Hi
I have this circuit. It charges a capacitor, and then discharges through a coil. All works, except that after the discharge there is a negative voltage of about 50V across the capacitor, which gradually dissipates to zero. I feel this isn't doing the capacitor any good. I guess this voltage is being produced by the coil, and the diodes are there to protect the thyristor, but what more can I do to protect the capacitor?

 

A diode between the capacitor and the coil.

 

I guess this voltage is being produced by the coil, and the diodes are there to protect the thyristor, but what more can I do to protect the capacitor?

Thyristor does not need protection, because discharge current not changes its direction.
Diodes should protect capacitor from reverse voltage.
If you see negative voltage on capacitor, then diodes are destroyed (because about 100 A current).
May be you will interested in this information:
https://forum.allaboutcircuits.com/...-gauge-for-pulses-current.189436/post-1768216
ADDED:
Instead 1N4007 may be used one diode MR760.

 

Ah OK thanks

 

A diode between the capacitor and the coil.

How can that prevent the reverse voltage on the capacitor?

The diodes to ground already there should do that, but they are apparently blown.

 

A diode in series would block reverse charging as ling as the reverse voltage is not higher than their breakdown. The existing diodes did not do that because they are positioned wrong, the should be directly across the coil.

Am I wrong?

 

Am I wrong?

Yes.
The inductor inductance causes the current to continue to flow in the same direction through the SCR after the voltage across it (and the capacitor) drops to zero, creating a reverse voltage on the capacitor.
Thus the diodes across the capacitor (and effectively the coil since the SCR is still on) should conduct that current and prevent the capacitor voltage form going more than 1V below ground.

 

The N channel MOSFET cannot work to charge the cap unless the charging signal is over 300V. It is hard to know what he really tested and what really happened. As you said, the diodes should have prevented the reverse charging, or at least limited it. And that did not happen.

Here is how I understand the operation, assuming he managed to get 300V on the capacitor somehow:

During the discharge (SCR on), those diodes are reverse biased and conduct no current since their reverse voltage is higher than 300V.

When the current falls below the the holding current of the SCR, it shuts off and the voltage on the inductor is reversed, but limited to 1 diode drop by the four, now forward biased, diodes. So the diodes only need to be able to conduct the minimum holding current of the SCR.

If this is right, the diodes should not have failed, and four of them was completely unnecessary, right?

If you agree, we still don't know why the cap became reverse charged.

 

When the current falls below the the holding current of the SCR, it shuts off and the voltage on the inductor is reversed, but limited to 1 diode drop by the four, now forward biased, diodes. So the diodes only need to be able to conduct the minimum holding current of the SCR.

With and without diode:
______

 

With and without diode:
View attachment 322814 ______View attachment 322809

Thank, Danko, now I get it.

 

When the current falls below the the holding current of the SCR, it shuts off and the voltage on the inductor is reversed

When the current falls below the holding current of the SCR there is very little inductive energy left and so little inductive kickback (which appears across the now turned-off SCR).

 

When the current falls below the holding current of the SCR there is very little inductive energy left and so little inductive kickback (which appears across the now turned-off SCR).

Checked and cured:
____

 

When the current falls below the holding current of the SCR there is very little inductive energy left and so little inductive kickback (which appears across the now turned-off SCR).

Yes, I had totally missed the fact that the current decreases as the cap discharges, which creates the negative voltage. Danko’s simulation cleared it up for me.

So the diodes do see a very large current, blowing them up and moving them parallel the inductor does not help.

So something is needed to dissipate the energy before the voltage becomes so negative.

 

OP here
Thanks for all the thought. Is is it agreed that a higher current rated diode across the capacitor will solve the issue?
(I think that originally the reverse voltage did not occur, which does suggest that the diodes were gradually destroyed)

 

Is this for a rail gun?

 

No. It's for a magic trick, in which something is made to jump "magically"

 

When the SCR is triggered the coil and capacitor are in paerallel, forming a resonant circuit that is excited by that current pulse. So until the SCR switches off the circuit will ring and only stop when the SCR switches off. Which is fast but not instant.
To pprotect the capacitor, if it needs protection, an SCR short circuiting it to trigger as soon as the voltage passes thru zero would work.Probably there will need to be a resistor in series, 100 ohms and a few watts.

 

Hi
I have this circuit. It charges a capacitor, and then discharges through a coil. All works, except that after the discharge there is a negative voltage of about 50V across the capacitor, which gradually dissipates to zero. I feel this isn't doing the capacitor any good. I guess this voltage is being produced by the coil, and the diodes are there to protect the thyristor, but what more can I do to protect the capacitor?
View attachment 322768

As drawn the diodes will clamp the voltage to some smallish (I need to simulate it) voltage. The forward conduction time of a 1N4007 is about a 1 us an there may be a short negative spike. 50 volts does seem excessive.

The circuit does want to swing the cap voltage negative and then ring because it is underdamped. To keep the voltage across C1 from swinging negative and/or ringing the circuit can be damped with a series resistance of 2 ohms, give or take. The resistor can be placed anywhere in the L-C-T1 current path.

R = (L/C)^0.5 R = the square root of L/C

 

Of course, I had never really thought about it oscillating, but I can see that it will. Will put in a resistor