chap. 5 field-effect transistors (fet)
DESCRIPTION
Chap. 5 Field-effect transistors (FET). Widely used in VLSI used in some analog amplifiers - output stage of power amplifers (may have good thermal characteristics if designed properly) n-channel or p-channel structure FET - voltage controlled device BJT - current controlled device. - PowerPoint PPT PresentationTRANSCRIPT
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 1
Chap. 5 Field-effect transistors (FET)
•Widely used in VLSI
•used in some analog amplifiers - output stage of power amplifers (may have good thermal characteristics if designed properly)
•n-channel or p-channel structure
•FET - voltage controlled device
•BJT - current controlled device
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 2
Physical structure of a n-channel device:
Typically L = 0.35 to 10 m, W = 2 to 500 m, and the thickness of the oxide layer is in the range of 0.02 to 0.1 m.
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 3
MOSFETs
•MOS - metal oxide semicondutor structure (original devices had metal gates, now they are silicon)
•NMOS - n-channel MOSFET
•PMOS - p-channel MOSFET
•CMOS - complementary MOS, both n-channel and p-channel devices used in conjuction with each other (most popular in IC’s)
•MESFET - metal semiconductor structure, used in high-speed GaAs devices
•JFET - junction FET, early type of FET
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 4
Cross section of a CMOS integrated circuit. Note that the PMOS transistor is formed in a separate n-type region, known as an n well.
CMOS
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 5
If VGS > VT (threshold voltage), an induced, conducting n-channel forms between the drain and source. The channel conductance is proportional to vGS - Vt.
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 6
Symbols and conventions
•n-channel
•several slightly different symbols
(source is often connected to the substrate which is usually grounded)
+VDS
-+ VGS
-
drain
source
gate
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 7
Symbols and conventions
•p-channel
•several slightly different symbols
(source is often connected to VDD)
+VDS
-+ VGS
-
drain
source
gate
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 8
An n-channel MOSFET with VGS and VDS applied and with the normal directions of current flow indicated.
Output characteristics (n-channel)
(linear)
+VDS
-
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 9
Input characteristics (n-channel)
+VDS
-
ID = K(VGS-VT)2
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 10
Summary of MOSFET behavior (n-channel)
•VGS > VT (threshold voltage) for the device to be on
•VDS > VGS - VT for device to be in saturation region
•ID = K(VGS-VT)2
•Enhancement mode device, VT > 0
•Depletion mode device, VT < 0 (conducts with VGS = 0)
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 11
Comparison of BJT and FET
FET•voltage controlled
•VGS > VT
for device to be on
•operates in saturation region (amplifier);VDS > VGS - VT
•ID = K(VGS-VT)2
BJT•current controlled
•VBE 0.7 V for device to be on
•operates in linear region (amplifier); BE junction forward biased, BC junction reversed biased
•IC = IB
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 12
MOSFET aspect ratio
ID = K(VGS-VT)2
K = transconductance parameter
K = 1/2 K' (W/L)
K' = nCox, where n is the mobility of electrons, and Cox
is the capacitance of the oxide
W/L is the aspect ratio, W is the width of the gate, L is the length of the gate.
ID W/L
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 13
Prob 5.41(a)
Given: VT = 2V, K = (1/2) .5 mA/V2
(a) Find V1
Use, ID = K(VGS-VT)2
10uA = (1/2) .5 (VGS - 2)2
Solve for VGS
VGS = 2.2V
V1 = - 2.2V
V1VGS -+
IDIG = 0
n channel
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 14
Prob 5.41(b)
Given: VT = 2V, K = (1/2) .5 mA/V2
(b) Find V2
Use, ID = K(VGS-VT)2
10uA = (1/2) .5 (VGS - 2)2
Solve for VGS
VGS = 2.2V
V2 = VGS = 2.2V
V2
VGS -+
ID
IG = 0
n channel
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 15
Prob 5.41(f)
Given: VT = 2V, K = (1/2) .5 mA/V2
(f) Find VGS
Equate current in load and transistor
Current in transistor: ID = K(VGS-VT)2
Current in resistor: I = (5 - VGS) /100K
Equate currents
(5 - VGS) /100K = (1/2) .5 (VGS - 2)2
Solve for VGS
VGS = 2.33V
VGS -+
ID
IG = 0
n channel
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 16
5.4 MOSFETS at DC
DC problemFind ID, and VGS, and VDS
VGS = 5VVGS > VT, so device is on
Assume device is in saturationID = K(VGS-VT)2
ID = (0.05 mA/V2)(5-1)2
ID = 0.8 mA
VDS = VDD - ID RD
VDS = 10 - (0.8)6VDS = 5.2V
VT = 1VK = 0.05 mA/V2
(typical values)
+ VGS
-
+VDS
-
ID
ID
IG = 0
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 17
General DC problem
+ VGS
-
+VDS
-
ID
IG = 0
DC problemFind ID, and VGS
Assume device is in saturationID = K(VGS-VT)2
ID = K(5 - ID RS -VT)2
18ID 2 - 25 ID + 8 = 0
Solve for ID, use quadratic formulaID = 0.89mA, 0.5mA, which is correct?
For ID = 0.89mA, VGS = 5 - (0.89)6 = - 0.34VFor ID = 0.5mA, VGS = 5 - (05)6 = 2V
Only for ID = 0.5mA, is transistor on! VT = 1V, K = 0.5 mA/V2
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 18
DC problem: two FETs in series
Find V
If devices are identical
ID
IG = 0
n channel
IG = 0 V
VDD = 5V
Ground
device
device
V =VDD/2 = 2.5V
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 19
5.5 MOSFET as an amplifier
g
d
s
n channel
ac model
d
s
g
vgs
+
-Ro = 1/slope of the output characteristics
. d
s
g
vgs
+
-
SPICE model
Ro
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 20
Transconductance
Useful relation: gm = 2 K ID
Transconductance = gm = dID/dVGS
= d [K(VGS-VT)2]/dVGS
= 2 K(VGS-VT)
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 21
Prob. 5.86
(a) Find the resistance of an enhancement load
Rin
Rin = resistance of current source || Ro
resistance of current source = voltage across current source / current in current source
resistance of current source = vgs / gmvgs = 1/gm
Replace current source by a resistor of resistance 1/gm
ac model
+
V
-
I gd
s
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 22
Prob. 5.86(a) Find the resistance of an enhancement load
Often,Ro >> 1/gm
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 23
Prob. 5.86
(b) To raise the resistance of the transistor by a factor of 3, what must be done?
R 1/gm
= 1 / 2 K ID
= [1/2 ] [1/K] [ 1/ ID]
= [1/2 ] [1/ 1/2 K W/L ] [ 1/ ID]
•Decrease ID by a factor of 9•Decrease W by a factor of 9•Increase L by a factor of 9
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 24
5.7 Integrated Circuit MOSFET amplifiers
•Resistors take up too much space on an integrated ciruit (IC)
•Use transistors as loads
Typical amplifier
DC analysis
Equate current in Q1 and load
I in Q1 = I in load
K(VGS-VT)2 = I in load
ID
ID
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 25
ac analysis of MOSFET amplifiers
ID
ID
ac circuit
Rin =
Rout = Rload || Ro
d
s
g
vgs
+
-Rin Rout
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 26
ac analysis of MOSFET amplifiers
Ai = iout / iin =
Av = vout/vin = -gmvgs(Ro || Rload) / vgs
= -gm(Ro || Rload)
d
s
g
vgs
+
-
vout
+
-
iin = 0 -gmvgs
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 27
Transistor loads: depletion load
I
V+
-
Depletion load
VGS = 0
R = Ro || resistance of current source with 0 magnitude
= Ro ||
= Ro
Ro = |VA| / I
Resistance iscurrent dependent
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 28
CMOS amp
Iref
I
•Q2 and Q3 form a p-channel current mirror load for Q1
•Q4 and Q3 establish Iref
I = Iref due to current mirror
Given:|VT| = 1V, |VA| = 50Vp-channel pCox = 20A/V2
n-channel nCox = 40A/V2
WQ1 = Wp = 100mWQ4 = 50mL = 10m
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 29
CMOS amp: power
Iref I
Given:|VT| = 1V, |VA| = 50Vp-channel pCox = 20A/V2
n-channel nCox = 40A/V2
WQ1 = Wp = 100mWQ4 = 50mL = 10m
Find Total power consumed
•Power consumed = 2IrefVDD
•Equate currents in Q3 and Q4 to find Iref
•IQ3 = IQ4 = K3(VGS-VT)2 = K4(VGS-VT)2
•Note that K’s are the same: K3 = (1/2)(20)(100/10) = K4 = (1/2)(40)(50/10)•Therefore, Q3 and Q4 behave the same, so VGS3 = VGS4 = 2.5V•Iref = K4(VGS-VT)2 = (1/2)(40)(50/10) (2.5 - 1)2 = 225A•Power consumed = (2) 5V 225A = 2.25mW
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 30
CMOS amp: DC analysis
Given:|VT| = 1V, |VA| = 50Vp-channel pCox = 20A/V2
n-channel nCox = 40A/V2
WQ1 = Wp = 100mWQ4 = 50mL = 10m
Find Vout •Consider current in Q1 or Q2•Using Q1, IQ1 = K1(VGS-VT)2
where VGS = Vout
225A = (1/2)(40)(100/10) (VGS - 1)2 •Solve for VGS, VGS = Vout = 1.75V
Iref
+Vout
-
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 31
CMOS amp: ac analysis
Given:|VT| = 1V, |VA| = 50Vp-channel pCox = 20A/V2
n-channel nCox = 40A/V2
WQ1 = Wp = 100mWQ4 = 50mL = 10m
Iref
+Vout
-
Find Av
Av = -gm1(Ro1 || Ro2)Ro1= Ro2 = 50/ 225A = 222Kgm = 2 K ID
= (2) [(1/2)(40)(100/10)] 1/2 225A = 300A/V
Av = -gm1(Ro1 || Ro2) = -300(.222/2) -33
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 32
CMOS multistage amp: ac analysis
DC circuit
ac circuit(neglects resistances of current sources)
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 33
CMOS multistage amp: ac analysis
Av of stage 1: Vout1/Vgs1 = -gm1Vgs1Ro1/Vgs1 = -gm1ro1
Av of stage 2: Vout2/Vgs2 = -gm2Vgs2Ro2/Vgs2 = -gm2ro2
Overall Av = (-gm1ro1) ( -gm2ro2) = gm1gm2ro1ro2
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Multistage CMOS amp: DC analysis
•Q3 and Q6 form a PMOS current mirror load for Q4•Q1 and Q5 form an NMOS current mirror load for Q2 •Q5 and Q6 establish the current in Q1,Q2,Q3 and Q4•The width of Q5 is adjusted to give a particular Iref
Iref
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ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002 35
Multistage CMOS amp: DC analysis•Equate currents in Q5 and Q6 •IQ5 = IQ6 = K5(VGS5-VT)2 = K6((VGS5 -
VDD)-VT)2
•Solve for VGS5, Use VGS5 to find Iref
•Other current s are multiples of Iref
•K3/K6 = IQ3/Iref
•K1/K5 = IQ1/Iref
•Find VD4, and VD1 = Vout from currents in those transistors
Iref
•Given KP = 80A/V2, KN = 100A/V2, |VT| = 1V, VDD = 9V100(VGS5 - 1)2 = 80((VGS5 - 9) - (- 1))2, VGS5 = 5.14V, 48.9VFind Iref, 100(5.14 - 1)2 = 1.7mAIQ3 = IQ4 = IQ2 = IQ1 because all KN’s and KP’s are equal