analog pi control of an old inverter

Discussion in 'General Electronics Chat' started by ak52, Sep 4, 2017.

  1. ak52

    Thread Starter Member

    Oct 15, 2014
    195
    5
    Hello everyone,

    I found this circuit in an old inverter ,which i tried to re-trace the circuit.
    Im confused at this circuit which i think is the a PI control loop.

    I tried to simulate it in ltspice but getting some wierd results.
    Firstly there is no DC gain present.Only when i give different AC values i see some difference in the output dutycycle.

    I check a few books and tried to derieve the transfer function and the co-efficients.
    Please correct me if i am wrong:

    Kp = R2/R3 = 5.62
    Ki = 1/(C1*R3) = 21276.5

    Will the output be (Vout) = -(R2/R3 + (1/C1*C3)) *Vin??


    Many thanks
    AK
     
    Last edited: Sep 4, 2017
  2. crutschow

    Expert

    Mar 14, 2008
    22,538
    6,614
    That circuit is an integrator with DC gain equal to the open-loop DC gain of the op amp.
    In real life, if not part of a feedback loop, the output will likely go to one of the rails due to input offset and stay there.
    I don't see any C3.
     
  3. ak52

    Thread Starter Member

    Oct 15, 2014
    195
    5
    Thanks for the reply.
    I made a typo in the formula.I meant (Vout) = -(R2/R3 + (1/C1*R3)) *Vin
     
  4. crutschow

    Expert

    Mar 14, 2008
    22,538
    6,614
    You formulas are incorrect because they do not have a frequency related S or jW term for the capacitive reactance.
    The impedance of a capacitor varies with the applied signal frequency and thus any expression for the transfer function of a circuit with capacitors must include a frequency term.
    Read this for more info.
     
Loading...