Revision as of 15:19, 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 $P={\frac {U^{2}}{R}}=20\ mWatt$

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

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

Note:

This method is not a replacement for a real calibration.

@@ Line 1: / Line 1: @@
+= Theory =
 [[Image:Current sensor Transfer impedance Prove.PNG]]
+The left impedance is the signal generator which is generating enough power for 1 ampere.
+This 1Amp. generates <math>P = I^2*R=50 \ Watt</math> 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 <math>P = \frac{U^2}{R} = 20 \ mWatt</math>
+= Reference measurement =
 [[Image:Current sensor Transfer impedance Ref measurement.PNG]]
+= Probe measurement =
 [[Image:Current sensor Transfer impedance final measurement.PNG]]
+= Calculation =
+<Math>Correction \ factor (dB)= P_{Measured} - P_{Reference} + 33.98</Math>
+<Math>P_{Measured}</Math> and <Math>P_{Reference}</Math> in dBm.
 {{note|This method is not a replacement for a real calibration.}}