I made this circuit and it works as open circuit.(power applied, led comes on) I want to make it work with a remote off/on switch, that is closed (powered) in its resting state; when the switch is depressed, then the led would come on. (along with a buzzer added that I haven't shown in this photo. Parts are: 555, 1K resistor, 1MK resistor, 100n cap, 2 (or less) µf cap and 220 ohm resistor for the led. I used the schematic in the 2nd photo as a starting point. Any ideas will be appreciated. I don't even know if it's possible to use this arrangement to work like I imagine.
Hi,
It would perhaps be easier to use a CD4049 Quad inverting buffer IC instead of the 555, in conjunction with a DC operated piezo-buzzer such as https://www.rapidonline.com/r-tech-350204-12v-pcb-piezo-sound-generator-3-20vdc-35-0204. The attached schematic shows the required connections. The buzzer has a built-in oscillator and only requires DC voltage to work. It consumes 15mA. The 4xganged inverters shown in the schematic will supply enough current for both the LED & buzzer. The schematic only shows the LED. To also connect the buzzer, connect its +ive terminal to the joined up outputs of the inverters and its -ve terminal to earth.
There are 6 inverters in the CD4049, and you can connect all 6 just as I have shown with 4 of the inverters (ie ganging up all 6 together)
EDIT: 08 Mar 2025. I had inciorrectly labelled the two schematics. I have now swapped over their labels and they are now correctly labelled
It is a hex buffer; the poster mentioned that later, the first mention incorrectly says Quad. Many 4000 series ICs with two-input gates are quad devices, so just a typo.