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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
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