Miller Integrator with Feedback (i.e. DC Stablization)

Discussion in 'Homework Help' started by jegues, Apr 22, 2011.

  1. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    735
    43
    I don't understand how to obtain the last portion of the question,

    Sketch output for a 0.1ms 1v pulse with the feedback resistor connected.

    I was able to do it without the resistor,

    \frac{-1}{RC} \int_{0} ^{0.1ms} 1d\tau + 0

    but I don't know how to do once the feedback resistor is implemented.

    Can someone explain?
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    Is the issue that you don't understand the solution provided?

    The unity frequency gain without the feedback resistance will be given by the condition ....

    1/(ωRC)=1 or when C=1/(ωR)=0.7958 nF

    When the feedback resistance Rf is added, the response will be a first order exponential function with a time constant determined by the feedback resistor and the capacitance.

    Given the input resistance will be 20kΩ the feedback resistor will be 2MΩ to give a DC gain of 40dB (100x).

    So the exponential time constant τ will be 0.7958nF*2MΩ=1.592ms

    The output for a constant DC input then has the general form

    Vo(t)=-k*100(1-exp(-at))

    where k=input DC voltage and a=1/τ=1/1.592ms

    This must be true because the DC gain has been set to 100x and at t=∞ the value vo(t)=-k*100. For +1V DC input there would be -100V DC output at t=∞.

    However, the 1V input lasts only for 0.1ms. So at 0.1 ms ...

    Vo(0.1ms)=-100(1-exp(-0.1/1.592))=-100(1-0.939)=-6.088V

    When the input drops back to 0V at 0.1ms, the output as a time function will then be

    Vo(t>0.1ms)=-6.088*exp(-a[t-0.1ms])

    A more rigorous proof requires you to solve a differential equation. The posted solution refers you to page 112 of the text, which presumably sets out the method of setting up & solving the DE.

    It's not hard to set up the equation ....

    If the input is Vi(t) then the input current is i(t)=Vi(t)/20k

    If the output is Vo(t) the parallel {Rf||C} feedback network must satisfy the condition

    i(t)=-\frac{V_o(t)}{Rf}-C\frac{dV_o(t)}{dt}

    or

    \frac{Vi(t)}{20k}=-\frac{V_o(t)}{Rf}-C\frac{dV_o(t)}{dt}

    In accord with this problem, one would solve this DE with Vi(t) a constant DC value.
     
  3. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    735
    43
    The first order DE is what I remember I just couldn't remember how to get it into that form.

    Since we have a constant input voltage between t1 and t2 the voltage will look like,

    V_{c}(t) = V_{c}(\infty) + [V_{c}(0) -V_{c}(\infty)] e^{\frac{-(t-t_{1})}{\tau}}

    correct?
     
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