sir,at @41:30 i(in) after entering R2 where does it go.dependent current source is open circuited as vpi=0.so the current has to go through R3 but if it so then Vout=i(in)R2.how could it be??
@Ali Hajimiri - Sir what if in the last example, current source of M1 is scaled instead of M2, it would suggest for a finite output voltage, gate voltage of M1 is zero and hence R1 is eliminated, can you point out where am I going wrong with this?
Good Question . If you assume vgs to be zero then you don't have any feedback right , so that cant 'be be right( as your analysis won't hold good anymore if there is no feedback) . This might suggest you to scale M2 instead of M1 . I cant think of a clearer solution with M1 being scaled . Also, M1 is in the feedback path and not in the forward path . We only scale the forward path gain 'A' by 'k' . There might be a different way to scale feedback paths .
The asymptotic formula H=H_infinity*T/(1+T) + H0/(1+T), still gives the right result taking M1 as reference. The new H_infinity is zero, T=gm1*R3*[R1/(R1+R2+R3) and H0=R3*(R1+R2)/(R1+R2+R3). The closed loop H is the same that the one with M2 as reference.
I think, when prof. defined the value of k. He placed the k in front of gain A, which is in the forward path. In the last example, M2 is in the forward path. Hence, it makes sense to scale the M2. M1, on the other hand, is in the feedback path, in which has nothing to do with k.
I don't understand how the example at 52:10 is a series output for feedback. Are we measuring a current and feeding that back? I guess so, since R_4 turns our output voltage into a current. But that is also the case in the example before that one. It seems arbitrary to me
gm depends on your quiescent conditions. so assuming gm=0 means you are assuming the circuit is biased at zero, or bias currents are zero, same thing. EDIT: Today I saw in Gray Meyer's book they assumed gm to be 0 in certain cases. Cant really say why. Good catch.
In 36:01, you were talking about being non-inverting or inverting. What video lecture did you go over that? Would like to understand why is that. Thank you
There are basically 3 configurations of single transistor amplifier. Common-source amplifier is inverting, Common-gate and Common-drain amplifiers are non-inverting.
@ 53.03 , since output is at drain of M2 its high output impedance node, so that means its sensing voltage right ? becaue voltmeter has high series resistance.. while current meter has low input impedance. here its opposite ? also here as output impedance is high ( M1 and M2 cascode ) its voltage sense.. but you used opposite logic in previous example.. you said we are sensing current because output impedance is high.. did not understand that
No, it is not because of the output resistance. It depends on the nature of the connection. If the sensing is done in parallel with the main signal measuring voltage, it is shunt; and if it is in series measuring current it is series. As discussed in lecture 163N, the shunt and series are often not "pure". The way to determine which one is more dominant you need to look at the change in the port impedance using Blackman equation. If the impedance is lowered at that port, it is a stronger shunt; and if it is raised it is stronger series. Watch: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-AdYTO5GimFo.html
@Ali Hajimiri Hello SIr.. At 48.44 time , for the output voltage/ input current feedback circuit , while calculating signs of feedback, why did you start with gate of the M2 rather than drain of the M2( Vout ) ? I did not understand that. please explain..
If you have a single feedback loop, you can calculate the sign of the feedback starting from anywhere in the loop as long as you come back to the same point. It allows you to start from the most convenient point.
Are there any resources/books that extensively talk about "Asymptotic Transfer Functions" of feedback? I found it to be really insightful, but i cannot find materials talking about this stuff. If anyone knows, please do write here. Will help the others too.
