Razavi Electronics2 Lec39: Application Examples of Feedback, More on Current-Voltage Feedback 

01:25 - Intro and Review 07:50 - Practical Problems and ISSCC Literature 16:50 - Current-Voltage Feedback: Output Impedance 27:16 - Transistor-Level Current-Voltage Feedback Example >> 33:13 - Formalization of Open-Loop Circuit Parameters >> 38:03 - Formalization of Closed-Loop Parameters: Loop-Gain >> 44:08 - Formalization of Closed-Loop Parameters: Output Impedance

01:16 Application of feedback in real-life and current-voltage feedback topology 04:16 Op-amp with voltage-current feedback has low input impedance. 09:30 Optical fibers have lower loss and can transmit data over long distances. 12:36 Optical communication example with a photodiode and transimpedance amplifier 17:31 Modeling amplifier with finite output resistance and understanding output impedance. 20:13 Measuring the output impedance of a feedback amplifier 25:33 Current-voltage feedback increases output impedance. 28:12 Feedback applied to improve quality of current source output. 34:58 Overview of Open-loop Amplifier Characteristics 37:40 Understanding open-loop and closed-loop parameters. 43:20 Calculation of closed loop output impedance is crucial. 45:56 Discussing output impedance and feedback topologies Crafted by Merlin AI.

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46:55 the resistance r (input resistance of k) and the load (if any) comes in series with Rout(1+kGm)

It depends where we cut the loop for k depending on the type of input accepted by the main amplifier and k

Thank You

35:59 l think l_out should point to the emitter, just as Professor Razavi pointed out at a later occasion

Was expecting the example of degenerated transistor for current voltage feedback. :)

3:22 - This configuration is something else, I would like more in-depth explanation. It utilizes voltage feed-forward high-gain amplifier with one input fixed (not connected to input or return signal). Also we don't have low impedance at that current summing node. Which means, each current source (input and return) is driving load sizable to its output impedance (when not looking at the effect of the feedback). Also input impedance of the amplifier is even bigger that output impedance of each current sources. Also voltage amplifier will try to lower its error input, which is voltage difference between in this case gnd and input node voltage. Not input node voltage is derived as voltage divider between amplifier output and input voltage... Combining all of this, we come to conclusion it acts as voltage-current feedback, but should we do that immediately at start? Does the circuit with that configuration result in same properties as plain voltage-current feedback amplifier? You are awesome Razave, for me even to come to this question, it's result of your great work :) Thank you immeasurably!

37:40 - Note: We are doing small-signal AC analysis, we presume bias is taken care of. 40:00 - Note: Output load is inserted in-between output current source and sensing network. 46:40 - Question: What about the r? Is the input impedanc of sensing network (K) in the case of returning current configuration going in series with open-loop output impedance, when finding closed-loop output impedance?

For last case k=r OLG=A0gm

26:52 - Why do we use current test source instead of voltage source? Because with the current source, the current is unchangeable, and writing equations with that in mind bring the right result for the output impedance of a feedback network. Is this right? If we wrote the equations as if the voltage was constant then the current trough R_out would be constant, and not I_x trough the entire feed-forward amplifier's output port. That might bring tatally different result for our feedback analysis.

op amp should not invert at 35:41

28:09