wiring 555 to disconnect itself from electricity

Question

i had idea for a circuit with 555 and here is the IDEA:

circuit is stationary somewhere, it is in "passive state" meaning its drawing no current from battery and pin 3 is set low, then someone comes by, presses the button and LED turns on, after 15 seconds LED turns off and circuit continues to draw no current and remains in this "passive state" untill next button press

QUESTION: this circuit doesn't work, and im wondering where did I make a mistake?

LINK: http://prntscr.com/5u1be2

PS: imagine LED connected to pin 3

Answer 1

Something like this will work. Timing and bypass components and load omitted for clarity.

^{simulate this circuit – Schematic created using CircuitLab}

When the output goes high, it turns on Q1 through R1, which turns on Q2 via R2. R2 should be low enough that Q2 stays reliably saturated with the highest load on the 555, plus the 555 supply current. R3 prevents leakage in Q1 from being amplified by Q2, so the total leakage in the passive state is merely the leakage in Q1 + Q2 (probably in the 10's of nA at room temperature), rather than tens or hundreds of times higher.

Edit: Expanding on the leakage- suppose the leakage of Q1 is 20nA, and gain of Q2 at 20nA is 40. All the leakage current from Q1 flows into the base of Q2. Then the collector current of Q2 will be 40 * 20nA = 800nA plus another 20nA for its leakage, so a total of 840nA. That's without R3.

Now add R3- the voltage across R3 will be 20nA * 20K = 400uV. That means virtually all the current will flow through the resistor and not the base (since the base-emitter junction doesn't really start to conduct until the voltage across it reaches hundreds of mV) so the leakage of Q2 is just the minimum 20nA, for a total of 40nA (21x better performance for a single resistor).

Edit2: Further to discussing why use this circuit as opposed to lifting the GND pin, here's the schematic of the original 555. It should be reliable as it is from the designer's (Hans Camenzind, RIP) book Designing Analog Chips.

Edit 2:

In order for the trigger to work properly in conjunction with the above-discussed power switching scheme- consider the below schematic. R6 may not be necessary.

^{simulate this circuit}

Answer 2

As Olin pointed out there's no way you can control a circuit power with an emitter follower from one of the circuit outputs itself, unless your circuit can provide an output voltage higher than the input, which is possible but is not the 555 case.

There actually is a way to achieve your goal:

^{simulate this circuit – Schematic created using CircuitLab}

When you press the button the circuit is powered, Q goes high and the bjt turns on, keeping the circuit powered when the button is released. After the time set by the 555 subcircuit Q goes low and turns off the bjt.

There are two things you should notice:

Since \$V_{CE}=0\$ the bjt might not turn on properly, you might try to include a resistor in series with the pushbutton (aim for some 2~200mV drop), or you need to replace the bjt with a relay.

Note that I connected the trigger to gnd through the pushbutton. Any other pin of that subcircuit that must be tied to gnd must be connected through the pushbutton, or your 555 might get some power from the protection diodes.

Answer 3

I looked at it only briefly since the 555 timer pins were just numbers without explanation. However, 2 things jumped out:

1. The pushbutton is permanently shorted. That of course would cause the circuit to ignore any pushbutton state changes.

2. It looks like you might be trying to power up the chip with a NPN emitter follower. That can't work. The output voltage of the emitter follower will always be less than what you are driving the base with. Since the chip can't produce a voltage higher than its own power voltage, you can't drive that power voltage from a output pin thru a emitter follower.