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Subject: Re: Audio noise in diff amps
X-Newsreader: Microsoft Outlook Express 5.50.4920.2300
Date: Wed, 18 Dec 2002 18:42:26 GMT
NNTP-Posting-Date: Wed, 18 Dec 2002 13:42:26 EST
Organization: Cox Communications
"Don Pearce" wrote in message
> Power is v squared / r.
> 2.4 uV in 22k ohms is -126dBm = (10 log((2.4e-6)^2 / 22000 + 30), or
But 2.4uV was obtained by taking sqrt(4kTRB).
What you've done is calculate 10*log(4kTRB/R) + 30 = 10*log(4kTB) + 30. Your
equation is missing a right parenthesis: it should have been written: (10
log((2.4e-6)^2 / 22000) + 30).
Assuming B = 16kHz, this is indeed -125.8, but you've removed R from the
equation. Given any resistor value, the noise power you calculate will
always be the same. While that would simplify the meter design - it would
read the same even if the probes weren't connected - I'm not sure it's very
I'm quite sure the meter is measuring something different...
-- Mike --
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