Instrumentation amplifier design with 5V power supply selection

[Weaver]

Homework Statement

Homework Equations

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CMRR = A_v/A_cm

A_cm= Δ/R , Δ = (2 x Tolerance of Resistor).R

The Attempt at a Solution

I have an issue with part e) and f) but here are all my workings

c) For this part, A_cm would be:
10 x10^-3
This makes 2.5mV -> 0.25 uV (which is the same magnitude as 0.22uV)

I am unsure for this part but would the CMRR in this case be 1/10x10^-3 = 10x10^3 ?

d) A_v = 1V/42mV = 23.81

e) Using this model of the instrumentation amplifier

A_cm = 23.8/(10 x10^3) = 2.4 x 10^-3

=> 0.12 % tolerance, which is closest to 0.1% tolerance

Using Rb = 9k and Ra = 100K, which get A_v = 23.2

One part I am not sure of is picking the 5V power supply. The only justification I can see, is that no clipping would occur on the output?

f)I'm really not sure of this part, but one possible reason is, it doesn't fix any problems. There is still the broadband noise on the signal, which will be amplified now too and an instrumentation op-amp will still have to be used to get the mic signal

 

[Tom.G]

You seem to have ignored the other contribution to CMMR.
See: https://masteringelectronicsdesign....-instrumentation-amplifier-transfer-function/