I want to make a capacitor discharge tool for use in an industrial application where the voltage may be as high as 400 volts and I have no idea what the maximum capacitance may be. I have been weighing the pros and cons of different resistor values. Then I thought, Why not ask some of the best engineers on the planet? If you have some insights for me it would be appreciated.
John
I'm not too good with FETs - when I started out they were a bit exotic and I've never had to use them for anything.
A constant-current configuration would be like this, wouldn't it?
The FET shuts down when the gate gets to below about -6V (below the source), so that would regulate the voltage across the resistor to about 6V and keep the current at around an amp.
It would need a reasonable heatsink - initial dissipation is 400W (on 400V), although it comes down quite rapidly. The 6 ohm resistor would need to be 10W or so.
That SiC part is very expensive though. You could build a whole constant-current load with a MOSFET for that (including a processor to give you constant-wattage rather than constant-current and other nice things).
Hi Doug,
I will take a look at the UJN1205K. I would like to keep the build simple and I would like to have some sort of indicator of the voltage. Here is an idea that came to mind. Ideally I will be able to find a sensitive LED that will indicate at low current levels. I think this will indicate voltage down to less than 10 volts though I am away from the shop so I can't actually build and test. Voltage above approximately 2 volts will cause current to flow through the shunt diodes D1 - D3. The bridge will make the polarity of the discharge probes irrelevant. R1 will limit discharge current to less than 1 amp at 400 volts. Watching the LED go from lit to off will encourage the operator to leave the discharge shunt in place long enough for voltages to drop to safe levels. I was already thinking of adding the non-slip tips into the build. The larger 1.2 mm size tips would be ideal for ensuring that the probe stay in place and make good contact with the capacitor leads. Thanks for you suggestions.
John
Hi Jon,
As promised I have run some experiments after returning home. I have taken a 2000 uF 200 Volt capacitor and charged it up. The formula for energy in a capacitor is 1/2 CV^2 which works out to about 40 Joules for this capacitor. I built the circuit in my schematic above using a 1/2 Watt 470 ohm series discharge resistor.
As you can see from the color of the resistor it sustained heat damage after three discharges. Here is a video of the action:
A second test using a 1 Watt 470 Ohm resistor was able to handle the discharge without any problem. The resistor did heat up but was able to be touched after the experiment. I would estimate it got to about 45 degrees C.
We were originally discussing a 100 uF 400V capacitor like one we might find in a small switching power supply. The energy storage for 100UF at 400 V is E=1/2CV^2 or 8 Joules. I have no easy way to put 400 volts on a 100 uF capacitor so I decided to simulate this by putting only 8 Joules of charge into the Capacitor in the above experiment. This works out using the formula V = SQRT(2E/C) or about 126 Volts. I will run this experiment using a fresh 1/2 watt resistor. Keep in mind this is an imperfect match as 400 volts from 100 uF would discharge the 8 Joules of energy much more quickly than our 126 volts from the 1000 uf capacitor. Here is a video of the test:
My conclusion is that the design in the schematic works well and while the 1/2 watt resistor would probably work for the 100 uF 400 V situation it would not be adequate for the 2000 uF 200 Volt capacitor. Jon's suggestion to use a higher wattage resistor is confirmed as the best way to go.
John
