I am planing to use an LF 510 -S / SP15LEM current transducer.

I am expected to sense -300A to 300A. What is the output of the Hall effect? Is it mA or mV?

If it is mA I will need a resistor to have a potential difference and then feed to the A/D or micro. Will this decrease the accuracy?

  • 1
    \$\begingroup\$ What even is this current transducer? It says it works for DC, but then it lists a "number of secondary turns" in the datasheet, a parameter that only makes any sense if it's a current transformer, which only works for AC. It also outputs a current instead of a voltage, which is pretty unusual for things that aren't current transformers. But then it requires a power supply, which is very unusual for things that are current transformers... What is this thing‽ \$\endgroup\$
    – Hearth
    May 11, 2022 at 15:43
  • \$\begingroup\$ @Hearth A closed loop Hall sensor \$\endgroup\$
    – devnull
    May 11, 2022 at 15:45

1 Answer 1


What is the output of the hall effect? Is it mA or mV?

There is no direct access to the Hall sensor with this device. The output signal is current (\$\pm 314\$ mA peak). The datasheet describes the device as a "Closed loop (compensated) current transducer" and based on details provided by the manufacturer, you can see that the cancellation current is the actual output (instead of simply amplifying the Hall sensor output):

enter image description here

Will need a resistor to have a potential difference and then feed to the A/D or micro?

If the ADC needs a voltage input, a current to voltage converter is needed. A resistor is the simplest one but more information is needed about the ADC to determine how the different current directions should be handled.

So will this decrease the accuracy?

Yes. Practically every single component, with its own tolerance, added to the signal path will add another source of uncertainty to calculate the overall precision of the circuit.


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