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Author Topic: HV power supply bleed resistor value  (Read 12162 times)
KB3TTP
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Posts: 32




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« on: January 28, 2012, 10:14:18 AM »

Hello All,

Installed in my HV PS. I have a 75K 100W bleeder resistor installed in front of the exhaust fan. The PS is separate from my amplifier.
 
Considering how many watts the transformer is rated for (665W), and the wattage of the resistor, it would seem I  may be taking to much of the power away from the transformer. It uses a full bridge rectifier circuit. (no doubler)

 
It generates a very large amount of heat. I do have 20K 10W resistors across each of the 16 capacitors which is 2 parallel banks of 8 Caps in series. 450V 220uF.

 
My question is: Should I go with a lower value resistor, such as 15 or 20K 50W , to improve performance, reduce heat and still be considered a good safety practice?  I have now seen this size used in some other PS.
 
The 2.7KV that the PS is set to now bleeds down to 10 volts in about 8 seconds after shutdown.
 
The 2.7KV at idle drops to about 2KV under max. load.  Valves being used are 572B's. They are Chinese valves that I put through HELL AND BACK after finishing the amp and tuning in of the tank coil key down. They are still at full output. I get about 750 W out of them running at these voltages.

Thanks for listening.

John
KB3TTP


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G3RZP
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« Reply #1 on: January 28, 2012, 10:51:09 AM »

My choice for that would be abou 200 K. I got relatively cheaply - well, for Surplus Sales! - 200k 200 watt resistor. It's about 14 inches long and 1.25 in diameter. Generally speaking, it would take a reasonable amount of time for the things to bleed down, but not as long as opening up the device to get at it! And you do use a chicken stick, don't you?
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W5FYI
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« Reply #2 on: January 28, 2012, 11:11:09 AM »

I'd have to see a schematic of your supply to see if you need the large bleeder, but if you have smaller resistors across each capacitor, I wouldn't think you would need it.

Or, you could keep it, but isolate it with relay. As long as the PS is powered, the relay's NC (normally closed) circuit keeps the resistor circuit open, thus it would not consume any power at all. When you turn off the PS power switch, the relay would release the armature and that would close the NC circuit, thus completing the resistor's circuit so that it can do its bleeding.

Alternatively, find a higher resistance bleeder resistor. If you use two 75K 100W resistors in series, they'll bleed the capacitors in twice the time, but consume only half as much power (48 watts), well within the power rating of the resistors.

It would be prudent to have a way to completely short the capacitors automatically whenever the access door is opened. As we all know, some capacitors can keep a lethal charge for days, or longer.  I've worked with HV photoflash capacitors, and always shorted their terminals with a "chicken stick" shorting bar, or alligator-clip connectors, before working on the circuits they were in.

Also, use eye protection. Sparks can fly and burn an eyeball (primarily the cornea) if they hit it, and ultraviolet energy from a spark flash could also damage the retina.

The old tech's rule, "one hand in your trouser's pocket," has probably saved several lives. Do what you can to engineer safety into any electronic circuit. GL
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AC5UP
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« Reply #3 on: January 28, 2012, 11:21:12 AM »

If I'm still handy with Ohms Law............. The 75k resistor currently installed draws 97 watts @ 2.7 kV.

Seems like a significant amount of wasted power overhead on a 665 Watt power transformer while an idle load of better than 100 Watts is a bit much in trade for the safety factor. A 300k resistor would draw 24.3 Watts, a 400k resistor would draw 18.23 Watts with both calculated across 2,700 Volts.

I see the 16 filter caps already have 20k balancing resistors across them and can't help but wonder if the separate bleeder is absolutely necessary. Yeah, the quicker the discharge the better, but how often do you work on the power supply and how willing are you to ignore your own caution label?

I'd pull the 75k resistor and install something near 400k @ 50 Watts, then clock the no-load discharge time and make a judgment call. If it takes three minutes to come down below 100 volts, I'd label it at five minutes and test before touch. Just because it's working OK today doesn't mean the bleeder will still be good the next time you open up the power supply... In other words: No matter what you do, assume the HV is HOT until proven NOT.  Cool
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KB3TTP
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« Reply #4 on: January 28, 2012, 11:32:45 AM »

Yup on the chicken stick.

Using the relay the way you mentioned is a good idea. I was thinking of a relay, but I didn't think of operating it automaticaly. I was thinking more of switching the relay manually.

Visit my QRZ page. The schematic of the amp and the PS is there.

Good advise you gave. Evan when we have heard it a hundred times we all need to be reminded of it. So hearing it 101 times rather than forgetting a procedure and risking your life is a good thing.

On the post about Ohms Law. Is there a smiley for hitting myself in the head. Great advise as well.

Thanks, 73
John KB3TTP
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AC5UP
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« Reply #5 on: January 28, 2012, 11:55:27 AM »

On the post about Ohms Law. Is there a smiley for hitting myself in the head. Great advise as well.

Just for the sake of the Peanut Gallery, I was tempted to mention that a 100 Watt resistor drawing 97 Watts continuously is not likely to live long and prosper, even though immediately inside the air flow of the fan is the best possible mounting location.

Which brings up another point to consider: If the purpose of the bleeder is a safety device then the reliability of the resistor should be considered along with the loading and bleed-down time of a given resistance value. Worst case example could be an owner who "knows" the bleeder will bring the power supply down to a safe value in less than 10 seconds and behaves accordingly, not realizing the bleeder went toes up six months ago and the HV now needs two minutes to drop below the 911 level.

Chicken Stick, HV Meter, whatever it takes to verify the filter caps are in the safe zone... Do It before you wish you had!
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KB3TTP
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« Reply #6 on: January 28, 2012, 12:19:10 PM »

I found a useful xls spread sheet calculator for computing bleed down time for HV caps with given values

2700V with a 200K resistor with my 55uF total capacitance = 121 second bleed down to 16V with 36 Watts or .0135 A used.

400K with same V and uF = 242 seconds bleed down to 16V with 18 W or .006 A used.

73 John

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KD8DEY
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« Reply #7 on: January 28, 2012, 12:38:08 PM »

I remember reading that for good regulation etc that the bleeder resistance should be calculated to consume 10% of the overall current draw.

so it would be a simple matter of applying Ohms/Watts law.

Say the power supply was 3000v@ 1A

so you would calculate the resistance
3000/.1= 30K ohm
divide this by the number of filter capacitors
Lets say 6 capacitors to be bridged by the resistors
so that would be no less than 5000 ohms per resistor.( you dont want to go below this number for this example)
The closest standard resistor value would be 5.1 K ohms for an overall value of 30.6k ohms.
So using 30.6 K ohms@ 3000v would give you a current draw of .098 amps. Close enough for gubmint work.

At 3000v x .098 amps, The resistors would consume just over 294 watts.
Divide that by 6 and you see that each resistor consumes just over 49 watts of power without even factoring in a safety margin.

I remember reading that in a lot of electronics designs that a  20% safety margin is usually employed at a minimum. So you are looking at resistors rated for at least 60 watts+.....

BUT I COULD BE WRONG
(I'm just a General Lite)
« Last Edit: January 28, 2012, 12:41:40 PM by KD8DEY » Logged
KD6KWZ
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« Reply #8 on: January 28, 2012, 10:11:55 PM »

It's been a while, but I remember on old ARRL book had a 1 meg bleeder resistor across the filter cap in a HV power supply. How long it should take to bleed off has to be balanced against the load it creates.

I found this, FWIW:

http://in.answers.yahoo.com/question/index?qid=20101227221150AAPmdBS

The function of the bleeder resistor is to discharge the reservoir capacitor (C) so that after switch-off, with no load connected, the PSU will return to a reasonably 'dead' and therefore safe state in a reasonable length of time. If we take a 'reasonably dead' state to mean that the voltage (V) on the reservoir cap is reduced to 1/e (about 1/3) of its normal working value (Vo) and the time to get to this state is 10 secs, then we can say that the RC time constant should be 10sec. So R (value of bleeder resistor in Ohms) = 10/C (C in Farads)

For a cap of 10,000uF this gives R = 10/10^-2 = 1000 Ohms (as an example)

Both my 'reasonably dead voltage' (Vt) and 'time to reach this state' (t) have been chosen arbitrarily - you might prefer different values. So the value of the bleeder resistor could be different according to how you choose these quantities. In particular, if the power dissipated in the bleeder seems excessive, you might prefer a higher value, and therefore a longer time for the PSU to become safe after switching off.

For a more general approach, you can use the formula for the discharge of a capacitor to calculate the value of R in relation to C and t -

Vt = Vo(1 - exp(-t/RC))

Vt is your chosen voltage at time t, Vo is the normal working voltage on cap C when fully charged. This transposes to give -

R = -t/(C*ln(1 - Vt/Vo))
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G3RZP
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« Reply #9 on: January 29, 2012, 03:37:09 AM »

The separate bleeder is a good idea, because if one of the equalising resistors opens, you still have some bleed. Any switch or relay will need at least a 3kV rating, so I wouldn't bother. Use a physically big resistor rated for 3 kV, too. One thing I find useful is something like a 200K resistor from the electrolytic at the bottom of the chain to small neon indicator lamp - they do them with internal resistors for 120 or 240 AC, or you can get them plain.

Another trick is to put something like a 0.1 microfarad across the lamp so that it is flashing all the time the HV is on. A flashing lamp will attract your attention more readily than a plain 'ON' light.
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KB3TTP
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« Reply #10 on: January 29, 2012, 07:38:17 AM »

Alot of good advise and some good ideas to incorporate into a project. Thanks!

Here is a link to a good excel spread sheet to calculate a bleed resistor.

http://nilno.com/docs/BleedRes.xls

73
John
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W6RMK
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« Reply #11 on: January 29, 2012, 07:52:02 AM »



Or, you could keep it, but isolate it with relay. As long as the PS is powered, the relay's NC (normally closed) circuit keeps the resistor circuit open, thus it would not consume any power at all. When you turn off the PS power switch, the relay would release the armature and that would close the NC circuit, thus completing the resistor's circuit so that it can do its bleeding.
Unless done very carefully, this approach is unsafe. If the relay sticks, you've lost the safety function of the bleeder.  What you might consider is a slow (high value R) that's permanently across the C, and the switched low value R to quickly discharge the C.
Quote

Alternatively, find a higher resistance bleeder resistor. If you use two 75K 100W resistors in series, they'll bleed the capacitors in twice the time, but consume only half as much power (48 watts), well within the power rating of the resistors.

It would be prudent to have a way to completely short the capacitors automatically whenever the access door is opened. As we all know, some capacitors can keep a lethal charge for days, or longer.  I've worked with HV photoflash capacitors, and always shorted their terminals with a "chicken stick" shorting bar, or alligator-clip connectors, before working on the circuits they were in
You want to have some resistance in that shorting stick, to limit the current to something like 10Amps.  Big bangs and rapid discharge is very hard on the capacitors (and your nerves, and little pieces of molten metal flying about are just not a great thing)
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W6RMK
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« Reply #12 on: January 29, 2012, 07:59:06 AM »



The function of the bleeder resistor is to discharge the reservoir capacitor (C) so that after switch-off, with no load connected, the PSU will return to a reasonably 'dead' and therefore safe state in a reasonable length of time. If we take a 'reasonably dead' state to mean that the voltage (V) on the reservoir cap is reduced to 1/e (about 1/3) of its normal working value (Vo) and the time to get to this state is 10 secs, then we can say that the RC time constant should be 10sec. So R (value of bleeder resistor in Ohms) = 10/C (C in Farads)

For a cap of 10,000uF this gives R = 10/10^-2 = 1000 Ohms (as an example)

Both my 'reasonably dead voltage' (Vt) and 'time to reach this state' (t) have been chosen arbitrarily - you might prefer different values.


The usual guideline is that you want the voltage to be <50V after some reasonable time. Reasonable is usually determined by how long it takes to get access to the HV parts.  If you have to remove 17 screws using 3 different screwdrivers, you can have a longer time than if it's flip a latch and open the cover. 

From 2kV to 50V is about 3.7 time constants  (ln(2000/50)), so if your RC time constant were 10 seconds, after 40 seconds, the system is "safe".

For systems where you want a long time constant (to save power dissipation in the bleeder) and where the door isn't opened much (e.g. you rarely take the covers off, but you want the thing to be safe within a reasonable time after turning off the switch) one strategy is to have a "automatic" mechanical shorting bar on the HV.  Normally, you turn off the power, wait a minute, then open the door.  But if somebody forgets to wait, or whatever, and open's the door, the shorting bar dumps the charge (with a bang and a flash, usually.)  Yeah, you might need to fix some broken parts, but at least nobody is dead.
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KB3HG
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« Reply #13 on: January 30, 2012, 08:22:56 AM »

Any one look up the NEC( National Electric Code )  Code lately? There is or was a mandated discharge time. I seem to remember 60 seconds and 60 volts were the numbers, but its been years since I could rattle off the details. Any how I 'd look it up. It could save a life and it could be yours. 2700 volts is nothing to fool with.

The suggestion of a relay sounds good but it has to be a high voltage relay with some size, between dv/dt and 1/2 cv^2 that's a lot of impulse power on contacts.

Tom Kb3hg
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W8JX
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« Reply #14 on: January 30, 2012, 11:13:29 AM »

Anybody that is smart enough to work on a amp is going to know to wait several minutes before opening it up and use a shorting stick too before removing finals. "Code" is not idiot proof nor is it meant to be. Also most amps will also show you the plate voltage as it decays too if it had real meters. I see no need to bleed it so heavy that hundreds of watts go to bleeder power and unnecessarily loading power supply and adding to heat generation in standby too. 
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