From: Don Pearce
Subject: Re: Audio noise in diff amps
Date: Thu, 19 Dec 2002 15:02:31 +0000
References: <email@example.com> <firstname.lastname@example.org>
NNTP-Posting-Host: host217-39-125-198.in-addr.btopenworld.com (220.127.116.11)
X-Newsreader: Forte Agent 1.91/32.564
On 19 Dec 2002 10:22:42 GMT, Rick Hellicar wrote:
>Don Pearce wrote:
>> On 19 Dec 2002 09:16:20 GMT, Rick wrote:
>> I know, we've been through all that. But when somebody presents you
>> with two figures, a voltage and a resistance - there is only one
>> calculation you can make to work out the power. That is what I did -
>> and that is where all the problems started.
>You've gone off at a tangent - I was trying to point out that
>you START with the available noise power, and from that compute
>the equivalent noise voltage, for which you need the value of R.
>This is directly against your assertion that...
>> 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"
>In other words,
>1. There's EVERY reason to suppose that 2.4uV contained R
>2. You don't use V and R to calculate the power, you *start* with
> KTB and compute V from it!
As far as I could make out, 2.4uV was a measured value, not the result
of a calculation of noise power, so that is where I started.