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

## Main Question or Discussion Point

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)

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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.

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.
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

The Electrician
Gold Member
Also when finding the Common Mode Input Resistance the circuit above excludes RE and includes ro, why ?
(finding input common mode resistance)
Apparently they are making some simplifying assumptions. What do you get for the common mode input resistance?

By the way, is this homework?

LvW
for my small signal analysis I would neglect REE because the voltage at x is zero and I would get the wrong equation
Why do you think the voltage at x would be zero? Can you verify this assumption?

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

By the way, is this homework?
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

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

The Electrician
Gold Member
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
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.

LvW
Because the symmetry ?
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.

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

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.
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 ?

LvW
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 ?
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:
The Electrician
Gold Member
Yes, it's the last circuit in the post. I'm not too sure how to analyse it without the small signal model
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?

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

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

The Electrician
Gold Member
View attachment 96070

Also when finding the Common Mode Input Resistance the circuit above excludes RE and includes ro, why ?
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:

LvW
Ok, why is it not true for common mode signals ?
Because common-mode signals will cause the same current changes in both transistors (same direction).

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):

View attachment 96185

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

View attachment 96186
They also neglected re, since it is much less than the parallel combination of 2REE and ro

Because common-mode signals will cause the same current changes in both transistors (same direction).
Is the voltage at x measured across REE ?

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

When finding the common mode input resistance why is it 2*Ricm

They go on to say that 2*Ricm = Vic/ib but isn't that just for one amplifier or just Ricm ?

LvW
When finding the common mode input resistance why is it 2*Ricm
They go on to say that 2*Ricm = Vic/ib but isn't that just for one amplifier or just Ricm ?