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Re: Negative Resistance Re: Gap Question



Original poster: Mddeming-at-aol-dot-com 

In a message dated 2/23/04 2:48:17 PM Eastern Standard Time, 
tesla-at-pupman-dot-com writes:

 > Holy cow that did throw a bit of a wrench in there.  My gears just
 > stopped.  That is hard to conceptualize based on im so used to the ohms

 > > law concept being so natural to me that it is very hard to imagine a
 > > negative resistance.  Not sure how to go about wrapping my head around
 > > that one.

If you think of Ohm's Law R=E/I it means that if you plot E vs I then R is 
the slope of the line. For most common metallic substances (and carbon) at 
"reasonable temperatures" this is just a straight line sloping up to the 
right and passing through (0,0). As the current increases, the voltage drop 
across the device increases if proportion. For plasmas and some other 
esoteric substances, a plot of E vs I over a certain range is reversed. It 
slopes up to the left. At any fixed point, if you measure E and I, they 
will be positive numbers and you can say R(static) =E/ I. But if you look 
at R(dynamic)=change_in_E / change_in_I then you get a negative number.

Example: If the voltage drop across the device is 100V when a current of 1 
Amp passes through it, R1(static)=100 Ohms. If increasing the current to 2 
Amps drops the voltage across the device to say 90 Volts then R2(static)=45 
Ohms.
But the slope, R (dynamic)= (E1-E2)/(I1-I2)=(100-90)/(1-2) = -10 Ohms, a 
"negative resistance".
In both cases, Ohm's Law holds at each point. If you are familiar with the 
Calculus,
R(static) = E/I while R(dynamic)= dE/dI. In VT circuits, R(dynamic) is 
usually positive but changes value continuously along the curve.
Hope this helps.

Matt D.