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Photoresistor and variable flux

  1. Mar 10, 2012 #1
    Before we jump into anything I want to emphasize that is isn't homework. This is me busting my head with random things!


    This is my circuit of interest.

    I want to find the output voltage on the capacitor if the flux is:

    [itex] \Phi(t)=\Phi _{0}(1+sin(\omega t+\phi)) [/itex]

    The resistance of the photoresistor is given by formula:

    [itex] R=R_0 e^{-A\Phi} [/itex]

    Where A is some constant, differs for various materials. In other words, doesn't matter.

    The problem comes up because the resistance of the photoresistor is not changing sinusoidally. It IS periodic, but its not sinusoidal.

    When writing differential equations, I get a mess:

    [itex]\Large \frac{dU_c (t)}{dt}+\frac{1}{\tau (t)}\cdot U_c (t)=\frac{E}{\tau (t)} [/itex]

    Where E is my input voltage, and it is constant DC voltage source.

    My time constant tau is:

    [itex] \tau (t) = R_0 e^{-A\Phi}C=R_0 e^{-A\Phi _{0}(1+sin(\omega t+\phi))}C [/itex]

    When I tried to solve this I came to conclusion that if I try to integrate the time constant(which I had to do), I couldn't. exp(sin(x)) is not integrable, at least not to my knowledge.

    I know it can be broken into infinite series and integrated piece by piece, but by all means that is my last resort.

    Can anybody give me a hint how to solve this, like without 10 pages of math?

    I am busting my head whole day, trying with replacing current sources etc. but I am not getting anywhere.

    My idea was that, the system of that resistor and input voltage I can "pack" into one subcircuit, and that subcircuit is generating some periodic voltage(because of variable resistor), which is not sinusoidal. Is this ok?

    I also thought of Fourier analysis. To brake this into harmonics and get an approximation.

    But I cannot find the coefficients because, the function exp(1+sin(x)) is not integrable...
  2. jcsd
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