Can I use any HV Probe for a Voltmeter/Ampmeter gauge?
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 2 days ago
Connor OlsenConnor Olsen
61
add a comment |
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 2 days ago
Connor OlsenConnor Olsen
61
add a comment |
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 2 days ago
Connor OlsenConnor Olsen
61
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 2 days ago
Connor OlsenConnor Olsen
61
asked 2 days ago
Connor OlsenConnor Olsen
61
asked 2 days ago
Connor OlsenConnor Olsen
61
asked 2 days ago
Connor OlsenConnor Olsen
61
asked 2 days ago
Connor OlsenConnor Olsen
61
61
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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 2 days ago
Spehro PefhanySpehro Pefhany
204k4150408
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 :-)
– winny
2 days ago
@winny but only once, I guess.
– Spehro Pefhany
2 days ago
How come? The probe division should take the majority of the voltage.
– winny
2 days ago
@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. ;-)
– Spehro Pefhany
yesterday
Of course! Hence correction factor!
– winny
yesterday
|
show 1 more comment
Only if the input impedances are the same to form a voltage divider.
answered 2 days ago
Sunnyskyguy EE75Sunnyskyguy EE75
63k22194
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
});
});
}, "mathjax-editing");
StackExchange.ifUsing("editor", function () {
return StackExchange.using("schematics", function () {
StackExchange.schematics.init();
});
}, "cicuitlab");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "135"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f415818%2fcan-i-use-any-hv-probe-for-a-voltmeter-ampmeter-gauge%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 2 days ago
Spehro PefhanySpehro Pefhany
204k4150408
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 :-)
– winny
2 days ago
@winny but only once, I guess.
– Spehro Pefhany
2 days ago
How come? The probe division should take the majority of the voltage.
– winny
2 days ago
@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. ;-)
– Spehro Pefhany
yesterday
Of course! Hence correction factor!
– winny
yesterday
|
show 1 more comment
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 2 days ago
Spehro PefhanySpehro Pefhany
204k4150408
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 :-)
– winny
2 days ago
@winny but only once, I guess.
– Spehro Pefhany
2 days ago
How come? The probe division should take the majority of the voltage.
– winny
2 days ago
@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. ;-)
– Spehro Pefhany
yesterday
Of course! Hence correction factor!
– winny
yesterday
|
show 1 more comment
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 2 days ago
Spehro PefhanySpehro Pefhany
204k4150408
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 2 days ago
Spehro PefhanySpehro Pefhany
204k4150408
answered 2 days ago
Spehro PefhanySpehro Pefhany
204k4150408
answered 2 days ago
Spehro PefhanySpehro Pefhany
204k4150408
answered 2 days ago
Spehro PefhanySpehro Pefhany
204k4150408
204k4150408
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 :-)
– winny
2 days ago
@winny but only once, I guess.
– Spehro Pefhany
2 days ago
How come? The probe division should take the majority of the voltage.
– winny
2 days ago
@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. ;-)
– Spehro Pefhany
yesterday
Of course! Hence correction factor!
– winny
yesterday
|
show 1 more comment
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 :-)
– winny
2 days ago
@winny but only once, I guess.
– Spehro Pefhany
2 days ago
How come? The probe division should take the majority of the voltage.
– winny
2 days ago
@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. ;-)
– Spehro Pefhany
yesterday
Of course! Hence correction factor!
– winny
yesterday
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 :-)
– winny
2 days ago
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 :-)
– winny
2 days ago
@winny but only once, I guess.
– Spehro Pefhany
2 days ago
@winny but only once, I guess.
– Spehro Pefhany
2 days ago
How come? The probe division should take the majority of the voltage.
– winny
2 days ago
How come? The probe division should take the majority of the voltage.
– winny
2 days ago
@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. ;-)
– Spehro Pefhany
yesterday
@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. ;-)
– Spehro Pefhany
yesterday
Of course! Hence correction factor!
– winny
yesterday
Of course! Hence correction factor!
– winny
yesterday
|
show 1 more comment
Only if the input impedances are the same to form a voltage divider.
answered 2 days ago
Sunnyskyguy EE75Sunnyskyguy EE75
63k22194
add a comment |
Only if the input impedances are the same to form a voltage divider.
answered 2 days ago
Sunnyskyguy EE75Sunnyskyguy EE75
63k22194
add a comment |
Only if the input impedances are the same to form a voltage divider.
answered 2 days ago
Sunnyskyguy EE75Sunnyskyguy EE75
63k22194
Only if the input impedances are the same to form a voltage divider.
answered 2 days ago
Sunnyskyguy EE75Sunnyskyguy EE75
63k22194
answered 2 days ago
Sunnyskyguy EE75Sunnyskyguy EE75
63k22194
answered 2 days ago
Sunnyskyguy EE75Sunnyskyguy EE75
63k22194
answered 2 days ago
Sunnyskyguy EE75Sunnyskyguy EE75
63k22194
63k22194
add a comment |
add a comment |
Thanks for contributing an answer to Electrical Engineering Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f415818%2fcan-i-use-any-hv-probe-for-a-voltmeter-ampmeter-gauge%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
