Electronic – Variable time alarm using 555 timer

Is the duty cycle supposed to switch from 50% to 0%, or decay from 50% to 0%? Is there any particular precision required on the 50%? Is there any requirement regarding quiescent current?

I would suggest that you start by using a 555 or similar chip to generate a symmetrical sawtooth-ish (doesn't have to have straight sides) wave, and then feed that into one side of a comparator. The other side of the comparator should be connected to a circuit which will generate a voltage that will start at half-rail when the button is first pushed, but will either rise toward VDD or fall toward ground after that. If you can use a SPDT pushbutton, a simple realization of such a circuit would be to connect the center pole of the button to a cap whose other leg is ground. Connect the normally-closed contact to VDD, and the normally-open contact to the comparator input. Connect a resistor from that comparator input to ground.

Conceptually, it would be nice if a single active device could be used to provide the output, without needing the separate comparator. It's possible for the 555 to provide an output whose duty cycle will be affected by an input voltage. Unfortunately, the frequency will also be affected by the input voltage; I'm not sure of any easy way to make a compensated circuit such that the decrease in high time will be balanced by an increase in the low time so as to yield a reasonably-constant frequency (as opposed to having a device generate a 1Khz reference signal which is then shaped by a second active device). A two-chip circuit doesn't seem unreasonable, though.

Here's a simple circuit which will illustrate the concept. For simulation purposes, it uses an op amp, but a comparator would work just as well. One slight note: many comparators ground the output when the + input is higher than the - input, so the polarity would be opposite that of an op amp.

I agree with Olin: ditch the 555 and get a microcontroller. The PIC10F200 is the first one which comes to mind if you only need a couple of I/Os and a simple program. Thanks to the internal oscillator it doesn't need any of the external components required by the LM555. The 10F200 is only a few cents more than the LM555 and can easily replace two of them. The LM556 is no competition either: it costs twice what you pay for the 10F200.

So Olin's answer is the right one? No :-). I'm going for the bonus! I recently found the Atmel ATTiny5, which is comparable with the PIC10F200: same SOT23 package, same 512 bytes program memory, internal oscillator. BUT! The ATTiny5 has an ADC, which the PIC doesn't have! Connect the potentiometer to it and the bonus is mine! :-)

(You can also make a crude ADC with the PIC, but it needs 2 pins which you can use as outputs, and since we already have two outputs and one I/O of the 10F200 is only input, you'll have to give up that buzzer output. Edit. On second thought, you can do it with one I/O, but you'll need two resistors an a capacitor instead of one resistor and a capacitor.)

The software will be very much the same as for the PIC: program an LFSR (Linear Feedback Shift Register):

An 8-bit LFSR can cycle through up to 255 combinations before it repeats, a 16-bit LFSR through 65535. Use the potentiometer reading to define the clipping values for the timer.