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Subject: Re: Audio noise in diff amps
X-Newsreader: Microsoft Outlook Express 5.50.4920.2300
Date: Thu, 19 Dec 2002 04:52:28 GMT
NNTP-Posting-Date: Wed, 18 Dec 2002 23:52:28 EST
Organization: Cox Communications
"Don Pearce" wrote in message
> On Wed, 18 Dec 2002 18:42:26 GMT, "Mike" wrote:
> >"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
> >> -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.
> >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
> >I'm quite sure the meter is measuring something different...
> >-- Mike --
> You're right - I did bungle my parentheses - I should have omitted the
> first one. But R is in there - 22000 is near the end of the equation.
> I didn't put in a bandwidth, because that was already implicit in the
> given voltage figure of 2.4 microvolts. I didn't calculate that 2.4 uV
> figure, you understand.
> Telecommunications consultant
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