Wiring LEDs in parallel with 5-pin LED on/off switch

Question

^{simulate this circuit – Schematic created using CircuitLab}

I'm creating a costume with some lights and am trying to run 10 LEDS and an on/off button with an LED from a 9V battery (so 11 LEDs in total). Due to the fact that it needs to be from a 9V for portability, I've opted to run the 10 individual LEDs in parallels of 2 each, with 100 Ω resistors at the front of each set, and then a 270 Ω resistor at the end of the series in front of the LED button. (Not gonna lie, I had to use the ledcalc website to guide me on how to wire them all up. I'm a novice when it comes to this sort of thing!)

I would prefer to have the switch at the opposite end of the circuit as the battery, since in the costume the battery pack will be at the top, the lines of 2 leds will be stacked down the middle, and then the button will be at the bottom. But I also know that I can make the wires as long or as short as needed in order to physically place everything where it needs to go so that's not as important. :)

My confusion/question comes from how to incorporate the on/off LED button into the circuit. I want the light to be off if the circuit isn't powered, but to come on when it is, but I can't have the LED part of the button accidentally complete the circuit. The details on the button say that the LED and the switch are separated, so they can be controlled independently, but I'm not not sure how to physically solder the wires correctly to make it work.

LED switch

Does anyone have insight into this 5-prong switch and could guide me on which prongs to attach to which? Thank you!

[edit] Added a schematic for visuals!

Answer 1

I should preface that I'm not 100% on this so be sure to test each of my hypothesis.

Your switch works like this: the middle three pins NC1 NO1 and C1 correspond to the switch itself whereas the 12VDC +/- corresponds to your LED. NC1 means "normally closed" and NO1 means "normally open". Normally open and close refer to which prong is connected to C1 when the button is in the On or Off position. (Off is, in my experience NC1) See this for more help: Can you clarify what an 1NO1NC switch is?

So thats cool to know but doesn't help your design. All you want to do is connect your LEDs to power when the switch is on and turn them off when the switch is off. In this case, you need to attach either the + or - of the battery to C1. Then you can attach your LED's to NO1 and they will be connected to the battery only when the switch is on. (If my earlier assumption that off = NC1 is wrong, just substitute N01 for NC1. The only difference between connecting your LEDs to NC or NO is what position of the switch will turn on your LEDs) Note: You will only use either NC or NO. The prong you don't use will have nothing attached to it.

Connect the free side of you LED's to the other terminal of the battery and you have a completed loop: Battery -> Switch -> LED -> Battery. (Make sure your LED's are connected in the correct direction. You want to see this pattern: Battery+ -> Switch -> +LED- -> -Battery OR Battery- -> Switch -> -LED+ -> +Battery)

As for the switch LED its just the same as your other LEDs. You're going one of the +,- prongs on the switch to the battery and the other prong to NC or NO that all your other LEDs are connected to.

Also, be sure to check what voltage the switch LED requires. Its concerning that its marked 12VDC. Its possible the LED for the switch already has a resistor and more worrying, 9Volts may not be enough to turn on the switch LED. You may need 12V with no external resistor.

^{simulate this circuit – Schematic created using CircuitLab}

Hope this helps! Sorry if its a bit unclear.

Answer 2

The general topology will look like the following diagram. I've broken it into two parts -- left and right -- with dashed boxes. The left side is your switch at the bottom of the costume (as I understand where you want things) and the right side is the battery system at the top of the costume and flowing down from there (as you mentioned you could do.)

^{simulate this circuit – Schematic created using CircuitLab}

As you can see, there are three wires which must go the entire distance: a NEGATIVE power wire, a POSITIVE power wire, and a SWITCHED POSITIVE power wire. You'll have to work out the details there. But you will need three conductors to make it down to the switch at the bottom of the costume.

(There are some other arrangements, but they all result in three wires because you say you want the battery supply on the opposite end of where the switch is at. If the battery and switch were together, then you could do with less than three.)

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One thing to worry about is that a \$9\:\textrm{V}\$ battery doesn't actually provide full voltage for long. It rapidly drops downward. You could plan on as little as \$4.8\:\textrm{V}\$, in fact, if you had some kind of circuit that would keep on providing the right current and voltage to your LEDs. But that's extra complexity and cost and I'm not going there. So you need to figure out what you are willing to accept and how long you expect all this to run.

You don't say what kind of LEDs you are using for \$D_1\$ through \$D_{10}\$. If these are white LEDs and need about \$3.2-3.6\:\textrm{V}\$ each, then you may be in a heap of trouble. Suppose you planned on a full, perfect \$9\:\textrm{V}\$ battery supply (never will happen; not even close) and decided to use \$R=\frac{9\:\textrm{V}-2\cdot 3.5\:\textrm{V}}{20\:\textrm{mA}}=100\:\Omega\$ as your current limiting resistor. Well, the battery voltage will quickly drop towards \$8\:\textrm{V}\$ (probably within a half-hour or so) and you will be supplying only \$10\:\textrm{mA}\$ (perhaps slightly more) then. Already, they are dimming down. By the time the battery gets down to \$7\:\textrm{V}\$ (where there is still some life left, too) your LEDs will be barely visible.

This is a serious difficulty in using a \$9\:\textrm{V}\$ alkaline battery. They don't hold their voltage for long. A circuit designed well for one of these should be able to operate down to some much lower voltage. At least down to \$7\:\textrm{V}\$ and perhaps even lower, towards \$6\:\textrm{V}\$.

But if pairing the LEDs will require all of that (and more, perhaps), then it is very difficult to make this work well with nothing more than a dropping resistor for each pair of LEDs.

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So. Disclose your exact LEDs (we need to know operating current and voltage.) Disclose how long you want all this to operate (this has a great deal to do with the battery supply itself as well as possible circuitry options to consider.) Disclose how tolerant you are willing to be about the LEDs dimming down over the period of operation you need.

The value for the resistors, including \$R_1\$, may depend upon all of the above. And, depending upon the details if and when you answer, the resistor idea may not even work well at all making the entire question moot.

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Thanks for the link. They say \$3.3\:\textrm{V}\$ provides about \$5\:\textrm{mA}\$ with \$100\:\Omega\$ in series. So that's \$2.8\:\textrm{V}\$, at that current. They also say they can operate up to \$6\:\textrm{V}\$, so in very round numbers I'd say you should plan \$3.0\:\textrm{V}\$ each (ignoring the resistor, assuming you are willing to bypass it -- short it out.) If you keep the \$100\:\Omega\$ resistor in place and want to operate these at full brightness, I'd say you should plan on about \$4.8-5.0\:\textrm{V}\$ each (which means you won't be putting them in series pairs.)

So. You should do some testing.

I'd recommend that you start out by considering the idea of bypassing the \$100\:\Omega\$ resistor that is included, as that may permit you to still use them in pairs. If you try that, consider putting two of them in series and bypassing (shorting out) just one of the two resistors in the pair. See how that works in terms of brightness and, if 10 of them running that way, longevity. Also set \$R_1=220\:\Omega\$, or thereabouts. Up or down, but in that neighborhood to start.

If that works for you, then go with that design. If not, you'll need to consider alternative ideas.