Re: HV Measurement - Back to Basics

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

Original poster: "David Speck by way of Terry Fritz <teslalist-at-qwest-dot-net>" <dave-at-davidspeckmd-dot-org>

Matthew,
The most important feature of the meter is its full scale current rating -- 
probably 500 uA, as indicated.  You could verify this with a simple circuit 
of a 9 volt transistor battery in series with a common 22 K ohm (red - red 
- orange ) resistor -- should give a scale reading of about 12.2 kV on your 
meter, for a current of 409 mA, approximately.

The 100 volt rating on the meter is really irrelevant if it is a true 
microammeter, as it is only the current flowing through it that determines 
the full scale deflection.  If anything, the 100 V rating might refer to 
the maximum terminal to case voltage allowed, probably not a problem if you 
will be running it with one end grounded.

The magnet symbol probably represents a schematic of a moving coil meter -- 
no big surprise, as they are probably the most common type out there. The 
other symbols don't correspond to anything I'm familiar with -- probably 
inspection, certification or acceptance stamps.

You are right on the resistance value and power dissipation of your divider 
string -- it will get warm.  In reality, this is probably not the best 
meter for this sort of application as its current sensitivity is relatively 
poor.  It's not hard to find meters with 50 or 100 microamp full scale 
movements at hamfests or on eBay.  These would dissipate only 0.75 watts, 
or 1.5 watts, respectively, in their dropping circuits.  Of course, with a 
MOT supply, you aren't going to miss an extra 7 watts anyway.  You would 
just have to use a dropping resistor, or series combination of smaller 
value resistors whose total power dissipation is more like 20 or 40 watts, 
so that heating will not dramatically affect the resistance of the 
chain.  30 seriesed 1 megohm 1 watt resistors would do the job 
inexpensively, and you could substitute a 1 meg trimmer pot for one of the 
resistors if you really want to tweak the circuit for maximum 
accuracy.  Just be sure to adjust it very carefully with a well insulated 
tool, and make it the resistor closest to the meter, and hence, ground.  Of 
course, I am assuming that you are measuring a DC voltage.  If you are 
measuring unrectified AC MOT output, you will have to add a precision 
rectifier circuit, a much more complicated issue.

HTH,
Dave

Tesla list wrote:

>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?
>
>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.)
>
>...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...
>
>In a word, help!
>
>Cheers
>
>M
>
>PS - FWIW, the meter is made by Ateliers Pekly of Paris.