0. 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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

1. 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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

2. 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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

3. 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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

4. 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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

5. Symbols and conventions
drain
n-channel
several slightly different symbols
(source is often connected to the
substrate which is usually grounded) +
VDS -
gate +
VGS -
source
ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

6. Symbols and conventions
drain
p-channel
several slightly different symbols
(source is often connected to VDD) +
VDS -
gate +
VGS -
source
ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

7. 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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

8. Input characteristics (n-channel)
ID = K(VGS-VT)2 +
VDS -
ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

9. 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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

10. Comparison of BJT and 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 = bIB
ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

11. MOSFET aspect ratio ID = K(VGS-VT)2 K = transconductance parameter
K = 1/2 K' (W/L)
K' = mnCox, where mn 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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

12. 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 ID
IG = 0 + V1
VGS -
n channel
ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

13. 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
IG = 0 +
VGS - ID
n channel
ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

14. 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
IG = 0 ID
n channel +
VGS -
ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

15. 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 ID
IG = 0 +
VDS - +
VGS - ID
VT = 1V
K = 0.05 mA/V2
(typical values)
ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

16. 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 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 = V
For ID = 0.5mA, VGS = 5 - (05)6 = 2V
Only for ID = 0.5mA, is transistor on!
IG = 0 +
VDS - +
VGS - ID
VT = 1V, K = 0.5 mA/V2
ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

17. DC problem: two FETs in series
Find V
If devices are identical
IG = 0
VDD = 5V
Ground
device
IG = 0 V
V =VDD/2 = 2.5V ID
device
n channel
ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

18. . 5.5 MOSFET as an amplifier ac model n channel SPICE model d g + Ro d
vgs g s -
ac model s g d
n channel +
vgs -
Ro = 1/slope of the output characteristics s
SPICE model
ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

19. Transconductance Transconductance = gm = dID/dVGS = 2 K(VGS-VT)
= d [K(VGS-VT)2]/dVGS
= 2 K(VGS-VT)
Useful relation: gm = 2 K  ID
ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

20. Prob. 5.86 ac model (a) Find the resistance of an enhancement load g I+ 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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

21. Prob. 5.86 (a) Find the resistance of an enhancement load Often,
Ro >> 1/gm
ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

22. 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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

23. 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
ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

24. ac analysis of MOSFET amplifiersg d + ID
vgs - s
Rin
Rout
ac circuit
Rin = 
Rout = Rload || Ro
ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

25. ac analysis of MOSFET amplifiers
iin = 0
-gmvgs g d + +
vout
vgs - - s
Ai = iout / iin = 
Av = vout/vin = -gmvgs(Ro || Rload) / vgs
= -gm(Ro || Rload)
ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

26. Transistor loads: depletion load+ V -
VGS = 0
Depletion
load
R = Ro || resistance of current source
with 0 magnitude
= Ro || 
= Ro
Ro = |VA| / I
Resistance is
current dependent
ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

27. 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
Given:
|VT| = 1V, |VA| = 50V
p-channel mpCox = 20mA/V2
n-channel mnCox = 40mA/V2
WQ1 = Wp = 100mm
WQ4 = 50mm
L = 10mm
Iref I
ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

28. CMOS amp: power Given: |VT| = 1V, |VA| = 50V p-channel mpCox = 20mA/V2
Iref I
Given:
|VT| = 1V, |VA| = 50V
p-channel mpCox = 20mA/V2
n-channel mnCox = 40mA/V2
WQ1 = Wp = 100mm
WQ4 = 50mm
L = 10mm
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 = 225mA
Power consumed = (2) 5V 225mA = 2.25mW
ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

29. CMOS amp: DC analysis Given: |VT| = 1V, |VA| = 50V
p-channel mpCox = 20mA/V2
n-channel mnCox = 40mA/V2
WQ1 = Wp = 100mm
WQ4 = 50mm
L = 10mm +
Vout -
Iref
Find Vout
Consider current in Q1 or Q2
Using Q1, IQ1 = K1(VGS-VT)2
where VGS = Vout
225mA = (1/2)(40)(100/10) (VGS - 1)2
Solve for VGS, VGS = Vout = 1.75V
ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

30. CMOS amp: ac analysis Given: |VT| = 1V, |VA| = 50V
p-channel mpCox = 20mA/V2
n-channel mnCox = 40mA/V2
WQ1 = Wp = 100mm
WQ4 = 50mm
L = 10mm
Iref +
Vout -
Find Av
Av = -gm1(Ro1 || Ro2)
Ro1= Ro2 = 50/ 225mA = 222KW
gm = 2 K  ID
= (2) [(1/2)(40)(100/10)] 1/2  225mA
= 300mA/V
Av = -gm1(Ro1 || Ro2) = -300(.222/2)  -33
ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

31. CMOS multistage amp: ac analysis
DC circuit
ac circuit
(neglects resistances of current sources)
ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

32. 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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

33. 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 Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002

34. 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 = 80mA/V2, KN = 100mA/V2, |VT| = 1V, VDD = 9V
100(VGS5 - 1)2 = 80((VGS5 - 9) - (- 1))2, VGS5 = 5.14V, 48.9V
Find Iref, 100( )2 = 1.7mA
IQ3 = IQ4 = IQ2 = IQ1 because all KN’s and KP’s are equal
ECE Electronics - Dr. S. Kozaitis- Florida Institute of Technology - Fall 2002