Subject: IFD and Johnson Noise
From: Alan Stummer <[email-masked]>
To: [email-masked]
Cc: Alan <[email-masked]>
Date: Fri, 06 Sep 2013 12:29:39 -0400

Hi;

IDEAS FOR DESIGN
Attached is an submission for IFD.  It is a solution for driving a 150V 
piezo motor.  Tried and true.

JOHNSON NOISE
Totally unrelated to IFD, but who knows, may become something.  I almost 
hate to ask because the answer is probably quite obvious.

Johnson noise <http://en.wikipedia.org/wiki/Johnson_noise> is always 
defined as the noise voltage appearing across a resistor with a defined 
bandwidth at a defined temperature.  Very annoying when high resistance 
is needed in the front end stage such as a transimpedance amp.  My 
problem with it is that Johnson noise is always defined as a voltage.  
The classic equation is:

    *Vn = sqrt(4KTRF)* where K is Boltzmann's constant, T is temperature
    in Kelvin, R in Ohms and F is bandwidth

Here is the dichotomy.  A literal interpretation of that equation would 
mean that an unconnected resistor on a table meets all requirements of 
temperature, resistance and any frequency. Therefore it has a voltage 
across it, albeit tiny but finite. Voltage across a resistance means 
power.  Power means heat.  Yet the resistor is obviously at room 
temperature.  Ergo, no voltage.

The simple solution is to say that there is only Johnson noise current, 
no such thing as Johnson noise voltage per se.  That would mean that 
shorting a resistor would cause it to heat up a tiny bit, yet it 
doesn't.  Ergo, no current...

In a circuit, Johnson noise is there.  We live with it.  However the 
equation for Johnson noise voltage does not always work, such as the 
resistor on a table example.  What are the conditions where this 
equation is true?  Is there another equation accounting for external 
impedances?

Cheers, Alan
-- 
*Alan Stummer*
University of Toronto
www.physics.utoronto.ca/~astummer