Capacitor Discharge Circuit with Energy Recuperation

[luky315]

I have another HV-Problem:
A capacitor (47µF) is charged to 100V. I need a circuit to discharge it in less than 5ms. My problem is that the circuit must be small and I can't simply dissipate the Power (for thermal reasons). The discharge Current must also be limited to minimize EMV-Problems. There are a lot of circuits for this problem which reuse the energy to recharge the capacitor (widely used for actuator drivers) but I don't need to recharge the capacitor, therefore I need a solution to use the energy for something else (powering the circuit or "putting it back" to the DC Power supply).
Any Ideas?

[Psi]

There aren't many options left if you eliminate heat.

motors/solenoids, light, sound,

Sound and motors/solenoids isn't going to be small and light isn't very efficient so you're back to heat.

[luky315]

I can't dissipate the heat on the PCB. The idea is to give the energy back to the DC input which powers the whole system including the charging circuit.

[jeroen74]

A buck converter with a slightly higher output voltage then the input voltage of the system.

[Simon]

I'm afraid we can't break physics for you, you can't "vaporize" power in 5mS and make no heat. What is your power limit ?, just shorting the cap with a mosfet or similar sounds like the best option to me. What are you actually doing ?

[mikeselectricstuff]

If you want to get exotic there are some phase-change materials that will absorb heat by turning from solid to liquid, and release it over a longer period of time.
However a single discharge of 47uf 100V - about a quarter of a joule - doesn't seem like too much heat to deal with if it doesn't happen very frequently . Compared to the volume of the capacitor, you wouldn't need much metal to store that & let it out more slowly - maybe a to220 sized MOSFET. Just need to make sure your heat-absorbing device has sufficiently low transient thermal   impedance.

Also, think about whether you actually have to discharge it all the way to zero in the 5ms.

[luky315]

I have to discare it frequently,  to 0V and I don't want to "vaporize" the power.
Just give it back to the main power rail.

[Simon]

what is the main power rail ?, by vaporize i mean dissapere without trace, which is what your almost asking, nothing can be destroyed, everything is transformed. Even if you can feed this "power" back to the input that won't discharge the capacitor to 0V

[jeroen74]

without some more details its impossible to help you. What is frequently? One per second, once per hour?  What is the main rail voltage? How much power does the device use? What exactly is this HV cap used for? Why charge it up, then discharge? That's like filling the bathtub with the tap running and filling a bucket at the same time from that same tap, and when the bucket is full dump it into the bathtub 

A similar application: old flash units that measure actual exposure use a very small flash tube to dump the excess energy in the main cap when enough light has been produced by main xenon tube. That will only reduce that cap voltage to 50V or so when the tube stops conducting, but at least you have lost 75% of the energy.

[mikeselectricstuff]

I have to discare it frequently,  to 0V and I don't want to "vaporize" the power.
Just give it back to the main power rail.

Will whatever is providing the main rail deal with it? What happens if it has a light loat - can it tolerate a high voltage spike?
if so then the only way would be a boost converter (could be a very crude, loosely regulated one). You will need significant space for an inductor though. Minimum hardware would be a switching transistor, diode, inductor and source of frequency to drive the transistor. You'd probably want to sweep the duty cycle to even out the current over the 5mS. You'll probably also want something to stop it if the rail voltage gets too high to prevent killing the transistor.

[G7PSK]

Sounds like a perpetual motion device.

[luky315]

The main power comes from a big 24VDC industrial Power supply with big output capacitors.
The discharge Frequency is 100Hz (5ms to charge, 5ms to discharge). So yes, heat is a problem.
And no, it's not a perpetual motion device but a test circuit for the charging unit. I tried with a simple circuit but it won't work for very long because of the thermal issues.

[Simon]

if it's a test circuit why is the heat a problem. use suitable parts and you will be ok

[luky315]

I have a very limited space and almost no air (vacuum!). The main power comes from the outside. I can't dissipate the heat inside the system.
I won't be asking for a circuit for energy recuperation if a simple Mosfet would solve the problem...

[Simon]

then take a pair of wires out of the box to discharge the cap from with a mosfet on the outside. remember what I said at the beginning ? you can't break physics. you are asking for the power to be made to vanish, can't be done. Do you need to discharge to 0V ?

[luky315]

I don't want the power to vanish. I want to "give it back" to the main supply rail.
I tried a lot of "simple" solutions but they don't work in my case.
I need a circuit to recover the energy. Nothing else will work in this case.
And I said already that I need to discharge it to 0V.

[Simon]

I don't want the power to vanish. I want to "give it back" to the main supply rail.



And I said already that I need to discharge it to 0V.

The two above statements make your requirement impossible, you have set a specific discharge time and want it to fully discharge the capacitor but you also want to "take the time" to reprocess the power and if your power input is 24V you won't get the capacitor down to 0V as there will be a minimum voltage for the buck/boost circuit.

I'm no expert of buck/boost circuits but i think you need to get it down to about 10-24V and then short the capacitor for the remaining voltage, you need to design a circuit to cut in the buck/boost converter at the right time and cut it off again. you also need to know that the power supply has enough load on it to use that power fast enough and your output voltage will need to be slightly higher than then main input voltage so that it takes over in order to discharge the capacitor fast enough.

Good luck

[jeroen74]

What supplies the energy to the cap?

[mikeselectricstuff]

I don't want the power to vanish. I want to "give it back" to the main supply rail.



And I said already that I need to discharge it to 0V.

The two above statements make your requirement impossible, you have set a specific discharge time and want it to fully discharge the capacitor but you also want to "take the time" to reprocess the power and if your power input is 24V you won't get the capacitor down to 0V as there will be a minimum voltage for the buck/boost circuit.

I'm no expert of buck/boost circuits but i think you need to get it down to about 10-24V and then short the capacitor for the remaining voltage, you need to design a circuit to cut in the buck/boost converter at the right time and cut it off again. you also need to know that the power supply has enough load on it to use that power fast enough and your output voltage will need to be slightly higher than then main input voltage so that it takes over in order to discharge the capacitor fast enough.

Good luck

A simple boost circuit could achieve 0V without additional hardware as you have a DC path through the inductor and switch. Remember you still have the incoming supply available to power the converter.

[Jovian]

If you want to put the energy from the capacitor into the 24V rail then you will need a buck converter. Then once you have the capacitor down to say 30V you can dissipate the rest with a resistor. But 5mS is not long, and you are going to have to dissipate about 50W on the 24V rail. Will the power supply mind that it is supplying a load that effectively pulses at 100Hz?

If you are testing a charger why use a capacitor as the load, unless you want to test high current impulses, which could easily be emulated with a power transistor (or multiple in parallel depending on the current). Could you use a resistor as the load, or put a resistor in series with the cap so that you have much less energy to deal with? (Remember the energy stored in a capacitor is proportional to the voltage squared.)

Regards,
Jovian.

[Simon]

The two above statements make your requirement impossible, you have set a specific discharge time and want it to fully discharge the capacitor but you also want to "take the time" to reprocess the power and if your power input is 24V you won't get the capacitor down to 0V as there will be a minimum voltage for the buck/boost circuit.

I'm no expert of buck/boost circuits but i think you need to get it down to about 10-24V and then short the capacitor for the remaining voltage, you need to design a circuit to cut in the buck/boost converter at the right time and cut it off again. you also need to know that the power supply has enough load on it to use that power fast enough and your output voltage will need to be slightly higher than then main input voltage so that it takes over in order to discharge the capacitor fast enough.

Good luck

A simple boost circuit could achieve 0V without additional hardware as you have a DC path through the inductor and switch. Remember you still have the incoming supply available to power the converter.

ok but surely at near 0V that will happen slowly ? can all this happen in 5mS ?

[G7PSK]

Use the heatsink of the switching transistor as the discharge resistance, paint it black and it will radiate the heat in a vacuum. Just give up on trying to recover all the energy,surely it cant be that important as it is just for testing purposes.

[Simon]

that is what I don't understand, if this is a test why all this fuss

[free_electron]

pumping energy back into the rail isone thing. it needs to be consumed there as well. if there is no load on the rail that can handle this energy than nothing happens....

if you are in vacuum your only option is to radiate it off as photons. dump the excess energy in a lightbulb... or a bunch of led's. the emitted light is effectively energy being wicked away...

[KedasProbe]

Does it have to be recovering the energy in every cycle or would it be OK to load to somewhere else and then later with a much lower frequency give it back to the rail.

Connecting a 10 times bigger capacitor parallel will drop the voltage to 1/11 without much loss of energy. The last bit of load on your cap (like 10%) you can decide to lose in heat to get to 0V.
You can mux between a few large caps and feed them later together to the rail or battery. (maybe connect them in serie)
basically a DC-DC converter.
Theoretically a variable capacitor would allow you to play with the voltage without changing the load.

Just an idea not really sure about the schematic.