ECE 595B CMOS Analog IC Design
Exam 1
ECE595B Exam 1 (Oct 12, 2007)
Purdue ID:
Full Name:
This is a closed book, closed notebook test. One single-sided letter size paper with equations is allowed. You are allowed to use a calculator. All work and answers should be written in the space provided.
State any assumption you make. It is recommended to keep your answer clear and concise.
At the end of test period (50min), turn in all test pages. You will be given a five minute warning prior to the end of the test period. Once the proctor leaves the exam room, no more test pages will be accepted.
Problem / Score1 / / 25
2 / / 25
3 / / 25
4 / / 25
Total / / 100
TIP: When two resistors are in parallel, you can use (R1 // R2) instead of R1R2/(R1+R2).
Use gmx and rOx for transconductance and output resistance, respectively.
x is a transistor number.
1. [25 points] A signal source with a source impedance RS is connected to the common source amplifier with a feedback resistor RF. Assume that the transistors are in saturation and ignore the channel length modulation and body effect.
(a) [10 points] Calculate the input impedance, Zin, of the amplifier. Ignore all capacitances.
(b) [15 points] Calculate the low-frequency voltage gain (uo/uin) of the amplifier.
2. [25 points] Calculate the low-frequency voltage gain (uo/uin) of the amplifier shown below. Assume that all transistors are in saturation, and ignore the body effect and channel length modulation.
3. [25 points] Calculate the low-frequency differential voltage gain [uo/uin = (uo+-uo-)/(uin+-uin-)] of the amplifier shown below. Assume all transistors are in saturation, the circuit and signals are fully symmetric, the signals are small enough to use small-signal models, and there is no body effect. (l ≠ 0).
4. [25 points] Calculate the low-frequency differential to single-ended voltage gain [uo/uin = (uo)/(uin+-uin-)] of the amplifier shown below. Assume all transistors are in saturation, the current source is ideal, the input signal is fully symmetric, and there is no body effect. Assume (W/L)1=(W/L)2, and (W/L)3=(W/L)4=(W/L)5=(W/L)6, and (W/L)7=(W/L)8. Hint: current mirror! You don’t need to calculate the exact current ratio of the mirror.
1/11
ECE595B