How can I build a 2:1 3 phase voltage attenuator from resistors

My datalogger is for 600VAC maximum, but sometimes my source is higher like 630V or so or maybe higher on malfunction or other problems. I would like to build a resistive divider to reduce the input to the logger by a 2:1 ratio. I can easily adjust the PT ratio in the software by whole numbers.

My thoughts are that if I arrange three 1 watt resistors in a delta configuration and then connect another three from each corner of the delta triangle to L1, L2, L3 then I will be able to read one half the source voltage at the corners of the triangle if,

the resistors in the triangle equal the source lead resistors divided by 1.73.

If this is correct, and I am not entirely sure it is, then if i chose a 1 MEG resister for each source resister, then I will need to select a 578.03 K resister for the triangle ones.

I am going to have to search hard to find combinations of two values to fit this equation so I wanted to check to see if my assumption of values is correct

Any related advice appreciated.

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Reply to
Frank White
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sampling dividers in high voltage transmitters are usually a stack of series resistors with a 1% tolerance.

even at low currents the I squared R value can sneek up on you when R is very large.

the resistive stack should take into account the load of the meter to insure calibration. often this is done with a trim pot in parallel with a resistor. remember the power rating of the pot is lessened when not at maximum resistance. a current limiting series resistor should be included in case the pot is adjusted to minimum.

wire with high voltage insulation should be used to connect to the test points.

the sampling stacks should be mounted on or in materials that are safe for high voltage. boards that are meant for HV rectifier stacks should work fine.

Reply to
TimPerry

Use potential transformers. You can get them with various ratios. It is much safer.

Ben Miller

Reply to
Ben Miller

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