Hello again, sorry for the delayed reply. There was some holiday activities involved. Anyway, back to the theory: I remembered Kirchhoff's Current Law: The sum of all currents, into and out of a point, is zero since we cannot store current in an ordinary junction point or a node if you will. So, I ran Mathematica using the formula: [tt]Solve[(0-v)/20+-v/1+(-5-v)/33==0, v][/tt] in order to find the voltage v at the filter input pin 2 on the CEM chip. This is for the case where the applied filter CV is zero volts. This is assuming that we stick with the resistors used in the data sheet (20K, 1K, 33K). Please try this for the OB values and you'll see what they most likely use in terms of voltage range. How did I arrive at the formula? ??? Assume that the filter CV voltage is v and that the applied CV is 0 volts. Observing the Kirchoff law we get: (0V-v)/20kOhm +-v/1kOhm+(-5V-v)/33kOhm = 0 :o Also, I used Ohm's law (V = R*I) => I=V/R in order to get an expression for the currents. Sometimes V is called U, E or emf depending on your application. Cheers!