Revision as of 15:26, 23 January 2009

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

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

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

Note:

This method is not a replacement for a real calibration.

@@ Line 10: / Line 10: @@
 The power in the lower 50 ohm impedance is <math>P = \frac{U^2}{R} = 20 \ mWatt</math>
-So 1 ohm: <math>10*^{10}log(\frac{0,02}{50}) \approx -33,98 dB</math>
+So 1 ohm: <math>10*^{10}log(\frac{0,02}{50}) \approx -33,98 dB = -20*^{10}log(50)</math>
 = Reference measurement =