circuit-designdelaylatchswitches
How would I make a switch that is controlled electronically, and has a ten second delay?
I have a circuit that will output either 3.3V or 0V. I would like to make a circuit that detects 3.3V, waits ten seconds, then closes a switch. The switch should remain closed even if the 3.3V drops back to 0V, possibly requiring manual reset.
Is this easily achievable?
You can use a latching relay. The relay would be set when a user presses a momentary switch which provides current to the coil which moves the contact closed. The relay would be reset when the electronics use transistors to provide current to the coil which opens the contact.
As you say, the obvious and simple way is to do this with a small microcontroller, like the PIC 10F200.
However, if you really want to do this with analog electronics, it would be simpler to use a transistor rather than trying to somehow fit the evil 666 555 timer into this role:
R4 and Q2 is basically the same thing you already have for switching the power. Q1 is similar to your M1 in that it activates Q2. The difference is that Q1 is activated directly from the switch, and turns off according to the decay time of the voltage on C1.
When the switch is pressed, C1 is charged up quickly thru R2. R2 is only there to avoid excessive current thru the switch when it would otherwise short a discharged capacitor across power.
R1 causes the voltage on C1 to decay exponentially towards 0. While that voltage is about 600 mV or more, Q1 is kept on enough to pull its collector low, and thereby turn on the power switch, Q2.
This circuit will turn on quickly, but fade off over a few 10s to 100s of milliseconds. If that is acceptable, then there is nothing further you need to do. If you need snap-action, then a little hysteresis is in order. That could take the form of some AC feedback from the drain of Q2 to the base of Q1.
The issue of the current this circuit would use was raised in a comment.
Look at the circuit carefully, and you will see that it uses very little current, especially compared to a 666 timer. When the switch (Q2) is on, the dominant current drain is thru R4. This should be obvious just from the values of the resistors. If V+ is 5 V, for example, then the current thru R4 will be less than 50 µA.
The current to keep Q1 on comes from the one-time pulse of current thru the pushbutton to charge up the timing cap. After the pushbutton is opened, no more supply current is used to keep the power switch on. Other than the initial inrush to charge C1, the steady state current with the pushbutton closed would be another 50 µA thru R1, and less than 1/10 of that thru R3.
Now compare that to the original proposal. The power voltage is kept across both R4 and R5 while the power switch is on, and this is before even considering the current to run the 666 timer.
In short, it should be quite obvious from even a cursory inspection that the circuit above draws considerably less current to keep the power switch on than the original circuit.
Best Answer
As is usually the case with a timer task like this, this would be an easy job for a small (6 or 8-pin) microcontroller such as a PIC10. But if you're not familiar using a microcontroller, that's a fairly steep learning curve.
So instead this circuit should do what you want.
Initially, both flip-flops are reset, and the \$\small \overline{\text{Q}}\$ output of the second flip-flop is high, so the 74HC03 pulls the EN line to ground (the 74HC03 is open collector, since the EN lead is pulled high by the Powerboost 500 module).
When the input goes high, the 74HC221 triggers. (The 74HC221 is a non-retriggerable version of the 74HC123.) \$\small \overline{\text{Q}}\$ of the '221 goes low, setting the first 74HC74 flip-flop. When the timer expires, the combination of the '221's \$\small \overline{\text{Q}}\$ high and the Q output of the first flip-flop high sets the second flip-flop, enabling the EN lead through its pullup on the Powerboost 500 board.
Pushing the reset button resets both flip-flops, allowing the sequence to repeat.
Because the circuit uses a 74HC221 instead of the retriggerable 74HC123, even if the input drops to 0 and then goes high again during the 10 second timeout, the timeout will not be extended. If that is not the behavior you want, substitute the 74HC123 -- it has the same pinout. Also, if the input stays high longer than 10 seconds, that won't affect the timeout either.
The circuit runs on 3.3V, assuming you use 74HC logic (not LS or HCT).