>When you are using a pot as a reference then you >need to account for the fact that the output >impedance of the pot varies with voltage. It should be possible to produce an expression that includes the variable resistance of the POT in calculating the output result. This is continuous allowing the original setting to be calculated. Brief example to try to clarify the above. Vcc=5 volt Imagine a 10K resistor from the pot wiper to the SD capacitor. Pot is 10K say Pot ends go to Vcc and ground. Micro drives SD cap also with a 10K resistor When pot is at centre point it looks like a 2.5V source behind a 5k resistor (Thevanin equivalent) The SD sees 2.5V behind 15K which LOOKS like 2.5*10/15 = 1.666V behind 10K so the SD reports a value of 1.666V (if it assumes that a 10K input resistor is used. With pot wiper at full supply the SD sees 5v behind 10K and reports 5V With pot wiper at ground the SD sees 0V behind 10K and reports 0V With wiper at 25% position from ground the pot looks like Vcc/4 = 1.25V behind 2.5K//7.5K = 1.875 K To the Sd this looks like 1.25V behind 11.875K which is the same as 1.053 volt behind 10K so it reports 1.053 volt. The result of all this is that the pot wiper reads as in the middle when it is in the middle but appears to approach the ends more rapidly than you would expect. Certainly reverse engineerable. Lesee Pot is set at fraction Position of Pot = X 0 <= X <= 1 Resistor from SD cap to pot wiper = R Pot Resistance whole track = P Pot connects to Vcc and ground 1. Pot effective resistance at wiper is the resistance of the two portions abvoive and below wiper in parallel = P.X//P(1-X) = P.X.P(1-X)/P = X.P(1-X) eg at 0.2 of full scale R effective = 0.2 * P(1-0.2) = 0.16 of full pot value ie a 10K pot set to 20% position will have a 1K6 effective resistance. 2. Pot open circuit voltage = Vcc * X 3. Equivalent resistance for Sd = R + Rpot = R+ X.P(1-P) 4. SD sees equivalent voltage of Vwiper x R/(R + Reffective-pot) *************************************************************** * SD voltage seen = Vcc * .X * R / (X * P * (1-X) + R) * *************************************************************** Whereas what it SHOULD see ideally is Vcc * X Solve for X (a quadratic) and you get X in terms of the SD voltage reported. Not pretty but workable. WARNING: I haven't checked my algebra - a slip somewhere is quite likely but anyone who will or can wade through this should be able to follow the argument well enough to spot any errors. ALSO - may have made a/some fundamental wrong assumptions :-) I have assumed that eg doubling the input resistance will halve the apparent voltage behind it (and as I recall this is correct but I haven't played with the theory recently) - if this is NOT correct (but I think it is) then NONE of the forgoing works ;-) Now, back to work !!! regards Russell McMahon _____________________________ - www.easttimor.com Updated regularly: 100,000 refugees STILL in West Timor face starvation! - www.sudan.com And you think Kosovo and Chechnya are bad! What can one man* do? Help the hungry at no cost to yourself! at http://www.thehungersite.com/ (* - or woman, child or internet enabled intelligent entity :-)) -----Original Message----- From: Walter Banks To: PICLIST@MITVMA.MIT.EDU Date: Tuesday, 18 April 2000 09:11 Subject: Re: [OT] What is a Sigma-Delta AtoD converter? [EE] >---------- >> From: Alice Campbell <1502amc@LO.SCSENG.COM> > >> And try as i might, a pot just doesnt seem to work well in >> this circuit. When i get a chance, i will investigate >> further, but i have not been successful in getting it to read >> a pot. It really likes to work with an opamp. > >The opamp isolates the circuit from the pot. >When you are using a pot as a reference then you >need to account for the fact that the output >impedance of the pot varies with voltage. > >As Scott said the input impedance is quite low >but we and many of our customers have used this >in commercial products. > >Walter Banks >