Difference between revisions of "Current sensor transfer impedance determination method"
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= Calculation = | = Calculation = | ||
− | <Math>Correction \ factor (dB)= P_{Measured} - P_{Reference} + 33.98</Math> | + | <Math>Correction \ factor (dB)= P_{Measured} - P_{Reference}+ 33.98</Math> |
<Math>P_{Measured}</Math> and <Math>P_{Reference}</Math> in dBm. | <Math>P_{Measured}</Math> and <Math>P_{Reference}</Math> in dBm. | ||
+ | |||
+ | = Example = | ||
+ | On 10 MHz we have the following information: | ||
+ | *Calibration: 0 dBm. | ||
+ | *Measurement: -27,96 dBm. | ||
+ | |||
+ | So: | ||
+ | <math>Imp.=-27,96-0.00+33.98=6,02 dBohm</math> | ||
+ | |||
+ | <math>Imp.= </math> | ||
{{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 15:45, 23 January 2009
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: