If the load current is under 50mA or so, then R5 is not needed, as it protects Q1's base from the current spike when C1 charges, via the load. The tiny leakage current through D1 will do no harm to Q1's base. If the power supply voltage is below 6V or so, D1 is not needed, as it protects Q1's base from negative voltages. Because the current through Q1's base and D1 is so low, one diode drop can be considered to be 0.5V, rather than the usual 0.6V, hence why two diode drops is said to be 1V. When M1 is off, C1 will charge via D1 and R5 again, returning the circuit back to the initial state. When M1 turns off, Q1 will start to turn on even more, this turning M1 off more, hence positive feedback. As Q1 turns on, it will connect M1's gate to 0V, thus turning it off. When the voltage on C1 reaches two diode drops above 0V, about 1V, Q1's base will start to conduct, which will turn it on. The circuit will remain in this state, until Q1's base starts conducting. With Q1 off, M1's gate will be held at +12V, via R2 and R3. Q1 will be off because no current will be flowing through its base. C1's negative plate now sits at the power supply voltage, minus two voltage drops, below the negative rail, which will be around -11V, with a 12V supply. C1's positive plate is connected to 0V, via M1's drain, which now sits at 0V. When S1 is closed, it connects M1's gate to +V, turning it on. The circuit will remain in this stable state until S1 is activated. R1 now keeps Q1 on, which holds M1's gate near 0V, via R3. When power is first applied, C1 charges via the load, R5, D1 and Q1's base. On the other hand it uses far more components, so could be considered to be overkill. The delay also starts the moment the button is pressed, rather than released and has no bearing on hold long the button is held for, which might not be an advantage, just different. You also might ask, why build the more complex circuit I posted, over the simpler one you posted? The advantages are: the delay is more predictable, less dependant on the supply voltage and completely independent of the MOSFET threshold voltage and the MOSFET turns on/off much faster. If this is undesired, the capacitor could be moved in parallel with R1. Also note, as drawn the MOSFET will turn on, when power is initially applied. The time delay will also be heavily dependant on both the supply voltage and the MOSFET'd threshold, although that's a non-issue if the supply is tightly regulated and a potentiometer is used for the timing resistor. Yes, that will work, just beware that it might shorten the life of the relay contacts a little, as they will not snap so sharply on, as the MOSFET will turn on very slowly, but it shouldn't be a problem if it's driving a solenoid directly. Some may have troubles programming the uC in which case the 555 could make sense but doesn't seem to be your case. If you leave space in the PCB for an optocoupler to add safe external control even more, and it will cost about the same than the 555 if not less.
Having a simple PCB with that hardware seems very useful to have in the drawer, it can solve many many problems using slightly different codes.
Simple delay timer transistor circuit code#
30s is fine for a 555, but if you now need a 5H timer? Grab the uC project and change one parameter in the code or start from ground up because the 555 isn't practicall anymore? Don't get me wrong, I like analog design much more than digital, but there are times it doesn't make sense, a switch with a timer, something a DIP 8 uC with minimal external parts can do perfectly and much more flexible than the alternative seems better to me. Then you can tweak the design in code after the PCB is settled, the HW is settled and you still have flexibility if you messed in something. That's the parts count, add a PS of your convinience, whatever between 3.3V and 5V will do no problem, likely wider range. UC, potentiomenter to set the time, one resistor, 2n3904, relay, switch.
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