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I'm constructing a boost converter, and I need to measure both the input current and the output current. Currents range anywhere from 25A to 200A, depending on the model. My controller is referenced to the negative rail of the converter. I've been focusing on hall effect sensors, but it occurs to me that I could use shunt resistors in the negative leg instead. What are the advantages and disadvantages of each approach?

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I was going to answer but I've read your bio and I bet that's a "Q&A style" question. Waiting for your answer here. –  Vladimir Cravero Jul 28 at 15:41
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I have some ideas, but I'm less confident in them than I have been on past questions! If you have an answer, please, post it! –  Stephen Collings Jul 28 at 16:03
    
Well I can add my two cents, I just feel I am way, way less experienced than you. I hope I can help you anyway. –  Vladimir Cravero Jul 28 at 16:09
    
Could we talk about sensitivity and noise? –  George Herold Jul 29 at 0:25
    
@GeorgeHerold that's out of my capabilities but I feel that much more might be added to this question, I'm starting a bounty on it. –  Vladimir Cravero Jul 31 at 14:19

1 Answer 1

up vote 15 down vote accepted

I am not an expert in the field but I can try to help jotting down some quick ideas.

Hall effect sensor
pros:

  • galvanic insulation between the measurement circuit and the circuit to be measured
  • they can be placed anywhere on the current path (voltage is not a problem), thus easyness of installation and eventually servicing
  • they nearly do not affect the measured current so they are great if this is a concern

cons:

  • cost: a high current, precise sensor can cost tens of bucks
  • bandwidth: the sensor and the sensed wire are coupled through a transformer, and of course it has its own frequecny response. A piece of copper (aka shunt resistor) is less affected by this problem.
  • magnetic fields: an external fixed magnetic field can cause an offset in the measurement that must be somehow taken into account

Shunt resistor
pros:

  • small and cheap, I bet that with a good pcb manufacturer you can make your shunt resistor on the pcb paying only for the increased size, but keep in mind that the copper resistivity depends on temperature, moreover the pcb outer layers thickness is not precise while inner layers are somewhat better.

  • You can get cheap SMD shunt resistors down to 1m\$\Omega\$ from ohmite

cons:

  • they can dissipate quite an amount of power, and a tradeoff exists between precision and dissipated power. They can get quite hot too.
  • they do affect the measured circuit, namely there's a voltage drop across them and that might not be acceptable for very low voltage, high current applications. You can't measure the current consumed by an array of cores that are powered with 1.8V with a shunt that drops some 100mV

That is just what comes to me from the top of my mind, I'd be very happy to integrate/correct this list reflecting any reasonable comment from below.

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I think the word you want is "affect", as in "they do not affect the measured circuit". Great answer! –  Stephen Collings Jul 28 at 16:26
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Bandwidth is sometimes an issue for hall effect devices. –  Andy aka Jul 28 at 16:44
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Paul J. Ste. Marie chimed in that "You can also get surface-mount 4-terminal shunt resistors from Ohmite, etc, down in the 1-10 milliohm range." [It was proposed as an edit. It's useful information, but not a valid edit. I took a liberty to convert it into a comment.] –  Nick Alexeev Jul 28 at 19:02
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Making a shunt resistor on the PCB turns out not to be a great idea. First, The tempco of copper stinks. (about 1/temperature, with T in degrees K) And for the outer layer of a pcb the copper thickness is very uncertain. (Inner layers are better.) –  George Herold Jul 28 at 20:05
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Also Hall effect devices can be susceptible to external or residual magnetic fields, which can cause offsets in measurement. (That's why there's a degauss button on current probes.) –  John D Jul 31 at 15:15

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