# differential amplifiers related: virtual ground confusion

Discussion in 'General Electronics Chat' started by acelectr, May 22, 2011.

1. ### acelectr Thread Starter Member

Aug 28, 2010
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0
Hi, I am trying to understand the basic concepts of the differential amplifiers topic and am having some difficulties in seeing some crucial points. I've attached this image, and as it's been told the voltage at the emitters is zero. Well I've tried a lot to get this zero result at the emitter terminal but I just can not get it. I've drawn the equivalent pi models and wrote couple of current voltage equations but could not get the voltage across REE exactly zero. I'll be glad if someone can write a good explanation for it. As far as I know there has to be a good one without any made assumptions on the circuit.
Another topic that I am having trouble with is the differential half circuits, well is there a specific technique in order to divide in to two a differential pair? What are the points that I need to focus and analys when trying to work on it?

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2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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The diagram that you post show small-signal AC model.
And you should treat this circuit as a CE+CB amplifier.
Then you will see that Ree resistor is parallel to emitter-base junction, so Ree is short by re resistance.
So the small-signal current flow through Ree will be very small.
So your book assume I_Ree = 0A

And from differential point of view (Vid not grounded) Ree has no influence on the circuit.
See the digram that I post.
And the KVL loop Vid = Vbe2+Vbe2

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• ###### aas_186.PNG
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Last edited: May 22, 2011
3. ### acelectr Thread Starter Member

Aug 28, 2010
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Well what do you mean CE CB? I dont see any CB here & well ofcourse vid is not grounded than how Ree and re becomse in parallel? one terminal of re's is connected to Vid while Ree is connected to ground. I am confused here

Last edited: May 23, 2011
4. ### Jony130 AAC Fanatic!

Feb 17, 2009
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The differential amplifier is a cascade connection of the two "basic" BJT amplifiers. The first one is CE (Common emitter) and the second stage is CB (Common base) stage.

When you apply differential AC signal to the input of the BJT differential amplifier. The AC input signal is 180° out of phase between the bases.
So in one BJT the Ie1 current is rising and in the second one the Ie2 current is decreasing. So AC signal current across Ree cancel each other,
becaues Ie1 and Ie2 are equal in magnitude but 180° out of phase.
And that's why AC voltage across Ree is equal 0V

5. ### acelectr Thread Starter Member

Aug 28, 2010
73
0
are there any equations that can prove tihs?

6. ### Jony130 AAC Fanatic!

Feb 17, 2009
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1,188
Well yes you can use this equation

$\frac{Ie1}{Ie2} = e^{\frac{Vbe1- Vbe2}{26mV}}$

And here you have the simulation result.
Simple if Ie1 increase due to input voltage increase (Vid = 10mV).
Of course Vbe1 also increase and Vbe2 will be decrease becaues Ve voltage will be more positive, and the base of the T1 is connect direct to GND.

ΔIe1 = 1.185mA - 1mA = 0.185mA
ΔIe2 = 0.817mA -1mA = - 0.183mA

So the AC sum of the emitters current Iee = ΔIe1+ΔIe2 will be equal almost 0A.
http://www.te.kmutnb.ac.th/~msn/224510diffamp.pdf

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