I am planning to design Rshunt basis current measuring. My current measuring range is 1A~2A. I need analog output of 0~5V analog with 1% accuracy. How can I do this?

I found conversion circuit here. Will it serve my application? I tried with these Hall effect current sensors from Sparkfun (1) (2).

But with these models, I found accuracy is not good; gives 5% error. Output is not stable even with load not connected. I tried with moving average and SD to filter, but distortion was greater.

Kindly give your suggestions for building shunt basis current sensor module.

Can I use this technique? Or a different one that uses an Allegro Microsystems ACS712?


4 Answers 4


Although other answers with a current sense IC were downvoted, I'll try with my favorite that I am currently successfully using in a product. The main difference with the LT6100 is that it has variable gain and 0.5% accuracy.

enter image description here

If you use it with a 100 mΩ resistor, a 2A load with drop 200 mV across the resistor. Using a gain of 20, this will give a 4v output. (You can also set the gain to 25, which would be right at the rail of the ADC.) Note that if you need to measure lower currents with higher accuracy, you can tie the programmable gain pins to the output of the microcontroller.

If you combine the LT6100 with a 100 mΩ 0.5% resistor, you will have 1% accuracy for the combination.

The LT6100 is available from Digi-Key for $2.36. I don't have any idea what it might cost in your area.


If you need the current sensor to work unisolated from your measurement equipment you can very easily use a 1%, 0.5% or 0.1% shunt resistor.

The easiest circuit is measuring the return path current, from the device into the ground node. Another trick, requiring only slightly more components is high-side measurement and level conversion. In both cases you need op-amps as well. Be aware, that every op-amp has different characteristics and your choice of one will be influenced by many details of the current signal you are not providing.

If there is a supply voltage above the highest input and output voltages and a negative supply under the lowest input and output voltage you enter the domain of very cheap op-amps with sufficient accuracy in spades. If you cannot provide these extended supply voltages you get stuck with rail-to-rail op-amps, these can have input and output voltages going from negative to positive supply. The basic R-R op-amp is affordable enough, but most all of them will give poor amplification performance in the extreme 500mV of the output signal, meaning fast-moving signals will be deformed and even slow-moving may have aberrations beyond your 1% specification in the last 250mV. These slow-moving aberrations are usually patterned, though, and might be solved with look-up tables in the digital domain. Another consideration is good input performance, but I'd expect most main stream modern op-amps to be sufficient on that respect.

I attach two example schematics, one for measuring the return current, one for measuring the high-side supply current. Be aware that getting 1% accurate components will give you worse than 1% absolute accuracy, as the absolute accuracies accumulate. I'd say, use at worst 0.5% for your shunt (not cheap) and 0.1% for all the normal resistances.

Set of example schematics drawn by hand

Note that the example assumes you have only Vin+ and Vin- as supply voltages. Standard cheap op-amps will be better off connected to (Vin+ + 3V) and (Vin- - 3V) for the high side measurement. The return current measurement can use a cheap op-amp if the Vin+ is at least 8V and there is a supply of (Vin- - 3V) or less.

Also note, this example is "the most basic". I do not know your exact requirements so I have drawn something sturdy with no filtering, so you get the best available current representation in your output signal.

Addition: So sorry for the delay, wanted to put the picture on a new permalink private domain do "keep them in personal maintenance", but as the DNS issues take long I forgot a while. Now just inserted it directly.


Several components will there for measuring the current in the circuit,out of that i will tel one of them which is used in my project Thermally Enhanced, Fully Integrated, Hall Effect-Based Linear Current Sensor.Part No:ACS758xCB

circuit diagram

The ACS758 outputs an analog signal, VOUT , that varies linearly with the uni- or bi-directional AC or DC primary sampled current, I P , within the range specified. C F is for optimal noise management, with values that depend on the application Allegro ACS758 Website

  • \$\begingroup\$ .v thanks for information. But This sensor is meant to measure 30A. How accurate you got your result.2) For measuring 2 A current it doesn't make any sense to purchase 30A sensor. up to 5A is ok for me.Even basic model ACS712 cost 250 RS in INDIA.I dont know how it cost me above sensor \$\endgroup\$
    – user50949
    Sep 22, 2014 at 6:08
  • \$\begingroup\$ Hi ACS758LCB-050U-PFF-T having Sensitivity 60 mv/A, if your ADC having more sensitivity,you can use this device for your application but this device cost is >480Rs. \$\endgroup\$
    – Praveen.v
    Sep 24, 2014 at 13:55

If you are looking for Hall effect current sensing at low cost, you can use the WCS2702/WCS2705. They are reasonably accurate and I've used them. However if you need 1% accuracy, Hall sensors won't be good enough. Your best bet is to use a current sense resistor of suitably low value. What are you trying to measure? AC or DC current? What is the voltage you are working with?

  • \$\begingroup\$ I am looking for DC one.I know these avialble in NSK.But these are current output. SO need to be amplified.If you already tried these sensor How accurate the sensor . How did to amplified to get 0~5v output.I usually use 0~5v analog but this time i am not getting proper sensor with analog out put. I tried with NSK ACS712 model . but they are more distortion.For Rshunt basis above circuit will work??? You have any other circuit please share me. \$\endgroup\$
    – user50949
    Sep 22, 2014 at 6:44
  • \$\begingroup\$ Will this circuit will workLink. I planning to use MCP602 ic for amplification?? \$\endgroup\$
    – user50949
    Sep 22, 2014 at 7:21
  • \$\begingroup\$ Some advice for a new user:It's best to use your answer to answer the question. Try not to use your answer to ask further questions. Use the OP comment section for that and update your answer based on the OP reply. \$\endgroup\$
    – akellyirl
    Sep 22, 2014 at 11:01
  • \$\begingroup\$ This is not the place to advertise goods or services. \$\endgroup\$
    – Dave Tweed
    Sep 22, 2014 at 18:56

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