I have a CT current sensor which, according to the datasheet, has a 62\$\Omega\$ burden resistor.
I have logged the output voltage for primary currents from 1mA to 10A. However, to obtain the lower currents I had to reduce the input voltage gradually to 0V since my variable load could not go lower than 200W at 230V.
My question is, does the input voltage matter as long as one is testing with the right primary current values? I need to use the sensor with a power strip connected to the mains. Can I use my test results in that scenario even though they have been obtained at different voltages?
Best Answer
In general, the CT will have no idea of the voltage of the circuit it is monitoring. All it can see is the current passing through its core.
What might matter is how you achieve the low currents. For example, if you were to use a dimmer switch and vary the trigger angle to reduce voltage and current you may find that a cheap CT does not handle and track the pulsed current very well. (It should be fine however.)
Figure 1. Triac trigger-angle control.
A simple solution for your 1 mA test is a 230 kΩ load. This will pass \$ \frac {230 V}{230k} = 1~mA\$. Power in the 230 kΩ resistor will be given by \$P = \frac {V^2}{R} = \frac {230^2}{230k} = 230~mW\$ (< 0.25 W).
simulate this circuit – Schematic created using CircuitLab
Figure 2. Low-current test circuit.
I've shown the 230 kΩ made up of a series combination of 100k + 100k + 33k to avoid full 230 V on a single resistor.
If you make a 1 mA current then 10 turns of that through the CT will give you 10 mA, etc. Similarly, a 100 W lamp should give you about 0.5 A. Add turns to generate 1 A, 1.5 A, etc.
Other tricks you can try are series connection of incandescent lamps, pull out the Christmas tree lights and use them as a low-current load, night-lights, etc., and calibrate your CT against a direct measurement with your multimeter.
simulate this circuit
Figure 3. Current divider.
Another possible method is to divide your current. Figure 3 shows four parallel conductors of equal wire size and length (and therefore resistance also). One quarter of the current will flow through your CT.