How to find common-mode voltage gain of a Differential Amp

AI Thread Summary
The discussion focuses on calculating the common-mode voltage gain (Acm) and differential voltage gain (Ad) of a differential amplifier, given parameters like beta, thermal voltage, and Vbe. Users clarify that with an infinite Early voltage (Va), Ro should also be considered infinite, leading to a common-mode gain of zero and an infinite common-mode rejection ratio (CMRR). There is confusion regarding two different values for Ad, with one participant correcting their own calculations for transconductance (g_m) and differential gain. The importance of consistent unit usage is emphasized to avoid errors in calculations. Overall, the conversation highlights the complexities of analyzing non-ideal conditions in transistor circuits.
DODGEVIPER13
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Homework Statement


Beta=100,Thermal Voltage=0.026 Volts, Va=infinity, and Vbe=0.75 Volts for the differential amplifier. Please look at the upload for a clearer picture.


Homework Equations


Ad=betaRc/2(rpi+Rb)
Acm=-gmRc/(1+(2(1+beta)Ro/(rpi+Rb)))

The Attempt at a Solution


I have uploaded my attempt but I am wondering if Ro is really zero I mean Va is infinity so I would think so but this would cause Acm=0 and CMRR= infinity this seems incorrect what am I doing wrong?
 

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I think what you did was correct.
The 'tail' in this diff amp is a pure current source (Ic3) and so not changed by a common mode input voltage.
Thus, the common mode gain, absent any transistor asymmetries of which none are specified, is indeed zero and the CMRR is infinity.

For Ad you seem to have two answers: "Ad = 73.169 = 28.846". What gives with that? Anyway, I got close to your the second number by approximating as
i = is exp(V/VT), ∂i/∂V = is (1/VT) exp(V/VT) = (1/VT) i = (1/.026)0.75e-3 = 34.7
so your number of 28.846 looks about right since my approx. will come in high.
 
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Well darn the 28.846 is actually my Gm the 73.169 is my gain I need to check that
 
I myself goofed in computing g_m. It should have been 0.029 A/V.
That times the 8K resistor gives A_d = 232 but that ignores beta, r_pi and the 2K base resistor so the actual gain will be somewhat less.
 
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Sweet thanks again man!
 
Hey I had one last question I had a friend bring up the point that the circuit is not ideal due to current biasing. So I resolved for R0 by running a kvl from 12 to Vec3 soI got 12-Ib(2)-0.7=Vec=IcRo and Ro=7.54 k ohms
 
DODGEVIPER13 said:
Hey I had one last question I had a friend bring up the point that the circuit is not ideal due to current biasing. So I resolved for R0 by running a kvl from 12 to Vec3 soI got 12-Ib(2)-0.7=Vec=IcRo and Ro=7.54 k ohms

I don't quite understand this. Ro is due to the Early voltage VA which is given to you as infinite, so Ro should also be infinite.

" the circuit is not ideal due to current biasing "?? Every transistor circuit is current biased, or maybe there is an interpretation here I don't know about. I realize the "tail" is a constant-current source = 1.5 mA.

Maybe you can elaborate on your equation.
 
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Alright man appreciate it well that kind of sums it up Ro=infinite
 
OK viper. One last suggestion - watch your units, like mixing mA and Ohms etc. It's probably best if you change everything to ohms, amps and volts before computing numbers.
 
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