I've designed the following schematic for my LED emergency lighting project;

I'm planning on building it soon, my knowledge in electronics is short right now (I started learning electronics by reading the book at "allaboutcircuits.com" two days ago).

I would like to know if my design is solid, correct and safe, should I add protection circuit, change the "Relay power OFF circuit", etc;

Do you have any suggestions on how can I make it better?

If changes are needed, could you please add a schematic for the change and how it should be connected to the existing circuit? remember that I'm a beginner.

LED emergency lighting schematics:

Purged delay circuit Revision change

Corrections, changed relay symbol to SPST Revision change

Changed parts designations Emergency LED lighting schematic

Former schematics
Deprecated-00 Former schematics Deprecated-01 Former schematics Deprecated-02 Former schematics

Design explanation:

  • The DC power supply is an old 5VDC-Output phone charger, it is going to stay connected to the mains 24/7.

  • It is essentially two separate circuits, one for the relay coil with power OFF delay and the other is to drive the LED's with 1.2V*4 AA battery pack.

  • Now the relay is connected directly to the 5VDC power supply as suggested; when the Mains power goes out, the relay will close the LED's driving circuit.

  • The switch S1 is when I don't need the light anymore while it is being used as a flashlight.

  • \$\begingroup\$ What is D4 for? It seams to be the wrong way. The C2 can be a problem as well (it will be huge ;-) ). \$\endgroup\$
    – Botnic
    Apr 7, 2015 at 13:00
  • 1
    \$\begingroup\$ If you have a 4.8V or 6.24V,... battery than you have 4.8V on one side and 0V on the other side. Not +-4.8V. I would put a switch on the high-side (+4.8V) to keep the circuit always connected to ground instead of positive voltage. \$\endgroup\$
    – Golaž
    Apr 7, 2015 at 13:00
  • \$\begingroup\$ Oops; About the switch position, shouldn't it be where all the Electrons are flowing from? what is the logic here? could you elaborate. \$\endgroup\$
    – Jon Tofly
    Apr 7, 2015 at 13:05
  • \$\begingroup\$ Switch: It works on both ways. But usualy it is connected on the positiv power side. \$\endgroup\$
    – Botnic
    Apr 7, 2015 at 13:06
  • 1
    \$\begingroup\$ About the switch position; Back in the stony ages, Benjamin Franklin thought that charge flowed from positive to negative, and by the time it was discovered that he'd made a wrong guess, it was too late. So, it got grandfathered in and we now call it "conventional current flow" in order to differentiate it from electron flow when the subject comes up. \$\endgroup\$
    – EM Fields
    Apr 7, 2015 at 14:45

2 Answers 2


1) You don't need the switch because if it's an emergency light Murphy will have turned it off at the most inopportune moment.

2) You don't need the delay because it's just adding a lot of unnecessary stuff to something that's supposed to just work and not have reliability problems.

3) You don't really need the 5 volt supply since you can just use a relay with an AC coil. However, it's probably a good idea to use low voltage where you can and minimize the possibility of someone getting zapped.

4) Make sure that, if it's a 5 volt relay, that 6.2 volt supply won't fry it.

Having said that, here's how I'd do it:

enter image description here


You basically have the right idea, but there are a few issues:

  1. That's one wierd relay symbol, so it's not clear how it's really hooked up. You are asking the relay to act like a SPST switch, and be open when the coil is energized. It's hard to tell if that's really how it's connected. Don't make up your own schematic symbols, at least not for common parts that have well established symbols.

  2. You talk about 5 V coming from a phone charger, which is how you detect AC power being present. However, you show 6.24 VDC coming from a separate battery. Huh?

  3. In two places you show both a + and - voltage, althought that is inconsistant with the battery voltages you show. It appears you meant 0 V in place of the negative voltages.

  4. I have no idea what you think C2 and D4 are doing for you.

  5. I think you mean 2N2222A, not 1N2222A.

Overall, I don't see the point in using a transistor to turn on the relay when power is present. Why not just connect the relay coil directly to the 5 V power you say is available? What logic is the transistor providing you?

In response to comments

If whatever software you are using doesn't have a acceptable symbol, make your own or use better software. There is no excuse. This still needs to be fixed.

The numbering of C2 and D4 is not what I was referring to in point 4 above. Your existing component designators were fine, as we could easily talk about individual components. Changing the designators after others have already referred to them is going to cause confusion. Anyway, the point was these components aren't doing anything useful. What exactly do you think their purpose is?

Again though, why isn't the bottom circuit simply the relay coil connected to the 5 V power supply without any other components?

  • \$\begingroup\$ 1. The relay is a Fritzing core component, I didn't find any SPST relay in the library. 2. I've corrected it, will upload shortly; the phone charger should keep the relay energized as long as there is power. 3. Corrected. 4. Changed the numbers, {C2 -> C1, D4 -> D1}. 5. Corrected. \$\endgroup\$
    – Jon Tofly
    Apr 7, 2015 at 13:22
  • \$\begingroup\$ I will correct the relay symbol. The logic of the capacitor is to keep the relay on for a second or two after the power is out, so that it will not close the LED circuit if the power is just out for a second; also for the joy of learning. \$\endgroup\$
    – Jon Tofly
    Apr 7, 2015 at 13:53
  • \$\begingroup\$ @Jon: But it's not going to do that. The capacitor has no way to get charged up in the first place. This is a emergency device. You want to keep it really simple to maximize reliability. What's the big deal if the emergency lights come on for a second when the power goes out for a second? This seems like a poor reason to make a fail-safe device more complex. \$\endgroup\$ Apr 7, 2015 at 14:06
  • \$\begingroup\$ Duly noted, I will change it the design. \$\endgroup\$
    – Jon Tofly
    Apr 7, 2015 at 14:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.