OP Amps

Discussion in 'Homework Help' started by Celostrophus, Jul 22, 2012.

  1. Celostrophus

    Thread Starter New Member

    Jul 22, 2012
    4
    0
    I'm attempting to solve this problem for Vout, but I keep getting an answer that makes no sense.

    Vout = 10*L*(1/1,000Ω)*dvin(t)/dt

    Any help?

    I know that the gain is 10 from G = (1 + 9000Ω/1000Ω) = 10

    I also know that Vout = VL(t) * G

    I guess I'm just having trouble determining VL(t) correctly.

    VL(t) = L * di(t)/dt

    iL(t) = Vin(t) / R , R = 1000Ω

    VL(t) = L * 1/R * dVin(t)/dt

    But since Vin(t) is a square wave this would make so sense at the places where the slope is infinity.
     
    Last edited: Jul 22, 2012
  2. #12

    Expert

    Nov 30, 2010
    16,298
    6,809
    You need to find the voltage wave shape at the inductor and just multiply that by 10. The voltage at the input to the amplifier will be the applied step voltage at time = 0 and then it will sag. At what rate?

    That is the essence of the problem, and it's not about sine waves or the response to a sine wave.
     
  3. Celostrophus

    Thread Starter New Member

    Jul 22, 2012
    4
    0
    So basically their are two equations that the output voltage is switching between depending upon the input voltage.

    Those two equations being:

    Vout(t) = 2.5e^(-t/τ) for 0 < t < 1, 2 < t < 3, ...
    Vout(t) = -2.5e^(-t/τ) for 1 < t < 2, 3 < t < 4, ...

    where τ = L/R = 2mH/1kΩ
     
    Last edited: Jul 22, 2012
  4. #12

    Expert

    Nov 30, 2010
    16,298
    6,809
    I actually don't know. I'm very good with op-amps but I suck at inductors.

    I can tell you that the inductor and the 1k resistor form a voltage divider. The apparent resistance of the inductor changes over time, from near infinity (because of the near zero rise time) to near zero after some time, but this frequency is so high that the inductor current never stabilizes.

    That's about all I have.
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Your second equation is incorrect.

    V_{in}(t)=i_LR+L\frac{di_L}{dt}

    It turns out

    V_L(t)=V_{in}exp^{ -\frac{R}{L}t}

    But that's only half the story - you have to deal with a transient condition with a square wave input rather than a simple DC voltage. That requires a little more careful thinking and the analysis to match.
     
    Last edited: Jul 22, 2012
    #12 likes this.
  6. WBahn

    Moderator

    Mar 31, 2012
    17,748
    4,796
    Here's your problem (and this is what t_n_k was focusing on). The current through a resistor is proportional to the voltage ACROSS the resistor. But Vin(t) is just the voltage on one side of the resistor. You have to subtract the voltage voltage on the other side of it in order to get the voltage across it.
     
    #12 likes this.
Loading...