MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Electrical Engineering and Computer Science
6.301 Solid State Circuits
Fall Term 2010
Problem Set 3
Issued : Sept. 17, 2010
Due : Friday, Sept. 24, 2010
Suggested Reading: Read as many of the following as you can. All of the recommended references are on
reserve at Barker Library.
1. Lundberg sections 4 and 5.
2. Grebene section 5.1.
3. Gray and Meyer section 3.5.
Problem 1: Find the mid-band gain for each circuit below. Assume that β = 400, VBE,ON = 0.6V, and
ignore ro .
(a) Circuit a
+5V
+5V
+5V
+5V
1.2k
1.3k
vo
1k
Q1
Q2
vi
1uF
1k
1.3k
1uF
2.6V
(b) Circuit b
+5V
+5V
2k
1k
+5V
1.4k
1.3k
1uF
vo
Q2
Q1
vi
+5V
Q3
3.0k
3k
2.4k
1uF
2.4k
1
1uF
Problem 2: Simulate the CE-EF-CE amplifier in Circuit 1b (above) with HSPICE. Use the following data
for your simulations: IS = 10−15 A, βF = 200, VA =100, τF = 0.1ns, cjeo = 10pF, and cjco = 2pF. Turn
in your HSPICE input file, and an Awaves plot showing the low and high frequency roll-offs.
Problem 3: For the differential amplifier shown below:
VCC
RC
RC
vo1
vo2
RL
vi1
Q1
vi2
Q2
RE
IEE
IEE
-VEE
(a) Find the differential voltage gain avd and the common mode voltage gain avc .
(b) Find both the differential and common mode input and output resistances (Rin,d , Rin,c , Rout,d ,
and Rout,c ).
2
Problem 4: For the single-ended differential amplifier shown below, assume that β = 200 and VBE,ON =
0.6V for both transistors, and that the DC common mode input voltage is zero. Neglect ro for this
problem.
VCC =15V
RL
RL
vo
vi1
Q1
Q2
vi2
RE
-VEE =-15.6V
(a) Express the differential voltage gain avd as a function of the voltage drop VL across RL .
(b) If VCE,SAT = 0.3V, what is the maximum avd possible?
(c) Select RL and RE so that Rin,d = 1MΩ and avd = 300. What is the common mode rejection ratio
(CMRR)?
3
Problem 5: For each circuit below, find the transfer function vo /vi , and draw the Bode plot (magnitude
and phase).
(a) Circuit a
+
−
vi
R1=100k
R2=100k
C1=0.1uF
C2=1uF
vo
(b) Circuit b
R1=100k
+
−
vi
C1=0.1uF
C2=1uF
vo
R2=100k
(c) Circuit c
C1=1uF
R1=10k
vi
+
−
R2=90k
4
vo
C2=4uF
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6.301 Solid-State Circuits
Fall 2010
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