This is really fantastic. Relating the feedback and sensing to sources and sensing is something that wasn't introduced by my classes, but it makes it much more intuitive. Thank you Dr. Razavi, for making this publicly available.
25:38 - Note: Because feedback with great A_1 will tend to minimize error, which is in this case difference between two currents, one the input one, and other the feedback current, the current that actually flows trough the input port of the amplifier is a lot lower and thus makes lower voltage drop then if whole input current would flow trough it, making it seem as amplifier with lower input impedance.
@enesog
4 жыл бұрын
Exact solution for the Example at 42:12 will be: K*A = - Vf / Vtest = gm2* (RD || [(1+gm1*r01)r02 +r01]). The Open Loop Gain is A = RD. --> Closed Loop Gain = A / (1 + k*A) = RD / (1+ {gm2* (RD || [(1+gm1*r01)r02 +r01])}) NOTE: gm2*RD || [(1+gm1*r01)r02 +r01] ~= gm2*RD like Razavi did in this example. More details on Lec 3 Cascode.
@tag1343
5 жыл бұрын
به عنوان یک ایرانی به وجود شما افتخار میکنم۰
@ELBAfzalMalik
Жыл бұрын
best lectures delivered by the best
@srinivasaprasanth
3 жыл бұрын
Best professor I have never seen.
@muratt6894
6 жыл бұрын
Thank You
@shanthoshkumarm6873
5 жыл бұрын
Sir,In 9:09 why current is subtracted?
@tag1343
4 жыл бұрын
Which current?
@vaisnava8224
4 жыл бұрын
Because it's a negative feedback amplifier.. The subtracted result only goes to the main amplifier.
@NaaJeevitham500
5 жыл бұрын
Why the current direction is inside to the K
@rahuldussa6927
5 жыл бұрын
to subtract and get i1-i2 ... for negative feedback
@shubhamnayak9369
5 жыл бұрын
We always take two port current and voltage notation for any two port system
@anku348
5 жыл бұрын
25:00 Shouldn't a good current source have high impedance?
@tag1343
4 жыл бұрын
He refers to A1. It is a block that wants to amplify a current so we wish all current enters A1 without losing any, This means in ideal case the input impedance shall be zero(like a current meter) so all current goes to A1. What you wrote is currect but it is not applice to the input of A1. If we had instead a block that wants to deliver a current at the output then yes you are right. It acts as a current source and must have very hight impedance in order to delvier all current to the load.
@zinhaboussi
6 ай бұрын
he said very good "current sensor" and not very good "current source"
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