chokes (and BIG Resistors) (power lost in RC)

From:  Jim Lux [SMTP:jimlux-at-earthlink-dot-net]
Sent:  Tuesday, May 26, 1998 11:21 AM
To:  Tesla List
Subject:  Re: chokes (and BIG Resistors) (power lost in RC)

Tesla List wrote:
> ----------
> From:  Gary Lau  25-May-1998 1526 [SMTP:lau-at-hdecad.ENET.dec-dot-com]
> Sent:  Monday, May 25, 1998 2:59 PM
> To:  tesla-at-pupman-dot-com
> Subject:  Re: chokes (and BIG Resistors)
> >From: Jim Lux <jimlux-at-earthlink-dot-net>
> >In a series RC charging system, the amount of energy dissipated in the R is
> >the same as the energy stored in the C. So, if you have a 450 VA NST
> >(typical), then you are going to be putting half that in the C and half
> >into the R, hence 200+ Watts dissipation in the R.  The fact that the NST
> >is really an inductive source actually reduces the dissipation in the R
> >somewhat, but this should give you the general idea.
> Hold on.  Not true.  "Half" of the energy goes into the R?  Shouldn't the
> value of the R factor into this somehow? 

nope...Here we go:

The power dissipated in the resistor is I^2* R.

	I = Echarge/R * EXP(-t/RC)

(Standard exponential RC charge, the initial current (when cap voltage
is zero) is E/R, and it tapers off to zero at infinite time, with a time
constant of RC)

So, instantaneous power is [E/R * EXP(-t/rc)]^2 * R, or, rearranging a
bit and collecting terms:

	P = E^2/R * EXP(-2 *t / RC)

Total energy is integral of power * dt (i.e. Joules is Watts times
Seconds): so,

	Energy = Integral [ P dt] = E^2/R integral [(exp(-2t/rc)dt)]
		(integrals from from t=0 to t=infinity)

integral[ exp(at) dt] = 1/a [exp(a * upperbound) - exp(a*lowerbound)]
for bounds = [0, infinity] this reduces to -1/a

	so, integral[exp(-2t/rc)dt] reduces to RC/2

substituting into the energy equation given above:

	Energy = E^2 / R * ( RC/2) = E^2 * C / 2

which is the energy stored in the capacitor = E^2 * C / 2

Oddly, the actual value of the resistor doesn't enter into it at all.
All the resistor does is set what the charging time constant is, and
therefore, the peak power. The energy is always the same.

 I think a first-order
> approximation of power would be I*I*R.  Assuming a 750 Ohm R and 60mA
> NST, that would be 2.7 Watts.  This could be doubled if using resonant
> charging.  Say 5.4 Watts.
Aha, now you are limiting the current using a NST, which makes the
constant voltage assumption above invalid.  If you truly have a constant
current (that is, the current limiting impedance is much, much greater
than the circuit impedance, which for a 60 mA transformer at 15 kV = 250
kOhms, it is), then the power lost in the resistor is:

	 Energy = I^2*R * charge time.

Assuming that charging stops exactly at Echarge, the charge time is
C*E/I. So, the energy lost in the resistor is:

	Energy = I * E * R * C

which for your example of 60 mA and 15 kV, and assuming a capacitor size
such that charging takes 1/4 60 Hz cycle( 4.166 mSec) =  .016 uF. The
total energy lost with your suggested 750 Ohm resistor would be

	= .06 * 15E3 * 750 *.016E-6 = .0112 Joules/charge cycle

multiplying by 120 charge cycles / second gives an average power in the
resistor of 1.35 Watts

The stored energy in the capacitor in this example is 15E3^2 * .016E-6 /
2 = 1.8 Joules, which multiplied by the same 120 cycles/second is 216
Watts. Looks like you are losing less than a percent in your resistor,
which isn't bad...

> A second-order effect is that, assuming one gap firing per half-cycle,
> the energy stored in the bypass caps is also dissipated in the resistor.
> Assuming a 15KV NST and a 500 pF bypass cap per side, charging up to
> 15KV*1.414/2 =10605 V, energy is .5 * C * V*V = .5 * 500E-12 * 10605 *
> 10605 = .028 Watt-seconds, times 120 bangs per second = 3.37 Watts.
> So, I can account for 5.4 plus 3.4 = 8.8 Watts per resistor.  This does
> not explain why my 750 Ohm, 50 Watt resistors are burning up though.

Probably because the NST isn't really a constant current source, nor is
it a constant voltage source, but somewhere in between. I'll have to do
some calculations allowing for the inductive component in the NST.