Small-signal transfer function from this differential input

AI Thread Summary
The discussion revolves around solving a small-signal transfer function problem related to a differential input circuit. Participants emphasize the importance of using given equations, such as BJT equations and Kirchhoff's laws, rather than relying on equivalent circuits. A key suggestion is to utilize the small-signal transconductance (gm) directly to simplify the solution. There is some confusion regarding the relationship between the exponential function and the problem, with hints provided to clarify its relevance. Overall, the focus is on finding an efficient method to derive the transfer function using the appropriate equations.
CorHawk
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Homework Statement


q1.png

2. Homework Equations [/B]
Small Signal Equivalent Circuit, Kirchhoff Current Law, and BJT equations mentioned in the question.

The Attempt at a Solution


q1a_1.png

It looks like I did something wrong. Can someone help me with fixing this?
 
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CorHawk said:

Homework Statement


View attachment 81419
2. Homework Equations [/B]
Small Signal Equivalent Circuit, Kirchhoff Current Law, and BJT equations mentioned in the question.

The Attempt at a Solution


View attachment 81420
It looks like I did something wrong. Can someone help me with fixing this?
I would not use equivalent circuits for this. You're given all the equations you need to solve the problem without resorting to equiv. ckts.
Hint: at some point use ex ≅ 1 + x for 0 ≤ x << 1.
OR you can do it with just gm.
 
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rude man said:
I would not use equivalent circuits for this. You're given all the equations you need to solve the problem without resorting to equiv. ckts.
Hint: at some point use ex ≅ 1 + x for 0 ≤ x << 1.
OR you can do it with just gm.

I don't understand how you want to get to the answer 1/2*gm by only using gm and nothing else. And I don't see how e^0=1 would be related to this.
 
Last edited:
CorHawk said:
I don't understand how you want to get to the answer 1/2*gm by only using gm.
I'll get you started:
gm = dI1/d(Vb1 - Ve) = -dI2/dVe
where Vb2 = 0 is assumed wlog ("without loss of generality"). Vb is base voltage, Ve is emitter voltage.
And I don't see how e^0=1 would be related to this.
That would come in handy if you choose to use I = Isexp(qVbe/kT) instead of gm directly.
I recommend using gm directly, it's shorter.
 
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