I have to measure current which is sine with 37Hz frequency 1350 mA(DC) + 190 mA(AC).

The voltage in circuit is generated using PWM 27V, 600 Hz, PWM dithering adds alternating part of current. I want as small voltage drop as it is possible. There is no chance of current going negative but I would like also reverse polarity protection. The measurement system is intended to measure many examples of circuits.

What will be the circuit for measuring current going into load (top side) and out of the load (bottom side)?

I have 0-10V input on my DAQ card.

Which solution will give more precise signal measurement: shunt resistor or some current transducer or you recommend something else?

  • \$\begingroup\$ Is the graph related to your problem or is just a decoration? \$\endgroup\$ Jun 18, 2016 at 11:34
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    \$\begingroup\$ What level of voltage drop can you tolerate on the measurement circuit? If you can tolerate up to 10 V it becomes very simple. If not then you will need to amplify the signal from the shunt. Add the information into your question. Also, is there any chance that the current could go negative? If so reverse polarity protection is required. \$\endgroup\$
    – Transistor
    Jun 18, 2016 at 18:24
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    \$\begingroup\$ On top of what transistor says, you should also mention if you want top side measure or bottom side measure. Is this part of a fixed circuit or is this intended to measure many different examples of circuits? \$\endgroup\$ Jun 20, 2016 at 16:08
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    \$\begingroup\$ "I do not want voltage drop. ... shunt resistor ..." Shunt resistors work by creating a voltage drop. You need to improve your question. \$\endgroup\$
    – Transistor
    Jun 20, 2016 at 16:58
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    \$\begingroup\$ Current going into the load (top side) or current coming out of the load (bottom side). \$\endgroup\$ Jun 20, 2016 at 18:14

3 Answers 3


This sounds like a pretty straightforward application of a current sense amplifier. These amplifiers basically measure the voltage across a sense resistor and output a corresponding current. You then feed this current into a resistor to GND so you get a corresponding ground-referenced voltage you can then feed into your ADC.

The nice thing about using a current sense amp is, once you understand the theory of operation, you can easily adjust the design to meet almost any criteria basically using Ohms law (V=IR) and the power equation (P=V^2/R, P=I^2*R). So you sound consider adding it to your bag of tricks.

Here's a simple one:

Typical simple current sense amplifier IC

Simply put, the amplifier drives current into the OUT pin and this same current goes through the 100 Ohm resistor. It will drive enough current so that the voltage across the 100 Ohm resistor is exactly the same as the voltage across the 0.02 Ohm resistor (in other words, so its +IN and -IN are at the same voltage).

So in this example, 1 Amp load current through the .02 Ohms is .02 Volts. So current across the 100 Ohms is .02/100 = 0.2 mA. Same current is pushed to the OUT pin, so voltage at the OUT pin is 0.2 mA * 1k Ohms = 2 Volts.

Power through your .02 Ohm sense resistor is 1A^2 * .02 = 20 milliWatt.

How low can you set the sense resistor? Well, it depends on how much error you can tolerate. Let's say the amp has 200 uV of offset error. This means the +IN and -IN pins might have 200 uV difference (instead of the ideal 0 Volts difference). The 1 Amp load generates .02 Volts, so error term is 200 uV/.02 Volts or +/-1% of error.

So let's say you want even more accuracy, you can 1) calibrate out the error on a per board basis (I actually do this), 2) use a larger sense resistor to generate a larger sense voltage at the expense of dissipating more power, 3) use a sense amp with even lower input offset voltage spec.

If you choose 1), make sure you check the temperature coefficient of the input offset voltage (how it varies over temperature), because even if you calibrate at room temp, it might shift enough at extreme temps to exceed your accuracy spec.

And if you want to filter the signal, you can just put a capacitor across the 1k resistor and it is just a simple RC equation.

So, yes, it is more work on your part, but as I mentioned, once you figure it out, you can adapt the same circuit over and over to many design projects.

One thing I found though, some manufacturers will specify tempco's of various parameters while others will not. The good ones also have better local FAEs (Field Application Engineers) that will actually help you design stuff like this and review your design especially from the manufacturers who are geared for lower volume, but more expensive, higher performance parts. Other manufacturers are geared for low cost parts, high volume, have sloppier performing parts that you'll need to add some design margin and you won't get as much design help. You will need to find what's right for your application.


Have a look at hall-effect current sensing chips, eg. ACS711 from Allegro. They have negligible voltage drop, enough bandwidth (100kHz). Depending on version they can measure current flowing in one or both ways.

The current sensing circuit is galvanically isolated from the rest of the chip so you can do both high- and low-side sensing.

  • \$\begingroup\$ ACS711 and others have relatively poor SNR. If accurate measurement is required, it should be done on resistor, while ADC is really close to it. \$\endgroup\$
    – user76844
    Jun 22, 2016 at 18:37

Because of the earlier thread you posted (and the data inside it), you are looking for an industrial signal conditioner.

Something like the http://www.dataforth.com/model.view.aspx?modelid=64


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