Subject: Re: Audio noise in diff amps
Date: 19 Dec 2002 09:16:20 GMT
User-Agent: tin/1.4.5-20010409 ("One More Nightmare") (UNIX) (CYGWIN_NT-4.0/1.3.12(0.54/3/2) (i686))
Don Pearce wrote:
> On Thu, 19 Dec 2002 06:00:15 GMT, "Mike" wrote:
>>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.
> 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?
Don - you never need to calculate the *available* power - it's always
K * T Watts/Hz. The source and load resistances just let you determine
how much of that power is delivered to the load.
Consider a noisy source resistor R1 loaded by a noisless resistor R2:
Noise generator voltage = sqrt(4*K*T*B*R1) Vrms
The two resistors form a divider, in the ratio R2/(R1 + R2),
so R1's noise voltage is divided down to:
V' = sqrt(4*K*T*B*R1) * R2
square this rms voltage (V') and divide by the load (R2):
Noise power into load = 4*K*T*B*R1*R2
(R1 + R2)^2
Note that if the load (R2) is infinite or zero, no power is delivered to
the load, which is consistent with expectations.
Note that under maximum power transfer R1 will be equal to R2, and the
power equation simplifies to KTB, as expected for thermal noise.
Of course, in reality both R1 and R2 would be noisy, so there should be
two noise equations, though you will observe that interchanging R1 and
R2 in the noise equation yields the same result - i.e. even if R1 and
R2 are unequal in value, they each deliver the same power to the other,
as Phil would have us require in order to maintain thermal equilibrium!