Is it possible to measure direct current without breaking the circuit? For AC parameters its no problem using current clamp. I would like to know if something like this is possible with direct current with sufficient precision (say 0.1%). Range of current lets say is 10 mA - 10 A.
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3\$\begingroup\$ No need to say, but if you have a resistive element (theoretically also the wire, but I wouldn't rely on that) in series with the point you want to measure, you can measure the drop over that. \$\endgroup\$– clabacchioMar 5, 2012 at 12:44
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2\$\begingroup\$ We need to know the range of current you are trying to measure to give you a good answer. I have experimented significant amount with Infineon Hall sensors for this purpose it is possible but precision depends on a few things. \$\endgroup\$– KtcMar 5, 2012 at 15:14
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2 Answers
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If you can turn the wire around a solenoid, also Hall effect measurements are possible for DC current. The precision depends by the magnitude of the current you want to measure and the number of windings (they amplify the signal) you can make.
See here
Update
Given the range of values: the range you want is quite big, but for the 1-10A measure should be feasible; for lower values, yes you can use more windings (but you need to know when the range is different, so you need to separate two different ranges) or amplify the signal. But in the latter case the accuracy will be probably lower.
Consider that you have the sensor in the link: for \$10\,A\$ you have \$0.33 \, V\$, which you can measure with almost every multimeter (better if benchtop) with a 4-digit accuracy (\$0.1\% - 0.33 \, mV \$ or \$ 330 \ \mu V \$). So you use 3 windings and obtain a full scale voltage of about \$1\,V\$. You obviously have to remove the \$4\,V\$ offset with an instrumentation (differential) amplifier to obtain the best from the instrument. Note that increasing the windings the offset is not multiplied.
Then you want to measure \$10 \,\, mA\$, which with the same configuration (3 windings) give a \$1\,mV\$ full scale output, of which the \$0.1\%\$ is \$1 \mu V \$. Depending on your instrument, you can have this accuracy or not. For sure you can't without removing the offset, especially in this case.
*Final note: Amplifying the output signal (you can do it after removing the offset) can help to use the range of the voltmeter, but note that doesn't effect the accuracy of the sensor, of which you have to check the specs.
answered Mar 5, 2012 at 12:56
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Yes, for example, TTi sell the I-Prober, using a a fluxgate magnetometer which measures down to DC, but not to 0.1%.
