Relative to the mil-std comparison: A level of 89 dBuA at 10 kHz is the
equivalent of 2000 Volts/meter.  Further, it assumes a plane wave illumination
over a 150 meter length of cabling.  It is difficult for me to imagine how
such a condition could exist.


on 6/14/07 12:35 PM, Bob Richards at [email protected] wrote:



Brian,
 

 
I am very familiar with 61000-4-6, and I think this is where the confusion is
coming from. 61000-4-6 specifies the 6dB adjustment not because of loss in the
fixture, but because the test level is specified as an open-circuit voltage. 
However, the calibration must be done in a terminated system, which will
reduce the measured voltage by half. There is a 9.6 dB associated with the
voltage divider network formed by the 150-50 ohm conversion.
 

 
GR1089 references 61000-4-6 for some reason. I think it should only reference
the mil std test, IMHO.  The test level is specified as a current level, and
the cal fixture is a simple loop wtih only one current path. So, I don't see
the justification of applying a 6dB reduction in the calibration level.
 

 
Side note: I gave a presentation recently on 61000-4-6, and in researching the
standard I was never able to find a good reason why the test level is
specified as an open circuit voltage.
 

 
Thanks,
 

 
Bob Richards, NCT.
 


"Kunde, Brian" <[email protected]> wrote:
 


   
Bob,
 

 
I'm not familiar with GR1089 but somewhat with EN61000-4-6 which is a
conducted immunity test standard.
 

 
When calibrating the CDNs or Injection clamps, you have to compensate for the
losses in the test fixture. For the CDN, it comprises of a 150 to 50 ohm
adaptor at the input and output of the CDN. The fixture acts as a voltage
divider, the the measured power levels during calibration would be much lower
than what is expected on the CDN output. So you have to account for this loss. 
 

 
In EN61000-4-6 section 6.4.1 "Setting of the output level at the EUT port of
the coupling device" it has an equation for calculating the measured power
level when you perform the calibration test:  
 

 
Umr = Uo / 6 (in linear quantities) or 
 
Umr = Uo - 15.6db (in logarithmic quantities where Uo is the test voltage from
table 1. 
 

 
There is a note at the bottom that says, 
 

 
"The factor 6 (15.6db) arises from the e.m.f. value specified for the test
level. The matched load level is half the e.m.f. level and the further 3:1
voltage division is caused by the 150 ohm to 50 ohm adaptor terminated by the
50 ohm measuring equipment".
 

 
For Current Clamps the equation is
 

 
Umr = Uo / 2 (in linear quantities) or
 
Umr = Uo - 6db (in logarithmic quantities).
 

 
I am assuming the test fixture acts like a voltage divider so you will only be
measuring a fraction of the true voltage at the CDN in the the calibration
setup. So yes, if you are not factoring in the fixture losses when you
calibrate, you would be testing at too high of level. 
 

 
I hope this is what you are referring to.
 

 
The Other Brian
 
 

  _____  


 


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