EE 435 Homework 4 Solutions Spring 2021
Problem 1
NMOS
PMOS
Problem 2
Part A
The current through M1 is π
ππ·π·βπππβ
1
2= 83.33ππ΄. So we can find π1 as follows:
π1 =2πΌπΏ
ππΆππ₯ππΈπ΅2 =
2 β 83.3π β 2π
112.8π β 0.22= 73.84ππ
Part B We know that ππΈπ΅1 = ππΊπ β πππ = 0.2π. Quiescently, ππΊ = 0π. So:
ππΊ β ππ β πππ = βππ β 0.79π = 0.2 β ππ = β0.99π
Part C Create an expression for ππππ divided by ππΌπ:
ππ1ππΌπ = ππππ (π πΆ1 +1
π 1)
π΄(π ) =ππ1
π πΆ1 + 1/π 1
Part D Start by finding the DC gain:
π΄0 = πππ 1 =2 β 83.33ππ΄
0.2β 50πΞ© = 41.66510 = 32.39ππ΅
Now find the corner frequency:
π πΆ1 +1
π 0= 0 β π = β
1
π 1πΆ1= 500π πππ/π ππ
Part E As found in Part D, the 3dB bandwidth is 500π πππ/π ππ.
Part F Increasing the power increases the gain-bandwidth, but not the bandwidth. The bandwidth is only dependent on π 1 and πΆ1.