I have a capacitor rated 15000 µF, 35V.

So, I charge my capacitor to 15V with no problems using a charging resistor. Now while discharging, I am using thyristor (2N6509G) as my switching device and a coil as my discharge resistor. Coil resistance is about 2 Ω.

Using the discharge equation of capacitor, I calculated the discharge time in this setup, which comes out to be around 81 ms for (assumed) final voltage of 1 V.

In my application, I wish to discharge the capacitor only for 20 ms. But when I do that, the capacitor voltage drops down to 0.9V in just this 20 ms duration instead of the calculated 7.7V (using discharge time calculators online).

If I am not doing anything wrong in the setup, what is it that I am not understanding conceptually? How do I go about discharging capacitors only to some fraction of the stored charge?

Edit: Thank you all for your comments, I am sorry for not being very clear with the details.

So, this is a coil gun model basically. Coil :- Copper magnet wire: AWG 20; no of turns: 650; Coil resistance: 2ohms

I actually used an online calculator for capacitor. Discharge Time Calculator

I used another calculator to get the inductance of this coil which came out to be 8.88mH. Coil Calculator

Discharging side of the capacitor

*Note: This circuit is not my original work.

  • 1
    \$\begingroup\$ Can you add the formula you used to calculate the discharge time? That would make answering this a lot easier. Also, what kind of coil? Why not simply a resistor? Are you aware of inductivity? Please add a schematic; I'm not 100% sure I'm on the same page as you regarding how you're using the thyristor. \$\endgroup\$ Commented Jul 26, 2017 at 21:26
  • \$\begingroup\$ 15000 uF is really 15 mF. Please use grown-up units. \$\endgroup\$ Commented Jul 26, 2017 at 21:52
  • 2
    \$\begingroup\$ With a thyristor you are not able to control the discharge, once you have triggered it, it has completely discharged the cap. \$\endgroup\$ Commented Jul 26, 2017 at 21:56
  • \$\begingroup\$ Voting to close as unclear since OP hasn't added a (direly needed) schematic nor any info on the coil to the question, though being asked. \$\endgroup\$ Commented Jul 26, 2017 at 22:14
  • \$\begingroup\$ she's probably having a hard time finding a 10W 9.0V Zener \$\endgroup\$ Commented Jul 26, 2017 at 23:30

4 Answers 4


I'm reasonably sure your real problem is that your expecting your thyristor to turn "OFF" when you remove your 20ms pulsed trigger signal. It doesn't, it will latch on until the energy stored in the capacitor is largely dissipated because the anode current from the capacitor holds it "ON".

With a coil you do have an LC circuit, or when accounting for the resistances in the circuit (such as the 2 Ohm coil resistance) an RLC circuit. That being said, with 15 mF, 2 Ohms, and a small enough inductance value (you don't give one) the resulting very high damping ratio will make treating it like a simple RC circuit an acceptable approximation. See below where the capacitor discharges to 7.7V at 20ms as previously noted. I’ve arbitrarily assumed a 100uH coil, it would have to be quite a bit larger to start to have an impact, and that would actually slow down the capacitors discharge, not speed it up. My assumption is the actual coil inductance is quite small, that assumption could certainly be incorrect. If large it would do a damped sinusoidal oscillation over a longer period of time as Olin mentioned.

enter image description here

I doubt you are getting a more rapid discharge at all, you are just actually discharging the capacitor for a lot longer than the 20ms trigger pulse because the thyristor latches “ON” until the capacitor is sufficiently discharged (it doesn’t turn off just because the trigger pulse ceased). Connected an oscilloscope to measure the voltage across the capacitor over time to check this and you would presumably see around 7.7V at around 20ms, but then the voltage continuing to drop from there until the current drops enough to finally turn “OFF” the thyristor.

If you want a 20ms pulse I wouldn’t use a thyristor, I would probably just use a simpler high current NPN transistor. That would “pulse” with the trigger, carry the large current, and not latch like a thyristor.

Another note of caution, if you abruptly cut off the current through the coil, any magnetic field built up in the coil will collapse and potentially cause a large voltage spike across the coil at that time, so usually some sort of a voltage clamp is recommended if this is happening.

  • \$\begingroup\$ Hi! Thanks for your answer! One more helpful addition!! I did mention later the inductance of the coil to be around 8.88mH, but I see your point about latching that even Olin mentioned. I do take the voltage down to zero at the gate when switching off, but I will definitely check with an oscilloscope and keep y'all posted. \$\endgroup\$
    – AmyM
    Commented Jul 27, 2017 at 17:45
  • \$\begingroup\$ @AmyM: Can you clarify that you understand that thyristors stay on until the anode to cathode current drops to zero. Turning off the gate (down to zero) does not turn off a thyristor. \$\endgroup\$
    – Transistor
    Commented Jul 27, 2017 at 19:31

It's not clear from your question, but it seems your are ignoring the inductance of your inductor, taking into account only its resistance.

A ideal capacitor "discharging" into a ideal inductor does NOT cause a simple exponential decay of the capacitor voltage, like would happen when discharging into a resistor. Instead, you get a sine voltage. With ideal components, the energy is continually sloshed back and forth between the capacitor and the inductor forever. The voltage across the two parts, and the current thru them are both sines, but 90° out of phase.

Discharging a capacitor into a inductor thru a SCR also gets interesting. Remember that the SCR stays on as long as there is current thru it. In the ideal capacitor/inductor case, the current builds up until the capacitor voltage is 0. That is actually the point of maximum current. The current then continues until the capacitor voltage is the negative of what it started out at. At that point, the current starts to build up in the opposite direction. This is where the SCR would turn off.

You actually have a capacitor, inductor, and resistor all in series. The pure capacitor/inductor case causes a infinite sine. Add a little resistance and you get a exponentially decaying sine. When the resistance dominates the inductance, you get mostly the R/C exponential decay you seem to be assuming, but with wrinkles. The resistor effectively adds damping. Whether you get ringing or not depends on how damped the system is.

You can't ignore the (probably) significant inductance of your "coil".

  • 2
    \$\begingroup\$ But, to be fair, without a schematic and coil data this is all idle speculation, isn't it? For all we know, the coil might be a 2 nH 20mA coil which through the Supercurrent Modification Of Komponent Entities (SMOKE) converts to a tubular carbon resistor with short-time central illumination. \$\endgroup\$ Commented Jul 26, 2017 at 22:37
  • 1
    \$\begingroup\$ @Marc: I'd say informed speculation would be more fair. Agreed that we don't know what we need to know to answer this clearly. I pointed out one thing for the OP to investigate and tried to explain some of the reason behind it. Maybe that ends up being useful, maybe not. A better answer requires a better question. I'll have to remember Supercurrent Modification Of Komponent Entities. Good one! \$\endgroup\$ Commented Jul 27, 2017 at 12:25

There are a few 2nd order effects in all electrolytic caps, just like batteries. Some more than others.

This means the ESR and C changes with pulse width due to the internal multiple effective caps in parallel each with different ESR*C time constants.

ESD and tan δ are inversely related, so you can compute one from the other. This affects the loss or dissipation Factor (DF) which is an older method but most common spec'd at 120Hz the 2f output for a bridge rectifier at 60Hz which does not need an ultra low ESR, so ESR is often never given in these parts.

You application needs good quality low ESR caps and a better spec to choose each part, including the thyristor, BEFORE you start the design.

So please give all the design requirements and not just a thumbnail sketch of your problem.

enter image description here

The simple approximation won't work using Ic=CdV/dt since it is not a simple as C. It is actually ESR1*C1//ESR2*C2 so what is more important here is the 1kHz ESR and the value of C at 1KHz , which is often not included.

Some brands are better than others for pulse discharging. Often the large C has a large ESR and the small C has a small ESR, so C effectively reduces for pulsed currents and increases for steady DC.

Choosing the right Cap for the job requires defining the rights specs for the job.

If you take a look at the product specs here for "Best-Cap" you will understand more.


simulate this circuit – Schematic created using CircuitLab

Remember this

It's all about Impedance Ratios when doing dynamic tests to consider the real and reactive elements as a function of rise time and load impedance to derive a better model.

We can call it Z(s) R+X(s) or work in the time domain.

YOu can prove this with a swept impedance plot with a sweep current source and small current sense R or use a fancy Vector IMpedance Analyzer, or compute the model from the RC for various pulse widths using any simulator like Falstad to help see the results.

It seems the cap you have has too much ESR1.

I highly recommend getting the RLC Nomograph to understand immediately the frequency vs impedance of every part in your model for starters. When you learn more about RF, they use Smith Charts.

But passive components are only ideal in school. In life they have warts and wrinkles. ( self capacitance, self inductance effective series resistance stray effects etc)


Well, even if you figured it out now, since I bothered to make the simulations anyway... The one on the right is basically what you have, with a thyristor modeled as the power PNP-NPN combo. The one on the left is with two NPNs, one signal and one power, connected as a Darlington pair (I assumed you might have current drive problems without the pair).

enter image description here

The bottom plots show the trigger signal, capacitor voltage, and capacitor current. I added the top plots to show the new severe voltage kick when the current is abruptly cutoff in the coil on the left schematic. I've also redrawn the left schematic to follow more standard conventions.

Also, 8.8 mH will have an appreciable impact on your discharge time, so using an RC approximation calculation isn't acceptable as Olin accurately assumed.

  • \$\begingroup\$ Thank you so much for your answers, and really appreciate you simulating all this. If you don't mind me asking , what software are you using? Also, you did mention that npn transistor can be an option, do you think a MOSFET or IGBT would be better in this application? \$\endgroup\$
    – AmyM
    Commented Jul 28, 2017 at 13:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.