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consider the following enter image description here

In normal circumstances, Vin1 and Vin2 change differentially, if Vin1 goes up by 10mV, Vin2 goes down by 10mV. Makes sense.

But I'm wondering, what would happen if Vin1 went up by 10mV, but Vin2 stayed the same and didn't go down by 10mV. The Ibias current source should then have a higher load current right, if it's implemented as an NMOS device?

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    \$\begingroup\$ Things don't "change differentially". Signals are read differentially in both your first case and second case. There is little difference between the two. The first case you have a difference of 20mV. The second case you have a difference of 10mV. The total current in both legs stay stays the same because it has to, but the voltage drops on the legs would change to accommodate so the same amount of Ibias current is steered to imbalance the legs differently. \$\endgroup\$
    – DKNguyen
    Oct 28 '19 at 23:08
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    \$\begingroup\$ The whole point of the differential amplifier is that the difference is amplified, and that the inputs don't have to move by opposite amounts. Just the difference between the two inputs count for the output current. (in theory – of course, this isn't a perfectly linear circuit in practice) \$\endgroup\$ Oct 28 '19 at 23:09
  • \$\begingroup\$ My apologies. @DKNguyen How would the current on the right branch change in the second case? It has the same Vgs on Q2 so by the square-law, it should stay the same. Or am I meant to assume channel-length modulation and thus Vds would actually decrease due to the lower current and hence lower drop across the resistor? \$\endgroup\$ Oct 28 '19 at 23:17
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    \$\begingroup\$ @AlfroJang80 Vgs does NOT stay the same. That current source adjusts it's terminal voltage to maintain the same level of current and one of those terminals is connected to both MOSFET source pins. Fiddling with the two Vins (which are referenced to GND and therefore do not move with changes in the source terminal voltage) makes Q1 and Q2 conduct more or less which changes the load seen by the current source. \$\endgroup\$
    – DKNguyen
    Oct 28 '19 at 23:23
  • \$\begingroup\$ @DKNguyen I see, so it has nothing to do with channel-length modulation? \$\endgroup\$ Oct 28 '19 at 23:29
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But I'm wondering, what would happen if Vin1 went up by 10mV, but Vin2 stayed the same and didn't go down by 10mV.

You need to work out the common mode and differential components of the input.

For example, your scenario is the same as both inputs going up by 5 mV, then Vin1 going up by another 5 mV while Vin2 goes down by 5 mV.

Roughly, the output would be the same as if you just had a 10 mV differential input. But if the amplifier is not well balanced you could also see an output change due to the 5 mV common mode input.

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  • \$\begingroup\$ I think this is a great answer to this question and what OP was asking about. So in the single ended input case the CMRR will come into effect which will make things worse than in fully differential input case(where there is no CM voltage and CM degradation). So it seems +5mV -5mV true differential input is better than +10mV 0V single ended input unless the CMRR is infinite. And it is never infinite because the symmetry is never perfect hence there will be some imbalance as you mentioned. \$\endgroup\$
    – floppy380
    Oct 29 '19 at 1:23
  • \$\begingroup\$ whenever the common-source(s) node moves, you'll have some CMRR effect \$\endgroup\$ Oct 29 '19 at 3:11
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In normal circumstances, Vin1 and Vin2 change differentially, if Vin1 goes up by 10mV, Vin2 goes down by 10mV. Makes sense.

No.

This is your specific input signal. That's not the "normal" case. It's the symmetric input special case.

But I'm wondering, what would happen if Vin1 went up by 10mV, but Vin2 stayed the same and didn't go down by 10mV.

The purpose of a differential amplifier is to amplify the difference between the two inputs.

So, that should, in an idealized differential amplifier, lead to half the output current compared to your symmetric case above. Nothing special here.

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