Integrator steady state error

Discussion in 'Homework Help' started by mentaaal, Oct 19, 2008.

  1. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
    451
    0
    Hey guys,

    I am doing a homework problem. We have to calculate the integrator steady state error for an integrator with:

    Rin = 10k
    Cint = 10nF
    Ib = 100nA
    Vio = 2mV
    Iio = 10nA

    The formula for integrator errrors is:

    1/RC∫Vio dt + 1/c∫Ib dt + Vio

    What i have done is this: (based on a previous example given to us)

    1/(10k*10nF)*2mv + 1/10nF*100nA + 2mv

    But when i do this my results is like in the tens of volts! because of the first multiplication.....

    Your thoughts?
     
  2. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    By assuming Ib and Vio to be constant the correct expression is that:

    1/(10k*10nF)*2mv*t + 1/10nF*100nA*t + 2mv

    t=time in seconds

    thus the error increases with time as the output does.
     
  3. mentaaal

    Thread Starter Senior Member

    Oct 17, 2005
    451
    0
    Thats what i was thinking but i was taking notes as gospel again and not trusting my own thoughts on it. Cheers mike.
     
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