1
\$\begingroup\$

I've looked around this forum but not found the answers I'm looking for. I've found some similar questions being asked, but this project is a bit specific so the answers were not really applicable to my situation. Also, I am very much a novice at this kind of thing. Maybe some of you experts will be able to help. Here's the background then...

I've got four decorative letters (one is actually a heart) which hang on the wall and are lit with white LEDs. Two have 10 LEDs, one has 11 and the other 12, so 43 white LEDs in total. Each 3V unit is powered by two 1.5V AAA batteries and each has its own on/off switch built into the battery compartment.

However, it's a real faff having to remove each unit separately from the wall twice a night, to switch them on and later to turn them off. The batteries don't last that long either and I seem to be constantly changing them.

So, in a moment of true innovation (for me anyway), I decided to mount them all permanently on a board that's painted the same colour as the wall and to wire the four 3V units in parallel to a 3V 1A mains power supply.

To complicate things slightly, I've also got four switches which are actually 12V car accessories (to switch on/off each of the units) and each switch also has a colored led in the rocker (1 x red, 1 x green, 1 x blue and 1 x yellow). I've tried connecting one switch to 2 x AA batteries as a test and the LED seemed to light up fine.

My questions are then, does this power supply sound like it would work okay with my LEDs and if so, how should I wire the units up? I'm guessing each 3V unit should be wired to the power supply in parallel. But if that is the case, should I also put some kind of resistor in the circuit for each unit and if so, what value?

Sorry, I'm a bit of a novice at stuff like this so thank you in advance for any advice you can give.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ Do you want a switch for each letter? Why not one switch for the whole thing? \$\endgroup\$
    – JYelton
    Commented Oct 21, 2019 at 18:13

3 Answers 3

1
\$\begingroup\$

Each letter most likely consists of a number of LEDs wired in parallel to the voltage source; itself two 1.5V cells in series for 3V total. Ideally, each LED will have its own current-limiting resistor, but sometimes they are omitted to reduce costs (such as with coin-cell powered devices).

AAA cells (like any alkaline cell) have internal resistance. Per the Energizer datasheet, this is 150-300 mΩ. In theory this means a maximum current, albeit for a very short time, of \$\frac{1.5V}{0.15Ω} = 10A\$, far more than an LED can withstand. This suggests to me that the product should have current-limiting resistors somewhere inside. You might want to inspect and post a photo to be sure.

schematic

simulate this circuit – Schematic created using CircuitLab

If this is the case, replacing the batteries with a line-powered 3V source should work. You could take some measurements with a multimeter to determine this more reliably.

To connect the letters together, you're correct that they should be in parallel:

schematic

simulate this circuit

The switches you have are likely for 12V use, and probably already have a current-limiting resistor for the internal LED. If they are lighting at all with 3V you can simply connect the two LED terminals on the switch in parallel to the letter it controls:

schematic

simulate this circuit

\$\endgroup\$
0
\$\begingroup\$

It does depend on the current draw of your LEDs, but given the draw of a typical(even extra bright) white LED, you should be alright. This LED uses a slightly higher voltage than you are supplying (3.1 V minimum), but even so only uses a maximum of 20 mA per LED. With the number of LEDs you have, that comes to approximately 860 mA.

Assuming your LEDs don't burn out at 3V (it sounds like you have already tried it!), you should be able to hook them up in parallel just fine without having any issues. As for wiring, it isn't necessary to have a switch for each letter, unless you want to be able to control them all independently.

Otherwise you just need to connect each letter's connections to the positive and negative sides of your power supply with the switch located close to your power supply so as to break the circuit when you want to turn it off. Western convention is to break the positive connection, which is what I would recommend, as it leaves less likelihood of an accidental ground being created.

The only change you would need to make if you wanted to operate them all independently is to have a separate positively wired connection for each of the letters with a switch on each connection.

\$\endgroup\$
0
\$\begingroup\$

photo of LED decorative letters

Thanks everyone, for your advice. As you can see from the picture, I have now finished my project with just one on/off switch. I used the 3V 1A power supply. I must say that although it all seems to be working fine at present, the LEDs are much brighter than they ever were when they were powered by batteries. I hope that doesn’t mean they will fail any time soon.

Anyway, thanks again for your help.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ They might fail sooner than they would have. While two AAA cells should put out 3V, in reality, the voltage drops if you draw a large current from them. The manufacturers of LED lights often allow for that in their design. Sometimes, they don't even bother with current limiting in their design, as the small batteries can only put out a limited current. \$\endgroup\$
    – Simon B
    Commented Oct 24, 2019 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.