Is it really just the RC time constant?

Discussion in 'General Electronics Chat' started by Management, Oct 29, 2009.

  1. Management

    Thread Starter Active Member

    Sep 18, 2007
    If you have a resisitor in series with a cap and a constant current source how long will it take to charge the capacitor?
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    A constant current source will be interesting, as it will raise the voltage in order to maintain its current output. Depending on the size of the resistor and the voltage rating of the capacitor, the results will be anything but roughly 5RC to charge the cap. The resistor could burn open (not likely), or the capacitor could have its voltage spec exceeded and explode.

    If you apply a regulated voltage to the capacitor through a fixed resistor, then you can use the rough figure of 5RC as the capacitor charge time. If you want a more accurate means of determining charge time, see our Ebook -
  3. Management

    Thread Starter Active Member

    Sep 18, 2007
    See I was getting the feeling that it didn't matter what the value resistor was so just RC wouldn't work.

    But thanks for the link. One would still use the current through a capacitor equation and if you want time then one has to just specify the change in voltage they one wants and you'll get the time.
  4. KL7AJ

    AAC Fanatic!

    Nov 4, 2008

    In theory, with a constant current source the capacitor would NEVER fully charge. Of course, a REAL capacitor would vaporize at some point. :)

  5. Ron H

    AAC Fanatic!

    Apr 14, 2005
    The voltage waveform on the cap will be a linear ramp. You have to define specify the voltage at which you consider the cap to be fully charged.
    The resistor in series just raises the voltage at the current source output by a constant I*R.
    This all assumes that the current source has sufficient headroom to keep it from saturating.
  6. AdrianN

    Active Member

    Apr 27, 2009
    To answer your question the charging time is

    t = C*V/I

    where C is the cap value, V is the voltage level at which the cap is considered charged, and I the constant current. So, the series resistor has no effect.

    However, this resistor starts being "seen" as the cap voltage aproaches the constant current source power supply level. From a linear charge behavior the cap voltage will start showing a curve as it approaches the current source power supply.
  7. electr

    Active Member

    May 23, 2009
    What happens when you connect a constant current source to an RL load? (a resistor in series with an inductor).

    The current source (a current mirror for example) is designed to immediately produce some positive value of current, but the inductor won't let that happen.

    So what would happen eventually in such case?
  8. SgtWookie


    Jul 17, 2007
    The current source would output at it's maximum voltage until the current through the inductor increased to the current sources' setting.