Re: HV Measurement - Back to Basics

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

 > 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.

This isn't entirely true.  When the voltage on the whole thing is low, the
"resistance" of the diode bridge is very high (i.e. the diodes haven't fully
turned on, and they are in their square law region), meaning that a larger
fraction of the total voltage is across the meter.  The circuit will act
very non-linearly, but not as a threshold circuit.

However,  given that most analog meters would be doing very, very well to
have 5% of full scale accuracy, and 5% of 15kV is 750V, some weirdness when
the voltage is 100 or so volts is insignificant.  You shouldn't really be
trusting this meter below 1/3 full scale anyway.

 >
 > 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.

The bridge will not contribute a 180 volt offset... When the diodes are
conducting, it is only a 1.5 volt drop (or thereabouts), which compared to
the 100V in the meter and the 14900V drop in the resistor string, is
insignificant.

I would think that the .707 is the RMS/Average correction... Depending on if
you put a capacitor across the meter, the RMS to meter reading depends a lot
on the mechanical dynamics, etc.  With a low current meter (i.e. 50 uA), and
a big capacitor, the meter tends to read peak voltage, in which case the
meter will read 1.4 times the RMS voltage.  With no capacitor (or a small
one), it's more like you'll be reading the average of the absolute value,
which is about .90 times the RMS value (or .64 times the peak value).

 >
 > Dave
 > Tesla list wrote: