Differential Mode Gain and 1/2 circuit

[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.

Homework Statement

Homework Equations

(the solution from the solution manual)[/B]

The Attempt at a Solution

I am getting stumped. Please help : /
Thanks for reading.

 

[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.

 

[perplexabot]

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.

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

 

[rude man]

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

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

 

[perplexabot]

I agree with Integral. This scheme involves no more than a power distribution network, and a very poor one at that.
To "charge" the Earth with any measurably useful non-local extracting force would require a placement of an earth-charging generation device likely to boggle the imagination, and likely far surpassing any ability we have.
Furthermore, what would be the point?
A considerable amount of energy would be required to charge the earth, for one, and the Earth's exceptionally poor conductivity would create current losses far exceeding the rationality of efficient distribution.
In my opinion, the idea is dead from the start.

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

 

[rude man]

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

?
Different thread.

 

[perplexabot]

?
Different thread.

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.

 

[rude man]

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.

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

 

[The Electrician]

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