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Home > Archive > Electrical Engineering > January 2007 > How can I build a 2:1 3 phase voltage attenuator from resistors
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How can I build a 2:1 3 phase voltage attenuator from resistors
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| Frank White 2007-01-29, 3:25 am |
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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.
Remove dashes "-----------" to e-mail.
delik2-------------@shaw.ca
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| TimPerry 2007-01-29, 9:25 am |
| Frank White wrote:
> 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.
>
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.
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| Ben Miller 2007-01-29, 1:25 pm |
| "Frank White" <delik2----------------@shaw.ca> wrote in message
news:9g5rr2l0ocv65o9d45654qli59407fldpu@4ax.com...
>
> 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.
>
Use potential transformers. You can get them with various ratios. It is much
safer.
Ben Miller
--
Benjamin D. Miller, PE
B. MILLER ENGINEERING
www.bmillerengineering.com
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