1. HKN ECE 342 Review Session 2
Anthony Li
Milan Shah

2. MOSFET’s
NMOS
PMOS

3. MOSFET Operating Point
Three regions of operation:
Cutoff (VG < VT): ID = 0
Linear/Triode (VG > VT, VDS < VGS - VT):
𝐼𝐷 = 𝜇𝑛𝐶𝑜𝑥( 𝑊 𝐿 )(( 𝑉 𝐺𝑆 − 𝑉 𝑇 ) 𝑉 𝐷𝑆 − 𝑉 𝐷𝑆 2 2 )
Saturation (VG > VT, VDS > VGS - VT):
𝐼𝐷 = 𝜇𝑛𝐶𝑜𝑥 𝑊 𝐿 ½ 𝑉𝐺𝑆 − 𝑉 𝑇 2 (1+ λ 𝑉 𝐷𝑆 )
Note: 𝜇 𝑛 𝐶 𝑜𝑥 𝑊 𝐿 = 𝑘 ′ 𝑊 𝐿 =𝑘

4. MOSFET Incremental Model
Transconductance: 𝑔 𝑚 = 2 𝐼 𝐷 𝑉 𝑂𝑉

5. Gain Calculation Av = -GMRout
GM = Small signal transconductance, ratio of iout to vin
ROUT = Equivalent incremental output resistance

6. Common Amplifier Topologies
Diode-tied Transistor
Common Source/Drain/Gate
Common Source with Degeneration
Common Drain with Modulation
Cascode
Diode Tied Transistor

7. Diode-Tied Transistor
ROUT = 1 𝑔 𝑚 || 𝑟 𝑑𝑠 ≈ 1 𝑔 𝑚 for 𝑔 𝑚 𝑟 𝑑𝑠 ≫1 or 𝑟 𝑑𝑠 →∞
Diode Tied Transistor

8. Common Source/Drain/Gate
𝑅𝑂𝑈𝑇 = 𝑅 𝑆 || 𝑟 𝑑𝑠 || 1 𝑔 𝑚
𝐺 𝑚 = 𝑔 𝑚
𝑅𝑂𝑈𝑇 = 𝑅 𝐷 || 𝑟 𝑑𝑠
𝐺 𝑚 = 𝑔 𝑚
𝑅𝑂𝑈𝑇 = 𝑅 𝐷 || 𝑟 𝑑𝑠
𝐺 𝑚 =− 𝑔 𝑚
Common Source. Your basic amplifier topology. Notice the infinite input resistance and inverted gain.
Common Drain. Also known as a level shifter. These make great buffers. Source follower?
Common Gate. good current buffer i believe.

9. Degeneration
When a resistance is “viewed” through the drain, it appears bigger by a factor related to the transconductance.
𝑅𝑂𝑈𝑇 = 𝑅 𝐷 || ( 𝑟 𝑑𝑠 + 𝑅 𝑆 + 𝑔 𝑚 𝑟 𝑑𝑠 𝑅 𝑠 )
𝐺𝑚 = 𝑔𝑚 1 + 𝑔 𝑚 𝑅 𝑆
Note that these formulas simplify to the common source ones if Rs is zero. Also remember that these equations rely on rds not being infinity.

10. Modulation Resistances seen through the source seem smaller:
𝑅𝐼𝑁 = 𝑅 𝑆 𝑔 𝑚 (1 + 𝑅 𝐷 𝑟 𝑑𝑠 ) for gmrds >> 1
𝑅 𝐼𝑁

11. Cascode 𝑅𝑂𝑈𝑇 = 𝑅 𝐷 || ( 𝑟 𝑑𝑠2 + 𝑅 1 + 𝑔 𝑚2 𝑟𝑑𝑠2 𝑅 1 )
𝑅𝑂𝑈𝑇 = 𝑅 𝐷 || ( 𝑟 𝑑𝑠2 + 𝑅 𝑔 𝑚2 𝑟𝑑𝑠2 𝑅 1 )
𝑅 1 =( 𝑟 𝑑𝑠1 + 𝑅 𝑆 + 𝑔 𝑚1 𝑟 𝑑𝑠1 𝑅𝑆)
𝐺𝑚 = 𝑔𝑚1 1 + 𝑔 𝑚1 𝑅 𝑆
𝑅 1

12. BJT

13. Regions of Operation Three regions of operation:
Cutoff: VE > VB < VC
Saturation: VE < VB > VC
Forward Active: VE > VB > VC
VT = kt/q
IC = ꞵIB
IE = IC + IB
Ꞵ = gmR𝜋

14. BJT Small Signal Model

15. Terminal Impedance RC = ro RB = Rℼ RE = 𝑅ℼ ꞵ+1 Diode-Tied = 𝑅ℼ ꞵ+1
Diode Tied Transistor

16. Common Emitter/Collector/Base
𝑅𝑂𝑈𝑇 = 𝑅𝜋+𝑅𝐵 ꞵ+1 ||𝑅𝐸
𝑅𝐼𝑁 = 𝑅𝐵+𝑅𝜋 +(ꞵ+1)𝑅𝐸
𝐺𝑚 = − ꞵ+1 𝑅𝜋+ 𝑅 𝐵
𝑅𝑂𝑈𝑇 = 𝑅𝑐 || 𝑟𝑜
𝑅𝐼𝑁 = 𝑅𝜋+𝑅𝐵 ꞵ+1
𝐺𝑚 = ꞵ 𝑅𝜋+𝑅𝐵
𝑅𝑂𝑈𝑇 = 𝑅𝑐 || 𝑟𝑜
𝑅𝐼𝑁 = 𝑅𝜋+𝑅𝐵
𝐺𝑚 = ꞵ 𝑅𝜋+𝑅𝐵

17. Degeneration
When a resistance is “viewed” through the collector, it appears bigger by a factor related to the transconductance.
𝐺 𝑚 = 1 𝑅 𝐵 +𝑅𝜋 ꞵ + ꞵ+1 ꞵ 𝑅 𝐸
RIN = RB+R𝜋 +(ꞵ+1)RE

18. Modulation Resistances seen through the Emitter seem smaller.
𝐺𝑚 = − ꞵ+1 𝑅𝐵+𝑅𝜋
𝑅𝐼𝑁 = 𝑅𝐵+𝑅𝜋 +(ꞵ+1)𝑅𝐸

19. Bode Plots Magnitude Pole: Roll down by 20 db/dec, 6 db/oct
Zero: Roll up by 20 db/dec, 6 db/oct
Phase: arctan(ω/ωp)
Usually -90° for poles, +90° for zeros
ωugf = 20log|An| * ωpn where n is the pole located before unity gain frequency

20. Miller Effect
𝐶 1 = 𝐶 𝑓 1+𝐴
𝐶 2 = 𝐶 𝑓 (1+ 1 𝐴 )

21. Problem 1: Midterm 3 Fa17

22. Problem 2: Midterm 2 Fa16

23. Problem 3: Midterm 2 Fa16
A. 𝐺 𝑚 = 𝑔 𝑚 𝑔 𝑚 2 𝑔 𝑚 𝑔 𝑚 𝑔 𝑚 1 𝑔 𝑚 3 , 𝑅 𝑜𝑢𝑡 = 1 𝑔 𝑚 𝑟 𝑑 𝑠 𝑔 𝑚 2 𝑔 𝑚 4 𝑟 𝑑 𝑠 2 || 1 𝑔 𝑚 𝑟 𝑑 𝑠 𝑔 𝑚 1 𝑔 𝑚 3 𝑟 𝑑 𝑠 1
B. Refer to solutions