My multimeter is rated to 600V. How can I readily (and safely) use it to measure higher voltages?
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3\$\begingroup\$ Having read about electrician Eddie Adams, and watched the video made by his colleagues, I would use a 600V multimeter to measure higher voltages by selling it and using the money to pay a very cautious experienced electrician with appropriate tools and appropriate arc-flash protective clothing/footwear/gloves/face-guard etc. But maybe I'm just a coward :-(. \$\endgroup\$– RedGrittyBrickMay 22, 2016 at 20:32
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\$\begingroup\$ @RedGrittyBrick - I assume somewhere around here there's a standard warning that "Electricity is dangerous. If you don't know what you're doing and take adequate precautions then please don't mess with it, since you could be injured or killed." \$\endgroup\$– feetwetMay 22, 2016 at 20:50
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\$\begingroup\$ What cat rating does it have? I believe quite a lot of the 20$ meters are rated to 600 V and you wouldn't trust them with my life. \$\endgroup\$– Oskar SkogMar 3, 2018 at 17:03
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1\$\begingroup\$ That movie is crazy ishn.com/articles/… \$\endgroup\$– Voltage Spike ♦May 21, 2018 at 22:31
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2 Answers
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The safest way is to buy a high voltage probe that is compatible with your meter.
I use a Fluke one that is rated at 40kVDC. Note carefully that this kind of probe is adequate for DC or mains-frequency. If you have higher frequency high voltage it is not necessarily safe.
Also, if you are using it on mains voltage make sure it is rated for the proper category for the situation or you may be at great personal danger from arc flash. Do not use an inadequately rated or unrated multimeter or probe in situations where high fault currents can occur.
answered May 22, 2016 at 3:54
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3\$\begingroup\$ Can you add details on the correct usage of such a probe? E.g., the one you have is specified for multimeters, but I can't tell what its three connectors are for. It also lists 1000:1 attenuation and a Cat I rating. So, presumably, it's good for up to 40kV DC on a multimeter with a Cat I rating to 40V DC? \$\endgroup\$– feetwetMay 22, 2016 at 4:43
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Use a voltage divider. In order to do this you first have to determine the resistance of your own multimeter. Set it to its maximum DC voltage metering range and then check the impedance through its probes with a second meter. This RM is typically 10MΩ for digital multimeters.
Now build a voltage divider using resistors.
simulate this circuit – Schematic created using CircuitLab
To get a D:1 divider (i.e., the multimeter will see and display the actual voltage divided by D) then
\$\frac{R_1}{(D-1)} = \frac{R_2 R_M}{R_2+R_M}\$
E.g., for a 10:1 divider on a typical 10MΩ multimeter:
\$\frac{R_1}{9} = \frac{R_2 \times 10M\Omega}{R_2+10M\Omega}\$, or \$R_2 = \frac{10M\Omega \times {}^{R_1}/_{9}}{10M\Omega-{}^{R_1}/_{9}}\$
So if we use a 10MΩ resistor for R1 then R2 should have resistance 1.25MΩ. Connect the multimeter to measure the voltage drop across R2, and the actual voltage being measured will be 10 times that displayed value.
Important note: Check the voltage capacity of your resistors! Standard ¼W carbon film resistors have a working limit of 350V. So if you're metering a voltage difference that could be 6kV you would need to break R1 into a series of at least 18 such resistors. E.g., with those components in that case I would use twenty 500kΩ resistors for R1, and three 417kΩ resistors in series for R2.
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4\$\begingroup\$ Just don't forget that your resistors need to be able to take the voltage! :) \$\endgroup\$ May 22, 2016 at 0:23
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1\$\begingroup\$ @ThreePhaseEel - Good catch! Just amended answer with details on that! \$\endgroup\$– feetwetMay 22, 2016 at 0:37
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\$\begingroup\$ Interesting and simple way to measure high voltage, but as I see it you are unnecessarily complicating things. I suggest you forget the impedance of the meter. If you use for instance 100 kohms resistors all the way, the impedance of the meter has no importance at all. You are just dividing the voltage as you wish. Agree that you have to make sure that the resistors can handle the voltage (and current) which means that you have to use many resistors. \$\endgroup\$– JornMar 3, 2018 at 16:09