# Testing a current transformer sensor - I had to reduce input voltage to get lower primary currents. Are my results valid?

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?

• Imagine the piece of wire that goes through the transformer as a resistor. Now think about what the voltage across that resistor is for different currents and how it changes with the voltage of your siggen(?) for the same current Mar 6, 2016 at 11:20
• @PlasmaHH: I'm an EE but I don't understand what your thought experiment is intended to demonstrate. Would you care to elaborate? Mar 6, 2016 at 14:49
• @transistor: ohms law. Resistance/impedance is fixed, current is fixed, so the voltage that resistor sees is fixed too Mar 6, 2016 at 20:37
• I think that might confuse the OP. He's specifically asking about testing over a range of currents for calibration. Mar 6, 2016 at 20:40

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.

My question is, does the input voltage matter as long as one is testing with the right primary current values?

Under extremes of frequency and primary voltages the relationship begins to gradually falter due to capacitive effects coupling primary wire to secondary. Also, at high frequencies the secondary leakage inductance will start to create a problem and you may also hit resonances which make the relationship between input and output currents tenuous.

However, the set up you have described should be just fine.