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When measuring the AC voltage across a DC battery (CR2032), and I am getting 6v. The DC voltage across it is 3.01 volts. I also tested on a 9 volt battery; again the same thing. I am reading about 20 v AC and 10v DC. I am using MASTECH MAS830L multi meter. I also used another multimeter, and I get the same result.

When I am putting the multimeter probes in the same polarity with the battery, I am getting the AC voltage, but none if I put the probes in reverse polarity.

Why is this happening?

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When I am putting the multimeter probes in the same polarity with the battery, I am getting the AC voltage, but none if I put the probes in reverse polarity.

This is the clue to what is happening - "nothing" when you connect in reverse means that the meter uses a simple precision half wave rectifier to convert AC to a rational DC voltage. I'm guessing the meter makes no claims to measure RMS AC voltage - it assumes the AC voltage is a sine wave and, the output from the precision rectifier is averaged (with a filter) to give a "steady value" that is representative of the AC RMS value.

The problem is that this "steady value" will be about 50% of the AC RMS value and, if the inputted voltage is always at a peak value (because it is DC), then the steady-value will also be the peak value hence, with a 3 V battery, the steady-value measured by the meter is 3 V (the meter doesn't know any better and indicates that it is measuring a 6 V AC voltage).

If it were a 6V AC source it would see about 3V.

I'm going to try and do a picture that might help visualize this so bear with me.

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  • \$\begingroup\$ Why is the "average value" 50% of the RMS value? Why is it no 50% of the peek value? \$\endgroup\$ – Arjob Mukherjee Mar 1 '14 at 17:32
  • \$\begingroup\$ @ArjobMukherjee I've added a picture - see also this link for background - en.wikipedia.org/wiki/Rectifier \$\endgroup\$ – Andy aka Mar 1 '14 at 17:37
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    \$\begingroup\$ @ArjobMukherjee to answer your question - the average value from a single diode rectifier (assuming zero volt drop across rectifier) - if the input waveform were a square wave then the average would be 50% of the peak value but the meter has to sensibly assume it is a sine wave because that is likely what most people use the meter for. \$\endgroup\$ – Andy aka Mar 1 '14 at 17:40
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Your multimeter uses a simple trick to measure AC voltages and it is pretty accurate provided that the input is pure sine wave. When you are measuring a DC source in AC mode, the applied correction factor is completely off because the input is not a sine wave. This is common behavior for somewhat cheaper range multimeters. The typical use case for these multimeters in AC mode is mains supply AC.

Only if you buy a true-RMS meter the display will be accurate, but these meters are much more expensive than one you have now. Perhaps starting from 100 or 150€.

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  • \$\begingroup\$ Yes, I was about to test a AC signal with a DC offset. If it reacts this odd with simple battery, I don't know what will happen with the signal! \$\endgroup\$ – Arjob Mukherjee Mar 1 '14 at 17:37
  • \$\begingroup\$ I do not recommend using DC ranges, especially low ones (i.e., 0.2vDC), on the AC line. The safer way to do it is set a high AC voltage range, and if that's near zero set a high DC range, and if that's near zero too, then go down a range or two and try both AC and DC again. This way you never get blind-sided by a large voltage that your meter could not see. \$\endgroup\$ – GR Tech Mar 1 '14 at 17:54
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First of all your meter it is not brocken, does not have internal DC blocking for the AC functions, and battery does not have any AC component.

In AC mode your meter runs the voltage input signal through a half wave diode rectifier (just like the old VOM), takes a time average of the result, and then multiplies by a constant factor. What is this factor? In case that is measure peak and then calculate the RMS the result will be 0.71 times the DC voltage. But in your case it is averages the waveform and multiply by

enter image description here

which is about 2.2 When connect a DC voltage at AC input, the meter does the same thing as for a sine wave, i.e. passes the signal through a diode, multiplies the time average of the result by 2.2, and displays that. But for a DC voltage, passing through a diode (in foreword direction passes quite nicely) does nothing, and the average of a constant voltage is just the voltage itself. So the meter displays 2.2 times the voltage.

You can verify the above doing this: Take a low amplitude sinewave signal from your function generator (namely 5~10V/50~60Hz) and measure with your DMM in AC scale. Next pass this through a diode and measure in DC scale, then multiply by 2.22.

If you want to block the DC component from AC signal, connect a 100nF capacitor in series with the red lead of your multimeter. The capacitor should be rated more than the peak-to-peak voltage you are going to measure. This capacitor in series with your meter input 10MΩ resistance will make a cutoff frequency from 20~30Hz (-3dB). But still you don't have an RMS multimeter as well as you you can't use this meter to read ripple on a DC supply.

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