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charging capacitor bank

Discussion in 'Electronic Design' started by Robert Morein, Jan 11, 2004.

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  1. I'm constructing a power supply that will have 56,000 ufd per positive and
    negative rail.
    The caps are cans with screw terminals, two 28,000 ufd units per rail.

    How many amps do I need to specify the bridge rectifier for to avoid blowout
    on slow start?

    I would prefer not to limit current with a resistor, though I guess I could
    use a relay bypass.
    But I've seen quite a few audio amplifiers that charge the bank without any
    extra current limiting component.
  2. James Meyer

    James Meyer Guest

    If you really mean "slow start", then the bridge current rating isn't a
    Even if you don't have a slow start mechanism the transformer will limit
    the surge current into the caps. I'd try a 50 amp bridge for starters.


  4. Key tidbits are missing from your post. I'm not referring to pineapple
    tidbits but rather informational tidbits. What are the relevant voltages
    involved? Are the capacitors being charged from the output of a
    transformer? If so, what is the transformer rated for. What kind of bridge
    rectifier did you have in mind. What is your load current? What kind of
    slow start system are you referring to?

    Generally speaking however inrush limiting is usually not mandatory (though
    in some cases desirable anyway) when charging even very large capacitors
    from the output of a transformer. Usually the transformer's leakage
    inductance and copper resistance will limit and absorb most of the transient
    stress upon power up. Since the transfomer is physically very large it has
    a lot of thermal inertia and can easily handle this extra stress without
    problems. If the transformer is massively overrated for the application
    then a larger percentage of the initial stress may be placed on the bridge
    rectifier, thus perhaps making inrush limiting more beneficial.
  5. James Meyer

    James Meyer Guest

    I can't think of any reason why that wouldn't work.

  6. Fritz Schlunder wrote...
    2 x 28000 uF implies moderate voltages of 170V or less. Robert, if this
    is a direct offline rectifier, inrush-current-limiting protection will
    be a wise addition. If you're using a power transformer, then as Fritz
    said, you may be able to live with the intrinsic current limiting that it
    provides. But 56mF at 170Vdc = 810J is a fairly-large apacitor-storage
    energy to charge quickly, so inrush limiting may well be mandatory. I've
    used a simple time-delay ac relay shunting a series input power resistor
    with good success for transformer-rectifier charging of capacitor banks.

    - Win

  7. Thanks. I think I can experiment using a Variac and a Tek P6021 current
    probe to see if the peak current exceeds 40 amps.
  8. James Meyer

    James Meyer Guest

    Don't forget that a nominal 40 amp diode will have a surge rating for a
    short time that is many times its continuous 40 amp rating.

  10. James Meyer

    James Meyer Guest

  11. James Meyer wrote...
    That's for only one half of a single 50 or 60Hz cycle.
    It may take much longer to charge a large capacitor bank.
    Some detailed calculations are in order.

    - Win

  12. Winfield Hill wrote...
    For example, a half-cycle of 400A peak half-sine charging
    will get Robert's capacitor bank voltage up to only about
    0.64 400A 8ms / 0.056F = 36V. This means many ac cycles
    would be required to complete the task, and an AC off-line
    directly-connected rectifier would be over stressed. If
    we assume 200W constant-power charging, it would take four
    seconds or 240 ac cycles (at 60Hz) to get the required 810J
    of energy into the caps.

    - Win

  13. Reg Edwards

    Reg Edwards Guest

    The switch-on surge current in a diode + big capacitor circut is limited by
    the internal impedance of the AC source, usually R+jwL. If a transformer is
    the source then leakage inductance and primary and secondary winding
    resistance must be taken into account.
  14. Reg Edwards wrote...
    In the case of the AC line, typially it can easily deliver 400A
    and do so cycle after cycle, creating a race between the circuit-
    breaker opening or the recifier dying. The capacitor's esr helps
    a bit, but not much for a 56mF bank of large electrolytics.
    Must? Haha, in Robert's scene, he should be very glad indeed to
    _get_ to take it into account. The transformer's high copper-wire
    heat capacity means it can be massively over-stressed for a few
    seconds without damage. But in a typical configuration it'll
    likely not lower charging currents enough to protect the rectifier.

    In my 250kW pulse generator I designed a series 200V 2A ac current
    limiter in parallel with a 25-ohm 50W resistor to solve the problem,
    but a timer to close a relay across the resistor could also work.

    Another approach is to manually slowly turn up an input variac,
    but one risks blowing stuff out if he ever forgets.

    - Win

  15. Hello Robert,
    here is something that may be of interest to you.
    There is the welding transformer, diodes and capacitors.
    A limiting inductor is used in series with the capacitor bank,
    four by 10,000 MFD capacitors but not in series with the
    output. 0 and 1.pdf
    (the circuit diagram of this mig welder is on page 15 so you
    don't have to download the whole booklet)

    I couldn't get any information out of the WIA company as to
    the value, and make up and exact purpose of the inductor.
    Very secretive they were. The use of a "limiting inductor"
    in your project, may reduce the initial surge current as the
    capacitors charge up when your power supply is first
    switched on. Just a thought.

    John Crighton
  16. [snip]
    Thanks John, and to everyone else.

    Does anyone have a time delay relay circuit handy?
  17. Here is a simple way to do it:

    I've never built it, or tried it out, so be careful. The current limit
    resistor for the charging phase must have a huge power rating, and the relay
    must be capable of handling the large charging currents when finally
    switched in without melting.

    Bob Monsen
  18. Rich Grise

    Rich Grise Guest

  19. James Meyer

    James Meyer Guest

    Can you give me a reason why they won't current share?

    Are you aware that a diode's forward voltage drop is proportional to the
    current it's passing? It's not a fixed 0.7 volts regardless of its current.

  20. James Meyer wrote...
    The diode has a traditional logarithmic component, which
    is both current and temperature sensitive, and a series
    resistive component, which follows ohms law. At high
    currents the resistive component by far dominates. If
    two high-current power diodes are paralleled, the current
    sharing will be largely determined by the relative values
    of their intrinsic series resistance. While for a given
    type of diode these resistance values will likely be the
    same, they are not guaranteed to be the same, and may in
    fact not be, especially if they have different date codes.
    The series-resistance component of diodes I have measured
    have varied by up to 50%.

    - Win

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