I am working on LEM HAIS 50P current sensor and I'm using it for measuring AC current. I have an offset of 2.5V, so at 0A the output of the sensor is 2.5v which is as expected but if I increase the current to 1A, technical there should be some changes in the output of the sensor but the output value of the sensor is still 2.5v and its the same for 2A too. I have followed the exact circuit as mention in the datasheet. I wanted to how exactly should I measure to get the desired output, i.e., if I increase the current from 0A to 1A there should be a change in the output of the sensor. Thanks https://www.lem.com/sites/default/files/products_datasheets/hais_50__400-p_and_50__150-tp.pdf +Uc=5V Uref=2.5V 0V= ground
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\$\begingroup\$ Hi there! A schematic or diagram would be great here. Even if you say you have copied the ciruit in that datasheet to the letter, some other part of your circuit could be wrong. \$\endgroup\$– MiNiMeCommented Nov 21, 2023 at 19:08
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\$\begingroup\$ Assuming you are using the model with the core hole, do you have a single conductor threaded through or are both wires threaded through? \$\endgroup\$– Spehro 'speff' PefhanyCommented Nov 21, 2023 at 19:09
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\$\begingroup\$ The datasheet shows a change in output voltage of 625mV for a sensed current change of 50A (Ipn). 1A would therefore make a change of 12.5mV, are you sure you're not interpreting this as noise? \$\endgroup\$– virCommented Nov 21, 2023 at 19:51
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\$\begingroup\$ You're not running two wires in the opening by any chance, are you? If so, it's possible that the current goes out one wire and comes back the other wire, canceling the magnetic field. \$\endgroup\$– Davide AndreaCommented Nov 21, 2023 at 20:53
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\$\begingroup\$ Just switch multimeter to AC. \$\endgroup\$– user263983Commented Nov 21, 2023 at 21:18
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1 Answer
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Measuring AC current
If you are connecting a multimeter to the output and, you are expecting to see a change when the AC current increases in amplitude, you are going to be disappointed. The device you are using produces a voltage waveform that is linked to the current waveform you are mentioning. You need to use an oscilloscope to see the change.
The circuit image from the data sheet should tell you what I'm talking about: -
Uout will site at half the DC power rails i.e. 2.5 volts and, the waveform will be superimposed on top of that.
answered Nov 21, 2023 at 20:04