# Differential Mode Gain and 1/2 circuit

1. May 31, 2015

### perplexabot

Hey all! I have been trying this problem for a while and can't seem to get the same answer as the solution. If someone can tell me where I am going wrong, that would be much appreciated. I am very close to the solution, but I am missing a term in the denominator. Everything is shown below! Thank you for your time.
1. The problem statement, all variables and given/known data

2. Relevant equations (the solution from the solution manual)

3. The attempt at a solution

2. Jun 2, 2015

### rude man

Since the two halves are coupled both at the emitters and collectors I would not try to use the "1/2 d.m. model" at all. Write the usual equations, I would go with kvl (actually I just sum currents to zero at every dependent junction which is basically the same set of equations.)

If you HAVE to use the d.m model I cannot be of help.

3. Jun 2, 2015

### perplexabot

Hmmm, ok. I will consider trying without half circuits. I don't have to use the 1/2 crkt model, I just thought it would be easier.

Thank you

4. Jun 2, 2015

### rude man

It would have been except for RL and RE.
I will keep the answers to myself until you come up with your answers, then let you know how they look.

5. Jun 2, 2015

### perplexabot

Ok, thank you. I'm in the middle of finals week, I will post an answer, just not too soon. I appreciate your help.

6. Jun 2, 2015

???

7. Jun 2, 2015

### perplexabot

Lol, I wonder where that quote came from?! That is weird, I was trying to quote your previous reply. My reply is still valid, except the quote is wrong. Let me try and remove it. I apologize for that.

8. Jun 2, 2015

### rude man

No prob.
Actually a pretty involved analysis. 7 equations, 7 unknowns.
For c.m. gain a hint: consider symmetry!

9. Jun 2, 2015

### The Electrician

The main problem I see is that you have calculated the voltage across RE/2 using the voltage divider formula applied to the input voltage Vi. But the current into RE/2 isn't just the current flowing down through rπ; you also have to include the current flowing down from the dependent current source, so you can't just use the voltage divider formula.