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Discharging Capicators Automatically

Discussion in 'Electronic Basics' started by Jonathan Mohn, Feb 8, 2004.

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  1. I recently built a variable regulated power supply (see It works great.

    When I turn the unit off, though, I still read a voltage for 5 or 6 seconds,
    and the LED stays lit for the same amount of time. I believe that happens
    because the 2200uF storage capacitor is discharging over this time period,
    supplying current both to the voltage regulator and to the LED.

    Is there a way to short the capacitor so that it discharges more rapidly
    when I turn the unit off? I think I could use a normally open push-button
    switch to short the capacitor -- I would simply press it after turning the
    unit off. I really don't like that idea, though. I would rather build
    something into the circuit that would automatically drain the capacitor once
    the power was switched off. Any ideas?

  2. Why? Why do you want to discharge the cap more rapidly?

  3. Jamie

    Jamie Guest

    you could use a switch/relay to short it how ever, the life of the
    contacts would not really last..
    the other way around is to use SCR as a trigure crow bar.
    your switch should have a DPDT so that you can use the off possition
    side to input voltage to the gate of the SCR which is tied accrossed the
    that should kill the charge fast enough and will auto release as soon
    as the charge gets down below the holding current of the SCR.
    if you don't have this type of switch then have a ralay in there
    using the NC side to bias the SCR.
    make sure you also use the same Relay on the other side to power the
    supply to insure you don't have a timing issue between powering up etc..
  4. CFoley1064

    CFoley1064 Guest

    Subject: Discharging Capicators Automatically
    Hi, Jonathan. Try using a DPDT switch to turn on the line voltage for 1/2 of
    the switch (normally open), and use the second half of the switch to provide a
    discharge resistor (normally closed). For anything less than 30V, a 220 ohm, 3
    Watt wirewound should work well for quickly discharging things. It would look
    something like this (use fixed font or M$ Notepad to view):


    | |
    L1 o-----o + o---------o-----------o
    o-----o--__ | | |
    o- | | |
    | | o
    Line In | |DC Out \
    | | \
    L2 | | o \o
    o---------------o | | |
    | | |
    | | .-.
    | | | | R
    | | | |
    | | '-'
    | | |
    | - o-------------o--------o
    | |

    If you're a newbie, be very careful to keep the line voltage and the DC
    separated. Use an ohmmeter to make sure you know what's going on with the
    switch before you wire it up, and use a big switch (like the hardware store
    kind -- don't use the minis available at Radio Shack) to make sure the wires
    don't even get close to each other. If you're not familiar with wiring line
    voltage, ask a friend who knows something to give you a hand. Make sure the
    wires are really secured on the switch before you install it in the box. An
    accident here could lead to electrical shock.

    Good luck

  5. I don't know that I do. I built this power supply to use not only as a
    tool, but as a project to help teach myself basic electronics. When I
    noticed the effect of the discharging capacitor, I figured there must be a
    method of handling it. I'm not sure whether I will actually build that
    capability into my power supply, but one of the great things about projects
    like this is that they invariably lead to new questions, and greater
  6. Thanks for the post and the schematic. I'll print it out but, because I am
    a newbie (as you correctly guessed), I'll get a little more experience under
    my built before I play around with this type of configuration. I'm a bit
    paranoid about electrocuting myself!

  7. OK. What is the voltage across the capacitor when the supply is on?
    Also, what is the voltage across the LED? If you replace R1 by a
    zener diode which takes up the better part of that voltage difference
    and a smaller resistor value to limit the current based on the
    difference between the present voltage drop and that drop minus the
    zener diode value, then the LED will go out much more quickly but the
    capacitor will stay charged longer. If the problem is just the LED
    then that might serve you. The charge on the capacitor itself
    shouldn't be a problem.


  8. Thanks for the post, Jeff.

    The voltage across the capacitor is about 29 volts. Right now, R1 is
    dropping about 27 V, and the LED is dropping about 2 V. I think I
    understand your idea -- if I put a 25V zener in series with the LED and,
    say, a 133 ohm resistor, then I should still get about 15 mA through the
    LED. Then, when I turn the power off and the capacitor discharges, as soon
    as the discharge voltage drops beneath around 25V, the LED should go out.

    That's really clever. The timing on that suggestion is perfect, too,
    because I've spent the last 30 minutes or so playing around with a zener
    diode, testing its break-down voltage, current, voltage drops, etc. These
    things are pretty useful little components. Thanks!

    The one thing I like about the LED remaining illuminated is that it tells me
    voltage is still being supplied to the voltage regulator and, hence,
    anything I have hooked up to the output. I haven't, yet, installed the
    in-line voltage meter shown in the schematic, so it is actually useful that
    the LED stays lit until the capacitor is nearly discharged. I guess that it
    is a pretty good reason to install the voltage meter, too. I had figured it
    was just for the convenience of not having to use my DMM.

  9. John Fields

    John Fields Guest

    You either have some sort of reading comprehension problem, or you don't
    bother to read what posters want before you start spewing your useless

    For instance, if you'd have gone to the site the OP cited, you'd have
    seen the schematic and the BOM for the power supply. With that
    information in hand, you would have been able to determine the voltage
    across the capacitor (well, maybe not _you_) and answered your own
    question. You might also have found out from the OP's post that he
    wants to discharge the cap quickly, which will automatically turn off
    the LED quickly without having to use your hare-brained Zener scheme
    which will, BTW, actually _exacerbate_ the discharge problem.

    In short, don't be offering useless advice when you don't know what
    you're talking about.
  10. John Fields

    John Fields Guest

    Yes. First off, it is definitely _not_ a good idea to short the cap with
    just a switch, since the only thing limiting the current through the
    switch would be the ESR of the cap and the inductance and resistance of
    the circuit, all of which would be very low. Because of that the
    possibility exists that you could weld the switch contacts closed and
    not know it, and then the next time you switched the supply on you'd
    start blowing fuses, best case.

    What I'd do would be to replace the ON-OFF switch with a DPDT switch
    wired to short the cap in the OFF position, like this:

    ON OFF
    120AC>--[FUSE]--->\ <-- -->\ <--O----+------>TO EVERYTHING ON
    \----------------\ | THE INPUT SIDE
    | | |
    | | [2200µF]
    [XFMR PRI] [100R] |
    | | |
    | | |
    120ac>--------------+ +----------+------>DC GROUND

    Notice that I placed the fuse _before_ the switch, which is the right
    way to do it. Your schematic shows it after the switch, which will
    leave the switch hot when the fuse blows. Not a good thing.

    The 100 ohm resistor will be discharging the cap at about 1/2 amp if
    it's charged up to 50V, but only for a short time, so something like a 1
    watt resistor would be fine. I just tried it with a 100 ohm 1/2 watter
    and a 3300µF cap charged up to about 40V, and I could feel the resistor
    warm up a little, so even that ought to be OK, but I'd use a 1 watter

    If you decide to try this, make sure you get a break-before-make switch.

    Good luck, and BE CAREFUL.
  11. Thanks, John. Someone else had recommended a DPDT switch, too, but your
    schematic makes it really clear how it would work (I wasn't 100% certain,
    before). That is great, and is exactly what I'm looking for.

    Thanks for the advice on the fuse. I wired as shown in the schematic. I'll
    rewire it as you suggest, and I think that, while I'm at it, I'll drop in
    the break-before-make DPDT switch, too.



  12. The capacitor will discharge quickly when you attach a load.

  13. I looked for some break-before-make DPDT toggle switches at my local Radio
    Shack today, but didn't see any specifically labeled as such. The store's
    tech guy didn't seem to know what I was talking about. They had 3-position
    DPDTs (on-off-on) and 2-position DPDTs (on-off). I picked up a 2-position
    (part #275-666 if that means anything to you). Is there any way I can
    determine whether it is break-before-make (I'm guessing that means, in this
    application, that the 120V AC circuit is turned off before the capacitor
    draining circuit is turned on). The RS tech told me that the switch should
    be okay, but I'd rather get your opinion.

    Also, since I got a couple of warnings about electrocution (which I do
    appreciate, BTW), would y'all mind commenting on my plans for wiring this
    switch? It has 6 pins on the back (see below), 1 and 4 are at the bottom
    (off position), 2 and 5 are in the middle, 3 and 6 are at the top (on
    position). When the switch is OFF, I get continuity between pins 1 and 2,
    and also between pins 4 and 5. In the ON position, I get continuity between
    pins 3 and 2, and also between pins 6 and 5. I was planning on wiring the
    120V AC through 6 & 5, and the capacitor drain through pins 1 and 2, as

    6 3

    5 2

    4 1

    *120 VAC (black wire) to Fuse to switch pin 6 (upper left pin, ON position)

    *Transformer Lead-1 to pin 5 (middle left pin)

    *Transformer Lead-2 to 120 VAC (white wire)

    Capacitor Drain:
    *Capacitor +lead to 100-ohm resistor to switch pin 1 (lower right pin, OFF

    *Switch pin 2 (middle right pin) to DC ground

    I did a test run with this wiring on a prototyping board, but with 25VDC
    instead of 120VAC, and it worked perfectly. Is there anything wrong with
    this set-up?

    Thanks for your time on this!!


  14. [questions/advice about power supply cap discharging elided]


    Although the switch idea is pretty good, one thing you could do which would
    be a simpler/cheaper thing would be to bypass the 2200uF cap with a
    resistor. If you used a 1k resistor, the cap would discharge very quickly
    (with no other loads, the voltage across it will decrease by 1/2 every 1.5
    seconds.) Since its already decreasing fairly quickly anyway, due to the
    regulator, this would probably be adequate to turn off the LED almost
    immediately. You could experiment with different resistor values to see
    which one turned off the led quickly enough.

    Note that at 30V, a 1k resistor will dissipate about 900mW (V^2/R), so it
    could get a bit hot. A 2W resistor (or even one of those big 5W sandstone
    resistors) might be better, since it wouldn't get as hot, due to the larger
    surface area. That will make it last longer, I believe.

    Bob Monsen
  15. John Fields

    John Fields Guest

    Break-before-make means that when you toggle the switch one contact is
    broken before the other one is made.

    Yes. The black wire should go to mains hot and the white to mains
    No, it should be capacitor + lead to pin 1, and pin 2 to one end of the
    resistor, with the other end of the resistor going to ground.
    Yes, the 100 ohm resistor must _not_ be in series with the capacitor.
    Here's the schematic with your switch terminals shown.

    BTW, the dashed line between the arms (the slashes) only indicates a
    mechanical connection between both switches in the package. Do _NOT_
    connect them electrically!!!

    ON OFF
    4 3
    120AC>-[FUSE]-O-->\ <--o o-->\ <--O----+------>TO EVERYTHING ON
    6 \----------------\ 1 | THE INPUT SIDE
    | | |+
    | | [2200µF]
    [XFMR PRI] [100R] |
    | | |
    | | |
    120ac>--------------+ +----------+------>DC GROUND
  16. Thanks for all the posts. Gradually, I'm beginning to see things more
    clearly! :)



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