I am a self taught beginner. I have put together a doorbell type circuit that I modified and messed around with til I got it to do what I wanted. My prototype works, but that doesn’t necessarily mean it’s correct, right?

Can anyone look at this and tell me if I have made any errors that might cause problems? before I solder it up.

Purpose of circuit: pressing a momentary pushbutton will activate either a flashing LED or, a flashing LED AND a beep from a buzzer depending on the position of the power switch.

Example here:

When the button is pressed momentarily, the LED flasher (a special flasher with built in resistor and flasher) will flash approximately 5 times, but will flash continuously when button is pressed continuously.

If the power switch selection is on LED/buzzer, the same as above will occur, but the buzzer will only beep one brief time whether the button is pressed once or continuously. The buzzer will reset after about 5 seconds so it can be triggered again.

Attached is the schematic.

 

The only thing that looks odd is the diode across the switch - it doesn't do anything, but it won't stop it working.
To do what you want I would expect the second '555 to be wired as a monostable too, triggered by the first one but, hey, if it does what you want go for it.

ETA: I can't see whether that diode is actually wired as it is shown on the schematic.

 

The diode across the switch is so when the switch is set for LED flashing only, it won't complete the buzzer part of the circuit.
I wasn't sure of any other way to accomplish that, but would love to learn!

The only thing that looks odd is the diode across the switch - it doesn't do anything, but it won't stop it working.

 

Just try removing the diode and see if it still works.
The switch connects and disconnects the buzzer.

EDIT: No hang on a minute. Does the switch have a centre off position?

 

You need a resistor in series with the LED. Since this is a battery powered circuit I would try 2.7k for the resistor, giving the LED a low current of about 2 ma.

The diode across the switch is so when the switch is set for LED flashing only, it won't complete the buzzer part of the circuit.
I wasn't sure of any other way to accomplish that, but would love to learn!

There are probably a million ways to do it a little bit better but the diode is the easiest and cheapest way I know of. For instance, I think that a MOS-FET connected as a diode would work but hardly worth the trouble.

 

You need a resistor in series with the LED. Since this is a battery powered circuit I would try 2.7k for the resistor, giving the LED a low current of about 2 ma.

The LED is one which has a built-in flasher circuit - it needs no external resistor.

 

Yes it is an on-off-on. But it's not on the prototype. I just switched where I was hooking the battery ground.

EDIT: No hang on a minute. Does the switch have a centre off position?[/QUOTE

 

The LED is one which has a built-in flasher circuit - it needs no external resistor.

Oops. I just assumed that the 555 was doing the flashing. I have never used a flashing LED. Do you have any feel for how much current they draw?

 

Yes it is an on-off-on. But it's not on the prototype. I just switched where I was hooking the battery ground.

EDIT: No hang on a minute. Does the switch have a centre off position?[/QUOTE

OK, the diode is fine. Get the soldering iron out

 

Oops. I just assumed that the 555 was doing the flashing. I have never used a flashing LED. Do you have any feel for how much current they draw?

I guess that is going to depend on size but is likely to be similar to normal LEDs.

 

If I remember correctly, 20 mA. I don't have the packaging handy so many might be wrong.