# BJT diff amp question

Discussion in 'General Electronics Chat' started by automagp68, Nov 18, 2015.

1. ### automagp68 Thread Starter Member

Nov 13, 2011
81
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Question for you gurus

Ive got this circuit here.
For the life of me when i look at the AC model for it i can not understand why they say RE is 125

AKA its divided by 2

Can someone explain simply why they are deviding RE/2 = 125

If Vin is some voltage and Vin2 is grounded then obviously we are talking about difference signal not common mode

AC Model

Dec 29, 2014
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3. ### automagp68 Thread Starter Member

Nov 13, 2011
81
0
Hello

I have seen that problem before. They work out to be the same answer
However its obvious to me how the other problem works. Its still not obvious to me how this one works

Thanks

Nov 13, 2011
81
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anyone?

5. ### Brownout Well-Known Member

Jan 10, 2012
2,375
998
For the differential model, you have vin/2 on the left input, and -vin/2 on the right input. (not vin and ground, that's not a differential input) So, split the 250 ohm resistor into two 125 ohm resistors. Electrically, this is the same circuit. Now, if you place an AC ground between the two 125 ohm resistors, you don't change the operation of the differential pair. Why? Because that point is already at AC ground potential. i.e., this voltage never changes even as vin changes. Now, you can split the two halves, and you get the resulting differential model.

6. ### automagp68 Thread Starter Member

Nov 13, 2011
81
0
Can you explain why the solution is incorporating the differential AC model If Vin and ground is not a differential signal?

I have bene fussing with this all day and do not understand.

I have seen Phd,s time and time again use the differential AC model while having the IB2 grounded

I understand for a differential model when you have two input 180 degrees out of phase you get a constant potential where the emitters meet.
For example Ie flowing from one side and -IE flowing from the other side in terms of AC and get a virtual ground so to speak.

This is not the case here.

7. ### Brownout Well-Known Member

Jan 10, 2012
2,375
998
The solution is modeled as a single amplifier because that is easier to analyze. It's just a simplification.

If you ground b2, then you use the differential model AND the common mode model, and add the results i.e. vo = 1/2vin*Av(diff) + 1/2vin*Av(common)

8. ### automagp68 Thread Starter Member

Nov 13, 2011
81
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Is there any chance you could explain in depth? You are really my last hope for the night and I've been trying to do this all day.

I think i understand what your saying but still can't see the picture

Could you draw a full picture of the AC model for me and then cross out the parts we are omitting for the simplified AC model?

That would be a huge help

9. ### Brownout Well-Known Member

Jan 10, 2012
2,375
998
See what I added to my previous post. What else do you need to have exapained?

10. ### automagp68 Thread Starter Member

Nov 13, 2011
81
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I need to see a picture of the models. I just can't understand how your using both models.
Why is it 1/2?

11. ### Brownout Well-Known Member

Jan 10, 2012
2,375
998
Did you read the part about splitting the 250 ohm resistor into two 125 ohm resistors. That works because you didn't change the value of the total resistance. So, after applying the ground, you can split the entire circuit, since the model for each side is a valid model. Sorry, I don't have a way to make a picture. Just draw what I've described.

12. ### automagp68 Thread Starter Member

Nov 13, 2011
81
0

I don't get how you can just split a resistor in two. Why not thirds, why nots 1/4,s what not 1/6,s

In the picture i drew above where is the current from the second dependent CS going? Is it going it through HIE to ground or is it going down to RE?
I could not possibly be more confused right now

13. ### Brownout Well-Known Member

Jan 10, 2012
2,375
998
You don't need more than two. You don't need the 2nd source because you split the circuit and only analyze half. Your drawing is wrong.

14. ### automagp68 Thread Starter Member

Nov 13, 2011
81
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I am at a loss then i guess. I have drawn 100 different pictures of this and still don't have a correct one.

There is no reason why there is an AC ground between a resistor that is a finite value. It just does not make electrical sense
If IB2 was of equal and opposite sigh i can understand why there is a virtual ground. However with IB2 connected to ground it just doesn't make any sense. There is not a constant potential at the connection of the emitters

15. ### Brownout Well-Known Member

Jan 10, 2012
2,375
998
I already told you, B2 is not connected to ground for the differential model. I don't care how many PHD's say otherwise.

16. ### automagp68 Thread Starter Member

Nov 13, 2011
81
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Im guessing this is what you want as the drawing. The round in the middle allows you to analyze just the right hand side.

However i can not understand for the life of me with IB2 grounded how you can say there is an AC ground splitting RE

17. ### automagp68 Thread Starter Member

Nov 13, 2011
81
0
Ok so if IB2 is not at ground what is it at?

The original problem shows it at ground?
So what is the voltage value at IB2?

18. ### Brownout Well-Known Member

Jan 10, 2012
2,375
998
Terrific! But B2 isn't connected to ground (see post #15)

Jan 10, 2012
2,375
998