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Reply-To: "Mike"
From: "Mike"
Newsgroups: sci.electronics.design
References: <4dc00voebmhialfsonrfvoqhog4qnu9hdo@4ax.com>
Subject: Re: Audio noise in diff amps
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Date: Wed, 18 Dec 2002 18:42:26 GMT
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"Don Pearce" wrote in message
news:4dc00voebmhialfsonrfvoqhog4qnu9hdo@4ax.com...
> Power is v squared / r.
>
> 2.4 uV in 22k ohms is -126dBm = (10 log((2.4e-6)^2 / 22000 + 30), or
> -125.8dBm.
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
useful.
I'm quite sure the meter is measuring something different...
-- Mike --
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