discharging caps

Well 200V and 200mF is 4kJ.

So not too much heat from, say, a kettle or hotplate element. Or an array of 20 incandescent lamps to add a nice warning display. Incandescents are a bit constant-powery too.

O O OOO OOO OOO O O O O O OOO O

Keeps it simple.

Motor generator? Ed

Have you ever shorted a kilojoule joule of capacitors with a screwdriver? It's an impressive explosion.

It's looking like we might actually have 0.12F at 150v, which is merely 1300J.

Incandescent is interesting. Running red hot, they would last forever.

One would have to consider startup, when they are cold. Don't want to hang up the power supply. So I guess we prefer constant-currrent, not so much constant-power.

AI says that an incandescent cold resistance might be 1/10 of hot. That might be OK.

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

Maybe not available in surface mount. And the heat still needs to go somewhere.

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

It sounds as if you want to detect when the AC has gone off, and only then discharge the capacitors.

Making the circuit too dumb wouldn't be a good idea

Putting a big power MOSFet in series with a big 100R resistor, and only turning the MOSFet on when the AC power has gone off, would be adequate. You'd need a battery to keep the MOSFet on for long enough to do the job.

Putting an inductor in series with the resistor could give you a faster discharge, but getting hold of an inductor that was big enough for the job could well be impractical.

When I was young I put together some linear power supplies for biggish arc lamps. We got some large metal cased power resistors and mounted them on big heat-sink extrusions and dissipated a couple of hundred watts indefinitely without even having to bother with fans.

Motor alternator - pump the energy back into the mains.

What sort of design needs 0.2 Farad cap, at 200 volts ?. If you are working at that level, put in a cheap relay and a rated heatsink wirewound resistor to dump the energy, when the power goes off. Cheap, and lossless as well. Motor drive inverters often have big resistors to brake the motor.

You could add some series resistance, of course.

It's a 1500 amp laser driver.

The input to our box is DC, from an external power supply.

I do want a circuit that's foolproof, that always discharges the caps but doen't often go up in flames.

A resistor makes an exponential decay which could be a very long time to get down to safe levels.

I thought the group might like a circuit design problem once in a while, a break from politics.

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

We get DC power and don't have access to the AC.

We want dumb. A smart circuit could make a mistake and not discharge, or might stay on and start a fire.

And it's fun to design dumb circuits. That's often more difficult than designing complex ones. It takes more thinking.

Size of my car maybe.

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

is a relay (or two) controlled by the AC not dumb enough?

maybe use a PTC self limiting heater to dump the energy in

We get DC power, and I wouldn't like the relay to fail anyhow and start a fire.

A PTC might work. It would sit there and get hot all the time and go sorta constant-power as the caps discharge. Maybe some PTCs and some series resistors, to not spike to a zillion amps at startup.

The DC power supply might not start up if it had a true constant-power load.

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

The load-dump resistor of a fairly large CNC milling machine was controlled by a power FET; twice in two years the FET failed and the resistor, which was in a cage on the top of the control cabinet, ran red hot. A temperature sensor inside the cabinet eventually shut the machine down.

Could you have a current-operated relay in the incoming supply with normally-closed contacts that bring in a contactor for the discharge current? The contactor could be supplied by the power it is discharging and would automatically drop out when the voltsge reached a safe level.

Big resistors are quite cheap to make from slate bars and resistance wire. - much cheaper than banks of metal-clad off-the-shelf devices and heat sinks.

Maybe. But any failure mode could start a fire.

Steady-state, the cap charging current can be zero.

We could mostly discharge the caps in two minutes by dumping a couple hundred mA, which we can do with maybe five 10-watt wirewound resistors. But an exponential decay can still leave bang-level charge in the caps for a long time. Big 'lytrics will also recharge themselves after you think they are discharged.

I think I have a circuit that will work, but I'd like to hear some other ideas.

John Larkin Highland Tech Glen Canyon Design Center Lunatic Fringe Electronics

When the mains fail, how can I discharge the caps?

Yeah I noticed that a long time ago.

I've used relays and big resistors to do it in the past. When the relay is energized the resistor is disconnected. I haven't read the entire thread so this has likely been covered.

If a stuck relay is a fire risk then maybe include a temperature sensor and a mosfet or another relay to disconnect when over temperature. What's the chance of both failing?

An audible warning may be useful if there's a fire risk. I'm hearing an AI generated voice saying "Maximum temperature exceeded, please unplug or turn off power now".

The traditional emergency discharge circuit for a supercon magnet like you used to make coil drivers for is two stud mounted diodes rated for the magnet current wired across from each other and mounted on a heatsink. Two diodes because you never know which way the last person to charge the thing connected the + and - leads :-). Switching was manual, of course, but it gave a constant power dump down to the diode cutoff.

Cheapest way is Darlington current source driven by a multiplier error amp taking voltage input from the cap and current input from the Darlington emitter current sense. The multiplier only needs to be one quadrant, eliminating the need for that overpriced ripoff AD633 and its equivalents. That gives you three options: transconductance variable gain amp, digital PWM multiplier using something like 4066, or a log-antilog using a quad opamp LM324 type. The quad opamp should be the least trouble. There are plenty of circuits in the old NatSemi app notes, nothing extreme by way of diode matching is necessary for this application.

When the voltage across the cap gets down to 10V, switch out the MOSFET and switch in a resistor. Current will be 2A by then.

The constant power components should come in at under $10 (small quantity). Those linear MOSFETs are another major ripoff too, and bipolar is perfectly adequate for this purpose, and cheap.

When I was in my "build huge hifi phase", I would put the power supply in one box (7U) and the amp in another.

The power supply needed something to ensure slow turnon (you don't want to apply 170V directly to a huge capacitor bank!). And, similarly, something to ensure the caps discharged and REMAINED discharged.

Can I introduce you to my lifelong friend, Murphy? :>

If the device is (and will always be) attended,this isn't usually too much of a problem. When the device is UNattended, then you have to be more aggressive in your protections.

I've met him but both a stuck relay and a shorted mosfet at the same time isn't usually his thing unless the mosfet/relay was underrated for the job.