on off timer

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steptoe

of course Im wrong, ask my wife™
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I have a project whereby when the on button is pushed its about 15 seconds before the user sees anything happen,

the problem is, if the 'on' button is pressed again in that 15 seconds it shuts down,

so, I need to have a secondary indication that the 'on' button has actually been pressed,

I would ideally like something that when it receives the intial voltage it will light a lamp/led or buzzer for perhaps 3 seconds then stop,

I know its do-able, just that my head is messed up with the simpliest method how to,

12v at [probably] <1w would be the maximum I think I need,

12v being as that is the easiest place for me to pick up the 'on' signal from.

<1w as I dont see the need for a buzzer or lamp/led needing more than that.

 
I would have thought it would be easier to just have an ON indicator light that illuminates when the button is pressed.

So that something has happened and someone does not try again.

If you wanted a signal lamp/buzzer that switches off after a few seconds then a 12V (if you can find one)  off delay timer relay would do the job so long as the signal stops after the power button is pressed.

Alternatively would it be possible to have the On button set up so that the system does not shut down on the subsequent presses?

 
I would have thought it would be easier to just have an ON indicator light that illuminates when the button is pressed.

So that something has happened and someone does not try again.

If you wanted a signal lamp/buzzer that switches off after a few seconds then a 12V (if you can find one)  off delay timer relay would do the job so long as the signal stops after the power button is pressed.

Alternatively would it be possible to have the On button set up so that the system does not shut down on the subsequent presses?


for the fact of 'authenticity' I'd like the lamp/indicator to go off while the system is actually running,

the 12v signal is constant once the power on button has been pressed, even when it has been pressed again to shutdown,

its actually for a homemade arcade machine using an RPi, I'll see if  can find the link to it,

my kids have a habit of not being patient enough to wait for the screen to start up, so press the button again that shuts the Pi down, but not the actual machine,.

 
so you need a timer for the LED that goes off after 15 seconds?
basically,

I was going to add a lamp/LED in the corner somewhere that would illuminate for 2 or 3 seconds just to show them that it has started and just to wait for the screen to light up

I think 15 seconds would be too long, but if that was an easy solution, then that would do.

 
what about something like a DP switch with one side charging a cap enough to keep an LED lit for a few seconds?
but wouldnt it keep charging the capacitor.?

thus keeping the LED liit.?

I think,,, and I may be wrong, I need something that can pulse the initial input,

charge the cap, and thats it,

 
yep. no idea what resistor or capacitor you would need, but you should be able to get it to last a few seconds
that really does seem the easiest solution

I had been on the mindset to turn the supply to the indicator off, but with that solution there is no need to, just divert it.

 
Ebay sell a cheap and cheerful 12Vdc delay off timer module (£7 UK, £4 china/bulgaria) that there should be enough room for; just power the unit and the relay will switch after your set time delay, connect the relay to your indicator LED /illuminated indicator and away you go. The unit will reset on power off.

 
I like the idea of completely killing the power so I think I'll order that link from @Ardet R

saves me messing about trying different sized capacitors and resistors, [ I havent got many different sizes anyway ]

 
I hope it works OK.

You will be powering the module all the time but not the indicator.

I have to ask because I am not so hot on electronics how the parallel capacitor and LED would work if you are using dc.

The way I see it the capacitor will quickly charge and then stop conducting; so the LED might come on at once if the capacitor were small but would be dim and would then brighten up as the capacitor stops conducting and would then stay on at full brightness.

If the LED were in series with a large capacitor then the LED would come on while the capacitor charges and then slowly dim as the capacitor reaches full charge and stops conducting, when the LED would go out.

Therefore I have missed something or failed in my comprehension of what would happen.

If someone could clarify that would help me, thank you.

 
I hope it works OK.

You will be powering the module all the time but not the indicator.

I have to ask because I am not so hot on electronics how the parallel capacitor and LED would work if you are using dc.

The way I see it the capacitor will quickly charge and then stop conducting; so the LED might come on at once if the capacitor were small but would be dim and would then brighten up as the capacitor stops conducting and would then stay on at full brightness.

If the LED were in series with a large capacitor then the LED would come on while the capacitor charges and then slowly dim as the capacitor reaches full charge and stops conducting, when the LED would go out.

Therefore I have missed something or failed in my comprehension of what would happen.

If someone could clarify that would help me, thank you.


you have an LED and resistor in series, and a cap in parallel

the LED lights while the cap is charging, once charged the cap provides a route of lower resistance for the current to flow, so the current no longer takes the 'harder' route through the resistor

 
............but what would happen when the switch is opened? The cap would be in parallel with the res and led, so would it not discharge through that causing the led to light (all be it a short time)

 
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