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I want make it so when the switch is turned off, the relay would be still on for about 15 to 30 seconds, and if it is switched on again in that time period it would just continue running. I am wondering if it would be enough to just put a capacitor connecting relay's coil, but I don't think I would achieve the desired time period and I don't really know how to calculate that time period. Maybe someone knows a better way to achieve this functionality or guide me in the right direction..

schematic

simulate this circuit – Schematic created using CircuitLab

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You should connect a 555 timer pin2 (trigger) to the line that feeds the relays signal .
And connect pin3(output) directly to the same line that feeds the relays signal. A high value resistor should be placed between this line and ground.

Look online for falling edge-triggered monostable delay 555 timer. With a simple capacitor and a resistor You should be able to specify the time you want the 555 timer to feed the relay for.

This way the 555 timer is triggered each time you switch off the switch. To run for let's say 15 seconds. That automatically resets if u switch it on and back off. And the relay will work normally if the switch was on.

Your only problem will be with this scenario. If you switched it off for let's say 5 seconds then switched it back on for 5 seconds then back off. What will happen is the timer will be triggered count 10 seconds while still sending a signal to the relay. Youd expect it to reset the timer when you switch back off finally But what is going to happen is that it will continue with the remaining 5 seconds finishing the first 15 seconds then start another 15 seconds. Which should be fine depending on your application.

Edit: Schematic for 15 second delay. enter image description here

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  • \$\begingroup\$ petervis.com/GCSE_Design_and_Technology_Electronic_Products/… This can help you setting up the 555 timer. \$\endgroup\$ – Malek May 27 at 12:18
  • \$\begingroup\$ of course with this guide you have to get rid of the Pull up resistor (because as I described earlier we added a pull-down resistor to the same line) \$\endgroup\$ – Malek May 27 at 12:23
  • \$\begingroup\$ Also do not forget to add diode parallel to the relay coil with the cathode to the positive end or you could damage the 555. \$\endgroup\$ – user9820366 May 27 at 14:48
  • \$\begingroup\$ Yes, that is very important, do that aswell. Thank you for the great save \$\endgroup\$ – Malek May 27 at 17:05
  • \$\begingroup\$ I have been trying to apply what you said, but I just can't figure a working example out... Can someone draw a schematic? \$\endgroup\$ – Edgaras Levinson May 30 at 9:41

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