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Question with RC Time Constant

Discussion in 'Electronic Basics' started by Panther, Dec 22, 2005.

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  1. Panther

    Panther Guest

    Sorry about these questions:

    Basically I've done a practical with an RC network and I've got 2 wires that
    touch to "complete the circuit" and provide current to the capacitor. I need
    to find out the amount of time they've touched by lookign tat the voltage in
    the capactiros. I have the capacitor charge/discharge curves.

    my question is can I work out the time using simply time constants? 5TCs =
    time to charge a capacitor (almost 100%, irrelevant). So because I have the
    capacitor charge curve, say it only had half the voltage o fthe input
    voltage, and i look up half on the y axis and find wher eit meets, and look
    up the x-axis' time and work out how long that is.

    or do i do this: http://www.play-hookey.com/dc_theory/rc_circuits.html

    basically use that formula. if i do use it i will get more ducks.

    are both these methods correct?
    thanks
     
  2. Bill Bowden

    Bill Bowden Guest

    are both these methods correct?

    Seems too complicated. Why not charge the capacitor with a constant
    current so the voltage change is linear? That way you would know the
    time to charge to 1 volt is 1/10 the time to charge to 10 volts.

    You can make a fairly constant current source with one transistor and 3
    resistors. But then you have the problem of discharging the cap. Maybe
    one more resistor could do that?

    -Bill
     
  3. You can only estimate a minimum charge time (must be greater than 5*
    R*C seconds) once the capacitor reaches essentially full charge
    voltage. In order to measure the charge time accurately from the
    voltage, the voltage has to still be measurably changing at the moment
    you open the switch.

    The capacitor charges to 1-(1/e) or about 63.2% of full charge (with e
    being the constant, 2.71828) when the charge time has been one R*C.

    The formula for voltage versus time, assuming the capacitor starts at
    zero, is, 1-e^(-time/R*C) times full charge voltage.

    To solve that for time versus voltage, time capacitor has been
    charging is -ln(1-V)*R*C seconds, with V being the fraction of full
    charge voltage, and R in ohms, C in farads.

    For other
     
  4. Bob Monsen

    Bob Monsen Guest

    Yes, you can do this, but you need to scale the output appropriately. A
    generic graph like that can have its time scale marked in time constants,
    and its voltage scale go from 0 to 1. In that case, using the y=0 axis as
    the initial voltage, you find the point where y = (end -
    initial)/(charging - initial). This allows you to determine how many time
    constants you've charged for. If a time constant in your circuit is equal
    to TC, then the answer is graphical time constants / TC.

    --
    Regards,
    Bob Monsen

    Mike is like the first man to discover fire. Fire was there all along
    after he showed them how, anybody could use it’¡Ä anybody with sense
    enough not to get burned with it.
     
  5. Pooh Bear

    Pooh Bear Guest

    You need this equation that's on that page.

    Vc = E(1 - e^(-t/RC))

    As suggested though, a constant current source makes it easier since theer will
    be a linear rise in the cap voltage voltage with time ( like a scope timebase ).

    Make sure the cap is fully discharged before commencing timing though.

    Graham
     
  6. Panther

    Panther Guest

    I can't change it though I've already done it. Thanks.
     
  7. Panther

    Panther Guest

    I can't, I've already done the practical. Thanks anyway.
     
  8. Panther

    Panther Guest

    I think it's best if I used the formula (more advanced).

    Thanksd
     
  9. Panther

    Panther Guest

    Thanks for the help!
     
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