In terms of power change in the system as a fuction of resistance (or scalor
of impedance),  then I expect it should be 10logR  "dB(Ohms)"

Why not use dB(ohms) if dB(V) is considered a valid unit?

Ralph McDiarmid, AScT 
Compliance Engineering Group 
Xantrex Technology Inc.


From: John Woodgate [mailto:[email protected]] 
Sent: December 18, 2003 2:58 PM
To: [email protected]
Subject: Re: CISPR 25 Conducted Emissions (Current Method)



I read in !emc-pstc that [email protected] wrote (in
<of227b3f0b.4367e65e-on85256e00.0056b9fa-85256e00.0056d...@jci.com>)
about 'CISPR 25 Conducted Emissions (Current Method)' on Thu, 18 Dec
2003:
>Please read the attached document that he assembled for the details.

Not relevant to the question, but I'm very disappointed to see 'dB(ohm)'.
The point is that this concept is mathematically nonsense.

Consider V = IR

20logV = 20log(IR) = 20log I + 20log R

Now consider

P = I^2R

10logP = 10log(I^2) + 10logR

So is the 'dB(ohm)' 20logR or 10logR?

The UK national EMC committee has pointed this out to CISPR committees on
several occasions, but the mistake keeps being made.
-- 
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk

Interested in professional sound reinforcement and distribution? Then go to 
http://www.isce.org.uk
PLEASE do NOT copy news posts to me by E-MAIL!


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