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The end goal is a resistance measurement for bulk semiconductor materials, we prefer AC measurement to avoid thermoelectric drifts. To validate my input current I am measuring AC amperage through a verified 1.01 ohm resistor with a Keithley 2000-20 multimeter.
The meter measures .94A AC during the sweep. If I send 1A without sweeping and read DC current, it measures 1.01A; as expected.
I'd like to better understand the difference in magnitude. How is the measurement impacted by not sending a sine wave, nor a bipolar square wave?
120 Hz AC is pretty quick for a 2400 to output, so yes it will definitely be tricky to get a good sine wave. The quickest you will likely be able to do would be about 1 point per ms so you could program a list sweep with about 8 points per cycle to get 120 Hz. It won't be a great sine wave, but it will be better than what you get with 2 points per cycle.
Since this is not a true sine wave, you cannot do a simple peak measurement and then divide by the square root of 2 to get the RMS value. Are you using a scope to measure waveform? Assuming you are using a TekScope, the RMS measurement on the scope makes a true RMS measurement, integrating the area under the curve, so you will get an accurate RMS measurement, whether it is a sine wave or not.
The graph above was AC voltage measured by a 500kS/sec DAQ unit. What I really want to know is how the Keithley multimeter 2000-20 arrives at its display number. Is it also integrating under the curve? It doesn't make sense to me why AC (0.94 amps) and DC (1.01 amps) numbers are so different...
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