Difference between revisions of "Current sensor transfer impedance determination method"
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<math>Imp.\approx 2 \ Ohm</math> | <math>Imp.\approx 2 \ Ohm</math> | ||
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{{note|This method is not a replacement for a real calibration.}} | {{note|This method is not a replacement for a real calibration.}} |
Revision as of 15:47, 23 January 2009
Contents
Theory
The left impedance is the signal generator which is generating enough power for 1 ampere.
This 1Amp. generates 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
So 1 ohm:
Reference measurement
Probe measurement
Calculation
and in dBm.
Example
On 10 MHz we have the following information:
- Calibration: 0 dBm.
- Measurement: -27,96 dBm.
So: