Bleeder Resistor Value & Wattage

In the old days I had a stick (Chinese Chopstick would do) with a copper nail pressed through the end, with a 10K resistor taped to it. One end went to the nail the other had a wire with a crock clip on it. But that was in the days of big solid carbon resistors. I don't know what power rating it was, but I never managed to damage it even putting it across a live circuit occasionally.

Any way I=E/R So 600/10,000=0.06 E*I=P So 600*0.06=36 The peak power would be 36 Watts. Since the voltage is falling, I reckon that the resistor would have been about a 5 Watt rating.

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Best Regards:
                     Baron.

A bleeder is generally always in place, low wattage, high resistance, may take quite a while to bleed down (RC of a minute or more), so that it won't load the circuit too much when running.

A shorting stick (applied to caps without a bleeder) is generally not "alligator clipped on both ends" since it is generally a poor idea to be quite that close, and is usually a lower value, high-wattage resistor. Exact figures depend on the caps in question - figure a reasonable time constant (ie, you'll hold the shorting stick on the capacitor for 5 or

10 seconds, and want it to be down to, say, 6 V - about 5 RC time constants, so a time constant of 1-2 seconds) for 600V on the largest typical capacitor you work with. One side has a clip and lead, the other side has some sort of stiff electrode, and an insulating stick handle leads back to the operator.

Remember that an easy method to up the power is to use multiple lower-resistance resistors in series - it may be much simpler to find 10

2 watt resistors than 1 20 watt resistor. You can also use a light bulb, (nice low resistance 100W [or whatever you buy] resistor, easily found) but in that case you need to check that it has not broken, and/or use a resistor in parallel with it so that you have some discharge even if the filament does break.

Once the cap is shorted, if there's some chance of stray charge pickup, a two-alligator clip lead (no need for a resistor, but there can be one if it makes you happy) can be installed.

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Cats, coffee, chocolate...vices to live by

I prefer to parallel higher resistance ones for more wattage ratings, like if I wanted a 10k 2 watt, I parallel 2 20k 1 watt resistors together. Which makes it easier as Dick Smith only sells 1/4,1 watt and 5 wa

I come from a high voltage background - in that context, the reduction of voltage across each resistor (when in series) is beneficial for inexpensive parts. At the voltages the OP is working with, makes no difference, either series or parallel would be fine.

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Cats, coffee, chocolate...vices to live by

For the job that you are wanting to do wire wound resistors would be just as good. Remember that the peak dissipation is only going to last for a fraction of a second. I just used something that came to hand.

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Best Regards:
                     Baron.

Something like a 10 watt, roughly 100 ohm wirewound resistor works fine.

John

I will do this, I have a couple of old bakelite bayonet lamp sockets, I can use one of them, what value resistor would I put in parallel? I regular 240 volt 100 watt light bulb tests at around 42 ohms.

Whatever's handy - probably a couple of hundred ohms. If you simply check (with an ohmmeter) that the filament is sound after you've discharged things, and before you reach in and grab stuff, you can do without it. Unless you drop it, the bulb should hold up for years in this application, as it never should get hot.

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Cats, coffee, chocolate...vices to live by

I'll just use a clear bulb, so the filament is easily visible. Think I may have one lying around somewhere, probably with those lamp sockets.

There's a "learn to solder SMT" QRP dummy load kit out there that's around 44 2k2 SMT resistors to make one 5W 50R load.

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Ben Jackson AD7GD

http://www.ben.com/