Differential Mode Gain: Positive Gain With -Vid/2?

AI Thread Summary
The discussion centers on the calculation of differential mode gain in a differential amplifier circuit. The user observes that using the second differential input (-Vid/2) yields a positive gain, while the first input (Vid/2) results in a negative gain, contrary to their expectations. A SPICE simulation supports the positive output for the second input, leading to confusion about the gain calculations. The basic operation of the circuit indicates that when V+ exceeds V-, the transistor Q2 should turn off, affecting the output gain. The user seeks clarification on the apparent discrepancy in expected versus calculated gains.
FOIWATER
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In the following diff amp, only one output is taken from collector two.

When I calculate the differential mode gain using the second differential input (-Vid/2) I get a positive gain. When I calculate the differential mode gain using the first differential input (Vid/2) I get a negative gain.

A spice simulation of this circuit shows that the output is positive.

I would then expect to get a negative gain for (-vid/2) and a positve gain for (vid/2)

Does someone know why that is not the case?
 

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Not sure how you are calculating gain. The basic operation is that when V+ is greater than V-, Q2 will try to turn off.
 
I made a small signal circuit out of the above using a differential mode half circuit. with the input as -Vid/2, I have -Vid/2 = ib(rpi + (1+b)*Ree/2) <-- (virtual ground between Ree/2 for diff mode.)

Vodm = -beta*ib*(Rc||Rcc/2) <--- another virtual ground between Rcc/2

Solving for adm = vodm/vid I get a positive number.

However, I would expect my negative input to be multiplied by a negative gain and give a positive output.

The same is true for Vid/2, I would expect it to be multiplied by a positive gain.

Am I missing things?

Thanks for the feedback
 
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