Discharging capacitor through potentiometer in ADSR circuit

Question

I'm trying to build a circuit for generating ADSR envelopes for use in an ananlog synthesizer, and a part of the circuit acting in a way I can't understand. Here is the part of the circuit that is giving me the problem (not the entire circuit):

^{simulate this circuit – Schematic created using CircuitLab}

(If this looks like a strange circuit, remember that this is only the part that is not working, stripped to its most simple form.)

The first op amp is outputting around 7 V DC. The gate voltage of the mosfet is 0 V. My problem is this: as I increase the resistance of R3, the voltage across C1 increases (it very quickly approaches 7 V as R3 is increased to some tens of kiloohms), and this only occurs if the mosfet is connected (gate voltage at 0). If I remove the mosfet from my breadboard, then R3 does not affect the voltage over C1, it simply remains discharged as I want it to.

I really have no idea why this is happening. Is there something I don't know?

Answer 1

Your MOSFET actually looks like this:

^{simulate this circuit – Schematic created using CircuitLab}

Every MOSFET has an internal diode so called body diode. It is a result of how the MOSFET is constructed. (the 3 bars in the symbol represent the sandwich of 3 differently doped silicon layers. n-p-n. Of which 2 are connected. The unconnected p-n pair results in this diode)

Since you have \$7V\$ comming from OA1, there will be a current through D2, R2, M1 and finally R3. This current will charge C1.

If the rest of your circuit allows it you could reverse M1. Then the body diode will not allow current to flow that way. Only if you make the M1 low impedance (switching it on) will current be allowed to flow that way.