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.
Electronic – Direct current measurement without circuit breaking
constant-currentcurrentcurrent measurement
Related Solutions
The opamp version of a TIA looks like this, (one of my fav. circuits.)
simulate this circuit – Schematic created using CircuitLab
In a nutshell, the first opamp and the transistor work together to draw a current through the 100Ω resistor such that the voltage drop across it matches the voltage drop across the 0.01Ω sense resistor. Since the former is 10000× the latter, the current through the transistor will be 1/10000 of the sensed current.
The current through the transistor is then fed through the two 1000Ω resistors, which creates a voltage drop of
$$2000\Omega \cdot \frac{I_{LOAD}}{10000} = 0.2 I_{LOAD}$$
The second opamp is just a voltage follower (buffer) for this voltage.
HOWEVER, the circuit cannot work as shown. The two inputs to the first opamp are at a value very close to V_PWR (24V), while the power supply for that opamp is at only ±12V. No opamp can work with its inputs that far outside the supply rails. You should use one of the dedicated current-sensing chips for this part of the circuit, which are specifically designed to deal with this situation.
In order to change the range of operation from 25A to 10A, the simplest thing would be to change the 2K resistance to 5K.
Best Answer
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.