LPF

This is an instrumentation amplifier implemented using two operational amplifiers. The general relation between input and output voltages is: Vout = (1 + Rf1/R1)Vin1 - (1 + Rf2/R2)( Rf1/R1)Vin2 When Rf1 = R2 and Rf2 = R1, Vout = (1 + R2/R1)(Vin1 - Vin2) For the circuit shown the gain is 1 + (20 kOhms) / (5 kOhms) = 5 The two input signals have equal peak of 1 V, they are at the same frequency but Vin2 is at a phase angle of 90 degrees. Subtraction of the two input signals is equivalent to inverting Vin2 and adding the result to Vin1. Inversion of Vin2 means adding 180 degrees to its phase, in this case the phase angle is simply negated to -90 degrees. Adding to Vin1 yields an amplitude of square root of 2 and a phase angle of -45 degrees, this is because both signals have equal amplitudes. Then Vout = (5) [sqrt(2) V @(-45 degrees)] Vout = 7.07 V at -45 degrees So the output signal has a peak value of 7.07 V. At time t=0 its value is Vout(t=0) = (5) [sqrt(2) V sin(-45 degrees)] Vout(t=0) = -5 V Notes: An offset voltage can be added at the bottom of R2. Making the gain adjustable is easier in the three-op-amp instrumentation amplifier.