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Why is this RC circuit not working in PSPICE?

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Discussion in 'General Electronics Discussion' started by alpha_centaur, Sep 9, 2010.

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  1. Sep 9, 2010 #1
    alpha_centaur

    alpha_centaur

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    Sep 9, 2010
    Hello all!

    I am really stuck up here. I have made this circuit and want to measure the voltage across the capacitor. The voltage should gradually increase as the capacitor charges, and then become constant. But, I only get a constant voltage of 6V across the capacitor.

    Infact, I have to measure the time constant. I am doing a transient analysis, and when I plot the voltage, it just shows a straight line, and no rising trend.

    Please help!

    The circuit image can be seen at:

    http://img641.imageshack.us/img641/8897/rccircuit.jpg
     
  2. Sep 9, 2010 #2
    (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    Is this your homework?

    Have you considered the starting conditions? I know that in the modelling tool I use, I can specify that it starts from either an equilibrium state or a zero state.
     
  3. Sep 9, 2010 #3
    alpha_centaur

    alpha_centaur

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    Sep 9, 2010
    I am using PSPICE. I dont know how to set starting conditions. If you can guide me it will be great.
     
  4. Sep 10, 2010 #4
    Mysteryname

    Mysteryname

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    Sep 6, 2010
    When setting transient analysis make sure you have a scale of time that allows you to see the charge of the capacitor.(Google a guide to PSPICE I know there is one out there cause I'm also learning how to use the program)

    The capacitor you are using is reasonably small. Calculate the theoretical time constant of the capacitor(I assume you already know how to do this) then use PSPICE to simulate the results.

    in the version of PSPICE I'm using the time scale is 1ms to 1s(defualt) at that time the capacitor is already charged.
     
  5. Sep 10, 2010 #5
    alpha_centaur

    alpha_centaur

    3
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    Sep 9, 2010
    Thanks for replying.

    My problem has been solved by setting the initial condition of the capacitor to 0 V.

    Cheers!
     
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