What's the reason for modifying these amplifier circuits?

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TheRedDevil18
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If I have a bjt amplifier with pure differential input signals like this:
Circuit schematic of a BJT differential pair with a dotted vertical line of symmetry; labels include VCC, RC, Q1, Q2, vC1, vC2, Vid/2 sources, RE, node x, IT, REE and -VEE.


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

Circuit schematic of a differential BJT amplifier with a vertical dashed line of symmetry; Q1 and Q2 each have RC to +VCC, RE to nodes x and y, and a tail network of 2REE with a current source IT/2 to -VEE; common-mode input sources vic are shown on both bases.


Also when finding the Common Mode Input Resistance the circuit above excludes RE and includes ro, why ?
Small-signal equivalent circuit for common-mode input resistance: input source v_ic with 2R_icm, base current i_b, emitter resistor r_e, controlled source α·i_e, tail 2R_EE, and output resistance r_o to ground.
(finding input common mode resistance)
 

Discussion

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.
 
LvW said:
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
 
TheRedDevil18 said:
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?
 
TheRedDevil18 said:
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?
 
The Electrician said:
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 don't understand certain things
 
LvW said:
Why do you think the voltage at x would be zero? Can you verify this assumption?

Because the symmetry ?
 
TheRedDevil18 said:
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 don't 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.
 
TheRedDevil18 said:
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.
 
The Electrician said:
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
 
LvW said:
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 ?
 
TheRedDevil18 said:
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:
TheRedDevil18 said:
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?
 
The Electrician said:
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)
 
LvW said:
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 ?
 
TheRedDevil18 said:
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):

Mathematica session solving for the common mode input impedance of a half circuit: a 3x3 admittance matrix Y, then Z = Inverse[Y], and ZinCom = FullSimplify[Z[[1, 1]]] giving a rational expression in re, Rc, ro, beta, RE and RG.


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

Mathematica notebook output simplifying an impedance expression by taking limits: ZNOre = Limit[ZinCom, re -> 0], ZN0RENOre with RE -> 0, Limit[ZNOre, Rc -> infinity] giving RE + 2 REE, and ZNORc = Collect[Limit[ZinCom, Rc -> infinity], re] giving re (1 + beta) + RE + 2 REE.
 
TheRedDevil18 said:
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).
 
The Electrician said:
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
 
LvW said:
Because common-mode signals will cause the same current changes in both transistors (same direction).

Is the voltage at x measured across REE ?
 
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 ?
 
TheRedDevil18 said:
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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