Electronic – Extending diode with a buzzer

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.

Without knowing what your power supply configuration is, I can offer a few suggestions with reasonable caveats.

• If you need to ensure that there is no overlap between your current two transistors and the third, then your circuit will be more complicated than a few more 555's. You're then looking for something called a "dead time generator" and you probably would need to define your minimum dead time requirements.

• If you are driving MOSFETs directly, and VDD for the 555 is the same as the load you are switching, you can use a PFET instead of an NFET, and that will work reasonably well. This will not work if your 555 is powered separately from your load. This is the simplest option and requires no additional parts. See schematic below:

^{simulate this circuit – Schematic created using CircuitLab}

• If you need to use all the same type of transistor (all NFETs or all PFETs) then the next best thing would probably be to use an inverter (or other integrated logic gate) that comes in a small package. This makes for a small circuit and easy to troubleshoot. If you're set on through-hole or breadboard-able parts, then there are a lot of chips with 6 inverters on them that will work great. Tons of options exist if you can live with some flavor of surface mount (you can have as few as 1 inverters in a package). This also has the advantage of working if your load is switched from a different VDD than your 555.

^{simulate this circuit}

• You could also use 555's to synthesize any logic function, such as an inverter.

^{simulate this circuit}