Diode to prevent shock from capacitor bank?

Discussion in 'General Electronics Chat' started by summersab, Feb 19, 2016.

  1. summersab

    Thread Starter Active Member

    Apr 8, 2010
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    0
    Say I have a capacitor bank. Like... a BIG capacitor bank. One that uses Maxwell supercapacitors and can jump start a car. Obviously, this is a scary-dangerous thing, and one touch to both poles is probably lethal. Is there a way to put a diode or other flow-preventing device between the positive pole of the capacitor bank and a car battery so that the bank will charge but the poles on the battery don't become deathtraps from current flowing out of the capacitors back toward the battery? I looked at high current diodes and have found some upwards of 70-200A, but a) I'm not sure if that's the right approach, and b) I figure the capacitors are capable of pulling in more than 200A during a charge and would fry the diode, right?

    Thanks!
     
  2. Lestraveled

    Well-Known Member

    May 19, 2014
    1,957
    1,215
    @summersab
    Sketch up a schematic of what you are thinking about and post it.
     
  3. AnalogKid

    Distinguished Member

    Aug 1, 2013
    4,546
    1,252
    A capacitor that is charged by a connection to a 12 V source charges up to ... wait for it ... 12 V. Enough current capacity to spot weld steel is not the same as enough voltage to overcome skin resistance or power a laptop. Except for patient-contact situations, UL considers DC voltages up to 32 V to be safe enough for non-protected handling.

    To be clear, 1000 A can be very dangerous if you're not paying attention. But until you drop a screw driver across the terminals, 12 V is just 12 V. The qualifying condition that makes your caps dangerous is not the voltage at the terminals, but the current being drawn by the external connections or devices or load.

    ak
     
    Lindembruck likes this.
  4. panic mode

    Senior Member

    Oct 10, 2011
    1,321
    304
    supercaps are polarized, you may want to make sure they are connected the right way. you can connect diode and resistor in parallel (diode oriented so that caps charge low voltage battery). diode can be used to allow rapid discharge and resistor to limit charging current when caps are empty. one touch to 12V can be lethal but not because of 12V being high, more because shorts can cause arcflash, or cause you to move away (perhaps into path of traffic).

    you can use circuit that only allows current flow when specific condition is met (correct, polarity, sufficient voltage etc.). but.... first step is to come up with some specs and description of what you wish to accomplish
     
  5. summersab

    Thread Starter Active Member

    Apr 8, 2010
    132
    0
    I suppose I should have followed up to this and marked it as closed instead of abandoning the thread. While browsing the Auto Zone app, I discovered that there are devices specifically designed for this. There are battery isolators (they use diodes) and high-power relays (often mislabeled as isolators). This should satisfy what I'm wanting to do.

    Thanks for the other info provided by those who posted, though - all good information!
     
  6. Lindembruck

    Member

    Feb 28, 2016
    32
    1
    The problem is not only direction flow but supply the power to the caps.
    To supply, you need to know what the caps will do, to you install a powerful and defined device, a diode.
    If the caps have a little to do, a mid sise diode with a low value resistor in serie will solve the problem but if you will need a high power from the caps, you will to need a very powerful source, with powerful diodes around doble power you need.~

    Examples:

    If you will need 500W output, a diode with 80A with a air heat-sink more a fan will solve

    If you will need 10.000W output, you will need a 1600A diode with maybe water heat-sink and this look be a problem to solve with a lot high power diodes in parallel more a very low resistor in serie with each one more, all power cords with the exactly same size more all power cords very heavy because ohm resistivity.

    The output power, is a calcule where will enter the power wave sise and its more low frequency.
     
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