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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
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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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.
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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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
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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CMOS
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.
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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If VGS > VT (threshold voltage), an induced, conducting n-channel forms
between the drain and source. The channel conductance is proportional to
vGS - Vt.
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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Symbols and conventions
drain
•n-channel
gate
+
VDS
-
+
VGS
-
•several slightly different symbols
source
(source is often connected to the
substrate which is usually grounded)
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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Symbols and conventions
drain
•p-channel
gate
+
VDS
-
+
VGS
-
•several slightly different symbols
source
(source is often connected to VDD)
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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Output characteristics (n-channel)
(linear)
+
VDS
-
An n-channel MOSFET with VGS and VDS applied and with the normal
directions of current flow indicated.
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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Input characteristics (n-channel)
+
VDS
-
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
ID = K(VGS-VT)2
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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)
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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Comparison of BJT and FET
BJT
•current controlled
FET
•voltage controlled
•VBE  0.7 V
for device to be on
•VGS > VT
for device to be on
•operates in linear region (amplifier); •operates in saturation region (amplifier);
VDS > VGS - VT
BE junction forward biased,
BC junction reversed biased
•IC = bIB
•ID = K(VGS-VT)2
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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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
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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Prob 5.41(a)
Given: VT = 2V, K = (1/2) .5 mA/V2
ID
IG = 0
(a) Find V1
Use,
+
VGS -
V1
ID = K(VGS-VT)2
10uA = (1/2) .5 (VGS - 2)2
Solve for VGS
n channel
VGS = 2.2V
V1 = - 2.2V
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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Prob 5.41(b)
Given: VT = 2V, K = (1/2) .5 mA/V2
(b) Find V2
V2
IG = 0
Use,
ID = K(VGS-VT)2
10uA = (1/2) .5 (VGS - 2)2
+
VGS -
n channel
ID
Solve for VGS
VGS = 2.2V
V2 = VGS = 2.2V
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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Prob 5.41(f)
Given: VT = 2V, K = (1/2) .5 mA/V2
(f) Find VGS
Equate current in load and transistor
IG = 0
ID
+
VGS -
n channel
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
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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5.4 MOSFETS at DC
DC problem
Find ID, and VGS, and VDS
VGS = 5V
VGS > VT, so device is on
Assume device is in saturation
ID = K(VGS-VT)2
ID = (0.05 mA/V2)(5-1)2
ID = 0.8 mA
VDS = VDD - ID RD
VDS = 10 - (0.8)6
VDS = 5.2V
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
ID
IG = 0
+
VDS
-
+
VGS
-
ID
VT = 1V
K = 0.05 mA/V2
(typical values)
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General DC problem
DC problem
Find ID, and VGS
Assume device is in saturation
ID = K(VGS-VT)2
ID = K(5 - ID RS -VT)2
18ID 2 - 25 ID + 8 = 0
Solve for ID, use quadratic formula
ID = 0.89mA, 0.5mA, which is correct?
For ID = 0.89mA, VGS = 5 - (0.89)6 = - 0.34V
For ID = 0.5mA, VGS = 5 - (05)6 = 2V
Only for ID = 0.5mA, is transistor on!
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
IG = 0
+
VDS
-
+
VGS
-
ID
VT = 1V, K = 0.5 mA/V2
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DC problem: two FETs in series
Find V
IG = 0
If devices are identical
VDD = 5V
IG = 0
V
device
V =VDD/2 = 2.5V
ID
device
Ground
n channel
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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5.5 MOSFET as an amplifier
g
.
d
d
+
Ro
vgs
g
-
s
ac model
s
g
n channel
d
+
vgs
Ro = 1/slope of the output characteristics
-
s
SPICE model
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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Transconductance
Transconductance = gm = dID/dVGS
= d [K(VGS-VT)2]/dVGS
= 2 K(VGS-VT)
Useful relation: gm = 2 K  ID
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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Prob. 5.86
(a) Find the resistance of an enhancement load
+
I
g
d
V
-
Rin
s
ac model
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
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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Prob. 5.86
(a) Find the resistance of an enhancement load
Often,
Ro >> 1/gm
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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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
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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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
ID
Equate current in Q1 and load
I in Q1 = I in load
K(VGS-VT)2 = I in load
ID
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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ac analysis of MOSFET amplifiers
g
ID
d
+
vgs
Rin
s
Rout
ac circuit
ID
Rin = 
Rout = Rload || Ro
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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ac analysis of MOSFET amplifiers
g
iin = 0
d
-gmvgs
+
vgs
-
s
+
vout
-
Ai = iout / iin = 
Av = vout/vin = -gmvgs(Ro || Rload) / vgs
= -gm(Ro || Rload)
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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Transistor loads: depletion load
VGS = 0
Depletion
load
I
+
V
-
R = Ro || resistance of current source
with 0 magnitude
= Ro || 
Ro = |VA| / I
Resistance is
current dependent
= Ro
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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CMOS amp
•Q2 and Q3 form a p-channel
current mirror load for Q1
•Q4 and Q3 establish Iref
I = Iref due to current mirror
Iref
I
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
Given:
|VT| = 1V, |VA| = 50V
p-channel pCox = 20A/V2
n-channel nCox = 40A/V2
WQ1 = Wp = 100m
WQ4 = 50m
L = 10m
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CMOS amp: power
Iref
I
Given:
|VT| = 1V, |VA| = 50V
p-channel pCox = 20A/V2
n-channel nCox = 40A/V2
WQ1 = Wp = 100m
WQ4 = 50m
L = 10m
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 = 225A
•Power consumed = (2) 5V 225A = 2.25mW
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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CMOS amp: DC analysis
Given:
|VT| = 1V, |VA| = 50V
p-channel pCox = 20A/V2
n-channel nCox = 40A/V2
WQ1 = Wp = 100m
WQ4 = 50m
L = 10m
+
Vout
Iref
-
Find Vout
•Consider current in Q1 or Q2
•Using Q1, IQ1 = K1(VGS-VT)2
where VGS = Vout
225A = (1/2)(40)(100/10) (VGS - 1)2
•Solve for VGS, VGS = Vout = 1.75V
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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CMOS amp: ac analysis
Given:
|VT| = 1V, |VA| = 50V
p-channel pCox = 20A/V2
n-channel nCox = 40A/V2
WQ1 = Wp = 100m
WQ4 = 50m
L = 10m
Iref
+
Vout
Find Av
Av = -gm1(Ro1 || Ro2)
-
Ro1= Ro2 = 50/ 225A = 222KW
gm = 2 K  ID
= (2) [(1/2)(40)(100/10)] 1/2  225A
= 300A/V
Av = -gm1(Ro1 || Ro2) = -300(.222/2)  -33
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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CMOS multistage amp: ac analysis
DC circuit
ac circuit
(neglects resistances of current sources)
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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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
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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Multistage CMOS amp: DC analysis
Iref
•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
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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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
Iref
•Find VD4, and VD1 = Vout from
currents in those transistors
•Given KP = 80A/V2, KN = 100A/V2, |VT| = 1V, VDD = 9V
100(VGS5 - 1)2 = 80((VGS5 - 9) - (- 1))2, VGS5 = 5.14V, 48.9V
Find Iref, 100(5.14 - 1)2 = 1.7mA
IQ3 = IQ4 = IQ2 = IQ1 because all KN’s and KP’s are equal
ECE 3111 - Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002
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