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# Calculate Input Voltage

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#### andy257

##### Member
Hi all,

Can anyone help me calculate the input voltage where I have indicated with the arrow.

The problem I am having is there are a lot of components after this junction that I believe will have an effect on the measurement. The only component I am disregarding is R5 as it is O/C.

Thank you

The signal level at the junction you show should be almost nothing because it is a "virtual ground" where the signal is almost completely cancelled by the negative feedback from R6. It will have signal only when the output of the opamp is clipping.

Hi Audio,

Thank you for the reply. It nice to see your still around the forums these days.

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.

Since V(b) is ~zero, then no current flows in R5, hence R5 and the the switch does nothing. Neither do D1 and D2 at the specified input voltage.

It appears to be a very long way around to make 170mV from 3.5V

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It is a Sallen-Key 2-pole low-pass filter, but again, the frequency response is the same switch open or closed, with or without the diodes.

"virtual ground"
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.
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.

It is not a Sallen Key 2-pole lowpass filter because it has only one lowpass pole, C2 the intergrator.
The diodes prevent the input of the opamp from damage when the signal level is so high that the output of the opamp is clipping like crazy and cannot clamp the input.

All of that still doesn't answer what you were trying to do? What is all of the monkey-motion with the switch and diodes? What is R3 supposed to do?

V(a) to a first approximation is just a voltage divider consisting of R1 and R2 with a DC blocking capacitor C1 on it's input.

Yes, it is a single-pole low-pass with -6db at 200Hz.

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.

Hi Mike,

Thanks for the Spice sim. Is there anyway to see what the exact peak value is of point A?

Thanks

Andy

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