Power engineering (pulse power) question 3

[Alue]

I have a simple circuit designed to discharge a large capacitor in a short amount of time. The latch I use to discharge the energy stored in the capacitor is robust and can handle well above the amount of current and voltage I am discharging. My problem is, when I connect a short load to the output the switching device is destroyed. I think this is caused by flyback voltage (the only thing between the switch and the load is an inductor). My question is how do you protect against this without reducing the amount of energy discharged or increasing the duration of the discharge too much?

 

[Marmite]

What do you mean by "short load"?
Regards
Marmite

 

[ScottyUK]

There's a duplicate thread over at thread248-292116. Can we pick one thread or the other and answer there?

Alue - please don't start multiple threads on the same topic. It fragments discussion and irritates everybody.


----------------------------------

If we learn from our mistakes I'm getting a great education!

 

[Skogsgurra]

If your switching device is destructed, I think it must be a semiconductor device of some sort. Mechanical switches usually aren't destroyed.

If you use a thyristor, there is an important parameter called current rate-of-rise. If your current rises faster than allowed, the conducting zone doesn't have time to spread over the whole semiconductor surface and the silicon will overheat and melt.

I do not think that you have a problem with inductive kick-back.

Scotty is right. It is a nuisance to have a discussion in more than one thread. Red-flag the other thread.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[ScottyUK]

Post a circuit diagram with part numbers and/or ratings. Include resistance of the inductor.


----------------------------------

If we learn from our mistakes I'm getting a great education!

 

[electricuwe]

Gunnar,

no problem to find capacitor banks that will easily destroy every mechanical switch and also you can design semiconductor switches to withstand quite severe discharges:

http://www.hzdr.de/db/Cms?pOid=21864&pNid=473

 

[Skogsgurra]

Right as usual, Uwe.

But if the guy thinks it is inductive kick-back, then I think he is using thyristors. And then, if he doesn't know about di/dt ratings, I thought it was a good idea to tell him.

Wanted to build up some interest first.


Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[Alue]

I'm sorry about the double thread, my accident.
Skogsgurra you are correct the switch is a thyristor. The peak di/dt of the switch is 30 kA/?Sec. The capacitor is 10?F charged to 3kV discharging into an inductance of 18mH. I don't remember the resistance, it is relatively low, but I will have to double check that. The inductor at the end is not imperative.

 

[Compositepro]

If there is no resistance in the circuit where will the energy of the capacitor be dissipated except the switch? Of course each component will have some resistance but the switch is likely to be the largest part. So is the switch capable of absorbing this amount of energy in a small fraction of a second?

 

[Skogsgurra]

Are you sure that the inductance is 18 mH? That's a very high L.

The maximum possible di/dt is no more than 17 kA/s, which is quite low. And very low compared to your spec (30 000 000 kA/s). The latter is, on the other hand, an extremely high value. Are you sure about the order of magnitude?

If the numbers are correct, then the current will probably continue through zero and you will have the reverse voltage that you mentioned in your first post.

A reverse diode parallel to the thyristor is what you need. With that high inductivity, I do not think that it needs to be very fast. If you see high reverse peak voltage also after adding the diode, a snubber can take care of that.

Yes, you need the resistor. If you leave that out, the thyristor will discharge the snubber capacitor and kill itself again.



Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[Alue]

Skogsgurra
I am sure about the maximum di/dt it's a Heavy Duty thyristor. The inductance I will have to double check on.
Excuse my dumb question, but where are you getting the 17 kA/s maximum possible di/dt from?
When I say 'the inductors at the end is not imperative', I mean that I can change the values or change it out as needed.
Thanks a lot for your help!

 

[PHovnanian]

Alue,
It sounds like (absent a schematic) what you've got once the thyristor fires, is an LC tank circuit. That's going to ring. Or at least it will want to until the thyristor turns off at the zero crossing. Then, several interesting failure modes could occur.

 

[Skogsgurra]

The maximum di/dt is the rate of rise when you connect 3000 V to an 18 mH inductor. Since E=-L*di/dt, you then have 3000=.018*di/dt, which makes initial di/dt equal to 3000/.018 and that is pretty close to 17 kA/s.

Re. the thyristor di/dt, I still wonder if your numbers are correct. This thyristor (

http://semtexinternational.com/SDiv/SPL/TST/HPSI/T36A7KV.pdf

) is a high power thyristor with a high di/dt rating. Still, it is limited to 300 000 kA/s. that is one thousandth of your rating. I think you need to have a closer look at your number of zeroes. Not that it is a problem - your 30 kA/us is more than you need. It is only a question of what is possible and what is not possible. The 30 kA/us is not possible.



Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[Skogsgurra]

Thanks! I have to reconsider what is possible.

Thanks.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[LionelHutz]

You need to read what PHovnanian has posted again. You can't dissipate the energy stored in the cap by discharging it into an inductor.

 

[electricuwe]

I only can agree to the recommendation by Skogsgurra:

Add a diode to protect the thyristor when the voltage reverses

 

[LionelHutz]

A diode doesn't allow the capacitor to stay discharged the first time. It would just allow the circuit to resonate and the cap would go through a few discharging and charging cycles before finally staying discharged. I believe you want to rid of most of the inductance in the circuit and use a suitable resistor.

 

[Skogsgurra]

That is not possible, Lionel.

A resistor does not limit du/dt. The inductor is probably the coil where the pulse is used to magnetize a workpiece. Or it may be the coil in an electric gun. Or some other use. Getting rid of the inductance is not a solution.

It is also not necessary to fully discharge the capacitor. The cycle is as follows: Charge capacitor (may take seconds), fire the thyristor (capacitor discharges in short time, 'millisecondish' or less), current in coil rises, falls and reverses and charges capacitor backwards 'somewhat'. The capacitor is then charged from the HV supply and the next cycle is ready to begin.

If the capacitor is polarized, it is an easy matter to add another diode so it will not charge in reverse.

Energy will be dissipated in resistance in thyristor, wires, coil, capacitor and diode.

Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[LionelHutz]

I can't read minds. The OP posted "a simple circuit designed to discharge a large capacitor in a short amount of time".

In a perfect L-C ciruit, the capacitor would be first discharged to zero voltage. At zero voltage, the current and the magnetic field in the inductor has both reached their peak levels. So, the magnetic field in the inductor will start to collapse which forces the current to continue to flow. This re-charged the capacitor to the same voltage level but opposite polarity. At this point, the magnetic field and current in the inductor are both zero and the thyristor will turn-off and not allow any reverse current to flow to discharge the capacitor. You have to remember, the thyristor will conduct until a current zero is reached, not until the voltage zero is reached. There can be a big difference between the two.

In reality, you have series resistances which creates a R-L-C circuit. Some energy is dissipated in the resistive element meaning the reverse voltage on the capacitor peaks at a lower level than the original starting voltage.

I don't see any "flyback" or "kick-back" issue requiring a diode here. There should be no energy left in the inductor to cause a destructive voltage surge. Still, the capacitor isn't left in a discharged state.

I have to question the gate signal the OP is using. He needs to hit the gate fairly hard to achieve those good di/dt capabilities. Try a gate circuit that is around 30V open circuit and will source 1A when shorted.

I also wonder if there is some stray capacitance in the circuit causing a higher frequency oscillation that rides on top of the discharge current. This could cause the thyristor to rapidly commutate on and off a number of times when it's first triggered and some thyristors don't take kindly to that kind of abuse.

 

[electricuwe]

Thyristors are not as simple as you believe, LionelHutz.

The main disadvantage of the circuit you suggest, is that after the pulse the capacitor is charged in the wrong direction.

However, if you want to apply it regardless of this disadvantage, you should take care that there is a snubber provided for the turn-off of the tyristor.

After the current has crossed zero the thyristor will exhibit a reverse recovery peak. This reverse recovery peak current will stop after recovery and you need the snubber to allow further current flow without destruction.

A freewheeling diode, either antiparallel to the thyristor or antiparallel to the capacitor will relief the thyristor from the reverse recovery.

Two recommendations to Alue:
- You would have avoided lot of guesswork, if you had posted your circuit and the components datasheet in the initial post
- There is lot of literature available on using thyristors in pulse power circuits, from IEEE Xplore library as well as from suppliers like ABB or Infineon (the two biggest players in the thyristor business)