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I'm trying to size a set of resistors for charging a capacitor. The source is a 500V 60Hz AC supply.

The initial instantaneous current is 0.5A sizing the resistors at 125W each. That seems too high because the current decays very rapidly. If the resistors only need charged every 3 minutes or so, how would I size their power requirements?

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Consider the pulse current rating of the bridge rectifier as well as the resistors. If you are feeding the bridge from a transformer, that may help by providing some current limiting depending on its characteristics. \$\endgroup\$ Sep 30, 2016 at 15:30
  • \$\begingroup\$ any questions @vini_i ? remember Joules =1/2 CV^2 = watt-seconds \$\endgroup\$ Sep 30, 2016 at 23:18

3 Answers 3

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Well if you were supplying the circuit from a DC supply it would take about 5 time constants before the cap could be regarded as fully charged. You could make the argument that an AC scenario would be no worse because at least the power comes intermittantly over a longer time period giving a little time for the resistor to cool so, stick with the DC circuit is my advice when analysing.

5TCs ia a total elapsed time period of about 17 seconds (on DC) so, you have to start digging around in the data sheets of power resistors to see how much their short term peak overload rating is.

For instance, a 1 watt rated resistor may be able to take 10 watts for 1 second or 100 watts for 0.1 seconds etc.. Regarding the current at any point in the charging process consider this graph: -

enter image description here

Along the base is time (measured in RC time constants). The rising graph tells you the capacitor voltage in terms of percentage fully charged voltage (500 V x 1.4142 and not 500 V).

The falling graph gives you percentage current and at 0.5 CR it is 61% of I max (which happens to be 0.7 A and not 0.5 amps as per the question). So, you could do a calculation based on approximate figures i.e. between t=0 and t=0.5RC you could assume the current to be 100% then, in the next half time constant assume it is 61% etc..

Or, you could do some digging around and develop the exponential formula to give a more precise figure for power.

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  • \$\begingroup\$ It's worth looking at datasheets for a few different kinds of resistor. e.g. wire wound resistors tend to have much better pulse withstanding characteristics than thin- or thick-film chip resistors, even if you look at so-called "pulse-rated" chip resistors. \$\endgroup\$ Sep 30, 2016 at 15:27
  • \$\begingroup\$ @pericynthion are you addressing me or the OP? If the op then do the normal @ thing. \$\endgroup\$
    – Andy aka
    Sep 30, 2016 at 15:44
  • \$\begingroup\$ OK @vini_i, you've done a good job so let's delete this mush under my answer.... \$\endgroup\$
    – Andy aka
    Sep 30, 2016 at 15:45
  • \$\begingroup\$ I guess it was addressed to you as a suggestion to add to your answer :) \$\endgroup\$ Sep 30, 2016 at 16:45
  • \$\begingroup\$ @pericynthion OK dude, cheers. \$\endgroup\$
    – Andy aka
    Sep 30, 2016 at 17:39
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One easy way is to let LTspice do the grunt work for you,

enter image description here

and here's the circuit list:

Version 4
SHEET 1 880 680
WIRE 320 -112 224 -112
WIRE 448 -112 384 -112
WIRE 272 -16 96 -16
WIRE 320 -16 272 -16
WIRE 448 -16 448 -112
WIRE 448 -16 384 -16
WIRE 448 16 448 -16
WIRE 96 128 96 -16
WIRE 448 128 448 96
WIRE 272 240 272 -16
WIRE 320 240 272 240
WIRE 448 240 448 192
WIRE 448 240 384 240
WIRE 96 336 96 208
WIRE 224 336 224 -112
WIRE 224 336 96 336
WIRE 320 336 224 336
WIRE 448 336 448 240
WIRE 448 336 384 336
WIRE 448 384 448 336
FLAG 448 384 0
SYMBOL voltage 96 112 R0
WINDOW 3 24 96 Invisible 2
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V1
SYMATTR Value SINE(0 707 60)
SYMBOL cap 432 128 R0
SYMATTR InstName C1
SYMATTR Value 3440µ
SYMBOL diode 320 -96 R270
WINDOW 0 32 32 VTop 2
WINDOW 3 0 32 VBottom 2
SYMATTR InstName D1
SYMATTR Value RFN5TF8S
SYMBOL res 432 0 R0
SYMATTR InstName R2
SYMATTR Value 1000
SYMBOL diode 320 0 R270
WINDOW 0 32 32 VTop 2
WINDOW 3 0 32 VBottom 2
SYMATTR InstName D2
SYMATTR Value RFN5TF8S
SYMBOL diode 384 256 M270
WINDOW 0 32 32 VTop 2
WINDOW 3 0 32 VBottom 2
SYMATTR InstName D3
SYMATTR Value RFN5TF8S
SYMBOL diode 384 352 M270
WINDOW 0 32 32 VTop 2
WINDOW 3 0 32 VBottom 2
SYMATTR InstName D4
SYMATTR Value RFN5TF8S
TEXT 414 408 Left 2 !.tran 180
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Use 3 or more 150W halogen tubes in series as charging R's and ensure Cap charging current is not exceeded. This will give the fastest charging rate.

The cold resistance will be 1/10 of hot.

If too much current for your powersupply reduce the bulb wattage.

3300mF 600V caps can handle 10Arms ripple current.

  • Cap ripple current is equiv to filtering a 5000W at 500V so three 150 W halogen lamps in series is a simple solution that will be easy to charge up in a few seconds or less and not stress the bulbs too much unless you plan on doing this forever, then use more bulbs in series in fuseholders. They can easily handle heat.
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