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Old 17th February 2014, 08:05 PM   #1
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Default Capacitor Drain for Tube Bench Power Supply ?

I am building a 1200VDC 200mA bench power supply.

It has tube rectifier. I placed the variac (voltage adjuster) on the primary side of the power transformer.

What is the best way to automatically drain/discharge the capacitors ?
There are 4uf, 30uf, and 30uf caps.

If using a resistor between B+ and ground is the best way, what would be a good resistance value ?

Also, are there any problem with putting the variac on the primary of the power transformer ?

Last edited by HP8903B; 17th February 2014 at 08:15 PM.
Old 17th February 2014, 09:56 PM   #2
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Beware that, as there are no rectifiers whit sufficient heater cathode voltage ratings, you will need an auxiliary transformer that has sufficiently high insulating properties.

Might I suggest insulating the transformer by putting a thin phenolic sheet between it and chassis.

I wouldn't go tube rectifier anyway. I would opt for multiple toroid s whit 230 vac out, you can find caps for 450V easily. Just stack the supplies. And add a choke on the low side of the supply.

George (Tubelab) wrote about that, I believe the transformers are called industrial control transformers.
Old 17th February 2014, 10:09 PM   #3
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I am using 2 6CJ3 damper diodes.
Old 17th February 2014, 11:10 PM   #4
DF96 is offline DF96  England
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You need to decide how quickly you want to discharge the caps, and how much power you are happy to waste in the resistor.

Be aware that a powerful PSU needs some serious design work if it is to be both safe and reliable.
Old 17th February 2014, 11:19 PM   #5
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I want a good balance between discharge speed and power waste.

Please make recomendation
Old 17th February 2014, 11:20 PM   #6
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Let's say around 10 seconds from 1000V to 0V
Old 18th February 2014, 12:27 AM   #7
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imo, if you have to ask this question, you shouldn't be fiddling with 1.2kv supplies.
"It may not be easy for some to not hear differences, even if they are not there." - Vacuphile,
Old 18th February 2014, 12:38 AM   #8
Keit is offline Keit  Australia
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I concur with DF96 - a 1200 VDC supply @ 200mA DOES need some serious COMPETENT thought in order to be safe. DC is a heck of a lot safer than AC, but 1200V can still kill. More likely, the kick will cause your muscles to violently react, causing you to knock yourself and anything on the bench flying. Further, you need to think about the consequences of shorted rectifiers, wiring shorts, flasher-overs, and the like. If the capacitors are series electrolytic strings, you could have explosions. If the capacitors are series strings (to get the voltage rating high enough, AND the ripple current rating high enough)) you need shunt resistors anyway, to ensure equal distribution of voltage. The electros in any one series string need to be matched. Otherwise - explosions!

In fact, if you need to ask a question like this, you don't have sufficient competence.

Having said all that, firstly, the primary side is the best place for the variac - for a start it means a variac that you can actually buy.

Requesting a shunt resistor value to go from "1000 V to 0V in 10 seconds" is the wrong question. It doesn't work that way. The voltage in CR circuits decays exponentially - fast at first, then slower and then more slower. It is generallly considered that less than around 70 V DC is human safe, but to protect tools and semiconductors, you need it much lower. But start with 70 V anyway. To get 30 uF (which will give you approx 700V ripple at 200mA, full wave rectification of 50 Hz) from 1000 V down to 70 V, you need approx 3 time-constants of decay, CxR ~ 3 sec, i.e., R = 100 kohm. Power in the resistor at 1200V (ignoring ripple) will be 14.4 watts. This sensibly low compared to the power supply output of 1200x0.2 = 240 watts. Choose resistors 2 to 3 times bigger than the calculated dissipation, as if ever the resistor fails, you've got a SERIOUSLY dangerous power supply.

But the ripple means that you will need MUCH more than 30 uF, and that will mean intollerable dissipation in the ressitor. What I do in cases like this is choose a relatively high value of shunt resistor, to get from 1200 V down to 1000V in one second, plus a circuit that detects abnormally high ripple or voltage drop (ie a drop in voltage equal to just larger than the calculated ripple), and when it occurs, trip a shutdown that 1) turns off the AC supply, 2) puts a very low value resistor across the output, and 3) turns on a lamp to tell you it's now safe. Than protects humans, tools, and the supply itself from excessive loads and shorts, inside or external. It's important that the safe lamp turns on, not off, otherwise you could hurt yourself just because the lamp failed.


Last edited by Keit; 18th February 2014 at 12:54 AM.
Old 18th February 2014, 12:51 AM   #9
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The smooth stage is CLCLC.
All caps are rated at 1500VDV.
The 4uf is paper in oil.
The 30uf are oil with no pcb.
Old 18th February 2014, 12:52 AM   #10
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I am using an external variac.

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