Can I use any HV Probe for a Voltmeter/Ampmeter gauge?












1














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?
Voltmeter/Ampmeter gauge






power-supply high-voltage multimeter probe







share|improve this question



























    1














    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?
    Voltmeter/Ampmeter gauge






    power-supply high-voltage multimeter probe







    share|improve this question

























      1












      1








      1







      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?
      Voltmeter/Ampmeter gauge






      power-supply high-voltage multimeter probe







      share|improve this question













      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?
      Voltmeter/Ampmeter gauge






      power-supply high-voltage multimeter probe




      power-supply high-voltage multimeter probe






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 2 days ago









      Connor OlsenConnor Olsen

      61




      61






















          2 Answers
          2






          active

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          4














          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.






          share|improve this answer





















          • 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



















          0














          Only if the input impedances are the same to form a voltage divider.






          share|improve this answer





















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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4














            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.






            share|improve this answer





















            • 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
















            4














            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.






            share|improve this answer





















            • 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














            4












            4








            4






            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.






            share|improve this answer












            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.







            share|improve this answer












            share|improve this answer



            share|improve this answer










            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


















            • 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













            0














            Only if the input impedances are the same to form a voltage divider.






            share|improve this answer


























              0














              Only if the input impedances are the same to form a voltage divider.






              share|improve this answer
























                0












                0








                0






                Only if the input impedances are the same to form a voltage divider.






                share|improve this answer












                Only if the input impedances are the same to form a voltage divider.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 2 days ago









                Sunnyskyguy EE75Sunnyskyguy EE75

                63k22194




                63k22194






























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