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I have this DC electronic load.

I am setting this electronic load to a constant resistance mode option - say 100 ohms - and turning the output ON.

When I measure the resistance across the electronic load terminals using a multimeter, I am not able to read the value as 100 ohms. I am getting high resistance values and it is showing as open circuit.

However, if I connect a power supply to the electronic load and ask it to consume a current equivalent of 100 ohms, the electronic load is consuming the required current from the power supply as if it was connected to a 100 ohm resistance across it.

  • Power supply was set to 10V.
  • DC constant resistance was set to 100 ohms.
  • Current consumption to the power supply was 100mA.

So, actually, the electronic load seems to be setting a constant resistance of 100 ohms when I set it as such, but I am not able to measure using the multimeter. Why is that?

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From the manual:

https://bkpmedia.s3.amazonaws.com/downloads/manuals/en-us/8540_manual.pdf

You need at least 1mA of input current for proper operation. Your DVM does not supply that much for resistance measurements.

Page 19:

enter image description here

How much current does your voltmeter supply? It's not readily found. However, I tried a simple setup here, and this is what I got for a 100 ohm resistor (0.3 mA):

enter image description here

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  • \$\begingroup\$ Thank you. So, you're saying that the input terminals of the DC Load would require atleast 1mA so as to read 100Ohms (which I have set) in the multimeter? And fluke.com/en-in/product/electrical-testing/digital-multimeters/… - This is the multimeter which I am using. How much current will it provide? \$\endgroup\$ – Newbie Oct 8 '20 at 14:49
  • \$\begingroup\$ Edited the answer above indicating that my meter is about 0.3 mA in resistance mode for 100 ohm load. YMMV. \$\endgroup\$ – Chris Knudsen Oct 8 '20 at 15:20
  • \$\begingroup\$ Thank you for your answer. Could you please clarify the first question in my comment \$\endgroup\$ – Newbie Oct 8 '20 at 15:30
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    \$\begingroup\$ Disclaimer: I do not own, and have never operated this unit. An electronic load (EL) is not a passive load. It's understandable that it would require some amount of drive signal in order to operate as a programmed resistance. It appears that your EL requires that the signal applied to its inputs to be able to deliver a 1 mA signal (minimum) for operation. \$\endgroup\$ – Chris Knudsen Oct 8 '20 at 15:52
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The electronic load would do something like read the voltage at the output and draw a constant current based on a calculation of that voltage * 1/Rset.

Your multimeter provides, by design, only a tiny current, which is too low for the circuitry in the electronic load to read accurately. There should be some indication of the resolution of voltage/current measurement in the datasheet or manual.

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