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How does a typical DMM measure AC current (circuit level)?
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replaysMike:
Hey all,
I've taken apart a few of my DMM's in trying to understand how they are measuring AC current, and scratching my head.
I see that they all seem to take the same approach using a shunt resistor (Constantan 0.1ohm). Looking at my Extech EX330 I see a TLC27M2C op amp that looks like its used for measuring the voltage drop on the shunt resistor. There are some other components before it gets there and I can't identify them all (Q4-Q7) so I don't know what they are doing before it hits the op amp.
The op amp has 5.9VDC at VCC, and I measured the default output of the op amp (on both outputs) and it's 1.5 VDC. When I run a 1amp AC current through it there is no change in output yet the meter is clearly measuring 1 amp. Perhaps this op-amp doesn't really deal with the current measurement but I can't see anything else that obviously does before it enters the MCU. And if it does, I don't understand how it works as the common mode voltage range is only 0.2-3.0V on that op amp and I'm measuring 120VAC. Because its a floating power supply it doesn't matter?
Anyone with more knowledge on the subject? I only half understand it.
Kleinstein:
The AC current measurement is usually just like DC, with a shunt and than measuring the small voltage. There may be some extra AC amplification with an OP or similar - however with battery operation this is more likely left out to save power.
For the AC voltage measurement there are different options: the cheap ones may use a kind of precision rectifier (e.g. an OP circuit with diodes), while the better ones tend to use an analog RMS to DC circuit to get true RMS readings. A few may also use a reasonable fast ADC to sample the AC waveform and than do the RMS calculation in software. With modern DMM chip sets much of this may be part of the big central chip.
For higher voltage (e.g. more than 0.2 V) there is a resistive divider at the input. For a cheap meter this could be the same divider also used for DC volts. The 1.5 V level seen could be a virtual ground connected to the common terminal. So relative to the signal ground the supply is more like -1.5 V and + 4.4 V.
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