# common-mode behavior of differential amplifier

Discussion in 'General Electronics Chat' started by screen1988, Jun 2, 2013.

1. ### screen1988 Thread Starter Member

Mar 7, 2013
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3
Can anyone shed some light on this short section from my book?
Here is common-mode behavior of differential amplifier:

What happens if VinCM = 0? Since the gate potential of M,and M2 is not more
positive than their source potential,both devices are off,yielding ID3=0.

This part is from the book. I want to ask a stupid question. How can you know that the gate potential of M1 and M2 is not more positive than than their source potential?
Is the smallest potential in this circuit possible only zero? There is no negative potential at any nodes in the circuit.

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2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Your understanding of how this circuit behaviour for Vcm = 0V is correct.
For Vcm = 0V M1 and M2 are OFF because even if M3 is in triode mode.
M1 and M2 source voltage is equal to 0V.
Vgs = Vg - Vs = 0V - 0V = 0V and Vgs<Vt ---->OFF

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3. ### screen1988 Thread Starter Member

Mar 7, 2013
310
3
That is not mine. It is Razavi's!
My confusion is how can you know that M2 and M3 are OFF.
I want to check the condition Vgs < Vth but in this case how can you know the voltage of source S is zero?
Is it possible that the voltage at the source S negative? Then Vgs can be larger than Vth. And M1 and M2 are ON.

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
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Simply assume the M3 is full ON. So Vds3 ≈ 0V

Do you see any negative voltage in the diagram?

5. ### screen1988 Thread Starter Member

Mar 7, 2013
310
3
Not sure about this. What do you mean full ON? When it is ON, it can be in saturation or triode region. But when I see the characteristic graph, if trans is in saturation its voltage is very large.
Nope, maybe it is only in my imagination.

6. ### kubeek AAC Fanatic!

Sep 20, 2005
4,686
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The voltage at point P can only be somehere between 0 and Vdd. So if Vin is at 0 then M1 and M2 will be off no matter what.