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Can I use any HV Probe for a Voltmeter/Ampmeter gauge?
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I purchased the multimeter pictured below for a high voltage power supply I'm working on. It's rated for 100V DC and 10A, but I hope to measure several thousand volts. Is it possible to use a HV voltage probe like I would use with a regular multimeter on this component, or no?
power-supply high-voltage multimeter probe
asked Jan 8 at 6:02
Connor OlsenConnor Olsen
61
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add a comment |
$begingroup$
I purchased the multimeter pictured below for a high voltage power supply I'm working on. It's rated for 100V DC and 10A, but I hope to measure several thousand volts. Is it possible to use a HV voltage probe like I would use with a regular multimeter on this component, or no?
power-supply high-voltage multimeter probe
asked Jan 8 at 6:02
Connor OlsenConnor Olsen
61
$endgroup$
add a comment |
$begingroup$
I purchased the multimeter pictured below for a high voltage power supply I'm working on. It's rated for 100V DC and 10A, but I hope to measure several thousand volts. Is it possible to use a HV voltage probe like I would use with a regular multimeter on this component, or no?
power-supply high-voltage multimeter probe
asked Jan 8 at 6:02
Connor OlsenConnor Olsen
61
$endgroup$
I purchased the multimeter pictured below for a high voltage power supply I'm working on. It's rated for 100V DC and 10A, but I hope to measure several thousand volts. Is it possible to use a HV voltage probe like I would use with a regular multimeter on this component, or no?
power-supply high-voltage multimeter probe
power-supply high-voltage multimeter probe
asked Jan 8 at 6:02
Connor OlsenConnor Olsen
61
asked Jan 8 at 6:02
Connor OlsenConnor Olsen
61
asked Jan 8 at 6:02
Connor OlsenConnor Olsen
61
asked Jan 8 at 6:02
Connor OlsenConnor Olsen
61
asked Jan 8 at 6:02
Connor OlsenConnor Olsen
61
61
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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No. Those cheap meters use a voltage divider adding up to around 280-290K and go directly into an 8-bit MCU (STM8 series).
Typical high voltage probes assume a reasonably accurate 10M$Omega$ input impedance for the multimeter- something like 999M in series with 1.11M (internal divider values), so it would read grossly low on this kind of an input.
answered Jan 8 at 7:15
Spehro PefhanySpehro Pefhany
206k5156412
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It would be highly impractical to manually account for a very uneven correction factor, but OP would be able to feed it higher voltages than before :-)
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– winny
Jan 8 at 7:35
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@winny but only once, I guess.
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– Spehro Pefhany
Jan 8 at 7:36
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How come? The probe division should take the majority of the voltage.
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– winny
Jan 8 at 8:37
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@winny If it's rated for 40kV but only outputs roughly 1/3 the voltage, connecting it to a 120kV source might not be wise. ;-)
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– Spehro Pefhany
Jan 8 at 12:36
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Of course! Hence correction factor!
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– winny
Jan 8 at 13:27
|
show 1 more comment
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Only if the input impedances are the same to form a voltage divider.
answered Jan 8 at 6:36
Sunnyskyguy EE75Sunnyskyguy EE75
65.4k22295
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No. Those cheap meters use a voltage divider adding up to around 280-290K and go directly into an 8-bit MCU (STM8 series).
Typical high voltage probes assume a reasonably accurate 10M$Omega$ input impedance for the multimeter- something like 999M in series with 1.11M (internal divider values), so it would read grossly low on this kind of an input.
answered Jan 8 at 7:15
Spehro PefhanySpehro Pefhany
206k5156412
$endgroup$
$begingroup$
It would be highly impractical to manually account for a very uneven correction factor, but OP would be able to feed it higher voltages than before :-)
$endgroup$
– winny
Jan 8 at 7:35
$begingroup$
@winny but only once, I guess.
$endgroup$
– Spehro Pefhany
Jan 8 at 7:36
$begingroup$
How come? The probe division should take the majority of the voltage.
$endgroup$
– winny
Jan 8 at 8:37
$begingroup$
@winny If it's rated for 40kV but only outputs roughly 1/3 the voltage, connecting it to a 120kV source might not be wise. ;-)
$endgroup$
– Spehro Pefhany
Jan 8 at 12:36
$begingroup$
Of course! Hence correction factor!
$endgroup$
– winny
Jan 8 at 13:27
|
show 1 more comment
$begingroup$
No. Those cheap meters use a voltage divider adding up to around 280-290K and go directly into an 8-bit MCU (STM8 series).
Typical high voltage probes assume a reasonably accurate 10M$Omega$ input impedance for the multimeter- something like 999M in series with 1.11M (internal divider values), so it would read grossly low on this kind of an input.
answered Jan 8 at 7:15
Spehro PefhanySpehro Pefhany
206k5156412
$endgroup$
$begingroup$
It would be highly impractical to manually account for a very uneven correction factor, but OP would be able to feed it higher voltages than before :-)
$endgroup$
– winny
Jan 8 at 7:35
$begingroup$
@winny but only once, I guess.
$endgroup$
– Spehro Pefhany
Jan 8 at 7:36
$begingroup$
How come? The probe division should take the majority of the voltage.
$endgroup$
– winny
Jan 8 at 8:37
$begingroup$
@winny If it's rated for 40kV but only outputs roughly 1/3 the voltage, connecting it to a 120kV source might not be wise. ;-)
$endgroup$
– Spehro Pefhany
Jan 8 at 12:36
$begingroup$
Of course! Hence correction factor!
$endgroup$
– winny
Jan 8 at 13:27
|
show 1 more comment
$begingroup$
No. Those cheap meters use a voltage divider adding up to around 280-290K and go directly into an 8-bit MCU (STM8 series).
Typical high voltage probes assume a reasonably accurate 10M$Omega$ input impedance for the multimeter- something like 999M in series with 1.11M (internal divider values), so it would read grossly low on this kind of an input.
answered Jan 8 at 7:15
Spehro PefhanySpehro Pefhany
206k5156412
$endgroup$
No. Those cheap meters use a voltage divider adding up to around 280-290K and go directly into an 8-bit MCU (STM8 series).
Typical high voltage probes assume a reasonably accurate 10M$Omega$ input impedance for the multimeter- something like 999M in series with 1.11M (internal divider values), so it would read grossly low on this kind of an input.
answered Jan 8 at 7:15
Spehro PefhanySpehro Pefhany
206k5156412
answered Jan 8 at 7:15
Spehro PefhanySpehro Pefhany
206k5156412
answered Jan 8 at 7:15
Spehro PefhanySpehro Pefhany
206k5156412
answered Jan 8 at 7:15
Spehro PefhanySpehro Pefhany
206k5156412
206k5156412
$begingroup$
It would be highly impractical to manually account for a very uneven correction factor, but OP would be able to feed it higher voltages than before :-)
$endgroup$
– winny
Jan 8 at 7:35
$begingroup$
@winny but only once, I guess.
$endgroup$
– Spehro Pefhany
Jan 8 at 7:36
$begingroup$
How come? The probe division should take the majority of the voltage.
$endgroup$
– winny
Jan 8 at 8:37
$begingroup$
@winny If it's rated for 40kV but only outputs roughly 1/3 the voltage, connecting it to a 120kV source might not be wise. ;-)
$endgroup$
– Spehro Pefhany
Jan 8 at 12:36
$begingroup$
Of course! Hence correction factor!
$endgroup$
– winny
Jan 8 at 13:27
|
show 1 more comment
$begingroup$
It would be highly impractical to manually account for a very uneven correction factor, but OP would be able to feed it higher voltages than before :-)
$endgroup$
– winny
Jan 8 at 7:35
$begingroup$
@winny but only once, I guess.
$endgroup$
– Spehro Pefhany
Jan 8 at 7:36
$begingroup$
How come? The probe division should take the majority of the voltage.
$endgroup$
– winny
Jan 8 at 8:37
$begingroup$
@winny If it's rated for 40kV but only outputs roughly 1/3 the voltage, connecting it to a 120kV source might not be wise. ;-)
$endgroup$
– Spehro Pefhany
Jan 8 at 12:36
$begingroup$
Of course! Hence correction factor!
$endgroup$
– winny
Jan 8 at 13:27
$begingroup$
It would be highly impractical to manually account for a very uneven correction factor, but OP would be able to feed it higher voltages than before :-)
$endgroup$
– winny
Jan 8 at 7:35
$begingroup$
It would be highly impractical to manually account for a very uneven correction factor, but OP would be able to feed it higher voltages than before :-)
$endgroup$
– winny
Jan 8 at 7:35
$begingroup$
@winny but only once, I guess.
$endgroup$
– Spehro Pefhany
Jan 8 at 7:36
$begingroup$
@winny but only once, I guess.
$endgroup$
– Spehro Pefhany
Jan 8 at 7:36
$begingroup$
How come? The probe division should take the majority of the voltage.
$endgroup$
– winny
Jan 8 at 8:37
$begingroup$
How come? The probe division should take the majority of the voltage.
$endgroup$
– winny
Jan 8 at 8:37
$begingroup$
@winny If it's rated for 40kV but only outputs roughly 1/3 the voltage, connecting it to a 120kV source might not be wise. ;-)
$endgroup$
– Spehro Pefhany
Jan 8 at 12:36
$begingroup$
@winny If it's rated for 40kV but only outputs roughly 1/3 the voltage, connecting it to a 120kV source might not be wise. ;-)
$endgroup$
– Spehro Pefhany
Jan 8 at 12:36
$begingroup$
Of course! Hence correction factor!
$endgroup$
– winny
Jan 8 at 13:27
$begingroup$
Of course! Hence correction factor!
$endgroup$
– winny
Jan 8 at 13:27
|
show 1 more comment
$begingroup$
Only if the input impedances are the same to form a voltage divider.
answered Jan 8 at 6:36
Sunnyskyguy EE75Sunnyskyguy EE75
65.4k22295
$endgroup$
add a comment |
$begingroup$
Only if the input impedances are the same to form a voltage divider.
answered Jan 8 at 6:36
Sunnyskyguy EE75Sunnyskyguy EE75
65.4k22295
$endgroup$
add a comment |
$begingroup$
Only if the input impedances are the same to form a voltage divider.
answered Jan 8 at 6:36
Sunnyskyguy EE75Sunnyskyguy EE75
65.4k22295
$endgroup$
Only if the input impedances are the same to form a voltage divider.
answered Jan 8 at 6:36
Sunnyskyguy EE75Sunnyskyguy EE75
65.4k22295
answered Jan 8 at 6:36
Sunnyskyguy EE75Sunnyskyguy EE75
65.4k22295
answered Jan 8 at 6:36
Sunnyskyguy EE75Sunnyskyguy EE75
65.4k22295
answered Jan 8 at 6:36
Sunnyskyguy EE75Sunnyskyguy EE75
65.4k22295
65.4k22295
add a comment |
add a comment |
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