# Currect measure through ADC and shunt

I spent some time already and I think I forget about something in my problem.

I have a motor, which single phase is R17 = 2 Ohm. I want to measure a current through it's circuit. I used a shunt with R11 = 5 mOhm. I measure voltages before and after the shunt, so I get the circuit:

R12 and R18 is R = 55 MOhm - those are ADC inputs of my AtMega8. I used two voltage dividers R13 & R10 and R15 & R14 to output the voltage to ADC-friendly level - in this case ~4 V. Measuring the ADC would output me almost the same result, because the voltages are different in the level of 0.01 V.

What should I do to output a nice, measureable results (like 4 V and 2 V is a good difference)?

EDIT 1:

I created another circuit with LT6106 as pipe suggested:

I think it worked like a charm. How do you think? Maybe I should add anything else?

EDIT 2:

I had a problem. I can't get fast enough any LT6106 and it's not a cheap one. I wanted to find any on my old boards(PCs, supplies) but I have none.

I have got few desoldered LM358 from ST. I managed to design a simple schematic that amplifies the differential voltage (dropout on R52).

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What do you think of it now? Is something wrong or something still lacks?

• Make sure to read the datasheet of the IC you end up using. The LT6106 datasheet tells you explicitly that your R35 (Rin) should not be larger than 500 ohm, and it has a section that tells you how to select Rin and Rout depending on your load.
– pipe
Oct 21, 2017 at 21:12
• Thank you again! Yes, you're right! I am doing many things in parallel and I just designed that circuit to see if it does work for me. I don't have LT6106 so yes - when I will choose mine IC - I will surely see it's datasheet for every requirement. Oct 21, 2017 at 21:17
• @pipe I edited again. I created another design, because I don't have any LT6106 or near that, but I do have LM358. It works in spice, but I should know if it really does before soldering. Oct 22, 2017 at 20:47

Use a high-side current sense amplifier.

These ICs are specifically designed to do this job: Amplify a tiny voltage across a tiny resistor placed on the "high side" of your circuit.

There are thousands of them, the first I found is the LT6106 from Linear, but there are many more. Here is a link from digikey to start with.

They have an integrated amplifier, and often use a clever and very simple feedback "trick" so that you just have to add two resistors to set the gain you want:

(From the LT6106 datasheet)

• I did as you suggested. You can see in my edit. It works, thanks! I didn't know it's so easy and those op-amps ICs are so wonderful. Oct 21, 2017 at 19:52
• @NikoValaday Good! Don't remember to mark an answer as accepted if it helped you. Sometimes it could be worth to wait for at least a full day to see if a better answer comes around.
– pipe
Oct 21, 2017 at 21:11

The answer is simple: Use a shunt with more resistance, and increase the supply voltage to account for the additional voltage drop.

simulate this circuit – Schematic created using CircuitLab

You can use a voltage regulator with a separate feedback input to keep the output voltage constant regardless of the shunt.

• Supply voltage has to be 12V for the purpose of my circuit. In this case more resistance means more Watts and Vdropout which I can't allow. I didn't tell you everything about the purpose of this circuit so it could mislead you, but thank you! Oct 21, 2017 at 21:22