# Re: [TCML] Pulse Capacitors

```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?
```
The dielectric will affect the dissipation factor of the cap and is 1/Q. If a poor choice of dielectric were used, Q will be affected. But when thinking of a pulse cap, consider the application as a pulse application and find a cap that will perform well in this application. You can certainly build your own cap for this purpose.
```
```
```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?
```
```Looks right.
It = Vp / R * (e^(-t/RC))
Q = C * Vp * (1-e^(-t/RC))

```
```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.

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?
```
Full discharge is 5 time constants. To fully discharge 9000V over a time span of 1ms, then t = 1ms/5 = 200us. I = (CV)/t = (.003F * 9000V)/200us = 135,000 amps, so a 0.067 ohm resistance.
```
```
You would need one hell of a cap and resistor for that! Let's get reasonable. A 0.003F charged to 9000Vp is a freakish energy to be discharged. Joules = 0.5*C*Vp^2 = 0.5 * .003 * 9000^2 = 121,000 joules. Hypothetically, your releasing 9000V across a .067 ohm resistance, so I^2R would dictate a 1.22 billion watt capability.
```
```
The problem is, your not discharging 10A in a thousandth of a second. If you want to limit the current to 10A, then time must be dramatically increased. Say 9000V/10A = 900 ohms. Then t = (CV)/I = (.003F * 9000V) / 900 ohms = 30ms, and 30ms * 5 = 150ms for full discharge. In this case, no, you would not really need a high energy cap because you have limited the current (this is totally unusable in TC duty). We want the cap to release high peak currents quickly into the primary impedance. 400A to 800A is typical and we do this at 100 pps and above. This is why a high energy cap is needed and designed for the pps were driving it at. So maybe look at 400A/8ms for di/dt as a reality check.
```
```
You can certainly build a glass, poly, bucket, salt water, etc. cap for TC duty. Even if the caps are designed for high energy and RF, there are also voltage standoff, corona losses, and mechanical aspects which is why I prefer to use an off the shelf cap.
```
Take care,
Bart

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