Bleeder resistors, What values?

[bowdown]

Hi everyone i am new to this site.looks like a pretty good site actually... Can somebody please help me with what value bleeder resistors i should use for my power supply and also how many i should use, i only have space for 5 watt resistors on my circuit board i have allowed for a max of 5 per rail as well so a total of 10. its running +/-92v and each rail has 50,000 uF of capacitance ( 5 x 10,000uF caps @ 100v) everyone keeps telling me different stuff about what value i should use. Im building a leach super amp. Oh just one other thing is 1x 1500VA transformer going to be ok for it? i am aiming for 350-400wrms @ 8ohms.I cant seem to find anything on the net on how to calculate what resistance values i should be using and also checked the threads.Any help would be extremely helpful. Thank you.

Regards
Bowdown

[Tube_Dude]

Hi bowdown!

everyone keeps telling me different stuff about what value i should use

That's normal ,because there are not a fixed rule to calculate the value of the bleeders.

I usually use 1 mA across the power suplly...so in your case 92 volts i will put 82k Ohms 1 W resistor across each rail.

Jorge

[janneman]

The bleeders are used to, err, bleed the charge off the reservoir caps when you switch off the amp. The 82k suggested above would work, but with your large caps it would take them many minutes to discharge. Ususally one would aim for 30 secs to discharge to half voltage. I'm just to lazy to do the calcs, but I would try 10K and go lower if necessary. You don't want to go too low because that generates more heat.

But there is nothing to stop you from building your amp and trying it out, because the quiescent current from the amp also works as a bleeder. In fact, the quiescent current is more than any bleeder would do, the problem is that if the supply voltage goes below a certain point, the amp will stop working properly, and from there on you need the bleeders. Your amp will work perfectly without the bleeders, so no need to commit yourself just now. If you do, you probably will decide to change it afterwards anyhow. I know, life's a b*tch.

Jan Didden

[jackinnj]

if you don't like the heat, use a DPDT relay, when the unit turns off the bleeder is connected into the system. you can thus use a bigger bleeder value -- although you might wish to put a Quencharc across the relay points to prevent arcing.

from a practical standpoint you want the bleeder to reduce the capacitor value to a low level in about the time it would take you to unscrew the cabinet.

make sure that the resistor is the proper wattage -- W > (E^2)/R

fwiw I use 50K in my tube amps -- and have been doing so since I built my first ham radio transmitter in the 1960's.

[sam9]

jackinnj,

I was planning to use a relay for this purpose on my next amp but was un aware that anyone else had done this, althouigh it seems pretty obvious. It's nice to know this has been before as is reasures me I'm not about to do something stupid.

Regarding the watt rating, it seems to me a lower watt rating (down to a point) could be used since the resistors are in-circuit only a few seconds rather than continuously. I figured I would stick a thermocouple on one the and swith the unit on and off several times, allowing evough time for full discharge. If the resistor stayed below 100deg C, all would be well. I would initially try for a resistance value that discharged the caps in 5-10 seconds.

[macboy]

I plan to use a circuit like the one attached. The Relay not only allows the bleeder resistor (1.8 K, 2 W) to be disconnected during normal operation, but also allows a soft-start resistor (27 R, 2 W)to be used to reduce the turn-on current surge that would otherwise occur. This can be a significant current with large capacitors and a large transformer, so limiting it is a good idea. The relay is activated after the power has been on for at least five time constants (about 5 seconds in this case) and is deactivated immediately upon removing power. The choice of reistor values was determined almost entirely by what I had in my parts box.

[jackinnj]

just think it through on a "fail-safe" basis -- the default condition should be "bleeder engaged" -- you can also put an LED with dropping resistor on to indicate whether the cap is charged.

i am getting pretty sticky on safety items -- only a week ago i accidentally came across a "line voltage" where I hadn't expected it.

[cpemma]

Formula for discharge is

ln(Vs) - ln(Vf) = t/RC

ln = natural log

Vs = Supply

Vf = Final level

t = time (sec)

R (kohms), C (mF)

So to drop from 60v to 24v would take 0.92 time constants, eg, with 20,000uF (20mF) smoothing and 2k bleeder, 37 seconds.

[bowdown]

Oh thanx everyone you guys are awesome.... Oh by the way will that transformer im using going to be sufficient???? thanx a million again.

Bowdown

[macboy]

This thread discusses Thermistor vs Soft Start, and I have posted a schematic there that works very well. It uses a power resistor instead of a thermistor, and a relay shorts it after ~ 1 second. It works very well and will soft start your transformer even if the power drops out only very briefly, something that a thermistor will not do. I built the entire thing including the power resistor and relay on a 1x2 inch PCB (sorry no eagle files to share as I do small boards by hand with a sharpie).

If you are convinced that you want to use a thermistor, you must consider the manufacturer's maximum capacitance rating. This capacitance (partially) determines the amount of energy which the thermistor will need to absorb when the power switched on. With too much capactiance, you will damage or blow the thermistor. Usually, these things are used in switching PSU's which recitify the mains directly, so the capacitance is specified assuming 120 VAC (~170 V rectified). Since energy stored in a cap is proportional to voltage squared, you can scale the capacitance up by the inverse-square of the reduction in voltage of your transformer's secondary. If the capacitance is specified for 240 VAC then adjust accordingly.
Example:
Dual 60 VAC secondaries: (120/60)^2 = up to 4 times the recommended maximum capacitance (total for both rails).
Or 40 VAC: (120/40)^2 = up to 9 times the rated capacitance.
Or +-35 VDC rails: (120/(35/1.414))^2: up to 23.5 times the rated capacitance.