[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

Re: [TCML] Pulse Capacitors



Hi Nicholas,

A well designed pulse capacitor will be very close to an "ideal" capacitor - no resistance, no inductance, and a perfect (lossless) dielectric. Although real capacitors can never reduce these parameters to zero, well designed pulse caps can minimize their effects by using foil plates, rolled construction with the extended foil plates (coming out on each end of the roll), sprayed metal to connect them all on each end, and low loss, defect-free dielectric film (typically polypropylene).

Furthermore, in Tesla coil use, the capacitors must be capable of withstanding voltage reversals (up to 100% - even more for DRSSTC's) and high currents (hundreds or thousands of amperes). Other pulsed power applications can demand discharge currents of hundreds of thousands of amperes/capacitor, often requiring special ultra-low inductance construction techniques.

Nicholas Goble wrote:
So for someone who is building their own capacitors, what do I have to do
differently to construct a pulse capacitor.  I mean a plate capacitor is a
plate capacitor.  Does the dielectric also determine the rate it can
discharge?

Unless the dielectric is very lossy, the overall inductance and resistance of the discharge loop (including the capacitor itself) determine the rate of discharge and whether the discharge will be oscillatory, exponential, or a combination.

My understanding of capacitors was that the discharging current,
with respect to time was determined by the following equation:

I(t) = (-Q/RC) * (e^(-t/RC))
Where e is the natural logarithmic value and Q = C * V.  Am I right or just
way off?

This would be correct only if there was no inductance in the circuit. Although all circuits contain inductance, the above is a useful approximation where circuit resistance dominates circuit behavior. Note that, at the time the switch closes (t=0+), the current will initially be V/R, and it will then decay exponentially towards 0 with a time constant of RC seconds. After about 5*RC seconds has elapsed, the current will be down to less than 1% of its initial value.


I guess I¹m just trying to get a feel for how pulse capacitors work and how
to design capacitors.  Take this hypothetical scenario:

If I wanted to discharge 10amps in a thousandth of a second.  t=.001 and
I(t)=10.  I¹d just pick a voltage such as 9kV, which would tell me that I¹d
need an the relationship between my resistance and capacitance would be the
following:

C = -.001 / (R * ln((10R)/9000))

So a suitable capacitance would be .003F with a resistance of .025Ohms.

The logic/math is incorrect. Plugging in the numbers for the scenario above, your initial voltage is 9000 volts and loop resistance is 0.025 ohms, with no inductance in the circuit (a bad assumption for this example). When you close the switch, a current of 9000/0.025 amperes will initially flow, and the circuit will discharge with a time constant of 0.003*0.025 seconds: 360,000 amperes at t=0+ decaying exponentially with a time constant of 7.5 usec. A BIG bang...


With this reasoning, if it¹s all correct, I¹d be able to discharge 10amps in
a thousandth of a second if I wired a .025Ohm resistor in series with a
.003F capacitor across a potential difference of 9kV, charged the capacitor
up, and discharged it.  Right?  So then why would I need a pulse capacitor
for that?

See the calculations above. In reality, for a low resistance circuit such as the above, the internal inductance of the capacitor and circuit wiring become an important limiting factor, and the circuit would actually have an oscillatory discharge that decays exponentially.

Tesla coils are designed so that the inductance of the primary winding has a much greater impact than the resistance, so the primary circuit oscillates. If there were no resistive losses, you could estimate the peak primary current by applying Conservation of Energy (COE). You know that at time t=0- the energy in the capacitor is 0.5CV^2. When we close the switch, the capacitor discharges into the primary inductor. As energy transfers from the capacitor to the inductor, the current begins to increase (sinusoidally), and eventually the capacitor voltage reaches 0 while the primary inductor current reaches its maximum value. At this point, all of the initial energy in the capacitor has been transferred into the inductor. Since the energy in an inductor is 0.5 LI^2 and we had no losses, then COE says that:

0.5*CV^2 = 0.5*L*I^2

Solving for primary current I:

I = V*Sqrt(C/L)

For typical small to medium spark gap Tesla Coils, peak primary current is several hundred amperes. Since our simple ideal LC circuit has no losses, it would continue to oscillate forever. Since real systems do have losses, it will actually decay exponentially at a rate determined by the circuit (and spark gap) resistance.

Pulse capacitors used in high voltage, high current, oscillatory circuits must be constructed to have low losses to prevent them from overheating and self destructing. They must also be designed to prevent degradation from corona in order to have acceptable lifetime. Most coilers no longer build their own tank capacitors, since there are commercial sources that are a more cost effective (or time effective) solution. Look up MMC capacitors in the Pupman archives for more information.

You may also find the following discussion useful for the transient response of RLC circuits:
http://www.physics.utoronto.ca/~phy225h/exercises/Currents_LRC/Transient%20voltages.pdf



Nicholas Goble

Bert
--
***************************************************
We specialize in UNIQUE items! Coins shrunk by huge
magnetic fields, Lichtenberg Figures (our "Captured
Lightning") and out of print technical Books. Visit
Stoneridge Engineering at http://www.teslamania.com
***************************************************
_______________________________________________
Tesla mailing list
Tesla@xxxxxxxxxxxxxx
http://www.pupman.com/mailman/listinfo/tesla