0
\$\begingroup\$

I'm trying to switch from reading my high-voltage unipolar square wave signal from a benchtop oscilloscope to a handheld multimeter, but I don't understand why my multimeter is giving the voltages it's giving.

My signal, according to the oscilloscope, is shown in the picture below as a 262 V peak, 500 Hz square wave (at least approximately), and the oscilloscope calculates that \$V_{RMS}\$ = 185 V. This matches my expectation, since for a unipolar square wave $$V_{RMS} = \frac{1}{\sqrt{2}} V_{peak} = 185.26 V$$

When I sample the wire (equal to the yellow curve in the oscilloscope plot) with a multimeter, however, I measure 123.79 V using the DC voltage mode, and 120.04 V using the AC voltage mode (plus, the multimeter reports that the frequency is 0 Hz). Both are around half of the peak voltage, but I don't really understand how the sensing circuits (e.g., this Stack Overflow post) respond to a unipolar square wave signal so I'm hesitant on completely relying on this multimeter. I'm using a Kaiheets HT118E TRMS multimeter, for reference.

From my understanding, the TRMS designation means that the multimeter should report true RMS values for bipolar signals, but I'm unsure how to properly measure the voltage of a unipolar square wave using a multimeter. Any insight would be appreciated!

Oscilloscope trace of a 262 V, 500 Hz square wave.

\$\endgroup\$

1 Answer 1

Reset to default
4
\$\begingroup\$

262V peak unipolar (0V/262V) square wave will have an RMS value of 262/\$\sqrt{2}\$ = 185V.

However the meter may actually be AC-coupling the input and measuring the RMS value of that. So you would get around the reading you are seeing (it's not a clean square wave, so a bit less than the peak value).

The meter manual shows a lower limit for true RMS readings of 40Hz (40Hz to 1kHz), so it does not go down to DC. On the DC range, it is reading the average, which (in this case) will be a similar reading.

Neither is the actual RMS value.

\$\endgroup\$
4
  • 2
    \$\begingroup\$ @asyndeton256 You might try a test: If the multimeter excludes the DC offset while doing a TRMS measurement, make your oscilloscope do the same...switch to AC-coupling rather than DC-coupling. Then see what your oscilloscope reports for RMS voltage. \$\endgroup\$
    – glen_geek
    Commented Jul 11 at 23:29
  • \$\begingroup\$ @glen_geek That was a great idea! Switching the oscilloscope to AC coupling mode, the oscilloscope now read $V_{max} = 142 V$ and $V_{RMS} = 130 V$. This is a lot closer to the multimeter readings, so Spehro-Pefhany's guess about AC coupling likely holds up. \$\endgroup\$
    – asyndeton256
    Commented Jul 12 at 23:27
  • \$\begingroup\$ From my understanding, I should be reading data from the multimeter as $$V_{peak,unipolar} = V_{DC,multimeter} + V_{AC,multimeter}$$ (since $$V_{RMS} = V_{peak}$$ for a bipolar square wave). i.e., the true peak voltage was 123.79+120.04=243.83 V, even though the oscilloscope read $$V_{peak} = 274 V$$. Is this correct? \$\endgroup\$
    – asyndeton256
    Commented Jul 12 at 23:29
  • 1
    \$\begingroup\$ I think it's correct, but only if the duty cycle is exactly 50%. For example, a 100V 10% duty cycle square wave will have an RMS voltage (AC coupled) of 30V and an average DC of 10V. (The actual RMS is 100/sqrt(10) = 31.6V) and the peak is 100V. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Jul 13 at 0:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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