Small-signal transfer function from this differential input

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CorHawk
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Homework Statement


Circuit schematic of a bipolar differential input stage with transistors T1 and T2, bias current source I_BIAS, supply V_S, inputs V_A and V_B, and output current labelled I_BIAS/2 + i_s; text above lists bipolar and MOS equations.

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

The Attempt at a Solution


Handwritten small-signal analysis of a differential BJT pair: left, a transistor pair with Va and Vb and currents I/2 minus ix and I/2 plus ix; middle, hybrid-pi equivalent circuit with gm Vb, r_pi, ro; bottom, KCL equation gm Va minus gm Vb minus 2ix = 0 and boxed result ix/(Va minus Vb) = -gm/2.

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