Tektronix Technical Forums are maintained by community involvement. Feel free to post questions or respond to questions by other members. Should you require a time-sensitive answer, please contact your local Tektronix support center here.
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...
Who is online
Users browsing this forum: No registered users and 1 guest