What's the reason for modifying these amplifier circuits?

  • #1

Main Question or Discussion Point

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


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

bjt common.PNG


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

Answers and Replies

  • #2
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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
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
 
  • #4
The Electrician
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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?
 
  • #5
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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?
 
  • #6
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
 
  • #7
Why do you think the voltage at x would be zero? Can you verify this assumption?
Because the symmetry ?
 
  • #8
The Electrician
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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.
 
  • #9
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Because the symmetry ?
Is there an ideal current source in the emitter leg? I don`t think so.
Hence, you have the classical common emiiter configuration with emitter feedback.
 
  • #10
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
 
  • #11
Is there an ideal current source in the emitter leg? I don`t 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 ?
 
  • #12
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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 can`t 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:
  • #13
The Electrician
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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?
 
  • #14
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)
 
  • #15
But his can`t 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 ?
 
  • #16
The Electrician
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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):

ComZ_1.png


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

ComZ_2.png
 
  • #17
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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).
 
  • #19
Because common-mode signals will cause the same current changes in both transistors (same direction).
Is the voltage at x measured across REE ?
 
  • #20
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Have a look into your diagram how and where Vx is defined.
 
  • #21
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 ?
 
  • #22
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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 ?
See my answer#17.
 

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