I teach an analog circuits (see http://sujan.hallikainen.org/cuesta/ , though I only had 2 students this semester, so it got cancelled). Anyway, for a quick analysis, I tell them to use the "theory of the happy op amp." Determine the voltage on the non-inverting input, assume the inverting input has the same voltage, then use Ohm's law from there. In this inverting amplifier case, the non-inverting input is grounded, so it has zero volts. Assume the inverting input has zero volts, then analyze from there. When you have another resistor from the inverting input to ground, if there is indeed zero volts on the inverting input, no current will flow through it, so it can be ignored. However, in reality, there is SOME voltage on the non-inverting input. The output voltage is the open loop gain times the differential input voltage, so if there is no differential input voltage, there would be no output voltage (we'll ignore input offset voltage, bias current, and offset current for now). However, the differential input voltage is usually very small, so it can be ignored USUALLY. When adding a second resistor to ground, you can combine this resistor with the input resistor and do a Thevenin equivalent. The Thevenin voltage would be Vin*(R1/(R1+R2)) where R1 is your original input resistor and R2 is the resistor to ground. This is driving your inverting amplifier. Next, the Thevenin resistance of the R1,R2 combination is R1//R2 = (R1*R2)/(R1+R2). Use this lower value in your gain formula (-Rf/Rin). You get the same gain as you had without the resistor to ground. But, the ACTUAL gain of an inverting amplifier is (-RF/Rin)*(A*B/(1+A*B)). A is the open loop gain of the amplifier. B is the "gain" of the feedback network (R1/(R1+R2)). With A being very large, the numerator and denominator of the AB term is very close to 1. Now, let's start decreasing that extra resistor to ground. Going back to the Thevenin equivalent of the input resistor and this extra resistor, we find we're driving the circuit with a smaller and smaller Thevenin voltage and a smaller and smaller input resistance. Ideally, the higher gain (-Rf/Rin) compensates for the lower input voltage. However, our B term (the gain of the feedback network) is ALSO getting smaller, so A*B is getting smaller. Evenutally, 1+(A*B) is substantially more than A*B, and the gain falls off. Summarizing, with an ideal op amp (infinite open loop gain), you could decrease the extra resistor to ground to zero without affecting the overall closed loop gain. With a less than infinite open loop gain, decreasing the "extra" resistor will decrease the closed loop gain once there is a substantial difference between AB and 1+AB. Harold -- FCC Rules Updated Daily at http://www.hallikainen.com - Advertising opportunities available! -- http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist