1. Recall Lecture 17 MOSFET DC Analysis
Using GS (SG) loop to calculate VGS
Remember that there is NO gate current!
Assume in saturation
Calculate ID using saturation equation
Find VDS (for NMOS) or VSD (for PMOS)
Using DS (SD) loop
Calculate VDS sat or VSD sat
Confirm that VDS > VDS sat or VSD > VSD sat
Confirm your assumption!

2. LOAD LINE, ID versus VDS
Common source configuration i.e source is grounded.
It is the linear equation of ID versus VDS
Use KVL
VDS = VDD – IDRD
ID = -VDS + VDD RD RD

3. ID (mA)
VDS (V)
VGS
VDS ID
Q-POINTS
y-intercept
x-intercept

4. DC Analysis where source is NOT GROUNDED
For the NMOS transistor in the circuit below, the parameters are VTN = 1V and Kn = 0.5 mA/V2.

5. Calculate the value of VGS KVL at GS loop:R D = 2 k S G
24 k
+ 5 V
1 V I
5 V
Assume the transistor is biased in the saturation region, the drain current:
Calculate the value of VGS
KVL at GS loop:
0 + VGS+ 1(ID) = 0
VGS = 4 - ID
1 k
ID = mA
VGS= V
Replace in VGS equation in step 1
VGS = 4 - ID
ID = mA
VGS = V
Why choose VGS = V ?
Because it is bigger than VTN

6. Calculate VDSsat = VGS – VTN = 2.646 – 1 = 1.646 VR D = 2 k S G
24 k
+ 5 V
1 V I
5 V
Use KVL at DS loop
Calculate VDSsat = VGS – VTN = – 1 = V
Confirm your assumption: VDS > VDS sat , our assumption is correct
IDRD + VDS + IDRS – 5 – 5 = 0
1.354 (2) + VDS – 10 = 0
VDS = 10 – – = V
1 k

7. Answers: R1 = 422 kΩ R2 = 248 kΩ RD = 2.84 kΩ
EXERCISE 2
Assume that the transistor parameters are VTN = 0.8 V and Kn = 0.80 mA/V2. Given VDD = 5V and R1+ R2 = 670 kΩ, design the circuit such that ID = 0.88 mA and VDS = 2.5 V. Confirm any assumptions you make during your analysis.
Answers:
R1 = 422 kΩ
R2 = 248 kΩ
RD = 2.84 kΩ

8. CHAPTER 7
Basic FET Amplifiers

9. For linear amplifier function, FET is normally biased in the saturation region.

10. Need to do DC Analysis first to find ID
AC PARAMETERS
where
Need to do DC Analysis first to find ID
vgs
gmvgs

11. The MOSFET Amplifier - COMMON SOURCE
The output is measured at the drain terminal
The gain is negative value
Three types of common source
source grounded
with source resistor, RS
with bypass capacitor, CS

12. Common Source - Source Grounded
A Basic Common-Source Configuration:
Assume that the transistor is biased in the saturation region by resistors R1 and R2, and the signal frequency is sufficiently large for the coupling capacitor to act essentially as a short circuit. +
vgs -
gmvgs vs

13. EXAMPLE The transistor parameters are:
VDD = 5V
RD = 10 k
520 k
320 k
0.5 k
The transistor parameters are:
VTN = 0.8V, Kn = 0.2mA/V2 and  = 0. vs
Voltage Divider biasing:
Change to Thevenin Equivalent
RTH = 198 k
VTH = V

14. gm = 0.442 mA/V DC ANALYSIS Calculate the value of VGS VGS – VTH = 0
Confirm your assumption: VDS > VDSsat, our assumption that the transistor is in saturation region is correct
Assume the transistor is biased in the saturation region, the drain current:
Use KVL at DS loop
IDRD + VDS – VDD = 0
VDS = VDD – IDRD = 2.56 V
Calculate the value of VGS
VGS – VTH = 0
VGS = V
Calculate VDSsat = VGS – VTN = – 0.8 = V
gm = mA/V

15. COMMON EMITTER GROUNDED
OUTPUT SIDE
Get the equivalent resistance at the output side, RO
Get the vo equation where vo = - gm vbeRO
INPUT SIDE
Calculate Ri
Get vbe in terms of vi
COMMON SOURCE GROUNDED
OUTPUT SIDE
Equivalent resistance at the output side, RO
Get the vo equation where vo = - gm vgs RO
____________________________________________
INPUT SIDE
Get vgs in terms of vi vs

16. The output resistance, Ro = RD The output voltage:
RTH
198.1 k
0.5 k
RD = 10 k
0.442 vgs +
vgs - + vi - vs
The output resistance, Ro = RD
The output voltage:
vo = - gmvgs (Ro) = - gmvgs (10) = vgs
Get vi in terms of vgs
vgs = vi

17. Equation of vo : vo = - gmvgs (Ro) = - gmvgs (10) = -4.42 vgs
vi = vgs
Av vi = vo  open circuit voltage
Avvi = vgs = vi
Av =  open circuit voltage gain

18. RSi = 0.5 kΩ vS vo
10 k
Ri = RTH = kΩ
To find new voltage gain, vo/vs with input signal voltage source, vs
vi in terms of vs  use voltage divider:
vi = [ Ri / ( Ri + Rs )] * vs = vs
vo = Avvi  because there is no load resistor
vo = ( vs)
vo/vs = -4.41

19. Type 2: With Source Resistor, RS
VTN = 1V, Kn = 1.0 mA /V2
RTH = k
VTH = ( 200 / 300 ) x 3 = 2 V

20. Calculate the value of VGS KVL at GS loop:
DC ANALYSIS
Assume the transistor is biased in the saturation region, the drain current:
Calculate the value of VGS
KVL at GS loop:
0 + VGS+ 3(ID) - 2 = 0
VGS = 2 - 3ID vs
VTN = 1V, Kn = 1.0 mA / V
ID = mA
Replace in VGS equation in step 1
VGS = 2 - 3ID
VGS= V
ID = mA
VGS = 1.43 V
Why choose VGS = 1.43 V ?
Because it is bigger than VTN

21. gm = 0.872 mA/V Use KVL at DS loop IDRD + VDS + IDRS – 3 = 0
Calculate VDSsat = VGS – VTN = 1.43 – 1 = 0.43V
Confirm your assumption: VDS > VDS sat , our assumption is correct
IDRD + VDS + IDRS – 3 = 0
VDS = V
gm = mA/V
 = 0
ro = 
VA = 

22. Equivalent resistance at the output side, RO
COMMON EMITTER with RE
OUTPUT SIDE
Get the equivalent resistance at the output side, Ro
Get the vo equation where vo = - gmvbeRo
INPUT SIDE
Calculate Rib = r + (1+ ) RE
Calculate Ri = Rib // RTH
vbe in terms of vi
COMMON SOURCE with RS
OUTPUT SIDE
Equivalent resistance at the output side, RO
Get the vo equation where vo = - gm vgs RO
________________________________________________________
INPUT SIDE
Get vi in terms of vgs : KVL

23. vi = vgs + gmvgs RS  vi = vgs(1 + 2.616) = 3.616 vgs
RTH
66.67 k
RD = 10 k
RS = 3 k + vi - vs
gm = mA/V
The output resistance, Ro = RD
The output voltage:
Get vi in terms of vgs : KVL
vi = vgs + gmvgs RS  vi = vgs( ) = vgs
vo = - gmvgsRD = ( vgs) (10) = vgs

24. Equation of vo : vo = - gmvgs (Ro) = - gmvgs (10) = -8.72 vgs
vi = vgs + gmvgs RS  vi = vgs( ) = vgs
Av vi = vo  open circuit voltage
Avvi = vgs = vi
Av =  open circuit voltage gain
3.616

25. 10 k
Ri = RTH = kΩ vS vo
To find new voltage gain, vo/vs with input signal voltage source, vs
vi in terms of vs
vi = vs  in parallel
vo = Avvi  because there is no load resistor
vo = (vs)
vo/vs = -2.41

26. Type 3: With Source Bypass Capacitor, CS
Circuit with Source Bypass Capacitor
An source bypass capacitor can be used to effectively create a short circuit path during ac analysis hence avoiding the effect RS
CS becomes a short circuit path – bypass RS; hence similar to Type 1

27. COMMON EMITTER with CE
OUTPUT SIDE
Get the equivalent resistance at the output side, RO
Get the vo equation where vo = - gm vbeRO
INPUT SIDE
Calculate Ri
Get vbe in terms of vi
COMMON SOURCE with CS
OUTPUT SIDE
Equivalent resistance at the output side, RO
Get the vo equation where vo = - gm vgs RO
____________________________________________
INPUT SIDE
Get vgs in terms of vi vs

28. IQ = 0.5 mA hence, ID = 0.5 mA ro =  gm = 1.414 mA/V + + vgs vi - -vs RG
200 k
RD = 7 k
1.414 vgs +
vgs - + vi - vs

29. The output resistance, Ro = RD = 7 k The output voltage: RG
200 k
RD = 7 k
1.414 vgs +
vgs - vi vs
The output resistance, Ro = RD = 7 k
The output voltage:
vo = - gmvgs (RD) = (7) vgs = vgs
3. The gate-to-source voltage:
vgs = vi

30. Equation of vo : vo = - gmvgs (RD) = -1.414 (7) vgs = - 9.898 vgs
vi = vgs
Av vi = vo  open circuit voltage
Avvi = vgs = vi
Av =  open circuit voltage gain

31. 7 k
Ri = RG = 200 kΩ vS vo
To find new voltage gain, vo/vs with input signal voltage source, vs
vi in terms of vs
vi = vs  in parallel
vo = Avvi  because there is no load resistor
vo = (vs)
vo/vs =