Most Power Saving Light Delay Circuit

There are a lot of circuits that once you switch on a light, the light will be turned off automatically after sometimes.

Most circuits do not consume too much power during light off. However the standby circuit still consumes little current.

This circuit uses a relay to cut the power completely after time out, such that no electricity is consumed during standby.

Step 1: The Circuit

Three parts in the diagram:
1. LED light
- LED in series with one 200ohm resistor

2. timer by 555
- the 1k, 33k resistors and 470uf capacitor will turn pin 3 to ground for around 10 seconds

3. relay
- when pin 3 is low, transistor will switch on so relay close and circuit is on
- when pin 3 is high, transistor will switch off
- hence relay will open and circuit is off

The flow logic is like this:
- when you press button to turn on the ciruit, LED is on and timer is on
- even when you release the button, as the timer is on, the relay remains close so electricity goes through relay to the LED
- when timer is off, relay losts power so relay is open and the circuit is disconnected from battery
- LED is then off

Step 2: The Material

LED and 200ohm resisitor

555, 1k and 33k resistors, 470uf capacitor
- choose different resistors nd capacitor for differnet timer setting

1k resistor, npn transistor (I used 9013), 3v or 5v relay

4.5v battery

Step 3: The Deliverable

As shown, you can put the circuit on breadboard as I did.

I don't have a button so I used a wire to touch the positive end of battery to switch on.

The circuit will keep light on for around 10 seconds then you will hear a tak sound from relay and the light will be off.

When off, zero power will be consumed.

That is all. Enjoy !



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    6 Discussions


    4 years ago

    This is astable. Check for for the formula. The config here will cause 10 seconds Low at pin 3 so relay will be on for 10 seconds. Note 555 timer may not be accurate if the duration is more than several minutes, eg over 5 minutes then inaccurate.

    4 replies

    Reply 4 years ago

    That website is very helpful, thanks. one question on your circuit, it looks like your switch bypasses the relay, is that intentional? I assume you switch circuit on, then press button to switch relay to turn on the load. switch just ensures it's completely off. That sounds correct?


    Reply 4 years ago

    Yes I international do that. But the flow is as below

    - when switch on, circuit goes via the switch into 555 so timer started and relay is triggered on
    - such that when you release switch, the electricity goes via relay to support 555 to keep relay on, at the same time electricity goes via relay to the LED
    - when 555 timer is time out, relay is off so no more electricity goes to 555, ie no electricity goes to LED


    Reply 4 years ago

    I understand now. Thanks for the explanation. Seems quite logical to me now I understand! Thanks for all the help, looking forward to making this


    Reply 4 years ago

    Sorry, I missed your reply here and was just checking things out once again. Thanks for the reply. I still haven't had a chance to build this, but will eventually. Thanks again


    4 years ago

    this is exactly what I've been looking for.Want to use it for sight light on my bow, I have a habit of leaving it on and draining batteries. Just wondering if this is a monostable out astable? Can never with that part out. just trying to find where I can work out what figures of resistor and cap will give me longer with the light on, and need to know what I'm searching for. this is great, thanks for posting it