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.
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.
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.
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.
01:25 - Intro and Review
03:20 - Voltage-Current Feedback Topologies
16:35 - Closed-Loop Trans-resistance
19:32 - Closed-Loop Input Impedance
25:42 - Closed-Loop Output Impedance
33:00 - Transistor-Level Example
41:10 - Closed-Loop Parameter Calculation
44:45 - Gain Improvement Preview
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.
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.
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Best professor I have never seen.
best lectures delivered by the best
Sir,In 9:09 why current is subtracted?
Which current?
Because it's a negative feedback amplifier.. The subtracted result only goes to the main amplifier.
Thank You
Why the current direction is inside to the K
to subtract and get i1-i2 ... for negative feedback
We always take two port current and voltage notation for any two port system
25:00 Shouldn't a good current source have high impedance?
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.
he said very good "current sensor" and not very good "current source"
19:52