All sensing and control applications use shunt-based current sensing which is prone to resistive loss, amplification noise, and high-side, low-side polarity limitations. Why can't we just use Hall-effect ICs which seem easier to use?

  • \$\begingroup\$ There's a large number of such sensors available that (so the manufacturers claim) are expressly designed for brushless DC motors. \$\endgroup\$ Commented Aug 5, 2023 at 6:08

2 Answers 2


I haven't looked lately but for all of my career the answer to that was:

  1. not good at low currents
  2. expensive
  3. susceptible to external magnetic interference

The one good use for those is in the feedback type of current sensors such as made by LEM, but those are very expensive.

  • 2
    \$\begingroup\$ I dropped a hall based current sensor from a motor control design specifically because of point 3. My bldc was just too close to the the board and I couldn’t keep the flux out of the sensor - my current measurement was crap. \$\endgroup\$
    – Bryan
    Commented Jan 17, 2023 at 0:05
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    \$\begingroup\$ Yes - the one thing you can be sure about motor systems is they're going to have large magnetic fields around, that's how they work \$\endgroup\$
    – pjc50
    Commented Jan 17, 2023 at 9:43
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    \$\begingroup\$ Those are very good point; I would slightly modify the first one as in "limited dynamic range", at least wrt a shunt + diffamp, and I would add bandwidth. You can reasonably build something with 1 MHz BW with a shunt resistor, I don't know if such an hall sensor is available though. \$\endgroup\$ Commented Jan 17, 2023 at 16:47
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    \$\begingroup\$ I completely agree with all these points. Additionally, I would mention the measurement noise. Typically the measurement noise is much lower when using a shunt-based measurement instead of a Hall sensor. An advantage of the Hall sensor is, that it is often isolated from the sensed wire and is very simple to integrate. I think Hall sensors are often used for sensing large (50A+) and low frequency signals, where some isolation is required and no high measurement accuracy is required. \$\endgroup\$
    – Mau5
    Commented Jan 17, 2023 at 20:09

Two major advantages of a series resistor over a Hall-Effect IC are simplicity and reliability.

Simplicity. The resistor is a very common part, readily available from a large number of suppliers at much lower cost.

Reliability. The resistor is very simple and thus a very reliable component. When it fails, the resistor will almost certainly go open circuit and disconnect the load. The HE IC will not.

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    \$\begingroup\$ Also, you can silkscreen the resistor value onto the PCB so that if it goes black and crispy on its way to being an open circuit then it's easy to tell what value to replace it with. \$\endgroup\$ Commented Jan 17, 2023 at 18:11
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    \$\begingroup\$ It might not be such a good idea to rely on a resistor failing open to disconnect a load. There is usually a lot of available power that will then destroy the measurement circuit and possibly other components. This can be addressed by using diodes across the shunt. And perhaps an optoisolator to provide an isolated shut-down signal. \$\endgroup\$
    – PStechPaul
    Commented Jan 17, 2023 at 21:46
  • \$\begingroup\$ @PStechPaul, yes, you're absolutely right that the resistor is not to be relied on to disconnect the load. If it's to be relied on then a deliberate circuit should be used, such as a fuse/polyfuse, over-current detect etc. But I described the component failure mode, most likely of which is open-circuit which disconnects the load after the resistor is destroyed. That failure mode improves the system reliability. Let me know if I'm not getting that across correctly and clearly, I can rephrase it. \$\endgroup\$
    – TonyM
    Commented Jan 18, 2023 at 23:18

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