For a student project, I need a Hall Effect current sensor. What I am going to do is to charge 8 rechargeable Li-Ion batteries of 3.7 volts in series with 34 volt supply and the buck circuit frequency is between 5KHz to 10KHz. My concern is finding the right hall-effect sensor for calculation of the input current to adjust the current flowing to the batteries.

The batteries are 4000-6000 mAh and I am going to use AVR to sense the output of current sensor. So, I need a sensor with 5 volt output. I am thinking about finding the right current sensor for this circuit. ACS712ELC-05B is an example however, it senses 5A which is too high for this circuit and reduces the sensitivity of my measurement. I need something in safe range but not too far. Could anybody help me with finding a better option?

  • \$\begingroup\$ How about a small resistor to measure current? \$\endgroup\$ – George Herold Apr 21 '15 at 12:12
  • \$\begingroup\$ @GeorgeHerold not acceptable. btw, on resistor, both pins a floating (and with high voltage). \$\endgroup\$ – barej Apr 21 '15 at 12:25
  • \$\begingroup\$ You could do a differential voltage measurement of the resistor with an opamp or instrument amp... 34 V is not what I would call high voltage. \$\endgroup\$ – George Herold Apr 21 '15 at 12:41
  • \$\begingroup\$ @GeorgeHerold it is still high for damaging AVR. \$\endgroup\$ – barej Apr 21 '15 at 13:05
  • 1
    \$\begingroup\$ Instrument amps have a reference output so you can put the output where ever you like... maybe an ina826? ti.com/lit/ds/symlink/ina826.pdf \$\endgroup\$ – George Herold Apr 21 '15 at 13:37

You should use a sense resistor and a current sense amplifier -- this will give you far superior accuracy to either a Hall effect sensor or an instrumentation amplifier.

The LTC6102 is an excellent current sense amplifier. Its maximum offset voltage of \$10\rm{\mu V}\$ permits great accuracy, but it can only sense current in one direction. If you need to sense bidirectional currents and are willing to live with a bit more error, the LT1787 is a good choice, with \$75\rm{\mu V}\$ of offset.

  • \$\begingroup\$ One direction is fine. But a few questions: can it be used alone in my case or it needs another amplifier. This image link made me worried to use another amplifier. Are those extra diodes and transistors really necessary? and is it required to remove voltage offset by a potentiometer? I do not like to use any variable resistor solution. \$\endgroup\$ – barej Apr 21 '15 at 23:53
  • 1
    \$\begingroup\$ @barej: A) The image in your link has a single amplifier. The LTC2433-1 is an Analog to Digital converter. You don't have to convert to digital if you don't want to. B) There are no diodes in the diagram, and the only transistor is inside the LTC6102. C) If you don't want to compensate for the offset, then you don't have to. In fact, that's why the offset is engineered to be so low: so that, under most cases, it's small enough that you won't need to correct for it. \$\endgroup\$ – Zulu Apr 22 '15 at 1:14
  • \$\begingroup\$ Thank you very much. For the case of my question, how much ohm resistance do you suggest me to use? \$\endgroup\$ – barej Apr 22 '15 at 12:32
  • \$\begingroup\$ This element is surface mounted. How to use it on prototype PCB boards? \$\endgroup\$ – barej Apr 22 '15 at 16:01
  • \$\begingroup\$ @barej, Page 12 of the LTC6102's datasheet provides guidance on selecting the sense resistor. Selection of the other components is covered in the following pages. \$\endgroup\$ – Zulu Apr 22 '15 at 16:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.