Evaluation of the loss gives 1.3x10-3 fs = 26.5 watts
27-8. a. Ds shorts out the snubber resistance during the BJT turn-off. Hence Cs1 is directly across the BJT as in problem 27-7a. Thus the loss reduction is the same as in problem 27-7a.
b. Equivalent circuit during BJT turn-off after free-wheeling diode reverse recovery is shown below.
Rs
sCI + tr rdi Cdt
vCE+
-vC
+
-v (0 ) = V+C d
vCE = CsRs dvCdt + vC ;; Cs
dvCdt = - Irr -
diCdt t
Combining equations and solving for vCE(t) yields
vCE(t) = Vd - Irr Rs - ÎÍÈ
˚˙˘
!IrrCs
!+!Rs!diCdt t -
diCdt
t22!Cs
At t = D T, vCE = 0 and turn-on is completed.
D T = - Irr!+!Rs!Cs!!
diCdt
!diCdt !
+ ÎÍÍÈ
˚˙˙˘
Irr!+!Rs!Cs!!diCdt
!diCdt !
2!+!!
2!CsdiCdt
![Vd!-!Irr!Rs]
D T goes to zero when Rs = VdIrr
. hence there is no increase in power dissipation
in the BJT due to the presence of Cs.
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27-9. a. Proposed snubber circuit configuration shown below.
1
23
4
L s
R s
C s
I o2 Vs
Equivalent circuit swith SCRs 3 & 4 on and 1&2 going off or vice-versa. Continuous flow of load current formces SCRs 1 & 2 to remain on past the time of natural commutation (when vs(t) goes through zero and becomes negative).
L s
R s
C s
2 Vs
3 or 4
1 or 2
I r r
With 3 & 4 on, 1 & 2 are off, and effectively in parallel with the Rs-Cs snubber. same is true when 1 & 2 are on and 3 & 4 are off. Thus the Rs-Cs snubber functions as a turn-off snubber.
b. w Ls = 0.05!Vs
Ia1 ; worst case situation (maximum reverse voltage across SCR
which is turning off) occurs when SCR which is turning on is triggered with a
delay angle of 90°. During reverse recovery of SCR1, Ls didt = 2 Vs and
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didt =
Irrtrr
. Solving for Irr yields
Irr = 2!w!trr!Ia1
!0.05 ; Cbase = Ls ÎÍÍÈ
˚˙˙˘
!Irr
! 2!Vs 2
= !w!!Ia1!trr
2
!0.05!Vs
Smallest overvoltage occurs when Cs = Cbase. Putting in numerical values
Cs = 0.9933 Ia1 mF
Rbase = 2!VsIrr
= !0.05!Vsw!trr!Ia1
= (0.05)!(230)
(377)!(10-5)!Ia1 =
3050Ia1
ohms
Rs,opt = 1.3 Rbase = 4000Ia1
ohms
c. Peak line voltage = 2 (230) = 322 V
Smallest overvoltage = (1.5)(322) = 483 V which occurs when Cs = Cbaseand Rs = 1.3 Rbase. For Ia1 = 100 A
Rs,opt = 1.3 Rbase = 4000100 = 40 ohms ; Cs = (0.0033)(100) = 0.33 mF
27-10. The resistor in the BJT/MOSFET snubber must be shorted out during the device turn-off so that the snubber capacitance is in parallel with the device. The uncharged capacitor delays the build-up of the large Vd voltage across the BJT/MOSFET until most of the current has been diverted from the switch. The snubber diode is forward biased during turn-off, thus providing the shorting of the snubber resistor as required.
The turn-off of the thyristor limits overvoltages arising from the interruption of current through the stray inductance in series with the thyristor. Lowest overvoltages are obtained when Rs = 1.3 Rbase and Cs = Cbase. Overvoltages are 70-% larger if Rs is zero. Hence a diode in parallel with the resistor is not desirable.
27-11. a. Check dvDS
dt at turn-off; dvDS
dt = Vdtrv
; trv < 0.75 microseconds
dvDS
dt > 700
7.5x10-7 = 930 V/ms > 800 V/ms limit so snubber is needed.
Not enough information available to check on power dissipation or overcurrents.
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didt =
Irrtrr
. Solving for Irr yields
Irr = 2!w!trr!Ia1
!0.05 ; Cbase = Ls ÎÍÍÈ
˚˙˙˘
!Irr
! 2!Vs 2
= !w!!Ia1!trr
2
!0.05!Vs
Smallest overvoltage occurs when Cs = Cbase. Putting in numerical values
Cs = 0.9933 Ia1 mF
Rbase = 2!VsIrr
= !0.05!Vsw!trr!Ia1
= (0.05)!(230)
(377)!(10-5)!Ia1 =
3050Ia1
ohms
Rs,opt = 1.3 Rbase = 4000Ia1
ohms
c. Peak line voltage = 2 (230) = 322 V
Smallest overvoltage = (1.5)(322) = 483 V which occurs when Cs = Cbaseand Rs = 1.3 Rbase. For Ia1 = 100 A
Rs,opt = 1.3 Rbase = 4000100 = 40 ohms ; Cs = (0.0033)(100) = 0.33 mF
27-10. The resistor in the BJT/MOSFET snubber must be shorted out during the device turn-off so that the snubber capacitance is in parallel with the device. The uncharged capacitor delays the build-up of the large Vd voltage across the BJT/MOSFET until most of the current has been diverted from the switch. The snubber diode is forward biased during turn-off, thus providing the shorting of the snubber resistor as required.
The turn-off of the thyristor limits overvoltages arising from the interruption of current through the stray inductance in series with the thyristor. Lowest overvoltages are obtained when Rs = 1.3 Rbase and Cs = Cbase. Overvoltages are 70-% larger if Rs is zero. Hence a diode in parallel with the resistor is not desirable.
27-11. a. Check dvDS
dt at turn-off; dvDS
dt = Vdtrv
; trv < 0.75 microseconds
dvDS
dt > 700
7.5x10-7 = 930 V/ms > 800 V/ms limit so snubber is needed.
Not enough information available to check on power dissipation or overcurrents.
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didt =
Irrtrr
. Solving for Irr yields
Irr = 2!w!trr!Ia1
!0.05 ; Cbase = Ls ÎÍÍÈ
˚˙˙˘
!Irr
! 2!Vs 2
= !w!!Ia1!trr
2
!0.05!Vs
Smallest overvoltage occurs when Cs = Cbase. Putting in numerical values
Cs = 0.9933 Ia1 mF
Rbase = 2!VsIrr
= !0.05!Vsw!trr!Ia1
= (0.05)!(230)
(377)!(10-5)!Ia1 =
3050Ia1
ohms
Rs,opt = 1.3 Rbase = 4000Ia1
ohms
c. Peak line voltage = 2 (230) = 322 V
Smallest overvoltage = (1.5)(322) = 483 V which occurs when Cs = Cbaseand Rs = 1.3 Rbase. For Ia1 = 100 A
Rs,opt = 1.3 Rbase = 4000100 = 40 ohms ; Cs = (0.0033)(100) = 0.33 mF
27-10. The resistor in the BJT/MOSFET snubber must be shorted out during the device turn-off so that the snubber capacitance is in parallel with the device. The uncharged capacitor delays the build-up of the large Vd voltage across the BJT/MOSFET until most of the current has been diverted from the switch. The snubber diode is forward biased during turn-off, thus providing the shorting of the snubber resistor as required.
The turn-off of the thyristor limits overvoltages arising from the interruption of current through the stray inductance in series with the thyristor. Lowest overvoltages are obtained when Rs = 1.3 Rbase and Cs = Cbase. Overvoltages are 70-% larger if Rs is zero. Hence a diode in parallel with the resistor is not desirable.
27-11. a. Check dvDS
dt at turn-off; dvDS
dt = Vdtrv
; trv < 0.75 microseconds
dvDS
dt > 700
7.5x10-7 = 930 V/ms > 800 V/ms limit so snubber is needed.
Not enough information available to check on power dissipation or overcurrents.
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b. dvDS
dt = IoCs
= 400 V/ms ; Cs = 100
4x108 = 0.25 mF
Choose Rs to limit total current ID to less than 150 A
150 A = 100 + 700Rs
; Rs = 700
!150!-!100 = 14 ohms
Check snubber recovery time = 2.3 Rs Cs = (2.3)(14)(2.5x10-7) = 8 ms
Off time of the IGBT is 10 microseconds which is greater than the snubber recovery time. Hence choice of Rs is fine.
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Chapater 28 Problem Solutions
28-1. Schematic of drive circuit shown below.
RG
VGG+
VGG-
VDS
100 A
100 V
+
-
vDS(t) waveform same as in problem 22-2.
During MOSFET turn-on dvDS
dt = Vdtfv
< 500 V/ms
From problem 22-2, Vdtfv
= ÎÍÈ
˚˙˘
VGG+!-!VGSth!-!!Iogm
RG!Cgd
During MOSFET turn-off, dvDS
dt = Vdtrv
< 500 V/ms
From problem 23-2, Vdtrv
= ÎÍÈ
˚˙˘
VGG-!+!VGSth!+!!Iogm
RG!Cgd
gm = Io
VGS!-!VGSth =
607!-!4 = 20 A/V
Estimate of RG for MOSFET turn-on:
5x108 V/sec > VGG+!-!4!-!
10020
(RG)(4x10-10) ; VGG+,min = VGSth +
Iogm
= 4 + 10020 = 9 V
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Choose VGG+ = 15 V to insure that MOSFET driven well into ohmic range to minimize on-state losses.
RG > 15!-!9
(4x10-10)(5x108) = 30 ohms
Estimate of RG at MOSFET turn-off:
5x108 V/sec > VGG-!+!4!+!
10020
(RG)(4x10-10)
Choose VGG- = -15 V to insure that MOSFET is held in off-state and to minimizeturn-off times.
RG > 15!+!9
(4x10-10)(5x108) = 115 ohms
Satisfy both turn-on and turn-joff requirements by choosing VGG+ = VGG-= 15 V and RG > 115 ohms.
28-2. a. Circuit diagram shown below.
R G1
RG2Q s
Tsw
Df+
-
I o= 200 A
Vd = 1000 V
When FCT is off we need VKG = (1.25) 100040 = 31.25 V = (1000)
RG2RG1!+!!RG2
Now RG1 + RG2 = 10-6 ohms so 31.25 = (1000) (10-6) RG2
RG2 = 31.25 kW and RG1 = 969 kW
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Choose VGG+ = 15 V to insure that MOSFET driven well into ohmic range to minimize on-state losses.
RG > 15!-!9
(4x10-10)(5x108) = 30 ohms
Estimate of RG at MOSFET turn-off:
5x108 V/sec > VGG-!+!4!+!
10020
(RG)(4x10-10)
Choose VGG- = -15 V to insure that MOSFET is held in off-state and to minimizeturn-off times.
RG > 15!+!9
(4x10-10)(5x108) = 115 ohms
Satisfy both turn-on and turn-joff requirements by choosing VGG+ = VGG-= 15 V and RG > 115 ohms.
28-2. a. Circuit diagram shown below.
R G1
RG2Q s
Tsw
Df+
-
I o= 200 A
Vd = 1000 V
When FCT is off we need VKG = (1.25) 100040 = 31.25 V = (1000)
RG2RG1!+!!RG2
Now RG1 + RG2 = 10-6 ohms so 31.25 = (1000) (10-6) RG2
RG2 = 31.25 kW and RG1 = 969 kW
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b. MOSFET characteristics
- large on-state current capability- low Ron- low BVDSS (BVDSS of 50-100 V should work)
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Chapter 29 Problem Solutions
29-1. Assume Ts = 120 °C and Ta = 20 °C.
Rq,rad ≈ 0.12A ; A in m2 ; Eq. (29-18)
Rq,conv ≈ 1
(1.34)(A) ÎÍÈ
˚˙˘dvert
DT 1/4
; Eq. (29-20)
Rq,conv ≈ 0.24A [dvert]
1/4 for DT = 100 °C.
A cube having a side of length dvert has a surface area A = 6 [dvert]2
Rq,conv ≈ 0.4
[dvert]1.75 ; Rq,rad ≈
0.02[dvert]
2
Rq,sa = net surface-to-ambient thermal resistance = Rq,conv!Rq,rad!
Rq,conv!!+!Rq,rad
Rq,sa = 0.04
![dvert]1.75!+!2![dvert]
2
Heat Sink # 1 2 3 5 6
Volume [m]3 7.6x10-5 10-4 1.8x10-4 2x10-4 3x10-4
dv = (vol.)1/3 [m] 0.042 0.046 0.057 0.058 0.067
A = 6 [dv]2 [m]2 0.011 0.013 0.019 0.002 0.027
dv1.75 0.004 0.046 0.0066 0.0069 0.0088
dv2 0.0018 0.0021 0.0032 0.0034 0.0045
Rq,sa [°C/W] 5.3 4.5 3.1 2.9 2.3
Rq,sa (measured) 3.2 2.3 2.2 2.1 1.7
Heat Sink # 7 8 9 10 11 12
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Volume [m]3 4.4x10-4 6.810-4 6.1x10-4 6.3x10-4 7x10-4 1.4x10-3
dv = (vol.)1/3 [m] 0.076 0.088 0.085 0.086 0.088 0.11
A = 6 [dv]2 [m]2 0.034 0.046 0.043 0.044 0.047 0.072
dv1.75 0.011 0.014 0.013 0.014 0.014 0.021
dv2 0.0058 0.0078 0.0071 0.0073 0.0078 0.012
Rq,sa [°C/W] 1.8 1.4 1.5 1.4 1.3 0.9
Rq,sa (measured) 1.3 1.3 1.25 1.2 0.8 0.65
Heat sink #9 is relatively large and cubical in shape with only a few cooling fins.Heat sink #9 is small and flat with much more surface area compared to its volume.Large surface-to-volume ratios give smaller values of Rq,sa.
29-2. Rq,conv ≈ 0.24A [dvert]
1/4 for DT = 100 °C ; From problem 29-1
Rq,conv ≈ 24 0 [dvert]1/4 for DT = 100 °C and A = 10 cm2 = 10-3 m2
dvert Rq,conv
1 cm 76 °C/W
5 cm 113 °C/W
12 cm 141 °C/W
20 cm 160 °C/W
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Volume [m]3 4.4x10-4 6.810-4 6.1x10-4 6.3x10-4 7x10-4 1.4x10-3
dv = (vol.)1/3 [m] 0.076 0.088 0.085 0.086 0.088 0.11
A = 6 [dv]2 [m]2 0.034 0.046 0.043 0.044 0.047 0.072
dv1.75 0.011 0.014 0.013 0.014 0.014 0.021
dv2 0.0058 0.0078 0.0071 0.0073 0.0078 0.012
Rq,sa [°C/W] 1.8 1.4 1.5 1.4 1.3 0.9
Rq,sa (measured) 1.3 1.3 1.25 1.2 0.8 0.65
Heat sink #9 is relatively large and cubical in shape with only a few cooling fins.Heat sink #9 is small and flat with much more surface area compared to its volume.Large surface-to-volume ratios give smaller values of Rq,sa.
29-2. Rq,conv ≈ 0.24A [dvert]
1/4 for DT = 100 °C ; From problem 29-1
Rq,conv ≈ 24 0 [dvert]1/4 for DT = 100 °C and A = 10 cm2 = 10-3 m2
dvert Rq,conv
1 cm 76 °C/W
5 cm 113 °C/W
12 cm 141 °C/W
20 cm 160 °C/W
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B
B
B
B
0
20
40
60
80
100
120
140
160
0 2 4 6 8 10 12 14 16 18 20
q
°C/W
dvert [cm]
R ,conv
29-3. Rq,conv ≈ 1
(1.34)(A) ÎÍÈ
˚˙˘dvert
DT 1/4
; Eq. (29-20)
Rq,conv ≈ 353 [DT]-.25 A = 10 cm2 and dvert = 5 cm
DT Rq,conv
60 °C 127 °C/W
80 °C 118 °C/W
100 °C 112 °C/W
120 °C 107 °C/W
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B
B
B
B
0
20
40
60
80
100
120
140
160
0 2 4 6 8 10 12 14 16 18 20
q
°C/W
dvert [cm]
R ,conv
29-3. Rq,conv ≈ 1
(1.34)(A) ÎÍÈ
˚˙˘dvert
DT 1/4
; Eq. (29-20)
Rq,conv ≈ 353 [DT]-.25 A = 10 cm2 and dvert = 5 cm
DT Rq,conv
60 °C 127 °C/W
80 °C 118 °C/W
100 °C 112 °C/W
120 °C 107 °C/W
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BB
B B
0
20
40
60
80
100
120
140
60 70 80 90 100 110 120
R ,convq
°C/W
DT [°C]
29-4. Rq,rad = !DT
5.1!A!ËÁÊ
¯˜!ËÁ
ʯ˜Ts
1004!-!ËÁ
ʯ˜Ta
1004!
; Eq. (29-17)
Rq,rad = 196 120!-!Ta(° C)
!ÎÍÈ
˚˙239!-!ÎÍ
È˚˙Ta(° K)
1004! ; A = 10 cm2 and Ts = 120 °C
Ta Rq,rad
0 °C 128 °C/W
10 °C 123 °C/W
20 °C 119 °C/W
40 °C 110 °C/W
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BB
B B
0
20
40
60
80
100
120
140
60 70 80 90 100 110 120
R ,convq
°C/W
DT [°C]
29-4. Rq,rad = !DT
5.1!A!ËÁÊ
¯˜!ËÁ
ʯ˜Ts
1004!-!ËÁ
ʯ˜Ta
1004!
; Eq. (29-17)
Rq,rad = 196 120!-!Ta(° C)
!ÎÍÈ
˚˙239!-!ÎÍ
È˚˙Ta(° K)
1004! ; A = 10 cm2 and Ts = 120 °C
Ta Rq,rad
0 °C 128 °C/W
10 °C 123 °C/W
20 °C 119 °C/W
40 °C 110 °C/W
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B B BB
0
20
40
60
80
100
120
140
0 5 10 15 20 25 30 35 40
Ta [°C ]
R ,rad
[ °C/W]
q
29-5. Rq,rad = !DT
5.1!A!ËÁÊ
¯˜!ËÁ
ʯ˜Ts
1004!-!ËÁ
ʯ˜Ta
1004!
; Eq. (29-17)
Rq,rad = 196 Ts(° C)!-!40
!ÎÍÈ
˚˙
ÎÍÈ
˚˙Ts(° K)
1004!
-!96 ; A = 10 cm2 and Ta= 40 °C
Ts Rq,rad
80 °C 114 °C/W
100 °C 120 °C/W
120 °C 110 °C/W
140 °C 101 °C/W
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B B BB
0
20
40
60
80
100
120
140
0 5 10 15 20 25 30 35 40
Ta [°C ]
R ,rad
[ °C/W]
q
29-5. Rq,rad = !DT
5.1!A!ËÁÊ
¯˜!ËÁ
ʯ˜Ts
1004!-!ËÁ
ʯ˜Ta
1004!
; Eq. (29-17)
Rq,rad = 196 Ts(° C)!-!40
!ÎÍÈ
˚˙
ÎÍÈ
˚˙Ts(° K)
1004!
-!96 ; A = 10 cm2 and Ta= 40 °C
Ts Rq,rad
80 °C 114 °C/W
100 °C 120 °C/W
120 °C 110 °C/W
140 °C 101 °C/W
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BB
BB
0
20
40
60
80
100
120
80 90 100 110 120 130 140
Ts [ °C ]
Rq,rad
[ °C/W]
29-6. PMOSFET,max = 150!° C!!-!50!° C!
!1!° C/W = 100 W ; 100 W = 50 + 10-3 fs
Solving for fs yields fs = 50 kHz
29-7. PMOSFET = 50 + 10-3 • 2.5x104 = 75 W
Rq,ja = Rq,jc + Rq,ca = 150!° C!!-!35!° C!
!75!W = 1.53 °C/W
Rq,ca = 1.53 - 1.00 = 0.53 °C/W
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BB
BB
0
20
40
60
80
100
120
80 90 100 110 120 130 140
Ts [ °C ]
Rq,rad
[ °C/W]
29-6. PMOSFET,max = 150!° C!!-!50!° C!
!1!° C/W = 100 W ; 100 W = 50 + 10-3 fs
Solving for fs yields fs = 50 kHz
29-7. PMOSFET = 50 + 10-3 • 2.5x104 = 75 W
Rq,ja = Rq,jc + Rq,ca = 150!° C!!-!35!° C!
!75!W = 1.53 °C/W
Rq,ca = 1.53 - 1.00 = 0.53 °C/W
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BB
BB
0
20
40
60
80
100
120
80 90 100 110 120 130 140
Ts [ °C ]
Rq,rad
[ °C/W]
29-6. PMOSFET,max = 150!° C!!-!50!° C!
!1!° C/W = 100 W ; 100 W = 50 + 10-3 fs
Solving for fs yields fs = 50 kHz
29-7. PMOSFET = 50 + 10-3 • 2.5x104 = 75 W
Rq,ja = Rq,jc + Rq,ca = 150!° C!!-!35!° C!
!75!W = 1.53 °C/W
Rq,ca = 1.53 - 1.00 = 0.53 °C/W