True RMS? NOT!

True RMS Voltage? NOT!

Modern Digital Multimeters Often Give an Incorrect Voltage Reading

By Doug Criner

New info: After first publishing this webpage, I received some additional information from Fluke. Scroll to the bottom of this webpage for this info.

My Fluke 179 DMM is a very nice, relatively expensive digital meter, and I’m very satisfied with it. It has many bells and whistles: auto ranging, temperature measurement, frequency, virtually drop-proof...the whole nine yards. The protection built into the meter is excellent–I've had hundreds of DC or AC volts connected to the leads with the meter on the ohms scale, mA scale, etc., without any problem whatsoever. Not one of those cheapo, generic, Asian knock-offs, no sir.

You might not be able to read the lettering above the LCD display in the above photo, but it’s labeled “True RMS Multimeter.” However, like most other such meters, it measures the true RMS of a periodic waveform only under special conditions–namely, if the waveform is symmetric about the zero axis.

By the way, why do we speak of true RMS? Can there be more than one RMS value of a voltage waveform, but only one of them is true? Apparently so.

To use simple Ohm’s law and power calculations for a periodic waveform, we must know its RMS, or root-mean-square value–which is the effective power-producing value. For a sinusoidal waveform, the RMS value is equal to the peak voltage divided by √2. What we call 120-VAC power is the RMS value of a sinusoid with positive and negative peaks of 170 V.

But built into my DMM’s firmware is the assumption that whatever periodic waveform it measures is symmetric about the zero axis, without a DC bias component.

Consider the square wave shown below in Figure 1. The RMS value is a. (I wonder why we call it a square wave when it's more like a rectangular wave, but I digress.)

Figure 1 - Square Wave With Amplitude a, RMS Value = a

Now, let's pass that square wave through a diode in series with a resistive load, clipping off one half, leaving the waveform shown in Figure 2. The RMS value of the new waveform is √(a²/2) = a/√2 = 0.707a. (If you don't believe this, then you can integrate the square of the voltage over one period, average it by dividing by the period, and then take the square root.)

Figure 2 - Square Wave With Amplitude a, One-Half Clipped by Diode. RMS Value = 0.707a (Not a/2)

But, my true-RMS meter will not give the correct value, 0.707a, for the waveform in Figure 2. Instead, the meter, in its infinite digital wisdom, conjures up the bogus waveform shown in Figure 3, symmetric about the zero axis, and calculates the RMS value for that. The RMS value of the waveform in Figure 3 is √[(a/2)²] = a/2 ≠ 0.707a; so the meter gives a/2 as the answer, which isn't close enough even for government work.

Figure 3 - Fictitious Wave Form Concocted by Meter from Figure 2. RMS Value = 0.5a (Incorrect)

Engineers, faced with a complex computational problem, often simplify and revise the problem until it becomes one that they can more readily solve. That's what my meter does–with reckless abandon.

Here is a practical problem which trips up some designers. Suppose that a resistance heater, connected to 120 VAC, is putting out too much heat. Instead of installing a step-down transformer or replacing the resistance heater, the suggestion is made: just put a diode in series with the load. That will reduce the RMS voltage by half, to 60 V, right? Wrong.

The diode will reduce the RMS voltage to only 0.707 x 120 = 85 V. True, the diode will drop the power by half, but not the voltage. It’s easy to fall into this trap, and an “untrue” RMS meter can fool you even more.

Although I'm not willing to pay the price, you can buy a “really, really true" RMS meter from Fluke or others. But until then, don't rely on your meter when measuring asymmetric waveforms.

Maybe Fluke gets off the hook because this problem applies to almost all manufacturers' DMMs, not just my Fluke 179. But in the interest of accuracy, this issue should be prominently explained in the owner's manual. I would also advise against using the term "true RMS" in any context.

Perhaps I should submit to Fluke a suggested new advertising slogan: If it works, it must be a Fluke. (Rim-shot!) Sorry, that's a very old joke, and I expect everybody at Fluke has heard it many times.

Additional Information

After posting the above webpage, I contacted Fluke to make sure I correctly understood how the true RMS function worked. Most of Fluke's (and their competitors') DMMs are AC coupled, which gives the RMS value of only the AC component of a waveform. It seems that the AC-coupled true RMS thing started many years ago in the electrical industry where a majority of the measurements were predominately sinusoidal with no dc offset – and perhaps the technical information hasn't been as widely publicized as could be.

Fluke offered the following work-around for measuring the RMS with an AC-coupled DMM. First, measure the RMS value of the AC component. Then, measure the waveform on the DC scale. Combine the AC and DC components by squaring each, adding, and then extracting the square root. Some of Fluke's higher end products have a function called AC+DC which essentially does the calculation for you.