# What's the reason for modifying these amplifier circuits?

1. Feb 18, 2016

### TheRedDevil18

If I have a bjt amplifier with pure differential input signals like this:

Now let's say I apply pure common mode signals to the same circuit, then why is the circuit above modified to this ?

Also when finding the Common Mode Input Resistance the circuit above excludes RE and includes ro, why ?
(finding input common mode resistance)

2. Feb 18, 2016

### LvW

For a pure common mode signal at both inputs both transistors are doing the same: Both will show slight increase of the collector currents.
OK - if they are doing the same, we can treat both transistor stages individually - however, in the first circuit the emittercurrent was split into two equal halfs. Therefore, each stage must have the same current IE as before: IT/2. And the same applies to REE. Putting both emitter nodes again together we gave 2REE||2REE=REE.

3. Feb 18, 2016

### TheRedDevil18

Ok yes I get that the two circuits are equivalent but then what if I analysed the circuit for the common mode signals without splitting the current source and REE (like first diagram), then for my small signal analysis I would neglect REE because the voltage at x is zero and I would get the wrong equation

4. Feb 18, 2016

### The Electrician

Apparently they are making some simplifying assumptions. What do you get for the common mode input resistance?

By the way, is this homework?

5. Feb 19, 2016

### LvW

Why do you think the voltage at x would be zero? Can you verify this assumption?

6. Feb 19, 2016

### TheRedDevil18

Ricm = Vic/ib
= (B+1)*ib*(2*REE//ro) / ib.................................// means parallel combination
= (B+1)*(2*REE//ro)

This is not homework. This stuff is from the textbook and I just dont understand certain things

7. Feb 19, 2016

### TheRedDevil18

Because the symmetry ?

8. Feb 19, 2016

### The Electrician

Are you just using the simplified circuit shown in the third image in post #1?

What do you get if you analyze half the circuit shown in the second image of post #1? If I perform a nodal analysis on that half circuit, I get a much different result.

9. Feb 19, 2016

### LvW

Is there an ideal current source in the emitter leg? I dont think so.
Hence, you have the classical common emiiter configuration with emitter feedback.

10. Feb 20, 2016

### TheRedDevil18

Yes, it's the last circuit in the post. I'm not too sure how to analyse it without the small signal model

11. Feb 20, 2016

### TheRedDevil18

In the book it says that the increase in signal current in one bjt is exactly matched by a decrease in the other hence the signal voltage at x is zero ?

12. Feb 20, 2016

### LvW

But his cant be true for a common mode signal.
If the increase in T1 is identical to the decrease in T2 we have a differential signal without any common mode portion. Remember: In post#3 you speak abouta common-mode signal.

Last edited: Feb 20, 2016
13. Feb 20, 2016

### The Electrician

So analyze the half circuit shown as the second image in your first post using the small signal model. What do you get if you do that?

14. Feb 20, 2016

### TheRedDevil18

It's the same answer that I posted above Ricm = (B+1)*(2*REE//ro)

15. Feb 20, 2016

### TheRedDevil18

Ok, why is it not true for common mode signals ?

16. Feb 20, 2016

### The Electrician

I can't tell you why they excluded RE, but when I solve for the common input Z, I get this (I included re in the model):

If I delete the effect of various components, I get:

17. Feb 21, 2016

### LvW

Because common-mode signals will cause the same current changes in both transistors (same direction).

18. Feb 22, 2016

### TheRedDevil18

They also neglected re, since it is much less than the parallel combination of 2REE and ro

19. Feb 22, 2016

### TheRedDevil18

Is the voltage at x measured across REE ?

20. Feb 22, 2016

### LvW

Have a look into your diagram how and where Vx is defined.