Discharge a Goldcap Capacitor

Discussion in 'General Electronics Chat' started by nicmon, Jun 30, 2011.

  1. nicmon

    Thread Starter Member

    Mar 18, 2011
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    I'm using a 1 uF, 5.5V Goldcap in my circuit which will be charged with a 3,3V current, Sometimes I would like to disconnect it from whole circuit, therefore I put a SPDT switch between 3,3V Source (comes from a LM1086) and the positive of Goldcap and a LED as indicator.

    the problem is once I change the switch to OFF position, the LED remains on for a while (I know, this is the normal behaviour of a Cap :D)

    referring to this article: http://www.allaboutcircuits.com/vol_6/chpt_3/17.html, can somebody help me and say which resistor can use there to discharge the Goldcap?

    Thank you.
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    If the 1uF cap is the only one in the circuit, then a 33k resistor connected across its' terminals would bring the cap voltage below 1.5v in about 30mS. To you, it would appear that the LED went off nearly instantly.

    While the circuit is powered, 3.3v across the 33k resistor is only a 0.1mA (100uA) load, so it won't be a significant power drain.
     
  3. nicmon

    Thread Starter Member

    Mar 18, 2011
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    Thank you very much, as I said, I'm using a SPDT to switch between with and without Capacitor modes in my circuit.
    Do I harm the Capacitor if I put its legs be connected to the resistor whole time that I'm in "without cap" mode?

    Sorry, for my noob questions.
     
  4. SgtWookie

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    Jul 17, 2007
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    The 33k resistor is so large that the current flow through it will be very small at 100uA, even at the full 3.3v voltage applied while the circuit is operating. That is 330 microwatts of power, or 0.00033 Watts; a very small amount of power.

    You will not damage the capacitor.
     
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  5. nicmon

    Thread Starter Member

    Mar 18, 2011
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    Thank you, really appreciate for your answers.
     
  6. strantor

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    Oct 3, 2010
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    SgtWookie, how did you get this answer? Is it a formula using RC time constant or from a chart in the datasheet or what? I would like to know for the future how to calculate how long it will take a capacitor to discharge for a given load.
    thanks
     
  7. shortbus

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    Sep 30, 2009
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    I too would like to know this formula :)
     
  8. Bernard

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    Aug 7, 2008
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    To drop from 3.3V down to 1.5V is about one RC time constant so T=RC; choose either a time or an R, & solve for the other. " Close enough for Gvt. work ".
     
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  9. #12

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    Nov 30, 2010
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    The formula is Ecap = dE (e^(-t/RC)) (I have it written on the wall behind my work bench.)

    In English,
    New voltage on the cap is the original voltage on the cap +/- [the change in voltage applied times e (2.7182818) to the (negative time over Resistance Capacitance)]

    Theoretically, the voltage never gets to zero, but like Bernard says...close enough for government work.

    People almost never use this formula bc it's easier to remember things like "10 time constants and it's empty" or, "one RC and its down two thirds". That's why the formula is written on the wall. I can never remember it.
     
    Last edited: Jun 30, 2011
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  10. Adjuster

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    Dec 26, 2010
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    If the LED is remaining on for a significant time at present, is the capacitor really only 1 microfarad? Perhaps it's really one of those 1F memory backup caps
     
  11. SgtWookie

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    You would be surprised how long an LED remains illuminated once lit, albeit dimly, by even a small cap across it, and how large a resistance you can use to cause the cap to discharge rapidly enough.
     
  12. strantor

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    There's one for my wall. thanks.
     
  13. #12

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    Nov 30, 2010
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    I never know which bits will be the most helpful, or who needs them, but it's always nice to get a "thank you".

    #12
     
  14. Adjuster

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    Dec 26, 2010
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    I believe that "Goldcap" normally refers to a Panasonic double-layer "supercap" product - they are not made in sizes as small as 1μF.

    Of course, the OP may have something else in mind, and in these days of Chinese copies tradenames can't be relied upon so much.

    http://panasonic.com/industrial/ele...arch.aspx?src=/www-ctlg/ctlg/qABC0000_AM.html
     
  15. SgtWookie

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    OK, our OP did say 1uF in his first post. If they really meant 1F instead, then it'll take about 8-1/3 hours for that same 1 time constant... :rolleyes:

    In that case, our OP could instead use a SPDT switch, the common being the input to charge the cap, and the two other terminals switching either to power or ground. Using a light bulb in series with the common to the cap (say like an automotive brake lamp) would limit the maximum current while charging or discharging the cap, but would have a very low resistance once the cap was charged or discharged.
     
  16. nicmon

    Thread Starter Member

    Mar 18, 2011
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    upps, they are 1.0F indeed (and not 1uF), so I must ask you again about the correct value of resistor.

    Can I use another value for resistor, e.g. 470K as I don't have a 330K in the hand?
     
  17. #12

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    Nov 30, 2010
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    I'd say you can use any value from 47 ohms 1/2 watt up.
    I'd also say I gave you the equasion to figure it out yourself in post #9.
     
  18. SgtWookie

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    I'd still suggest that you use an incandescent lamp in the charge and discharge path for the cap. Incandescent bulbs have a non-linear resistance; when they are cold, the resistance is very low. As they heat up, the resistance increases a great deal.

    If your cap is completely discharged, the regulator on your input will be very stressed due to the high current required to charge the capacitor.

    Here's one way to connect it up:
    [​IMG]

    I don't know why you need such a large capacitor, but perhaps you'll explain that.
     
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