Source: http://www.amb.org/audio/epsilon24/
I have been trying to understand the operation of the above circuit. To simplify, I am ignoring K1, Q1, R4 and D2. I also assume there is a jumper across T1 and T2. Since I assume there is no K1, I assume that R3 is connected to 12V rail and the LEDB junction is connected to the drain of Q3.
As I understand, this is a circuit which is used to power up a device from mains (through a 12V relay) using a momentary switch. When the circuit powers up C3 will charge up to 12V. Q3's drain is also at 12V. Q3's gate, because the capacitor charges up to 12V, should be at almost 0V and therefore it should be off.
What I don't understand is what happens when the switch is pressed: I assumed that Q3's drain is at 12V and the cap is also at 12V. But if that is the case, nothing should happen when the switch is pressed because both points are at 12V.
So I must be wrong with some of my assumptions.
Best Answer
When power is first applied to this circuit, voltage is applied simultaneously to the top of R1 and to the top of R5 (via the output relay). However, Q3 has a significant amount of gate capacitance, so Q2 will turn on first, shorting the gate of Q3 (and the current coming from R1) to ground. This will also prevent C3 from charging. The circuit will be in the "off" state.
The first time the button is pressed, C3, which is discharged, will pull the base of Q2 low enough to cut it off. This will allow Q3 to turn on, which pulls its drain low, turning on the relay and cutting off Q2. The circuit will be in the "on" state. Now C3 can charge up via R1 and R2.
The next time the button is pressed, the voltage on C3 will turn on Q2, which will short the gate of Q3 to ground, turning it off. This will also discharge C3, but Q2 will be held on by the voltage on the drain of Q3, via R5. The circuit is again in the "off" state.
Obviously, if you simply hold the button down, it will slowly toggle between the two states.