# checking HV transformer

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#### chris54

##### New Member

I have a HV tranformer that I need to check. The secondary output is 1800 volts AC But I do not have a meter that goes that high.

How would I be able to test this

I use a 110 VAC to 11 VAC transformer to test HV transformers. I apply 11VAC to the primary of the HV transformer and measure the secondary at 10:1. That is much safer. In your case the output will be 180 volts.

If you need to measure the transformer at full voltage them do this: A typical multimeter has a 10MΩ input resistance. If you connect nine 10MΩ resistors in series and measure the voltage through them the voltage will be reduced by a factor of ten, so the meter will read 180V.

Note: Be very careful with such a high voltage since it's decidedly lethal.

If you need to measure the transformer at full voltage them do this: A typical multimeter has a 10MΩ input resistance. If you connect nine 10MΩ resistors in series and measure the voltage through them the voltage will be reduced by a factor of ten, so the meter will read 180V.

Note: Be very careful with such a high voltage since it's decidedly lethal.

I would never do that without HV probes.

make a voltage divider using few series resistors. calculate the value depends on maximum out put current, or to draw few mA, measure the voltage between one of the series resistor that woulod give a range in your multimeter.

rather than giving a low input and measure, measuring at rated voltage would give a good result about its conditon. all depends on the purpose of your test such as to find turn ratio, or etc.

high voltage probe can be the best option if you can effort.

razeen

razeen

I would never do that without HV probes.

So make a HV probe by putting the nine 10MegΩ resistors in a straight line inside a glass tube. Handle it from the end connected to the DMM only.

10MegΩ is too high and would make drop when DMM is connected, it can be in KΩ range,
just take more care, you can measure it

I still think that Ron's idea is the best.

On Ron's idea: You can use any low voltage transformer. A 3 or 6 volt (AC) wal-wort works well. This gives you a current limit just in case things go wrong. (high voltage can go wrong)

10MegΩ is too high and would make drop when DMM is connected, it can be in KΩ range,
just take more care, you can measure it
You misunderstand the circuit. The 10MegΩ resistors are connected in series with the 10MegΩ multimeter input impedance. The meter resistor is part of the voltage divider circuit. The drop provided by the DMM is desired (indeed it's a required part of the circuit).

Resistors in the kΩ range would dissipate a lot of power

You misunderstand the circuit. The 10MegΩ resistors are connected in series with the 10MegΩ multimeter input impedance. The meter resistor is part of the voltage divider circuit. The drop provided by the DMM is desired (indeed it's a required part of the circuit).

Resistors in the kΩ range would dissipate a lot of power

Agreed.

1800V + 100k resistor = 32.4W

>Watch that you resistors can handle the voltage. Some resistors are only good for 100 volts.
>Not all meters are made the same.
>>Some meters have a low impedance in AC mode. less than 1M ohm
>>Some meters in AC mode have a DC blocking capacitor. If you try to measure a DC voltage the capacitor will cause the divide down to not work.
>If the meter is unplugged the divide will not work.
>I have several HV probes and they have a resistor across the meter so even if the meter is unplugged or turned off the divide down will work. Example; use a 100k across the meter so the probe will divide down even if the meter is unplugged or switched to AC mode.

Draw an arc across the output leads of the HV transformer. The voltage will be approximately 10KV/cm, if I remember correctly. Don't be near the fixture while the circuit is powered up.

what about 100k x10 numbers in series? power discipation would drop to 3.24 W?
also each resistor discipates 0.324?
connect the meter across one of them and measure?

what about the accuracy of the result when 10M ohms are in series with the meter? ( input impedence of the DMM is so accurate?)

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what about 100k x10 numbers in series? power discipation would drop to 3.24 W?
also each resistor discipates 0.324?
connect the meter across one of them and measure?

what about the accuracy of the result when 10M ohms are in series with the meter? ( input impedence of the DMM is so accurate?)

I don't see any advantage on hookin 10x 100k in series.

Yes, the input impedance of a DMM is very accurate.

Accurate? Or High? It's very very high but I doubt the manufacturer is overly concerned about accuracy of the input impedance since the meter was not designed to measure resistances anywhere near the input impedance.

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It has to be High Accuracy, otherwise voltage measurements would be wrong.

Usually it's 10 MOhm +- 1%.

for the voltage measurement to be accurate it doesnt require accurate impedence, instaead requires high impedence only. it should not load the source, its the main concern for the manufacturers.

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