Hello, I'm having some issues with small-signal analysis of BJT differential amplifiers.(adsbygoogle = window.adsbygoogle || []).push({});

Let's say I have this BJT differential pair and suppose that Q1 and Q2 perfectly match: http://i50.tinypic.com/14ahdgn.jpg

Suppose a small differential input is applied and consider the small–signal equivalent circuit:

Having two opposite-phase inputs with the same amplitude ([itex]\frac{v_d}{2} , - \frac{v_d}{2}[/itex]) makes the potential of the emitter E stuck at its initial value.

My book says that:

- If the potential of E is stuck at its value, then E can be considered as a ground point for the small-signal circuit;

- If E is grounded, then every point the two circuits have in common is grounded; this means that the two circuits become totally independent one from each other and I can solve the small-signal circuit of Q1 and Q2 separately;

Can anyone please explain me the reason for these two statements?

One last question: both small-signal circuits of Q1 and Q2 have an input impedance [itex]r_\pi[/itex] and an output impedance Rc || [itex]r_o[/itex].

What's the input/output impedance of the whole differential pair?

Thanks in advance for your time.

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# BJT differential amplifier, small signal analysis

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