UC Berkeley EECS Midterm 1 (Fall 1994) - Electrical Engineering Problems, Exams of Microcomputers

The take-home midterm exam for the electrical engineering and computer sciences (eecs) 145l course at the university of california, berkeley from fall 1994. The exam covers problems related to differential amplifiers, instrumentation amplifiers, amplifier noise, and filter design.

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UNIVERSITY OF CALIFORNIA
Electrical Engineering and Computer Sciences
145L MIDTERM #1 (take-home)
September 19, 1994
Due Monday, September 26, 1994
(100 points total, 3 points deducted for each school day late)
(no credit after graded midterms have been returned to students)
PROBLEM 1 (16 points)
For the differential amplifier circuit shown in Course Reader Figure 2.4, and assum-
ing that the open loop gain A is infinite, do the following:
a. Derive the expression for the output V0 as a function of the four variables R1, R2,
R3, and R4. Your result should be of the form V0 = aV2bV1.
b. Derive expressions for the differential gain G± and the common mode gain Gc in
terms of R1, R2, R3, R4, using the expression from part a. and the following:
V0 = aV2bV1 = (a + b)(V2V1)/2 + (ab)(V2 + V1)/2
= G± (V2V1) + Gc Vc, where Vc = (V1 + V2)/2
c. Using your expressions from step b., derive an expression for the CMRR in terms
of R1, R2, R3,R4.
d. Under what conditions does Gc = 0?
e. When Gc = 0 is satisfied, what does your expression for G± reduce to?
f. For G± = 1 and G± = 1000, first set R2/R1 = R4/R3 = G± and then find the percent-
age variation in R3 that causes CMR 120 dB. Note that G± changes very little.
Comment on the resistor accuracy required for a good CMRR at the two differ-
ential gains.
Note: G± is primarily determined by R2/R1 and R3 (or R4) can be used to “fine tune”
Gc.
PROBLEM 2 (16 points)
You are given an instrumentation amplifier with a gain that is adjustable from 1 to
1000. At a gain of 1, the bandwidth is 106 Hz. At a gain of 1000, the bandwidth is 104
Hz.
a. Both input terminals are connected to ground with 5-M resistors. If the input
leakage currents on the two inputs are 0.5 nA and 1.5 nA, what is resulting output
offset at a gain of 1000? At a gain of 1?
b. What is the output noise in the 104-Hz bandwidth at a gain of 1000 due only to the
room-temperature Johnson noise in the 5-M resistors? (Hint: If two uncorre-
lated noise sources are added, the rms noises combine as the square root of the
sum of their squares.)
c. When the inputs are connected directly to ground, the output-voltage noise with
a gain of 1000 is 1 mV rms in the 104 Hz bandwidth. When the gain is reduced to
EECS 145L Midterm #1 page 1 Due Sept. 26, 1994
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UNIVERSITY OF CALIFORNIA

Electrical Engineering and Computer Sciences

145L MIDTERM #1 (take-home) September 19, 1994

Due Monday, September 26, 1994 (100 points total, 3 points deducted for each school day late) (no credit after graded midterms have been returned to students)

PROBLEM 1 (16 points)

For the differential amplifier circuit shown in Course Reader Figure 2.4, and assum- ing that the open loop gain A is infinite, do the following:

a. Derive the expression for the output V 0 as a function of the four variables R 1 , R 2 , R 3 , and R 4. Your result should be of the form V 0 = aV 2 – bV 1.

b. Derive expressions for the differential gain G ± and the common mode gain Gc in terms of R 1 , R 2 , R 3 , R 4 , using the expression from part a. and the following: V 0 = aV 2 – bV 1 = ( a + b )( V 2 – V 1 )/2 + ( ab )( V 2 + V 1 )/ = G ± ( V 2 – V 1 ) + Gc Vc , where Vc = ( V 1 + V 2 )/

c. Using your expressions from step b. , derive an expression for the CMRR in terms of R 1 , R 2 , R 3 ,R 4.

d. Under what conditions does Gc = 0?

e. When Gc = 0 is satisfied, what does your expression for G± reduce to?

f. For G± = 1 and G± = 1000, first set R 2 /R 1 = R 4 /R 3 = G± and then find the percent- age variation in R 3 that causes CMR ≈ 120 dB. Note that G± changes very little. Comment on the resistor accuracy required for a good CMRR at the two differ- ential gains.

Note: G± is primarily determined by R 2 /R 1 and R 3 (or R 4 ) can be used to “fine tune” Gc.

PROBLEM 2 (16 points)

You are given an instrumentation amplifier with a gain that is adjustable from 1 to

  1. At a gain of 1, the bandwidth is 10^6 Hz. At a gain of 1000, the bandwidth is 10^4 Hz.

a. Both input terminals are connected to ground with 5-MΩ resistors. If the input leakage currents on the two inputs are 0.5 nA and 1.5 nA, what is resulting output offset at a gain of 1000? At a gain of 1?

b. What is the output noise in the 10^4 -Hz bandwidth at a gain of 1000 due only to the room-temperature Johnson noise in the 5-MΩ resistors? ( Hint: If two uncorre- lated noise sources are added, the rms noises combine as the square root of the sum of their squares.)

c. When the inputs are connected directly to ground, the output-voltage noise with a gain of 1000 is 1 mV rms in the 10^4 Hz bandwidth. When the gain is reduced to

1, the output-voltage noise is 0.1 mV rms in the 10^6 Hz bandwidth. What is the amplifier noise with respect to the input ( D 1 ) and the output ( D 0 )? (Express the noise in units of nV Hz–1/2.)

PROBLEM 3 (10 points)

The classic instrumentation amplifier circuit is shown in figure 2.13 of the course reader (page 82).

Assume the following:

  • R 1 = 100 Ω, R 2 = 5 kΩ, R 3 = 1 kΩ, R 4 = 10 kΩ.
  • Input V+ = +1 volt d.c. plus 1 mV p-p (peak-to-peak) sine wave at 1 kHz
  • Input V– = +1 volt d.c. plus 1 mV p-p sine wave at 1 kHz
  • Differential input (V+ – V–) = 2 mV p-p sine wave at 1 kHz
  • Power supply voltages are –10V and + 10V

a. What are the amplitudes of the d.c. and 1 kHz components of V 3?

b. What are the amplitudes of the d.c. and 1 kHz components of V 4?

c. What are the amplitudes of the d.c. and 1 kHz components of V 4 – V 3?

d. What are the amplitudes of the d.c. and 1 kHz components of V 0?

PROBLEM 4 (16 points)

A new instrumentation amplifier circuit has been proposed, as shown below:

V (^1) R 2

V 2^ R 2

R 3

R 3

V 3 V 5

V (^4) V 6 R 4

R 4

V 0

V

V +

R 1 / 2

R 1 / 2

Assume the following (same values as Problem 3):

  • R 1 /2 = 50 Ω, R 2 = 5 kΩ, R 3 = 1 kΩ, R 4 = 10 kΩ.
  • Input V+ = +1 volt d.c. plus 1 mV p-p sine wave at 1 kHz
  • Input V– = +1 volt d.c. plus 1 mV p-p sine wave at 1 kHz
  • Differential input (V+ – V–) = 2 mV p-p sine wave at 1 kHz
  • Power supply voltages are –10V and + 10V

Answer the following: