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I was trying to use a DM-4100A multimeter to verify a power supply, and I'm not sure I'm using the multimeter correctly. The multimeter is >10 years old too, so maybe it is not working correctly.

When I set the meter to measure AC voltage, with a 500V or 200V max, it reports 170-175V. I was expected more around 120V (I am in the USA). Is 170V an expected voltage, or would that suggest I'm doing it wrong?

UPDATE: It sounds like it may be reporting peak voltage instead of RMS. The multimeter does not seem to have an RMS setting. Can I just take the voltage and divide by 1.414 to get the RMS voltage? I want to do this for the purposes of verifying whether power adapter is still producing the correct voltage.

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    \$\begingroup\$ Many digital multimeters will give wrong readings if the battery is weak. Try measuring 9 volt and 1.5 volt batteries, as another reference. I have an older meter that started reading incorrectly for no apparent reason, and a fresh battery didn't help. \$\endgroup\$ May 16, 2013 at 22:33
  • \$\begingroup\$ The document strongly suggests that the instrument is supposed to measures RMS AC. \$\endgroup\$
    – Kaz
    May 17, 2013 at 1:31
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    \$\begingroup\$ Warning! This could also be a loose neutral picking up another phase! Measure between one leg and ground and you should get ~ 120v, and another leg and ground should get very near 0v. If you have a substantial non-zero voltage you need to get an electrician out immediately. \$\endgroup\$ May 17, 2013 at 18:49
  • \$\begingroup\$ When using a multimeter I frequently reverse the leads to verify a reading. If I see 170 volts AC then I switch the multimeter's black and red wires and see if it still reports 170 volts. As 170v is not expected I would disconnect that power supply ASAP and then try the multimeter in several AC outlets and perhaps in another building to see if the meter is consistently wrong or your power supply is unusual or failed. The power supply may be generating an unusual waveform that the multimeter interprets as 170 volts. As it's supposed to be 120 volts I would discard that power supply as failed \$\endgroup\$ May 20, 2021 at 17:27

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The manual you linked to, on page 3 says "Average responding, calibrated in RMS of sine wave", so it should not display the peak value of the AC voltage. I would not trust it, or "adjust the reading to RMS" by dividing by 1.4.

I suggest you get another meter, as this one appears to be broken in some way.

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At 120 \$V_{rms}\$ your peak voltage should be \$\sqrt{2}*120=169.7\$ -> ~\$170\$ so you must have a peak hold function on that DMM.

It's either that or all the appliance you have plugged into that circuit are exploding around you ... becasue 175 \$V_{rms}\$ represents a 2X factor in power, most appliances would have problems with that.

update: after posting the manual, it's apparent that this DMM does not have a \$V_{peak}\$ ability. The natural conclusion is that the DMM is not functioning properly.

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170V would be approximately the peak voltage reading of a 120V (rms) AC source. If you multiply the rms voltage of 120v by 1.414 you will get 170v. Does your meter suggest that it is measuring Peak voltage? Does it have another setting which measures RMS voltage?

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UPDATE: It sounds like it may be reporting peak voltage instead of RMS. The multimeter does not seem to have an RMS setting. Can I just take the voltage and divide by 1.414 to get the RMS voltage? I want to do this for the purposes of verifying whether power adapter is still producing the correct voltage.

If this is indeed correct, then YES.

$$V_{rms} = \frac{V_{peak}}{\sqrt{2} \approx 1.414}$$

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  • \$\begingroup\$ Just want to mention that this isn't wrong, but it's true if, and only if, you know that it's a proper sine wave AC and that you're actually measuring the peak. If either of these things is in question, assuming that you're measuring the peak of a sine wave may put you in danger. As other answers said: read your meter's documentation and get a different one if you're not sure. Don't make assumptions when it comes to mains voltages. \$\endgroup\$
    – Sarah G
    Apr 8, 2018 at 4:48

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