Suppose you have designed the circuit above with R_a = R_c = 100 ohm, and chosen

Question

Suppose you have designed the circuit above with R_a = R_c = 100 ohm, and chosen R_b and R_d such that where, A_dm is the difference-mode gain, and the common mode gain is zero (i.e. CMRR is infinity). However, the resistors you've used to build the circuit are not precise. Each of them has an x% error (therefore, a resistor with a nominal value of 100 ohms will actually have a resistance of 99 or 101 ohms). Write a program to find the worst possible CMRR for a given value of x and A_dm. Plot the CMRR (in dB) vs. x for x = 1,2,... 10. Obtain three graphs (and display in the same window) for A_dm = 1,10,100.

Explanation / Answer

An op-amp differential amplier is built using four identical resistors, each having a tolerance of
±5%

CMRR:-Common Mode Rejection Ratio.

Reference has the equations for the gain.
Vo = ((R1+Rf)/R1) (Rg/(Rg+R2)) (V2) – (Rf/R1) (V1)

Set the two inputs equal, which should give you zero output. Set inputs at 1 volt for easier calculation.
Vo = ((R1+Rf)/R1) (Rg/(Rg+R2)) – (Rf/R1)

Now figure out which combination of out of tolerance resistors gives you the highest Vo. You may have to try several combinations and do the calculations for each. When you get the highest Vo, the CMRR = 20 log Vo

for example, my first guess would be to make Rf and R2 high, the others low. this would be the same if you made Rf and R2 10% high and the others nominal.
Vo = ((R1+Rf)/R1) (Rg/(Rg+R2)) – (Rf/R1)
Vo = ((1+1.1)/1) (1/(1+1.1)) – (1.1/1)
Vo = (2.1)(1/2.1) – 1.1 = 1 – 1.1 = 0.1
in this case, CMRR = 20 log 0.1 = –20 dB

Second guess would make R1 and Rg high
Vo = ((1.1+1)/1.1) (1.1/(1.1+1)) – (1/1.1)
Vo = 1– 0.91 = 0.09
CMRR = 20 log 0.09 = –21dB

looks like 21 dB is the answer

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