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Device selection to discharge HV capacitor

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Zak28

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Trying to find a means to dump 20juels @ 400v out from HV capacitor immediately after the 5th time constant. I insist on some sort of semiconductor it seems unlikely however I find relays most sensible to incorporate. Perhaps there are capable thyristors which are not huge railroad modules?
 
Last edited:

Hi,

Is this homework?
I've never seen such a specification.

Especially "after the 5th time constant". This is a relative expression. It can be microseconds or hours.

Time constant says you need a resistor, that dissipates the energy. Is the resistor value known?

Tell us more..What's behind this...

Klaus
 

Conventional definition says that a capacitor reaches full charge (99.3%) after five RC time constants.
Click to expand...
Or discharge to 0.7 % of initial voltage. That's clear, but the OP didn't say what the time constant is.
 

KlausST said:
Hi,

Is this homework?
I've never seen such a specification.

Especially "after the 5th time constant". This is a relative expression. It can be microseconds or hours.

Time constant says you need a resistor, that dissipates the energy. Is the resistor value known?

Tell us more..What's behind this...

Klaus
Click to expand...

Put simply to dump a cap rated 400v right about when its terminal voltage reaches ~99% rated voltage hence I wrote 5th time constant. Seems like triacs are capable although their ability to immediately dump 30A @ 2uS pulse-width seems unlikely.
 

A relay is about the worst choice for capacitive switching, due to contact fusing. If the peak current is just 30A, then a MOSFET or IGBT should work fine.

What is the capacitor ESR?
 
Last edited:

Hi,

I´d say about every 400V rated relay could be used.

Lets say one with 10mA rating. Add a 1M resistor in series and it discharges the capacitor to less than 1% with 5 time constant. (with only 0.4mA of current)

****
The question simply is not elaborated.

No device without additional circuitry (or manual connection) will work.
Neither a relaycontact nor a triac is made to dissipate 20 joules of energy.


Klaus
 

Seems like triacs are capable although their ability to immediately dump 30A @ 2uS pulse-width seems unlikely.
Click to expand...
Guess you didn't read datasheets. Pulse handling of moderate size triacs and SCR is much higher.

The required discharge time is still unclear. Where did you get the 2 µs? Why not 2 ms or 20 ms? Why 30 A?

20 Ws at 400 V is 250 µF. It would be discharged with 250 µs time constant through a 1 ohm resistor, peak current is 250 A. Respectively the peak current with 2.5 µs discharge time is 25000 A, presumed the capacitor ESR is low enough to provide it and the capacitor doesn't explode during discharge.

So again what is the required discharge time, what's the intended discharge current? What's the capacitor ESR and pulse current capability?
 

Hi

Some calculations:

400V and 30A means 13.3 Ohms. With a peak power of 12000W. (This is more than a standard 0207 THM resistor will withstand)
20J = 20WS = 250uF @ 400V

So now the time constant is 13.3 Ohms x 250uF = 3.33ms.
Now select a typical Triac. Let´s say BTA08-600.
Hold current of 10mA worst case.
10mA with a peak of 30A means 1/3000 = 0.033%.
Discharge to 0.033% of initial current means 8 time constant
8 x 3.3ms = 27ms.

The calculation shows: Once the triac is triggered you have to wait at least 27ms for the triac to safely relase.
If you power up the capacitor within the 27ms .. and the triac has not yet released.. funny things happen.

Lowering the current = increasing resistance = increasing time constant = longer time for the triac to release. (0.4A --> 1kOhm --> 925ms)

Klaus
 

KlausST said:
Hi

Some calculations:

400V and 30A means 13.3 Ohms. With a peak power of 12000W. (This is more than a standard 0207 THM resistor will withstand)
20J = 20WS = 250uF @ 400V

So now the time constant is 13.3 Ohms x 250uF = 3.33ms.
Now select a typical Triac. Let´s say BTA08-600.
Hold current of 10mA worst case.
10mA with a peak of 30A means 1/3000 = 0.033%.
Discharge to 0.033% of initial current means 8 time constant
8 x 3.3ms = 27ms.

The calculation shows: Once the triac is triggered you have to wait at least 27ms for the triac to safely relase.
If you power up the capacitor within the 27ms .. and the triac has not yet released.. funny things happen.

Lowering the current = increasing resistance = increasing time constant = longer time for the triac to release. (0.4A --> 1kOhm --> 925ms)

Klaus
Click to expand...

Load being an inductor ~0.9 Henry @ 0.4Ω discharging cap into coil will be 400v 1mF myself sees an issue since because 1st constant yields current ~0.3A however simply the sheer size of the cap causes me to believe semiconductor will explode or incur damage. The whole object this circuit has to do is discharge cap thru load immediately after cap has regained ~99% its voltage rating.
 
Last edited:

Hi,

Another little piece of information.
For me: Still far away from knowing what the OP wants to achieve. It seems we can wait another countless posts until the information is complete. I don't want to assume, guess, ask, try, calculate ... that long. If the OP isn't interested in giving full information...he isn't interested in answers. Therefore I'm out of here.

Klaus
 

Load being an inductor ~0.9 Henry @ 0.4Ω discharging cap into coil will be 400v 1mF
Click to expand...
So stored energy is 80 rather than 20 Ws.

Looks like you never analyzed the expectable voltage and current waveforms. The 1mF 0.9H 0.4 ohm circuit is a resonant circuit with little damping. Discharging the capacitor through a SCR generates a sine halfwave current which stops at the current zero crossing and a capacitor voltage of nearly -400V.

discharge.png

Currents are moderate can be well handled by semiconductor switches. Of course the capacitor must be a bipolar foil capacitor, quite bulky with 1 mF.
 

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