Take Razavi's Electronics 1 & apply a feedback loop around it. The result is: we have gained a deeper understanding, but sacrificed viewers by a factor of ~10. We've come a long way guys, cheers!
@jaykamat1927
4 жыл бұрын
Good one !
@coolwinder
3 жыл бұрын
01:25 - Intro and Review 03:20 - Improving the Gain in Voltage-Current Configuration >> 10:00 - Input voltage swing due to input current is low as TIA has small input impedance, especially compared to the output voltage that is amplified >> 12:15 - Resistor can approximate current source if it's input voltage is large and impedance that it is connecting to is small >> 13:30 - Closed-Loop Parameters >> 21:00 - Sign of the Feedback K 22:09 - Non-Inverting Amplifier Construction - Changing Current Amplifier to Voltage Amplifier >> 27:20 - Open-Loop Parameters - Problems, differing to later >> 28:38 - Finding Closed-Loop Gain Directly >> 31:05 - Finding Closed-Loop Input Impedance Directly - Utilizing Miller Theorem (as we know voltage swing across only resistance) >> 35:12 - Finding Closed-Loop Gain by Breaking it 37:28 - Current-Voltage/Series-Series Feedback Topology >> 43:05 - Finding Closed-Loop Gain Directly >> 44:59 - Finding Closed-Loop Input Impedance
@DJTrancenergy
2 жыл бұрын
The easiest way to think about the inductive behavior of the input and output impedances of this shunt-shunt feedback scheme is to know that: 1) A shunt connection will reduce our impedance at that point 2) Obviously, due to the low-pass behavior of our controller/nullor/op-amp, this is not going to hold true for a long range of frequencies 3) Therefore, the opposite will happen. The impedance at the shunt connected node will start increasing -> inductive behavior.
@coolwinder
3 жыл бұрын
02:55 - Because we are minimizing current error, input impedance will be lowered for voltage-current feedback; Because we are minimizing voltage error, input impedance will be increased for voltage-voltage feedback. Everywhere in the loop due to negative feedback changes would be harder to make...
@tag1343
5 жыл бұрын
به عنوان یک ایرانی به وجود شما افتخار میکنم۰
@hardikjain-brb
7 ай бұрын
12:09 we could give Vout to the source and Vin to gate for neg fdbk but in that case it would heavily load M3
@zinhaboussi
6 ай бұрын
01:16 Examples of Voltage-Current and Current-Voltage Feedback 03:34 Addition of stages to increase open-loop gain 08:47 High gain leads to large output voltage swing 11:36 Voltage-current feedback network with a resistor 16:45 Understanding the concept of K in voltage-current feedback circuits. 19:13 Feedback is negative 24:26 Erklärung des Aufbaus und der Funktionsweise eines invertierenden Verstärkers 26:34 Understanding the concept of a Transimpedance Amplifier (TI) 31:09 Finding the input impedance of the circuit using Miller's theorem. 33:14 Evaluating different perspectives on voltage-current feedback. 39:02 Setting up feedback network for current measurement 41:06 Importance of input impedance in voltage and current measurements 45:13 Der Eingangsabstand zur gesamten Schaltung ist entscheidend. Crafted by Merlin AI.
@yishanliu422
5 жыл бұрын
Thank you so much professor! Very good teacher!
@evivoulgari4569
4 жыл бұрын
Thanks a lot for the courses! They are pretty cool! Will you record also Analog Design I and II courses? By the way it is a pity that the microphone makes a lot of noise :)
@coolwinder
3 жыл бұрын
24:10 - Don't we need to take our current source output impedance if it were non-ideal into account? Either way, doesn't our assumption of K=-1/R_F fails now as we change amplifier's input impedance to infinity or some big value? 31:05 & 35:12 - K*A_transimpedance = A_voltage
@erwinshad9143
4 жыл бұрын
The A0 of a transimpedance has a unit of (R). It is different from voltage gain which doesn't have a unit and is used used in 32:30 in Millers theorem. I don't understand why professor Razavi mentioned A0 as VOLTAGE gain of the amplifer while he was talking about transimpedance amplifiers.
@許祐嘉-u1c
3 жыл бұрын
Me too... But I think A0 is the gain of OP amp, while A1 is the gain of the Transimpedence amp.
@muratt6894
6 жыл бұрын
Thank You
@amithpandit4043
5 жыл бұрын
What would be the closed loop output impedance for that final inverting amplifier circuit? How would you derive that?
@coolwinder
3 жыл бұрын
17:00 - Do we expel feedback K from the circuit when calculating Input and Output impedances, if not, where do we break the circuit as it may impact resistance seen on input/output ports?
@justuschiang4710
2 жыл бұрын
Break at Vout right to the R(F). And in this case the KA equals: 1/RF * RF gm1 / (1+RF gm1) * RD1 * gm3 RD2. The only difference is there's no RF gm1/(1+RF gm1) in this video. Actually, when we calculate the K by inspection, we also see an input impedance of "1/gm1" from the left of R(F) which has been neglected in this video. Hence K should equals the (negative) sum of R(F) and 1/gm1.
@MohamedMahmoud-df6uc
5 жыл бұрын
Thank You Dr.Razavi
@hardikjain-brb
7 ай бұрын
For op amp k=-1/RF Closed loop gain by KCL KVL is -RFA0/1+A0 By this G is -RFA0 and K=-1/RF and LG=A0 Way to get G : Break feedback connect RF from negative input terminal to ground (opening loop properly) and get Vout/Iin = -RFA0 as above as G.
@surendra3335
3 жыл бұрын
13:17 how the resistor in feedback sense the output voltage it is not in parallel .... can any one help out this ?
@許祐嘉-u1c
3 жыл бұрын
The Resistor is in parallel with Vout. Because Vout is measured from the output node to GND.
@hardikjain-brb
7 ай бұрын
Look at it this way -> the sensing is done directly it takes Vout and gives Vout bw its right terminal to ground There is just a wire to do that job so they are in parallel
@bitvision-lg9cl
Жыл бұрын
what is the open loop input impedance? kzitem.info/news/bejne/wmiFm22wnX-Kq2k
Пікірлер: 24