BJT differential amplifier, small signal analysis

[eliotsbowe]

Hello, I'm having some issues with small-signal analysis of BJT differential amplifiers.

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 makes the potential of the emitter E stuck at its initial value.

At this point, 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 r_pi and an output impedance Rc || r_o .
What's the input/output impedance of the whole differential pair?

Thanks in advance for your time.

 

[(*steve*)]

Hello, I'm having some issues with small-signal analysis of BJT differential amplifiers.

Thanks for placing this is the homework section

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

That's the normal theoretical start position.

Having two opposite-phase inputs with the same amplitude makes the potential of the emitter E stuck at its initial value.

OK, remember what you said here. Do you understand that?

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

Sure, I'll try

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

You recall analysis of simpler (linear) circuits using (say) just resistors and voltage and/or current sources. In those circuits, you could make any node the ground node. You may have been taught that there were some nodes which, if chosen as a ground node, made analysis easier.

This statement is simply a rewording of the statement above (that I asked you if you understood), and an observation that because the potential is essentially fixed, we can easily compare all the other voltages in the circuit to it (or at least as easily as we can any other node that remains at a fixed potential).

This point is chosen though, because it helps offer insight into the operation of the 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;

And here is the insight. If we assume this is a ground, we can simplify our small signal analysis. Being able to treat both halves of the circuit independently (even though we know they're co-dependant) simplifies the analysis enormously.

One last question: both small-signal circuits of Q1 and Q2 have an input impedance r_pi and an output impedance Rc || r_o .
What's the input/output impedance of the whole differential pair?

I guess this is the question you have to answer

Google is your friend. Here are some lecture notes dealing with this.

When you've got it figured out, come back and explain it to me. If you can explain it then (a) it means you have a good understanding, and (b) it means someone can check your explanation.

 

[eliotsbowe]

Thanks for replying.

I liked the point you made about the pro sides of E being the ground node.

I've figured out that E is forced to be grounded if I apply two opposite-phase input with the same amplitude, because the two sides of the circuit are symmetric and KVLs give two opposite voltages between E and the ground, which is impossible as long as the potential of E is nonzero.

I do think google is my friend (that's how I found this forum ) but it wasn't friendly about my last question.
I mean, I can find the input/output impedance for the each (symmetric) side of the small signal equivalent circuit, but what role is that impedance going to have when I deal with the whole differential couple?
Looking at the picture, I'd say:
Q1's side input impedance = input impedance of the differential couple;
Q2's side output impedance = output impedance of the differential couple;

But if an input is applied between the base of Q1 and the base of Q2, how can I state that it is going to "see" the impedance of the Q1's circuit only?

 

[(*steve*)]

At this point, I have exhausted all my book learnin'.

Whilst I could possibly fumble through the small signal stuff to get an answer, I am not sufficiently confident that I could try to provide pointers.

I recommend you look at Wikipedia, which gives you an answer with some reasoning.

That should give you insight as to how to approach the small signal analysis.

 

[eliotsbowe]

No problem, Steve, thanks for trying anyway!