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From: Don Pearce
Subject: Re: Audio noise in diff amps
Date: Wed, 18 Dec 2002 23:24:25 +0000
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On Thu, 19 Dec 2002 09:41:28 +1100, "Phil Allison"
>"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.
> ** No, Mr Pearce calculated a compleley wrong one instead.
> Then tried to justify it - and could not.
> Same with his dangerous advice on appliance grounding.
> .............. Phil
Phil do please give this up - you are looking increasingly silly and
desperate. If you cannot manage a calculation as simple as
10 log((2.4e-6)^2 / 22000) + 30
without getting the answer 16dB in error, then there is really not a
great deal of hope for you.
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