Current sensor transfer impedance determination method

Theory

The left impedance is the signal generator which is generating enough power for 1 ampere.

This 1Amp. generates $P=I^{2}*R=50\ Watt$ in the right impedance.

The current sensor has 1 ohm transfer impedance, this means 1 ampere generates 1 Volt on the measuring part below.

The power in the lower 50 ohm impedance is $P={\frac {U^{2}}{R}}=20\ mWatt$

So 1 ohm: $10*^{10}log({\frac {0,02}{50}})\approx -33,98dB=-20*^{10}log(50)$

Reference measurement

Probe measurement

Calculation

$Correction\ factor(dB)=P_{Measured}-P_{Reference}+33.98$

$P_{Measured}$ and $P_{Reference}$ in dBm.

Example

On 10 MHz we have the following information:

• Calibration: 0 dBm.
• Measurement: -27,96 dBm.

So: $Imp.=-27,96-0.00+33.98=6,02dBOhm$

$Imp.\approx 2\ Ohm$

Verification

The left impedance is the signal generator which is generating enough power for 1 ampere.

This 1 Ampere generates $P=I^{2}*R=50\ Watt$ in the right impedance.

The current sensor has 2 ohm transfer impedance, this means 1 ampere generates 2 Volt on the measuring part below.

The power in the lower 50 ohm impedance is $P={\frac {U^{2}}{R}}=80\ mWatt$

So 2 ohm: $10*^{10}log({\frac {0,08}{50}})\approx -27,96dB$

The difference is $-27,96-(-33,98)=6,02dB$

Conclusion

Correction to a 1 ohm impedance is $20*^{10}log(R_{probe})$

Note: This method is not a replacement for a real calibration.