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From: Don Pearce
Subject: Re: Audio noise in diff amps
Date: Wed, 18 Dec 2002 19:35:20 +0000
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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
>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 --
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.
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