# Output voltage of MOSFET differential pair with current mirror load

How are the drain voltages equal?? I can't seem to figure it out, it's not even symmetric?

Vd3 = Vsply+ - Vsg = Vsply+ + VGS

I can see that but what about Vd4??It can't be due to symmetry. Q3 forces its drain and gate to be equal with a short, but Q4 doesn't have that.

• Try read this answer electronics.stackexchange.com/questions/353604/… Did this answer help?
– G36
Oct 17 '18 at 14:12
• How are the drain voltages equal?? You're right, usually they aren't equal. Only in the purely theoretical case where the transistors are exactly identical and $I_{bias}$ divides exactly half-half between Q1 and Q2 and the output resistances of Q1, Q2 and Q3, Q4 exactly match then the drain voltages could be the same. But even in a simulator they will not be the same, for start Q3 has a different Voltage to current relation than Q4 because Q3 has local feedback and Q4 does not. Anyone who claims that the voltages at the drains of Q1, Q2 are equal still has a lot to learn! Oct 17 '18 at 14:13
• @G36 i still don't know how the drain voltages are equal. But I realized that just as in the BJT version, there needs to be a wire to sink out the current difference. The current mirror forces both sides to have equal currents, but the differential voltage forces different currents. The current difference must be sinked out by a wire connected to either side of the differential pair. Oct 17 '18 at 15:26
• In theory, if Vin1 = Vin2 and all transistor are perfectly matched you can get the same voltage at the output.
– G36
Oct 17 '18 at 15:46
• @G36 not sure if im stupid but, if the input voltages are equal and all the transistors are matched, then all the transistors have the same currents. But if they are all in saturation, their drain voltages could be anything. if there was no feedback in M3, then I could believe that they are symmetric but m3 does have feedback and m4 doesn't. I'm not sure how to come up with an exact value/equation for Vout. Oct 17 '18 at 16:20