Potentiometer Theory

[Jan Olligs]

Originally posted by Ulibrium
Ok, I understand now how Vg = 0(I have a sophomoric understanding of circuits). But... what is I1 and I2? It's Kirchoff's loop rule or something. I'm just curious.

Just follow the circuit (electrical flow, not physical): The electrons ome from +5V and go through the first part of the poti. Then they are at a "junction": they can go either through the connected circuit, in this case the analog input of the controller, or through the second part of the poti to ground. This way you get two different currents: I1 through the controller and I2 to ground. Because of the knot rule, they add up to the current that enters the circuit, I. Thus, I1 + I2 = I.
You get the different voltages because of Ohm's Law. Look at the resistance in the upper part (X) as R1 and the resistance in the lower part (1/( (1/(R-X)) + (1/R_controller) )) as R2 (because R_controller is pretty high, 1/R_controller is about 0 and thus, R2 is about R-X). Because the same current flows through both resistors (no knots), you can calculate the voltages over the top and the bottom part via Ohm's Law, U = R * I. This way you get U1 = R1 * I, U2 = R2 * I. Now, solving for I you get U1 / R1 = I = U2 / R2. Multiplying by R1 / U2 you get U1 / U2 = R1 / R2. Because R1 equals X and R2 equals approximately (R - X), you get (approximately):

U2   R2       U2     X                     X
-- = --  <=> ---- = ---  <=>  U2 = (U-U2)*----  <=>
U1   R1      U-U2   R-X                   R-X


        X        X           /     X  \       X
U2 = U*--- - U2*---  <=> U2*( 1 + ---  ) = U*---  <=>
       R-X      R-X          \    R-X /      R-X


    / R-X    X  \       X            R       X
U2*(  --- + ---  ) = U*---  <=>  U2*--- = U*---  <=>
    \ R-X   R-X /      R-X          R-X     R-X


       X
U2 = U*-
       R

I would advise you to verify that, but that is what I remembered from Physics; I hope I didn't make a mistake. I hope that helps you a bit.