I was referring to eetech's example.

There is no -12 V in #12 and #14. Am I missing something?

ak

 

it will quickly discharge thru the load, or, as AK states, an additional resistor can be used to discharge C1.

No need for it in your circuit. The load will discharge C1 down to 0.7 V, after which its own leakage current will continue to discharge it (more slowly).

ak

 

If the button is held down past the timing period, then the output will remain energized.

That might be a significant difference to the TS. However, since he clearly wants a latch function, the normal case probably is that the button is released before the circuit times out.

Another difference is that his circuit is not self-resetting. That is, with his approach (the latch is a separate function from the outut driver), you will not get another timed output cycle if you push the button again before power is cycled, With your circuit, once C1 is fully charged the circuit becomes a momentary action power switch - power is passed to the output whenever PB is pressed, for as long as PB is pressed.

ak

 

There is no -12 V in #12 and #14. Am I missing something?

ak

I had a brain flop and confused charge with voltage.