From: Don Pearce
Subject: Re: Audio noise in diff amps
Date: Thu, 19 Dec 2002 08:22:42 +0000
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On Thu, 19 Dec 2002 06:00:15 GMT, "Mike" wrote:
>"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.
>Well, gosh, Don, if you look closely, you'll notice that R is also in my
>equation. It's in the numerator and the denominator, just like it's in
>yours. Even if you didn't calculate the 2.4uV number, you know where it came
>from, and that it contains R.
>-- Mike --
Oops - just seen that last phrase. There is no reason at all to
suppose that the 2.4uV figure contains R - indeed if you do the maths,
it can't. If it did, how would you then use R to calculate the power?
No, that 2.4uV can only contain B, which is what I have assumed
And of course if it did contain R, what value would you use? Phil told
us 22000 ohms, but then used 600 ohms for his calculation (inherent in
the adoption of 0.775 volts as a voltage level for 0dBm). So what
would you say is the correct way to interpret this?