Re: HV Measurement - Back to Basics

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Jim Lux by way of Terry Fritz <teslalist-at-qwest-dot-net>" <jimlux-at-earthlink-dot-net>


 > Original poster: "Matthew Smith by way of Terry Fritz
<teslalist-at-qwest-dot-net>" <matt-at-kbc-dot-net.au>
 >
 > Hi All
 >
 > Could some kind soul give me a hand with this little problem?
 >
 > I bought (blind) a 15kV voltmeter, which I planned to sit on the end of my
 > MOT-based power supply.  When I first saw it and discovered that the
 > terminals are about 8mm apart, I decided that this is just a meter with a
 > 15kV *scale*, not a meter than can be connected to and measure up to
 > 15kV...  Never fear, I thought, it's just the question of sizing an
 > appropriate resistor/resistor network.
 >
 > Looking at the base of the scale, I see some small symbols; the first
 > appears to be an underscore - possibly this is a moving coil (DC) meter
 > (terminals are also marked + and - which would tend to confirm this). The
 > second symbol is a star with a 2 in it - goodness knows what this
 > means.  The third symbol is an upside-down capital T with 1.5 above
 > it.  The fourth symbol appears to be a horseshoe magenet, pointed
 > downwards, with something between the poles.  The fourth symbol is a
 > standard Euro resistor symbol with a very helpful R in it.
 >
 > If anyone can shed any light on the above, I'd be interested, but the
 > imporant bits followed: 500uA 100V.  Now, I'd read that as being 500uA
FSD,
 > and a maximum voltage rating of 100V.  (A bit less than 15kV, eh?)
 >
 > I canna remember how I'm supposed to wire this up!  I'm fumbling with
 > this:  if FSD is 500uA, I would need a series resistance of:
 >
 > R = 15,000V/500uA = 30Mw (where w represents capital Omega)
 >
 > This, however, doesn't sound right because then the whole thing would be
 > dissipating:
 >
 > 15,000V x 15,000V / 30Mw = 7.5W  Wouldn't this be getting a bit warm?

yep, that's the problem... But, a 30 Meg resistor that can handle 15 kV is
going to be fairly large physically, so the 7W dissipation shouldn't be a
horrible problem.

Say you make your resistor string out of 15  pieces of 2 Meg, 2 watt
resistors... Each resistor will have only 1000 V across (nicely) and will
dissipate about  1/2 watt, so they won't get too hot (which would change
their value).

In reality, the series resistance of the meter itself will throw things off
a bit.  AND, you should put a resistor across the meter so that if the meter
goes "open" it doesn't float up to the HV (unless you've got the case
insulated from ground and user).

If you're looking at measuring AC, you need a bridge rectifier and (maybe) a
filter capacitor across the meter as well.


 >
 > I assume that I'd have to have a potential divider somewhere around here
to
 > make sure that the meter never sees more than 100V across it (if, indeed,
 > that is it's rating.)

Unlikely that the meter could handle 100 V across the terminals.. that would
imply that it's internal resistance is quite high, and that's unlikely for a
0.5 mA FS movement (I'd expect a few kOhm at most)

A bit of testing might be called for... get a 1.5 V battery or a  5V power
supply, and some resistors.  A 10K resistor and 5V will be 0.5mA, so hook
that up with the meter and see what it reads.  It will read lower (because
the total resistance includes that of the meter), but that will allow you to
calculate what the meter's internal resistance is.


 >
 > ...and that's where I've come to a grinding halt.  I don't know whether I
 > started off going the wrong way or if I've just got the math wrong
 > somewhere.  Thought it was just basic Ohm's Law...
 >