I am analyzing two Op-Amp circuits to calculate the gain. Circuit 1: Two input voltages are connected to the inverting lead of an Op-Amp through a resistor on each input voltage which also has negative feedback across some other resistor. The non-inverting lead of the op-amp is connected to ground through a resistor. Here is a circuit diagram showing the circuit, http://www.fileden.com/files/2006/8/19/175303/Circuit/6.wmf Don’t pay any attention to the numbers on the circuit…I just had to include those so the program I used to make the circuit would allow me to place the components; they have on relevance to this question. V1 = I1 * (R1 + R4) V2 = I2 * (R2 + R4) Where I can solve for I1 and I2 (which are the currents coming from each of the inputs) as, I1 = V1 / (R1 + R4) I2 = V2 / (R2 + R4) The sum of these two input currents is the total current, I_T = V1 / (R1 + R4) + V2 / (R2 + R4), Which is equal to the oppotite of the current flowing around from Vout (V3 on the diagram). V3 = - I_T * (R3 + R4) I_T = -V3 / (R3 + R4) For simplicity sake, lets say that R1, R2, and R3 are all equal to some value which we will call R, so we can simplify the equations to be, I_T = (V1 + V2) / (R + R4) I_T = -V3 / (R + R4) So setting the two equations for I_T equal to each other, (V1 + V2) / (R + R4) = -V3 / (R + R4) And solving for V3, V3 = -(R + R4) * (V1 + V2) / (R + R4) Which is equal to, V3 = - (V1 + V2) What bothers me about this answer I have just obtained is that it seems that no mater the value of R4, it doesn’t seem to affect the results since it will get canceled out in the end. This seems strange to me. I have seen numerous examples on how to do these types of problems (inverting summing amplifiers), but they never have an R4 in the circuit, it always leads directly to ground. If my above calculations are correct, and R4 doesn’t matter, then this would make sense why the examples I have seen never bother to include an R4. Circuit 2: http://www.fileden.com/files/2006/8/19/175303/Circuit/7.wmf This is a very similar circuit to the one above, only the feedback resistor has been replaced by a capacitor. My intuition tells me that (because of the capacitor) it should integrate the input signal….or in this case, the sum of the input signals. Following much the same process as before, I get…. (Letting R1 = R2 = R) (V1 + V2) / R = -V3 / Z_C Where Z_C is the impedance of the capacitor. Solving for V3, I get, V3 = -(Z_C / R)(V1 + V2) And then substituting in Z_C = C * dv_c / dt Where C is the capacitance of the capacitor, v_c is the voltage across the capacitor (which is equal to V3), I get, -(1 / (RC)) * (V1 + V2) = dV3 / dt The integrating both sides I get, V3 = -1/RC * integral of (V1 + V2) So do these calculations look correct? I have not done enough of these / seen enough examples to have much confidence in my answers as of yet, but assuming these are correct then I think I might have ‘figured it out’ so to speak.