Re: HV Measurement - Back to Basics

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

Matthew,
Interesting -- the meter has a built in dropping resistor, not the most 
common arrangement, but as you have pointed out, certainly one that is 
surmountable.

My suggestion is that you use only 1M2 resistors in you chain.  If they are 
half watt resistors, they you will have a total available power dissipation 
of about 14 watts, so each individual resistor will be dissipating about 
half of its rating.  If you use 1 watt resistors as you mentioned, then 
each would dissipate only a quarter of its rating, so you would have less 
resistance change from heating.

As previously mentioned on the list, IIRC, the max voltage rating on 
typical film resistors is about 1000 volts, so an 1100 volt drop across 
your 2M2 resistors might be pushing things a little.  You might get away 
with it, but if there is a single resistor failure, then the whole string 
might fail catastrophically.  Best to check the specs on the particular 
brand of resistors that you plan to use.  Remember that the TC environment 
is full of spikes and transients which can easily be double or triple the 
"rated" values of the circuit, so it's a good idea to over engineer the 
circuits.  It's not the steady state situation that destroys things, but 
the sneaky transients that come up to bite you.

If you put your rectifier bridge at the low end of your divider string, 
remember that diodes are nonlinear devices.  They will not conduct at all, 
and your meter will show a zero deflection, until the voltage across the 
bridge is about 1.2 volts.  This means that the supply will be delivering 
15 kV * (1.2V / 100V) or 180 volts before the needle even begins to move.

You can view the system as a current mode operation after the diode bridge 
begins to conduct, so you would not invoke the 70.1V correction that you 
proposed in your last posting.  The bridge will make a -180 volt 
subtractive offset across the entire range of the meter, rather than a 
0.701 multiplication effect of the scale.  That's not enough to seriously 
affect you scale accuracy, but you will have to remember that "0" on the 
meter does not necessarily mean 0 volts from the supply.

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 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?
>
>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)