esercizi su np-pn hetero junction

8/3/2019 Esercizi Su Np-pn Hetero Junction

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Exercise n. 1

Consider a pn heterojunction between Al0.4Ga0.6As (hereafter material A), with NAA =51017 cm3 and Al0.15Ga0.85As (hereafter material B), with NDB = 210

16 cm3. Assumealloys properties as obtained by Vegard law for AlxGa1xAs:

Eg = 1.4 + 0.9 xq = 4.07 1.06 x

r = 12.9 2.84 x

and

NCA = 7.961017 cm3; NV A = 1.4910

19 cm3

NCB = 5.621017 cm3; NV B = 1.3610

19 cm3

1) Sketch the energy band diagram at thermal equilibrium, using the the affinity rulefor the evaluation of valence and conduction band discontinuity.

2) Evaluate the electric field distribution.

3) Evaluate the extension of the depletion region in each side of the junction andcalculate the peak electric field.

4) Calculate the excess of minority carriers injected at the boundary of the neutralregion under an applied forward bias of 1 V.

5) [advanced] Compare the extension of the depleted region at equilibrium with the De-bye length of the majority on both sides. Draw a detailled behavior of the depleted

region.

Solution

1) Band diagram: From Vegard law we can evaluate all relevant paramter:

Al0.4Ga0.6As : EGA = 1.760 eV, qA = 3.646 eV, rA = 11.764

Al0.15Ga0.85As : EGB = 1.535 eV, qB = 3.911 eV, rB = 12.474

Band edge discontinuities:

EC = q = qB + qA = 0.265 eV

EV = q EG = 0.04 eV

Type II heterojunction, see Fig.1 (left).

Work-functions and built-in potential:

qSA = qA + EGA KBT log(NV ANAA

) = 5.3177 eV

qSB = qB + KBT log(

NCB

NDB ) = 3.9977 eV


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q Vbi = U0B U0A = qSB qSA = 1.32 eV

Notice that our convention is to measure the built-in potential (energy) as thedifference of the vacuum level from the material on the right to the material on theleft: this may cause the built-in potential to become negative in some cases (like thisone). With the same hypothesis the built-in voltage is also the voltage differenceof the material on the left with respect to the material of the right (VA VB).The students need to be careful to maintain this convention in all the followingcalculations.

Type IInp

U0

EvB

EcB

EvA

EcA

EFB

EFA

0.265 eV

0.04 eV

U0

Ev

Ec

EF

0.265 eV

0.04 eV

1.32 eV

Figura 1: Band diagrams of material A and B separately and equilibrium band diagram

Electron transfer from material with the lower workfunction towards the one withthe higher (or holes in the opposite way). A space charge region builds up acrossthe heterojunction, like in p n homojunctions, see Fig. 2 (left). The bands benddownwards if the charge is negative and upwards if positive. The final equilibriumband diagram is skeched in Fig. 1 (right).

r

q NDB

-q NAA

-xp xnE-xp xn

Figura 2: Charge distribution and electric field

2) Electric Field: The analytic form of the charge distribution of Fig. 2 (left) may bewritten as:


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(x) =

0 x < xp

q NAA xp < x < 0

+ q NDB 0 < x < xn

0 x > xn

Enforcing charge neutrality, i.e. the same total negative charge in the p-side andpositive chare in the n-side, we have:

0

(x) dx =

0

(x) dx

q NDB xn = q NAA xp

Starting from the charge distribution profile and integrating Gauss law dEdx

= itis possible to derive the electric field profile. This will resemble the one found in the

homojunction case, except for a discontinuity at x = 0 (which corresponds to theheterojunction), due to the difference in the materials relative dielectric constants.Notice however that at x = 0 the continuity of the displacement vector: AE(x =0) = BE(x = 0

+) must be in any case be verified (Note: A = 0r,A and B = 0r,Bwith 0 = 8.8510

14 F/cm ).

The analytic expression of the electric field is:

E(x) =

0 x < xp

qNAA

A(x + xp) xp < x < 0

qNDBB

(x xn) 0 < x < xn

0 x > xn

where the integration constant have been found enforcing the continuity of theelectric filed for x = xp and x = xn. For x = 0 we have a discontinuity of theelectric field as expected:

E(x = 0) = qNAA

Axp

E(x = 0+) = qNDB

Bxn

but the continuity of the electric displacement vector E is verified due to the chargeneutality condition. In fact:

AE(x = 0) = BE(x = 0

+)

i.e.:qNAAxp = qNDB xn

Since A < B, the peak electric field will be located at x = 0, i.e. in the p-side of

the junction, see Fig. 2 (right).


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3) Depletion region widths:

Integrating the electric field we obtain the electrostatic potential fromddx =

E .

(x) =

0 x < xpqNAA

2A(x + xp)

2 xp < x < 0

qNDB

2B(x xn)

2 + |Vbi| 0 < x < xn

|Vbi| x > xn

where the continuity of the electrostatic potential in x = 0 yields:

|Vbi| =qNAA

2Ax2p +

qNDB2B

x2n

Solving the system: |Vbi| =

qNAA2A

x2p + + qNDB

2Bx2n

q NDB xn = q NAA xp

xp =

2(N)eq

qN2AA|Vbi| = 11.8 nm

xn =

2(N)eqqN2DB

|Vbi| = 295.5 nm

where

(N)eq = (ANAA)//(B NDB ) =ABNAANDB

ANAA + B NDB= 2.119104 Fcm

Hence the peak electric field:

|Emax| =qNAA

Axp = 91.115 kV/cm.

4 ) Minority carrier concentration at the border of the depleted region. First of all weneed to recall the relevant relationship in the equilibrium condition.

n side : ECA = U0A qA

p side : EV B = U0B qB EGB

so thatECA EV B = U0A qA U0B + qB + EGB

Recalling the definition of built-in potential

q Vbi = U0B U0A


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and that EC = q = qB + qA and EV = q EG, we have thetwo equivalent forms:

ECA EV B = q Vbi EC + EGB

ECA EV B = q Vbi EV + EGA (1)

The minority carrier concentration in the equilibrium condition condition at theborder of the depleted region can be expressed as:

p0(xn) = NV B exp

EF EV BkBT

n0(xp) = NCA exp

ECA EFkBT

The product of the minority carriers at the border of the depleted region on the twosides of the junction is:

p0(xn) n0(xp) = NCA NV B expECA EV B

kBT

(2)

while in the same side of the junction we have:

p0(xn) n0(xn) = p0(xn) NDB = NV B NCB exp

EGBkBT

and

p0(xp) n0(xp) = n0(xp) NAA = NV ANCA exp

EGAkBT

i.e.:p0(xn) =

NV B NCBNDB

exp

EGBkBT

and

n0(xp) =NV ANCA

NAAexp

EGAkBT

Substituting the above expressions in Eq. (2) and recalling Eq. (1)

p0(xn) = NAANV BNV A

exp(qVbi EV

kBT) = 8.757106 cm3.

n0(xp) = NDB NCA

NCBexp(qVbi EC

kBT) = 9.4831011 cm3.

The above results hold in equilibrium. 1 Outside equilibrium the externally appliedvoltage modifies the amount of voltage drop on the junction with respect to the

1The Fermi energy on the two sides

n side : EFB = U0B qB kBT log

NCB

NDB

p side : EFA = U0A qA EGA + kBT logNVANAA


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equilibrium value Vbi. As already noticed, the built-in voltage is the voltage diffe-rence of the material on the left with respect to the material of the right ( VAVB).On the other hand, any externally applied voltage is conventionally applied to thep-side of the junction with respect to the n-side. Hence in our case the externalapplied voltage will add to the built-in voltage:

Vbi Vbi + V

Under forward bias V = 1 V:

p(xn) = p0(xn) (exp(V /VT) 1) = 4.441011 cm3.

n(xp) = n0(xp) (exp(V /VT) 1) = 4.79106 cm3.

It is worth noticing that the ratio p(xn)/n(xp) 105 is much larger than the one

expected for a homojunction. In fact, in the homojunction this ratio depends only

on the doping levels and we would expect p(xn)/n(xp) = NA/ND = 25.5 ) The Debye lengths in each side of the heterojunction are:

LDA =

AkB T

q2NAA= 5.8 nm

LDB =

B kBT

q2NDB= 29.9 nm

Hence, the Debye length in the p-side is comparable to the extension of the depleted

region while in the n-side it is negligible. The depleted region on the p-side cannot be considered as abrupt (like in the full depletion approximation) but it has thebehavior shown in Fig. 3. Note that the value of xp must be recalculated with thenew accurate charge distribution by solving the Poisson equation.

The analytic form of the charge distribution may now written as:

must be equal at equilibrium, hence:

U0B qB kBT log

NCB

NDB

= U0A qA EGA + kBT log

NVA

NAA

q Vbi = EC EGA kBT logNVANCB

NAANDB

q Vbi = EV EGB kBT logNVANCB

NAANDB

and therefore we verify that

p0(xn) = NAANV B

NVAexp(

qVbi EVkBT

) =NV BNCB exp(

EGBkBT

)

NDB=

n2iBNDB

as expected in equilibrium. Analogously for the expression of n0(xp).


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rq NDB

-q NAA

-xp xn

LDA

Figura 3: Detailled charge distribution considering the Debye length of the majoritycarriers in the p-side

(x) =

q NAA exp

x + xp

LDA

x < xp

q NAA xp < x < 0

+ q NDB 0 < x < xn

0 x > xn


0

(x) dx =

0

(x) dx

0

(x) dx =

xp

(x) dx

0

xp

(x) dx

i.e. q NDB xn = q NAA LDA + q NAA xp

Definingxp = xp + LDA

we haveq NDB xn = q NAA x

p

Turning to the electric field, integrating the Gauss equation dEdx

= we obtain

E(x) =

qNAA

ALDA exp

x + xpLDA x < xp

qNAA

A

x + xp

xp < x < 0

qNDBB

(x xn) 0 < x < xn

0 x > xn

where the integration constant have been found by enforcing the continuity of theelectric filed for x = xp and x = xn and the conditionlimxE(x) = 0. Noticethat for x = 0 we have a discontinuity of the electric field:

E(x = 0

) =

qNAA

A x

p


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E(x = 0+) = qNDB

Bxn


AE(x = 0) = BE(x = 0

+)

i.e.:qNAAx

p = qNDB xn

Integrating the electric field we obtain the electrostatic potential fromddx =

E .

(x) =

qNAAA

L2DA exp

x + xp

LDA

x < xp

qNAA2A

x + xp

2+ L2DA

xp < x < 0

qNDB2B (x x

n)2

+ |Vbi| 0 < x < xn

|Vbi| x > xn


|Vbi| =qNAA

2A

xp

2+ L2DA

+

qNDB2B

(xn)2

which reduces to the standard expression for LDA 0.

Solving the system: |Vbi| =

qNAA2A

xp

2+ L2DA

+

qNDB2B

(xn)2

q NDB xn = q NAA x

p

Defining:

|Vbi| = |Vbi| qNAA

2AL2DA = 1.307 V

we have

xp = 2(N)eq

qN2AA

|Vbi| = 11.765 nm

xn =NAANDB

xp = 294.1 nm

and finally:xp = x

p LDA = 5.948 nm


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Exercise 2 (solved in class)

Consider a pn heterojunction between Al0.3Ga0.7As (hereafter material A), with NAA =11016 cm3 and GaAs (hereafter material B), with NDB = 510

16 cm3. Assume alloysproperties as obtained by Vegard law for AlxGa1xAs:

Eg = 1.41 + 0.87 xq = 4.07 1.06 x

r = 12.9 2.84 x

and

NCA = 6.991017 cm3; NV A = 1.4310

19 cm3

NCB = 4.71017 cm3; NV B = 710

18 cm3





5) Calculate the depletion capacitance for an applied voltage V = 5 V (voltages areapplied to the p-side with respect to the n-side).

Solution

1) Band diagram

From Vegard law we can evaluate all relevant paramter:


Band edge discontinuities:

EC = q = qB + qA = 0.32 eV

EV = q EG = 0.06 eV

Type II heterojunction. Work-functions and built-in potential:

qSA = qA + EGA KBT log(NV ANAA

) = 5.2311 eV

qSB = qB + KBT log(NCBNDB

) = 4.1283 eV


The qualitative behavior of the junction is like Exercise n. 1 (see Fig. 1)


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2-3) The qualitative behavior of the charge and electric field is like Exercise n. 1 (seeFig. 2).

(N)eq = (ANAA)//(BNDB ) =AB NAANDB

ANAA + BNDB= 8.9856103 Fcm

xp =

2(N)eqqN2AA

|Vbi| = 351.95 nm

xn =

2(N)eqqN2DB

|Vbi| = 70.39 nm


|Emax| =qNAA

Axp = 52.8 kV/cm.

Since A < B, the peak electric field will be located at x = 0, i.e. in the p-side of

the junction.4 ) Minority carrier concentration at the border of the depleted region.

p0(xn) = NAANV BNV A

exp(qVbi EV

kBT) = 1.8454104 cm3.

n0(xp) = NDBNCANCB

exp(qVbi EC

kBT) = 1.2728107 cm3.

The above results hold in equilibrium. Outside equilibrium the externally appliedvoltage modifies the amount of voltage drop on the junction with respect to theequilibrium value Vbi. As already noticed, the built-in voltage is the voltage diffe-

rence of the material on the left with respect to the material of the right ( VAVB).On the other hand, any externally applied voltage is conventionally applied to thep-side of the junction with respect to the n-side. Hence in our case the externalapplied voltage will add to the built-in voltage:

Vbi Vbi + V


p(xn) = p0(xn) (exp(V /VT) 1) = 9.32691012 cm3.

n(xp) = n0(xp) (exp(V /VT) 1) = 6.4325109 cm3.

5) Depletion capacitance.

The charge stored in the depleted region (e.g. in the n-side) is:

Q = qNAAxp =

2q(N)eq(|Vbi + V|)

and the depletion capacitance is

Cdepl =dQ

dV=

q(N)eq

2(|Vbi + V|)

For V = 5 V: Cdepl = 1.0853108 F/cm.


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Exercise n. 3

Consider a np heterojunction between Al0.3Ga0.7As (hereafter material A), with NDA =1017 cm3 and GaAs (hereafter material B), with NAB = 210

18 cm3. Assume alloysproperties as obtained by Vegard law for AlxGa1xAs:

Eg = 1.4 + 0.9 xq = 4.07 1.06 x

r = 12.9 2.84 x

and

NCA = 6.991017 cm3;NV A = 1.4310

19 cm3

NCB = 4.71017 cm3;NV B = 710

18 cm3





5) [advanced] Compare the extension of the depleted region at equilibrium with the De-bye length of the majority on both sides. Draw a detailled behavior of the depletedregion.

Solution

From Vegard law we can evaluate all relevant paramter:


GaAs : EGB = 1.4 eV, qB = 4.07 eV, rB = 12.9

1) Band diagram:

EC = q = qB + qA = 0.32 eV

EV = q EG = 0.05 eV

Type II heterojunction, see Fig.4 (left).


qSA = qA + KB T log(NCANDA

) = 3.8 eV

qSB = qB + EGB KBT log(

NV B

NAB ) = 5.4374 eV


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Notice that our convention is to measure the built-in potential (energy) as thedifference of the vacuum level from the material on the right to the material on theleft. With the same hypothesis the built-in voltage is also the voltage differenceof the material on the left with respect to the material of the right (VA VB).The students need to be careful in maintaining this convention in all the followingcalculations.

Type IIp

GaAs

nAlGaAs

U0

EvB

EcB

EvA

EcA

EFB

EFA

0.32 eV

0.05 eV

U0

Ev

Ec

EF

0.32 eV

0.05 eV

1.738 eV

Figura 4: Band diagrams of material A and B separately and equilibrium band diagram

A space charge region builds up across the heterojunction, as in np homojunctions,see Fig. 5 (left). The bands bend downwards if the charge is negative and upwardsif positive. The final equilibrium band diagram is skeched in Fig. 4 (right).

r q NDA

-q NAB

-xnxp

E

xp-xnFigura 5: Charge distribution and electric field

2) Electric Field: The analytic form of the charge distribution of Fig. 5 (left) may bewritten as:

(x) =

0 x < xp

+ q NDA xn < x < 0

q NAB 0 < x < xp

0 x > xn


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0

(x) dx =

0

(x) dx

q NDAxn = q NAB xp

Starting from the charge distribution profile and integrating Gauss law dEdx

= it

is possible to derive the electric field profile. This will resemble the one found in thehomojunction case, except for a discontinuity at x = 0 (which corresponds to theheterojunction), due to the difference in the materials relative dielectric constants.Notice however that at x = 0 the continuity of the displacement vector: AE(x =0) = BE(x = 0

+) must be in any case verified (Note: A = 0r,A and B = 0r,Bwith 0 = 8.8510

14 F/cm ).

The analytic expression of the electric field is:

E(x) =

0 x < xnqNDA

A(x + xn) xn < x < 0

qNAB

B(x xp) 0 < x < xp

0 x > xp

where the integration constant have been found enforcing the continuity of theelectric filed for x = xn and x = xp. For x = 0 we have a discontinuity of the

electric field as expected:

E(x = 0) =qNDA

Axn

E(x = 0+) =qNAB

Bxp


AE(x = 0) = BE(x = 0

+)

i.e.:qNDAxn = qNABxp

Since SA < SB , the peak electric field will be located at x = 0, i.e. in the n-side

of the junction, see Fig. 5 (right).

3) Depletion region widths:


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Integrating the electric field we obtain the electrostatic potential from ddx

= E .

(x) =

0 x < xn

qNDA

2A(x + xn)

2 xn < x < 0

qNAB

2B (x xp)

2

Vbi 0 < x < xp

Vbi x > xp


Vbi =qNDA

2Ax2n +

qNAB2B

x2p

Solving the system:

Vbi =qNDA

2Ax2n +

qNAB2B

x2p

qNDAxn = qNABxp

xp =

2(N)eq

qN2ABVbi = 6.48 nm

xn =

2(N)eq

qN2DAVbi = 129.66 nm

where

(N)eq = (ANDA)//(BNAB ) =AB NDANAB

ANDA + BNAB= 1.0188105 Fcm


|Emax| =qNDA

Axn = 194.53 kV/cm.

4 ) Minority carrier concentration at the border of the depleted region.

First of all we need to recall the relevant relationship in the equilibrium condition.2

n side : EV A = U0A qA EGA

p side : ECB = U0B qB

2The Fermi energy on the two sides

n side : EFA = U0A qA kBT log

NCA

NDA

p side : EFB = U0B qB EGB + kBT log

NV B

NAB

must be equal at equilibrium, hence:

U0A qA kBT logNCA

NDA

= U0B qB EGB + kBT logNVB

NAB


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so thatECB EV A = U0B qB U0A + qA + EGA

Recalling the definition of built-in potential:

q Vbi = U0B U0A

we haveECB EV A = q Vbi + EC + EGA (3)

The minority carrier concentration in the equilibrium condition condition at theborder of the depleted region can be expressed as:

p0(xn) = NV A exp

EF A EV AkBT

n0(xp) = NCB exp

ECB EF BkBT

The product of the minority carriers at the border of the depleted region on the twosides of the junction is:

p0(xn) n0(xp) = NV ANCB exp

ECB EV AkBT

(4)

while in the same side of the junction we have:

p0(xn) n0(xn) = p0(xn) NDA = NV ANCA exp

EGAkBT

andp0(xp) n0(xp) = n0(xp) NAB = NV B NCB exp

EGBkBT

i.e.:

p0(xn) =NV ANCA

NDAexp

EGAkBT

and

n0(xp) =NV B NCB

NABexp

EGBkBT

q Vbi = EC + EGB kBT logNVBNCA

NABNDA

q Vbi = EV + EGA kBT logNVBNCA

NABNDA

therefore we verify that :

p0(xn) = NABNV A

NV Bexp(

qVbi + EVkBT

) =NVANCA exp(

EGAkBT

)

NDA=

n2iANDA

as expected in equilibrium. Analogously for the expression of n0(xp).


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Substituting the above expressions in Eq. (4) and recalling Eq. (3)

p0(xn) = NABNV ANV B

exp(qVbi + EV

kBT) = 2.61010 cm3.

n0(xp) = NDANCBNCA

exp(qVbi + EC

kBT) = 1.387107 cm3.

Outside equilibrium the externally applied voltage modifies the amount of voltagedrop on the junction with respect to the equilibrium value Vbi. As already noticed,the built-in voltage is the voltage difference of the material on the left with respectto the material of the right (VA VB). On the other hand, any externally appliedvoltage V is conventionally applied to the p-side of the junction with respect to then-side. Hence in our case the external applied voltage will be subtracted from thebuilt-in voltage:

Vbi Vbi V


p(xn) = p0(xn) (exp(V /VT) 1) = 1.3107 cm3.

n(xp) = n0(xp) (exp(V /VT) 1) = 7109 cm3.

It is worth to note that the ratio n(xp)/p(xn) 5102 is much larger than the

one expected for a homojunction. In fact, in the homojunction this ratio dependsonly on the doping levels and we would expect n(xp)/p(xn) = ND/NA = 510

2.

5 ) Proceed like explained in Exercise 1.


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Exercise n. 4

Consider a np heterojunction between Al0.4Ga0.6As (hereafter material A), with NDA =51016 cm3 and GaAs (hereafter material B), with NAB = 10

17 cm3. The materialparameters to be used (please, discard Vegard rule in this case) are as follows:


NCA = 7.96321017 cm3, NV A = 1.489810

19 cm3

GaAs : EGB = 1.42 eV, qB = 4.07 eV, rB = 13.1

NCB = 4.71017 cm3, NV B = 910

18 cm3

(a) Draw the qualitative band diagram in thermodynamic equilibrium, by exploitingthe affinity rule;

(b) Draw the charge density and electric field distributions at thermodynamic equili-

brium, and calculate the depleted region widths in each material and the maximumvalue of the electric field;

(c) Draw the qualitative behavior of the depletion capacitance versus the applied biasvoltage and calculate its value (per unit area) for V = 0 V and V = 2 V. Externalvoltage is applied on the p-side with respect to the n-side.

Solution

(a) Band diagram

Band edge discontinuities

EC = q = qB + qA = 0.4240 eV

EV = q EG = 0.0840 eV

Type II heterostructure. The qualitative behavior of the junction is as inExercise n. 3.

qSA = qA + KBT log(NCANDA

) = 3.7180 eV

qSB = qB + EGB KBT log(

NV B

NAB ) = 5.3730 eV


(b) Chrage, electric fielded regions.

The qualitative behavior of the junction is as in Exercise n. 3. The peak electricfield will be located at x = 0, i.e. in the n-side of the junction.




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xp =

2(N)eq

qN2ABVbi = 86.21 nm

xn =

2(N)eq

qN2DAVbi = 172.42 nm


|Emax| =qNDA

Axn = 132.49 kV/cm.

Monority carriers at equilibrium:

p0(xn) = 9.48091010 cm3

n0(xp) = 8.0757105 cm3

(c) Depletion capacitance Out of equilibrium:Vbi Vbi V

The charge stored in the depleted region (e.g. in the n-side) is:

Q = qNDAxn =

2q(N)eq(Vbi V)

and the depletion capacitance is (see Fig.6)

Cdepl =

dQ

dV =q(N)eq

2(Vbi V)

Vbi

4.16

2.80

C F/cmdepl

Figura 6: Depletion Capacitance

For V = 0: Cdepl = 4.1672108 F/cm. For V = 2 V: Cdepl = 2.804110

8 F/cm.


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Exercise n. 5

Consider a n p heterojunction between InP (hereafter material A), with NDA = 71016 cm3 and In0.53Ga0.47As (hereafter material B), with NAB = 710

17 cm3. Assumealloys properties as listed below:

InP : EGA = 1.35 eV, qA = 4.2 eV, rA = 12.3,

NCA = 81017 cm3, NV A = 2.510

19 cm3

In0.53Ga0.47As : EGB = 0.8 eV, qB = 4.5 eV, rB = 14.0,

NCB = 61017 cm3, NV B = 910

18 cm3

1) Draw the energy band diagram at thermal equilibrium, using the the affinity rulefor the evaluation of valence and conduction band discontinuity.



4) Calculate the excess of minority carriers injected at the boundary of the neutralregion under an applied forward bias of 0.6 V (voltages are applied to the p-sidewith respect to the n-side).

5) Calculate the depletion capacitance for an applied voltage V = 1 V.

Solution

1) Band Diagram.

EC = q = qB + qA = 0.3 eV

EV = q EG = 0.25 eV

Type I heterojunction.


qSA = qA + KBT log(NCANDA ) = 4.2721 eV

qSB = qB + EGB KBT log(NV BNAB

) = 5.1830 eV


Electron move from material A to material B. A space charge region builds up acrossthe heterojunction, see Fig. 7 (right). The bands bend downwards if the charge isnegative and upwards if positive. The final equilibrium band diagram is skeched in

Fig. ?? (left).


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U0

Ev

Ec

EF

0.3 eV

0.25 eV

0.91 eV

InP In Ga As0.53 0.47

E

xp-xn

r

q NDA

-q NAB

-xn

xp

Figura 7: Band diagram, charge distribution and electric field

2) Electric Field.

Integrating the Gauss law, the analytic expression of the electric field is:

E(x) =

0 x < xnqNDA

A(x + xn) xn < x < 0

qNAB

B(x xp) 0 < x < xp

0 x > xp

For x = 0 we have a discontinuity of the electric field as expected:

E(x = 0) =qNDA

Axn

E(x = 0+) =qNAB

Bxp

but the continuity of the electric displacement vector E due to the charge neutalitycondition qNDAxn = qNABxp. Since SA < SB , the peak electric field will be locatedat x = 0, i.e. in the n-side of the junction, see Fig. 7 (right).

3) Depletion region widths.



xp =

2(N)eq

qN2ABVbi = 4.1884 nm

xn = 2(N)eqqN2DA

Vbi = 418.84 nm


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|Emax| =qNDA

Axn = 43.094 kV/cm.



exp(qVbi + EV

kBT) = 0.0805 cm3.

n0(xp) = NDANCBNCA

exp(qVbi + EC

kBT) = 3.3449105 cm3.

Outside equilibrium:Vbi Vbi V

Under forward bias V = 0.6 V:

p

(xn) = p0(xn) (exp(V /VT) 1) = 3.1183108

cm3

.

n(xp) = n0(xp) (exp(V /VT) 1) = 1.29501015 cm3.

5 ) Depletion capacitance.

Cdepl =dQ

dV=

q(N)eq

2(Vbi V)

For V = 1 V: Cdepl = 1.4409108

F/cm.


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Exercise n. 6

Consider a np heterojunction between In0.53Ga0.47As (hereafter material A), with NDA =71017 cm3 and InP (hereafter material B), with NAB = 710

16 cm3. Assume alloysproperties as listed below:

In0.53Ga0.47As : EGA = 0.8 eV, qA = 4.5 eV, rA = 14.0,

NCA = 61017 cm3, NV A = 910

18 cm3

InP : EGB = 1.35 eV, qB = 4.2 eV, rB = 12.3,

NCB = 81017 cm3, NV B = 2.510

19 cm3

1) Draw the energy band diagram at thermal equilibrium, using the the affinity rulefor the evaluation of valence and conduction band discontinuity.



4) Calculate the excess of minority carriers injected at the boundary of the neutralregion under an applied forward bias of 0.6 V (voltages are applied to the p-sidewith respect to the n-side).

5) Calculate the depletion capacitance for an applied voltage V = 10 V.

6) Compare the extension of the depleted region at equilibrium with the Debye lengthof the majority on both sides.

Solution

1) Band Diagram.

EC = q = qB + qA = 0.3 eV

EV = q EG = 0.25 eV

Type I heterojunction.


qSA = qA + KBT log(NCANDA

) = 4.4960 eV

qSB = qB + EGB KBT log(NV BNAB

) = 5.3373 eV


A space charge region builds up across the heterojunction, see Fig. 8 (right). Thebands bend downwards if the charge is negative and upwards if positive. The final

equilibrium band diagram is skeched in Fig. 8 (left).


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U0

Ev

Ec

EF

0.3 eV

0.25 eV

0.91 eV

InPIn Ga As0.53 0.47

E

xp-xn

r

q NDA

-q NAB

-xn

xp

Figura 8: Band diagram, charge distribution and electric field

2) Electric Field.

Integrating the Gauss law, the analytic expression of the electric field is:

E(x) =

0 x < xnqNDA

A(x + xn) xn < x < 0

qNAB

B(x xp) 0 < x < xp

0 x > xp

For x = 0 we have a discontinuity of the electric field as expected:

E(x = 0) =qNDA

Axn

E(x = 0+) =qNAB

Bxp

but the continuity of the electric displacement vector E due to the charge neutalitycondition qNDAxn = qNABxp. Since SA > SB , the peak electric field will be located

at x = 0+

, i.e. in the p-side of the junction, see Fig. 8 (right).

3) Depletion region widths.



xp =

2(N)eq

qN2ABVbi = 402.63 nm

xn = 2(N)eqqN2DA

Vbi = 4.0263 nm


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|Emax| =qNAB

Bxp = 41.426 kV/cm.



exp(qVbi + EV

kBT) = 3.3449105 cm3.

n0(xp) = NDANCBNCA

exp(qVbi + EC

kBT) = 0.0805 cm3.

Outside equilibrium:Vbi Vbi V

Under forward bias V = 0.6 V:

p

(xn) = p0(xn) (exp(V /VT) 1) = 1.29501015

cm3

.

n(xp) = n0(xp) (exp(V /VT) 1) = 3.1183108 cm3.

5 ) Depletion capacitance.

Cdepl =dQ

dV=

q(N)eq

2(Vbi V)

For V = 10 V: Cdepl = 7.4658109

F/cm.

6 ) Debye lenght. The Debye lengths in each side of the heterojunction are:

LDA =

AkBT

q2NDA= 0.38308 nm xn

LDB =

BkBT

q2NAB= 4.0869 nm xp

Both lenghts are negligible with respect to the extension of the depletion regions.

No further analysis is required.

esercizi su np-pn hetero junction

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