Can I Measure the Output of an Unknown High Voltage Transformer?

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The discussion revolves around measuring the output of an unknown high voltage transformer, specifically an old furnace ignition transformer without specifications. The user initially plugs it into 240V, producing a 10mm arc, and seeks advice on safely measuring its output and current. Suggestions include applying lower voltages to determine the turns ratio and using a moving coil meter for current measurement, while cautioning against the dangers of high voltage. A successful test at 12V yielded an output of approximately 190V, suggesting a potential output of 3800V at 240V. Ultimately, the thread was closed due to safety concerns regarding high voltage discussions.
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Hi, I have an old high voltage furnace ignition transformer with no specification plate on it, When plugged to 240v it gives about 10mm arc length; with no proper equipment to measure such high voltage I was wondering if I can plug it 50v and measure the output to calculate the ratio.
I'm also stuck in calculating the current output at 240v.
I suppose that this kind of transformers are made to work in short circuit mode so the nominal current equals the shortcut current?

Any help would be a real relief because my brain is having a huge heat loss.
 
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I recall a figure of 30000 volts per centimeter. So your arc is being caused by about 30000 volts.

I doubt that it would be rated to do this constantly as a furnace would be started with a spark and then the spark would be turned off.

You could probably measure the AC current in the secondary circuit if you are careful to insulate the meter, but you would need a moving coil meter to do this and not an expensive digital multimeter.

Although the meter would give a reading, it may not mean much because conduction would only be for a short time at the positive and negative peaks of the incoming sinewave. So the actual current will be a lot higher than the indicated current.
 
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I suppose that this kind of transformers are made to work in short circuit mode so the nominal current equals the shortcut current?

I would guess it's made like a ballast. Open circuit voltage is quite high but as soon as current starts to flow voltage drops radically.
Magnetic shunts give that characteristic by allowing flux to bypass secondary winding when current starts to flow and it's making secondary amp-turns.

If you're just characterizing the transformer perhaps you could power it as a step down.
Apply 240V to high voltage secondary winding and see how many millivolts you get on primary. That should give an idea as to turns ratio without need to handle high voltage. You can easily fry an expensive meter with just a couple kilovolts.

Then , if first test says you can expect a safely measurabe secondary voltage with 6V on primary, maybe you could put 6VAC on primary from a filament transformer.


old jim
 
Thanks for the help, I applied 12V on the primary and I got about 190V, So on 240v I would have 3800v.
Maybe an idea to get the maximum amps through the secondary?
 
Hohenheim said:
Hi, I have an old high voltage furnace ignition transformer with no specification plate on it, When plugged to 240v it gives about 10mm arc length; with no proper equipment to measure such high voltage I was wondering if I can plug it 50v and measure the output to calculate the ratio.
I'm also stuck in calculating the current output at 240v.
I suppose that this kind of transformers are made to work in short circuit mode so the nominal current equals the shortcut current?

Any help would be a real relief because my brain is having a huge heat loss.

Hohenheim said:
Thanks for the help, I applied 12V on the primary and I got about 190V, So on 240v I would have 3800v.
Maybe an idea to get the maximum amps through the secondary?

Sorry, this is too dangerous to be discussed here. Thread closed.
 
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