# Differential Amplifier: Differential Input

Discussion in 'Homework Help' started by blah2222, Nov 27, 2011.

1. ### blah2222 Thread Starter Distinguished Member

May 3, 2010
582
38
Hello all, just a question regarding the input when a differential signal is passed to a matched differential amplifier.

$V_{id} = V_{1} - V_{2}$ (1)

Then they describe V1 and V2 in terms of their differential and common-mode terms:

$V_{1} = \frac{V_{id}}{2} + V_{ic}$ (2)
$V_{2} = -\frac{V_{id}}{2} + V_{ic}$ (3)

I just don't understand how they determine that half of the differential signal will be sent to the bases of both transistors, with opposite polarities. I do understand how they got equation (1), but the half Vid is confusing me.

Thanks!

2. ### crutschow Expert

Mar 14, 2008
22,210
6,469
A normal differential input circuit has the same input impedance on each input so any differential voltage applied to the inputs will be equally divided between the two, thus the factor of 1/2. And by definition, each half of the differential signal carries 1/2 the voltage. Obviously you can't have the full signal appearing across each input at the same time.

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3. ### blah2222 Thread Starter Distinguished Member

May 3, 2010
582
38
Thanks for the response!

Also, if given a differential BJT pair that is biased in the emitter by a current source, how would you be able to find out the operating Q-point of each transistor if they have different beta values?

Thanks again!

4. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,349
731
It might help if you work backwards from this equation:

$V_{out} = A_{sig}(V_{in}^+ - V_{in}^-) + A_{com}(\frac{V_{in}^+ + V_{in}^-}{2})$

A is gain.

The left half is the desired signal to be amplified, And $A_{sig}$ is the gain for that signal.

The right half is the common mode signal, and $A_{com}$ is the gain for the common signal (same as $A_{sig}$, except for the following).

The common mode signal is applied to both inputs, which is greatly reduced after the subtraction. Then, as the two desired signal inputs form a sort of "virtual floating ground" at the zero point halfway between them, the common signal is reduced by half as well.

I could be overthinking this. Your book is using signs on the common mode (before amp signal) since it is showing a - sign one of the common modes is that if you had +1V common mode on both pins, the end result would be 2V, so they wave the majik wand and the equation is formed with the differential voltage divided by 2.

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5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
Another perspective

• ###### Vcm etc.png
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