held high logic to shorter duration logic ?

hello : )

i have been trying different ways to turn a logic held high signal (constant 5v trigger signal)
into a shorter duration ( 7 secs) logic output.

The only way ive found so far, is to use a 555 timer at a low frequency, to pulse a decade counter and have the decade counters second Qoutput reset the circuit after the first Qoutput has sounded a buzzer and led for 7 seconds.

Although this works, the footprint seems very large just to sound a buzzer for 7 secs.. I tried to use a 555 timer by itself; and although i can make a 555 timer output for 7 secs and then stop by itself using a momentary switch as a trigger...as soon as i connect this constant logic trigger signal to the 555 timer, the 555 timer goes around in a seven second loop and never stops.

i just wonder if there is a better soultion?

thanks for your help : )

In your post, I didn't get what you meant by:

as soon as i connect this constant logic trigger signal to the 555 timer, the 555 timer goes around in a seven second loop and never stops.

Click to expand...

Isn't your trigger signal a single low-going pulse?

However, you can use this circuit:



If you momentarily press the switch and let go, you get the required ~7 seconds delay. If you keep the switch latched, you still get the required ~7 seconds delay.

Hope this helps.
Tahmid.

Do you want the buzzer to sound for 7 seconds only at power-up?

If so, then the schematic below can provide a high signal on power-up. Then it gradually drops.

At some point on the slope, your buzzer would stop. Or you can use a buffer device to switch it off cleanly.



The capacitor retains its charge as long as power is on. The buzzer is off.

When power is shut off, (1) the diode is needed to prevent backflow of current from the capacitor, and (2) the capacitor discharges slowly through the high resistance (in parallel).

Tahmid said:

In your post, I didn't get what you meant by:


Isn't your trigger signal a single low-going pulse?

However, you can use this circuit:



If you momentarily press the switch and let go, you get the required ~7 seconds delay. If you keep the switch latched, you still get the required ~7 seconds delay.

Hope this helps.
Tahmid.

Click to expand...

Thank you very much Tahmid . i just tested it out and it works , yes its just right , 7 secs if momentary, or 7 secs if the logic is held on. : )

i shall see how small a footprint i can get now ,

thanks once again.

have a great week : )

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BradtheRad said:

Do you want the buzzer to sound for 7 seconds only at power-up?

If so, then the schematic below can provide a high signal on power-up. Then it gradually drops.

At some point on the slope, your buzzer would stop. Or you can use a buffer device to switch it off cleanly.



The capacitor retains its charge as long as power is on. The buzzer is off.

When power is shut off, (1) the diode is needed to prevent backflow of current from the capacitor, and (2) the capacitor discharges slowly through the high resistance (in parallel).

Click to expand...

thanks brad the rad for your idea. Thamid managed to sort it out for me.

i really appreciate your post. very clever that it can hold its energy, for when its is needed : )

thanks once again.