Razavi Electronics2 Lec38: Examples of Voltage-Current Feeback, Current-Voltage-Feedback

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  • เผยแพร่เมื่อ 2 ม.ค. 2025

ความคิดเห็น • 25

  • @polarjsapkota2484
    @polarjsapkota2484 4 ปีที่แล้ว +21

    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!

  • @coolwinder
    @coolwinder 3 ปีที่แล้ว +3

    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

  • @tag1343
    @tag1343 6 ปีที่แล้ว

    به عنوان یک ایرانی به وجود شما افتخار میکنم۰

  • @zinhaboussi
    @zinhaboussi 8 หลายเดือนก่อน

    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
    @yishanliu422 5 ปีที่แล้ว

    Thank you so much professor! Very good teacher!

  • @coolwinder
    @coolwinder 3 ปีที่แล้ว +1

    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...

  • @AmirhosseinBozorgzadeh-r7b
    @AmirhosseinBozorgzadeh-r7b 5 วันที่ผ่านมา +1

    32:59 but what about Rout

  • @hardikjain-brb
    @hardikjain-brb 9 หลายเดือนก่อน

    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

  • @coolwinder
    @coolwinder 3 ปีที่แล้ว

    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

  • @coolwinder
    @coolwinder 3 ปีที่แล้ว

    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
      @justuschiang4710 3 ปีที่แล้ว +3

      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.

  • @muratt6894
    @muratt6894 6 ปีที่แล้ว +3

    Thank You

  • @erwinshad9143
    @erwinshad9143 4 ปีที่แล้ว

    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
      @許祐嘉-u1c 3 ปีที่แล้ว

      Me too... But I think A0 is the gain of OP amp, while A1 is the gain of the Transimpedence amp.

  • @DJTrancenergy
    @DJTrancenergy 2 ปีที่แล้ว +1

    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.

  • @MohamedMahmoud-df6uc
    @MohamedMahmoud-df6uc 6 ปีที่แล้ว +2

    Thank You Dr.Razavi

  • @surendra3335
    @surendra3335 3 ปีที่แล้ว

    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
      @許祐嘉-u1c 3 ปีที่แล้ว +3

      The Resistor is in parallel with Vout. Because Vout is measured from the output node to GND.

    • @hardikjain-brb
      @hardikjain-brb 9 หลายเดือนก่อน

      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

  • @amithpandit4043
    @amithpandit4043 5 ปีที่แล้ว

    What would be the closed loop output impedance for that final inverting amplifier circuit? How would you derive that?

  • @evivoulgari4569
    @evivoulgari4569 4 ปีที่แล้ว +2

    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 :)

  • @hardikjain-brb
    @hardikjain-brb 9 หลายเดือนก่อน

    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.

  • @mrpossible5696
    @mrpossible5696 5 ปีที่แล้ว

    14:07

  • @bitvision-lg9cl
    @bitvision-lg9cl ปีที่แล้ว

    what is the open loop input impedance? th-cam.com/video/a2O59zdMYx4/w-d-xo.html