Re: HV Measurement - Back to Basics

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Original poster: "Matthew Smith by way of Terry Fritz <teslalist-at-qwest-dot-net>" <matt-at-kbc-dot-net.au>

Hi Folks

Thanks for your answers!

I hooked the meter up to a battery and resistor and got only the smallest 
deflection.  Since the unit is obviously unused (no traces of solder on 
solder pots, manufacturer's label still attached) I doubted that it had 
"blown".  I checked the resistance across it using my trusty Wavetek and lo 
and behold, the series resistance was 200kw*, so the 100V on the label was 
actually correct.  (500uA x 200kw = 100V)

So, taking your advice and looking to standard resistor series, I calculated:
Vtest=15kV
Ifsd=500uA
V=IR
.'. Rtotal = 15kV/500uA = 30Mw

Rmeter=200kw
.'. Rload = 30Mw - 200kw = 29.8Mw

Using standard series, a string of 13 x 2M2 and 1 x 1M2 gives this 
total.  The greatest volt drop we see over any of the components is 1100V 
on each of the 2M2's, which works out at:
P=VV/R
.'. P = 15kV * 15kV / 2.2Mw = 0.55W

... so 1W or greater resistors should be OK - agree?

It really is convenient to have a calculator that actually allows one to 
enter k, M, u, etc!  Wish I'd had it at college...

NB - As far as I'm concerned, HV meters are "no touch" when energised, so I 
haven't included a shunt resistor in case of meter failure.

Now all I have to do is re-calculate the whole lot to allow for the fact 
that the meter should be rated at 70.1V rather than 100V - it's going to be 
on the end of a bridge rectifier.  (100V x SQRT(2) / 2)

Cheers

M

* I use w to represent Ohms, due to lack of capital omega in this 
characterset ;-)


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Matthew Smith            | Business: http://www.kbc-dot-net.au
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