@5:15 how du you conclude that Acm=0 out of vo1-vo2=0? The common mode gain is zero only in the case of an ideal tail current source, therefore there is no small signal (common mode) current through RD. I think what u mean is that there is no common mode to differential mode conversion. Nevertheless your explanations are very good and helped me alot, thank you!
Thanks for the feedback. I am glad that you found the video useful. Going over the videos with practical examples on the differential pair are probably the best way to understand it
Actually, she starts with assuming Vg2 = 0, then says Vs2 needs to be -Vt in order to keep M2 in on-mode. I = (1/2)*kn*Vov1^2 I/2 = (1/2)*kn*Vov2^2 After making side by side division, you will see that actually our Vov that steers all the current to one side equals to sqrt(2) of that causes equal currents at both branches. More than that voltage will cause no change in current, ergo, voltage as in graph 10:17. Therefore, this will be the maximum voltage that keeps us in the linear zone. Since we assume Vg2 = 0, Vidmax = Vg1 = Vg1 - Vs + Vs = Vgs1 + Vs
Notice that Vov is not that of Id = I case. I = (1/2)*kn*Vov1^2 I/2 = (1/2)*kn*Vov2^2 In that case, you will see that Vov that causes a drain current of I is equal to sqrt(2) times of that causes a drain current of I halves. Vov1 = sqrt(2)*Vov2
At 20:16, did you mean to say "we want to still keep M2 turned on", instead of in saturation?
@5:15 how du you conclude that Acm=0 out of vo1-vo2=0?
The common mode gain is zero only in the case of an ideal tail current source, therefore there is no small signal (common mode) current through RD.
I think what u mean is that there is no common mode to differential mode conversion.
Nevertheless your explanations are very good and helped me alot, thank you!
great explanation. I had a struggle with understanding differential pair. Now I totally understand it.
Thanks for the feedback. I am glad that you found the video useful. Going over the videos with practical examples on the differential pair are probably the best way to understand it
Love this!
very confusing at the end.
How you get Vov from I ?!
why Vidmax = Vgs1 + Vs??
Actually, she starts with assuming Vg2 = 0, then says Vs2 needs to be -Vt in order to keep M2 in on-mode.
I = (1/2)*kn*Vov1^2
I/2 = (1/2)*kn*Vov2^2
After making side by side division, you will see that actually our Vov that steers all the current to one side equals to sqrt(2) of that causes equal currents at both branches. More than that voltage will cause no change in current, ergo, voltage as in graph 10:17. Therefore, this will be the maximum voltage that keeps us in the linear zone.
Since we assume Vg2 = 0, Vidmax = Vg1 = Vg1 - Vs + Vs = Vgs1 + Vs
nice explanation,It help me during examination
Thanks! Very helpful.
Is easy to understand. Gracias.
nice explanation, I guess u should become a professor at my college lol.....
at 22:24 it should be Vgs1 = Vt + Vov and not sqrt(2)Vov.
Notice that Vov is not that of Id = I case.
I = (1/2)*kn*Vov1^2
I/2 = (1/2)*kn*Vov2^2
In that case, you will see that Vov that causes a drain current of I is equal to sqrt(2) times of that causes a drain current of I halves.
Vov1 = sqrt(2)*Vov2
Thank you