# Differential amp single-ended

Discussion in 'Homework Help' started by Rit.aux, Sep 30, 2017.

1. ### Rit.aux Thread Starter New Member

Jul 29, 2015
2
0
Tell me please. I'm dealing with a differential amplifier. And he got confused.
I solve the examples and question 9 in the spoiler - the answer is there and the formula. I do not understand where it comes from.
Question 9

How do I understand the operation of a differential amplifier with one input signal:
Transistor Q1 is an amplifier with a common emitter. The amplification of Rc / (re + RE || re). Since RE >> re, the gain is approximately A = Rc / 2re.
As I understand it, if Q2 is in a mode with a common base, then its input resistance is Z = re + RE || re.
And since A = Vout / Vin = (i * Rc) / (i * Z) => Rc / 2re.

I understand correctly?

2. ### MrAl AAC Fanatic!

Jun 17, 2014
6,383
1,382
Hi,

I guess for this question you are allowed to consider the two transistors as being two separate amplifiers connected together. Otherwise you can do a full analysis such as a Nodal analysis.
If no one else helps more here i can help more later tonight or tomorrow sometime.

3. ### MrAl AAC Fanatic!

Jun 17, 2014
6,383
1,382
Hello again,

A complete Nodal analysis turns up almost the same formula. It's the same formula but multiplied by the factor:
G=(RE+re)/RE)=1+re/RE

so when 're' is small they are very much the same. This is also when approximating B+1 as just B as is common in these kinds of calculations, otherwise we have to include the Beta B in the calculation.

The difference with Rc=1000, RE=100, and re=10 is one formula gives Av=48 and one gives Av=53, although we may want to use some more realistic values.

Last edited: Oct 2, 2017
4. ### Rit.aux Thread Starter New Member

Jul 29, 2015
2
0
And what is this coefficient? Where does it come from? I did not understand.
And the second question. The output of the circuit is the collector of the transistor, which is switched on by a common base. How does Rc / (re + RE || re) get there? After all, on the Q2 emitter, the voltage will be Vin-0.7, because Q1 here works (for Q2), as an emitter follower

5. ### MrAl AAC Fanatic!

Jun 17, 2014
6,383
1,382
Hi again,

The 'coefficient' you are talking about is the factor i mentioned? If so, it comes from first doing a complete Nodal analysis (or whatever you want to use to analyze it fully) and then comparing the Nodal analysis to the 'book' analysis result. The only difference is that factor.

If you are only hung up because of the 0.7v then you may be happy to know that you can call that zero rather than 0.7. This is common in small signal analysis. So just assume the base emtter drop is 0.000 volts.
Another approximation you may need is the B+1=B approximation where we assume B is high enough so that B+1 is almost the same value and only affects the final outcome by a small fraction and after all the exact value of B in a real life circuit is never known exactly anyway, so the error from the approximation gets absorbed into the normal error anyway.

Try it with this information and see if you can work it out. If not we'll have to go over it step by step i guess.