Bleed resistor

[appicnc]

Hello Group

I am building a new power supply and have lost the formula for the bleed resistor in the power supply, can someone post the formula, any thing to be careful of with the new supply ( other thank life and limb ) the supply should supply 70v DC at up to 30amps, the cap is rated at 49000mfd so I guessing that you don't wwant to come into contact with any part of either side of the transformer!.

Thanks

Ed

[vger]

The cap should not discharge back through the transformer with the diodes/birdge in place. The choice of bleeder will depend on how quickly you want the cap to discharge once power is removed, and how much current you want to draw with the supply turned on. The RC time constant comes into play here. T (in seconds) = R (in ohms) * C (in farads)
That is to get the cap discharged to 36.8% of it's original charge level. Double the time and you get to 13.5%, triple it and get to 4.9%.

Assuming a 10k resistor across a 49,000 microfarad, gives you 490 seconds, or about 8 minutes 10 seconds to get the voltage down to about 25 volts. The 10K resistor will draw only about 7 ma at 70 volts and that is about 1/2 watt so I would use a 1 watt resistor.

Decrease the resistance and your time decreases accordingly, the current goes up though and wattage increases. Tradeoffs abound

Steve

[Aleks]

There are several approaches. Theoretically the capacitor will never discharge to zero volts.
Approach 1)
Decide what is a "safe" voltage (Vd)
Decide what is the maximum acceptable time (t) to discharge to VD.
Calculate R:
Let Vc = the normal charged voltage.
R = t*/{C*[ln(Vc) - ln(Vd)]}
Where ln(x) is the natural logarithm of x.
Calculate the wattage according to Paul Taylor's advice.
~
Approach 2)
Decide what is a "safe" voltage (Vd).
Decide on the minimum value resistance that you can use, based on
dissipation.
Calculate the time required to discharge to Vd.
t = R*C*[ln(Vc)-ln(Vd)] so> R=t/C*[ln(Vc)-ln(Vd)]

*****************************************************
The amount of energy stored in the capacitor is given by:

0.5CV^2 joules

A joule dissipated over a second is one watt so an 1/8th watt resistor
should not be expected to dissipate more than one joule every 8 seconds, a
quarter watt resistor in 4 seconds, half watt in two seconds and one watt
resistor in one second. Work out your joules and decide how quickly you
want to dissipate the energy and that will give you the resistor wattage.
Remember that this resistor will dissipate this power when the supply is on
as well so it would be better to over rate it e.g. use a 1watt type where
the dissipation is half a watt.

>From the decided dissipation and the supply volts you can calculate the
resistor value W = V^2/R so R = V^2/W

[UKVoyager]

With that amount of power I would like a visible indicator that its safe to play with. What about a lamp across the cap switched on by a relay normally closed contact pair, with the coil connected before the bridge. When the lamp stops glowing the volts should be really low.

[Aleks]

It's always good idea to use visible indicator but keep in mined... what if you'll have a bad light? To use regular voltmeter is much safely

[apache405]

I would use both a visible light and a voltmeter. That way you can glance at the supply from across the room and see if it is close to being discharged, then double check the voltmeter.

[rdpzycho]

I normally use a quick discharge path in case of really high voltage (used to work with 400V, but 70V can already be harmful if not fatal) and really high energy storage cap (I've worked with 300J to 2400J).

the path is consist of a power PNP (relays would easily get weld in very high currents) and a low resistance resistor. another resistor is tied to base and the positive rail, with a button from base to ground. a push on the button will quickly discharge the capacitor if the resistance is low enough and the transistor can handle the discharge pulse (way back in school I would normally short the capacitors with wires and enjoy the sparks around the place, of course it is not a safe practice).

[vger]

Many moons ago.... Electricians would lay/throw a crobar across a circuit they were working on to protect them while working on the equipment. If the power switch was in a different location (sometimes miles away) and an unknowing person re-energised the circuit, the "crobar" shorting the circuit would protect them by blowing the fuse. Now days in electonic circuits (power supplies) the "crowbar" is generally replaced by an SCR that is designed to have a very high peak current. Most often seen as a protection circuit to kill the power supply in case of overvoltage malfunction. The advantage of the SCR is that once triggered it will continue to conduct untill the current through it reaches a VERY low threshold, even with the trigger removed.
No "lack of power" indicator is totally foolproof. Lamps burn out, voltmeters can as well. A voltmeter is the best check, but, just to be on the safe side, after checking with the voltmeter seeing the voltage is low, "crowbar" that cap before you touch it.

Steve

[apache405]

Originally Posted by vger Many moons ago.... Electricians would lay/throw a crobar across a circuit they were working on to protect them while working on the equipment. If the power switch was in a different location (sometimes miles away) and an unknowing person re-energised the circuit, the "crobar" shorting the circuit would protect them by blowing the fuse. Now days in electonic circuits (power supplies) the "crowbar" is generally replaced by an SCR that is designed to have a very high peak current. Most often seen as a protection circuit to kill the power supply in case of overvoltage malfunction. The advantage of the SCR is that once triggered it will continue to conduct untill the current through it reaches a VERY low threshold, even with the trigger removed. No "lack of power" indicator is totally foolproof. Lamps burn out, voltmeters can as well. A voltmeter is the best check, but, just to be on the safe side, after checking with the voltmeter seeing the voltage is low, "crowbar" that cap before you touch it. Steve

So basically you are proposing the use of an SCR to control the discharge of cap, and then a shorting bar be switched in after the SCR has finished?

[pastera]

For the bleeder pick either the power level or bleed down time to calculate the rest.

I would pick a bleed down time of 5 minutes and label the equipment to allow 20 minutes after power down before servicing. So 3 time constants would be five minutes, therefore, your resistance would be 750 ohms (closest 5% in stock).
Power would be 6.5 watts so anything over a 10 watt would work reliably.

Vishay
HL02506Z750R0JJ - $2.37

Huntington Electric
FVTS10-750 - $3.00


Aaron

[rdpzycho]

SCRs are better (most capacitor discharge lamps have them shorting the caps after a power failure), and they are used in resistance welders..it just happened that I have lots of PNPs lying around..

actually, I always do short the caps after getting a low enough reading (and keep the shorting wires attached while servicing the unit)..a few years ago I was using a homebrew ESD gun with capacitor voltage multipliers..all the caps are already reading low at 5V, but touching the caps will give you the sensation that it still has a few hundred volts (it was 1kV per capacitor bank for 8kV, energy per bank is roughly 0.8J which is, IIRC, way above the standard used for ESD testing)..

don't use your pliers or any other tools when shorting high energy caps...the caps can make the tool useless (had a plier permanently welded by shorting a 300J cap)..

[apache405]

Originally Posted by rdpzycho don't use your pliers or any other tools when shorting high energy caps...the caps can make the tool useless (had a plier permanently welded by shorting a 300J cap)..

your work area must have smelled *wonderful* after that. also kinda reminds me of what happened when i was checking a 120V light switch and shorted hot to ground with my meter probe. (new new pair of meter probes later...)

rdpzycho-
i gather from your previous posts that the SCR is used as a switch that directs the current flow through the resistor being used to discharge the cap. the SCR would direct current into the resistor until the voltage is low enough (.6V i'm assuming) to change the state of the SCR. am i on the right track?