From: Jim Lux [SMTP:jimlux-at-earthlink-dot-net]
Sent: Monday, June 01, 1998 4:28 PM
To: Tesla List
Subject: Re: Capacitor discharge
> From: Marco Denicolai [SMTP:marco-at-vistacom.fi]
> Sent: Monday, June 01, 1998 1:11 AM
> To: tesla-at-pupman-dot-com
> Subject: Capacitor discharge
> Because tuning a TC is a sequence of power-on / check / power-off /
> wires I would like to make sure all capacitors are discharged so that I
> don't get any electrical shock (that could be my last one...).
> 1. A 10 Mohm HV resistor in parallel with each capacitor. This is what is
> in the capacitors employed in microwave ovens (they have got this
> INSIDE the cap can). But how that will affect the HV capacitor
This works great. However, check your time constant on the RC combination.
Figure out how long it will take to discharge to less than 50V. If your cap
is charged up to 15 kV (for example) it will take almost 6 time constants
to discharge down to 50 V ( ln(15E3/50) = 5.7). If you have a .01 uF
capacitor, the time constant with a 10 Meg resistor is 0.1 seconds, so
you'll be down to 50 Volts in less than a second, which seems reasonable.
Check the power dissipation on that resistor: 22.5 Watts.... Also, make
sure the resistor can stand the voltage. If the time constant of the RC
combination is "long" compared to the 60 Hz or the tesla frequency, it
won't have any effect. For your example, you might even want to go to a 50
Meg resistor, which will make it take 3 or 4 seconds to discharge (with
0.01 uF), but will dissipate only 2.25 watts total. A series of 15
resistors 3.3 Meg, 2 Watt should work nicely. You want the 2 watt kind
because their larger body creates a longer "creepage path" for flashovers.
Or, use smaller resistors and stick them in oil. The oil insulates them
electrically, and helps conduct the heat away. (of course, watch the oil
temp, if you put it all in a PVC pipe, the PVC acts as a good thermal
insulator) Life is a series of tradeoffs, of course.
> 2. A kind of delayed-relay battery that will short all caps through, say,
> 100 Kohm resistors. But then you will have HV wires running through your
> TC assembly, which is not very nice. And HV relays will be needed too
> (do they exist?).
HV relays do exist, they are expensive (several hundred $) (Check out
http://www.rossengineering-dot-com/ for example). The usual scheme is to use a
normally closed relay, and power the relay coil from the same power as runs
your transformer (except before the variac, if you have one). Then, when
the power is turned off (or unplugged, or fails, or....) the relay closes,
discharging the capacitors.
A series resistor that limits the discharge current to a few amps is nice.
It gives the stored energy in the cap somewhere to go (other than melting
the relay contacts), and reduces the "bang" when they discharge. It can be
a few hundred ohms for instance. A water resistor made with copper sulfate
solution is a handy energy dump. You don't care about the exact resistance,
and the water has huge thermal mass to absorb the energy without heating
up much, and they are really cheap to build. (see
http://home.earthlink-dot-net/~jimlux/hv/rwater.htm for more info). Make sure
you design the resistor big enough to absorb the full transformer output
for a few seconds, in case the relay doesn't open when you put power to the
transformers. In the case of 1kW microwave oven transformers, that means
the resistor needs to take the full kW and then some.
A combination of a bleeder resistor of a few Megs and a shorting relay
would be the safest scheme overall (and is what I use on my HV power
supplies). That way, even if one fails, the other will make sure that the
circuit is really and truly dead when the power is off. (And, I ALWAYS
check it with a meter, too.)
Finally, in most voltage multiplier schemes, you actually have several caps
in series. You need to discharge them individually. Shorting the whole
string at once doesn't necessarily discharge the individual caps, if there
are any capacitance variations. All shorting the string does is make sure
that the sum of the cap voltages is zero. Imagine that you had two caps,
each charged to 10 kV. Now hook them up so that the positive side is
connected to the positive side. You've shorted the entire string, but they
still have 10 kV on them.