Hmm,
That's a bit tricky- it depends on lots of variables. The voltage doubler circuit doesn't actually double your rms AC voltage, and it also depends on the load you are feeding, the transformer secondary, and the value of the caps in the doubler.
Probably the best thing to do is to build the circuit from the mains transformer up to the voltage doubler. Then miss out R16, and just connect your voltage divider- i.e. the four resistors R1-R4. Make sure the circuit is off and the caps are well discharged, and connect up a DMM set to its highest DC range (1000V or greater) across the resistor chain, so that you're measuring the DC voltage at the output of the doubler. The resistors are there as a temporary load, to roughly simulate the actual circuit conditions.
[Tip- when measuring a high voltage (above 200V or so)
don't use the standard "prods" supplied with the DMM. Make up a set of test leads with the standard red and black 4mm plugs on one end, and then solder insulated crocodile clips (the ones with plastic "boots") to the other end. This way you can clip
both leads onto the points/terminals where you want to measure, and then when they are both secure, switch on the power. This way, there is no chance of you slipping or accidentally touching exposed live terminals, or touching a leaky probe. The absolute worst thing to do is to wade into an HT circuit with the two DMM probes, one held in each hand when the power is on :shock: A half measure (and how I usually work) is to have the 0V lead as a croc clip lead connected to the 0V point, and then have the +Ve lead as a standard probe, so that you can at least work "one handed". This is much safer as there is less likelihood of HT across the heart...]
End of safety lecture.
Switch on the circuit, and measure the voltage. Switch off once you've got a stable reading. You can then work out the size of R16 by the following:
1. Add together the values of the 4 resistors R1 to R4. The voltage across these needs to be 500V.
2. Find the current through these resistors when there is 500V across them. I = V/R, so I = 500/R1+R2+R3+R4
3. This current will also flow through R16. So the voltage drop will be V = I * R. You know the current, and you'll now know the voltage you want to drop: Vacross R16 = Vmeasured - 500V
4. So drop these values into R = V/I to find R16:
R = Vacross R16/Ithrough R1+R2+R3+R4
Find the nearest preferred value, and you're sorted.
5. Before you select the resistor, you'll need to find its power rating: P = I^2 * R
Drop I and R into the above, and multiply it by 1.5 to 2 to be on the safe side.
Neon- I have a feelling you need a "naked" neon- i.e. one without a resistor- I think M@plin sell them. The reason is that it's used as a "blinker"- check this link:
neon blinkers Neons trigger and glow at ~90V, and when you strap a cap across them and feed a voltage via a resistor, they act as an oscillator (as the cap charges to 90V, the neon flashes, discharges the cap, and the cycle starts again...)
:thumb:
Mark
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