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I'm trying to measure the current through a 1 milli ohm shunt resitor, I need a resolution of at least 250mA but I only have some op 07, I'm from Venezuela an here its pretty hard to find components, the best I could find was the op07 and the 1milli ohm shunt, it's feasible to do wthat I need? the resistence of any wire is bigger than the resistence of the shunt so I can't use the op as an non inverting, the wire resistence will give me an offset of what I have any control, I'm thinking to use three amplifiers to get a diferential input. There is a better way to approach this problem? I would do it with regular 5% resistance, I can allow an error of about 10-15% I have already tested the circuit suggested by Spehro Pefhany, I have to use potentiometers to get the values of resistors.

EDIT

I have already tested the circuit suggested by Spehro Pefhany, I have to use potentiometers to get the values of resistors.

The first two values are quite far from each other , I'm thinking in take the median and used to calculate the current, any better sugestion?

The resistance is supposed to be of 1miliohm

enter image description here

Edit:

I have made more measurements and get very different values, the red ones are with a small load connected between the shunt and ground, I think there is a problem with common mode voltage.

enter image description here

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  • \$\begingroup\$ AC or DC makes a difference. \$\endgroup\$ – Andy aka Jul 7 '15 at 7:07
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0.25A * 0.001 ohm is 250uV, which isn't too bad.

Vos of the E version is 75uV maximum, and you can null that out. The more difficult one is TCVos which is maximum 1.3uV/°C. If you allow 20°C change that's 26uV, which is 26mA on your shunt.

The OP07 is a very old op-amp but not a bad precision op-amp especially if you have the dual supplies available. To get much better you have to use a chopper type of amplifier, which can have other issues.

You should be fine. Make sure to follow the data sheet requirements on supplies (especially common mode range) and (very important) make sure you use a proper Kelvin connection to the shunt.

Edit: Here is what I mean by Kelvin connection- this amplifier has a gain of 50, so 5.0V out for 100A through the shunt. R2-R5 represent connection and leadwire resistances, not real resistors.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Thanks for the answer that That reassures me, isn't that four wires connection for resistance measurement? I can used to for measure the current? \$\endgroup\$ – Luis Ramon Ramirez Rodriguez Jul 7 '15 at 1:01
  • \$\begingroup\$ It's the same concept. The heavy current goes through the "outer" connections ("force" for resistance measurement), the "inner" connections ("sense" for resistance measurement) go to a differential amplifier (which can be made with your OP-07). \$\endgroup\$ – Spehro Pefhany Jul 7 '15 at 1:26
  • \$\begingroup\$ There isn't a problem with the voltage of the inverter input been so close to ground? that was the connection that I had tried, but with 10k resistors, I had got a quite big nonlinear error, for currents between 50 and 300 mA, now I have to try with a bigger current. I had make the test in a breadboard maybe thats why I get that error. If it works well I will like to use it to measure the inductor current in a dc dc converter, other wise (if works well, but not that well) just to measure the current in the load. \$\endgroup\$ – Luis Ramon Ramirez Rodriguez Jul 7 '15 at 2:44
  • \$\begingroup\$ Not at all- but you do need a big enough negative supply. At least -5V is good. \$\endgroup\$ – Spehro Pefhany Jul 7 '15 at 2:58
  • \$\begingroup\$ I have made 4 more measurements, and get differences up to 50%, did you now how can I calculate the common mode voltage of the circuit? \$\endgroup\$ – Luis Ramon Ramirez Rodriguez Jul 13 '15 at 21:05

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