I am starting to think I’ve drawn my circuit wrong as I’ve had this circuit on a board and I measure a signal, approx 0.6V ac.
Thank you Mike. I just moved to a computer with SPICE to prove the point has about half a volt on it, and there it is in your picture below. If the spice in my head works.It is a Sallen-Key 2-pole low-pass filter
Thank you Mike. I just moved to a computer with SPICE to prove the point has about half a volt on it, and there it is in your picture below. If the spice in my head works.
The Guru is almost right about virtual ground but....... Point B only.
Point A is a virtual ground at DC but at AC (depends on frequency ) there will be a AC voltage.
..We know the (-) input of U1 is at ground. "virtual ground"
..For "Out" to change voltage, current must flow through C2. (charge up C2)
..C2 current come from R4, R1, C1, V1. Current in R4 causes point A to move away from ground.
Another way to think about it is: R4 & C2 with U1 makes an integrator. Lift up on A, B=0V, this causes Out to go down.
A third way to think about it. Replace C2 with a resistor that is the impedance of C2 at 500hz. Solve.