6
\$\begingroup\$

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.

\$\endgroup\$
  • 1
    \$\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\$ – Peter Bennett May 16 '13 at 22:33
  • \$\begingroup\$ The document strongly suggests that the instrument is supposed to measures RMS AC. \$\endgroup\$ – Kaz May 17 '13 at 1:31
  • 1
    \$\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\$ – Bryan Boettcher May 17 '13 at 18:49
5
\$\begingroup\$

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.

| improve this answer | |
\$\endgroup\$
11
\$\begingroup\$

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.

| improve this answer | |
\$\endgroup\$
5
\$\begingroup\$

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?

| improve this answer | |
\$\endgroup\$
0
\$\begingroup\$

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.

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

| improve this answer | |
\$\endgroup\$
  • \$\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 '18 at 4:48

Not the answer you're looking for? Browse other questions tagged or ask your own question.