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schematic

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 LED switch schematic

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!

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  • \$\begingroup\$ I don't really understand – Where do you want your switch to be? I'm also not 100% sure I understand your wiring. Could you maybe edit your question and use the built-in schematic editor to draw your schematic? That would solve all confusion and would give us a common base of discussion! \$\endgroup\$ – Marcus Müller Jan 18 '17 at 20:39
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    \$\begingroup\$ Oh, I didn't even realize there was a schematic editor! I will definitely create a mock-up and post it, I'm sure it would help clarify a lot of things. Thanks! \$\endgroup\$ – Alister M Jan 18 '17 at 20:46
  • \$\begingroup\$ @MarcusMüller I've inserted a schematic of what I had in mind! \$\endgroup\$ – Alister M Jan 18 '17 at 21:23
  • \$\begingroup\$ What piece of lint vtc as repair? \$\endgroup\$ – Passerby Jan 19 '17 at 0:29
  • \$\begingroup\$ Three "Wise men " voted for this ?! \$\endgroup\$ – Autistic Jan 21 '17 at 23:51
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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.

schematic

simulate this circuit – Schematic created using CircuitLab

Hope this helps! Sorry if its a bit unclear.

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  • \$\begingroup\$ Ahh I think I was having trouble visualizing the path the current takes through the button, but somehow you explained it in a way that helped me lay it out in a straight line. So I CAN connect one prong to another, I wasn't sure if that was an option or not. The description of the switch LED says that you can connect 3V-6V to the LED and it will light because there is a built-in resistor (I didn't see anywhere that said how much it was, though) and that if you go up in voltage, to add a 470 resistor. \$\endgroup\$ – Alister M Jan 18 '17 at 21:26
  • \$\begingroup\$ Ah, if it is specified in documentation 3V - 6V, then you should be good. The 12V may be a max voltage or pertain to a slightly different switch. \$\endgroup\$ – Mark Schmidt Jan 18 '17 at 21:40
  • \$\begingroup\$ I can't comment above (reputation limits) so I'll comment here. Are you sure in your circuit diagram that you want the 270 in line with all the LED's? I believe the 270 should be drawn in parallel with the other resistors and you need to add your switch LED to that diagram next to the 270 resistor. The 270 where you have it will be opposing the combined current of all the LEDs. Remember, your switch LED should be wired almost identically to all your other LEDs. (Except for a different resistor and only one LED) \$\endgroup\$ – Mark Schmidt Jan 18 '17 at 21:44
  • \$\begingroup\$ This might just be a misunderstanding on my part of how parallels work, but wouldn't putting the switch in parallel with the other LEDs then allow the circuit to close? Because wouldn't the current be able to flow through the other LEDs and resistors back to the battery and bypass the switch? \$\endgroup\$ – Alister M Jan 18 '17 at 22:10
  • \$\begingroup\$ I added a diagram. Let me know if that doesn't help clear things up. \$\endgroup\$ – Mark Schmidt Jan 19 '17 at 20:29
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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.)

schematic

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.)


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.


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.


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.

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  • \$\begingroup\$ The exact LEDs I'm using are these and they do say they need about 3.4V each. And I've also realized after the fact that they actually have the 100 resistors built IN to the board. As for the time to make the lights last, probably not more than about three hours, although I would definitely prefer them to last longer if possible. I'm more than willing to change my circuit/battery options if that's what it takes, like I said I'm really just beginning to understand the complexities at work so my base knowledge needs to be expanded on. \$\endgroup\$ – Alister M Jan 19 '17 at 3:34
  • \$\begingroup\$ @AlisterM See added note at bottom of answer. \$\endgroup\$ – jonk Jan 19 '17 at 8:32
  • \$\begingroup\$ I'm doing my best to follow along, so please bear with me. Initially I was assuming I wanted to run them at 25mA each for max brightness, but it seems like dropping that down to 20mA might be best, which I'm on board with doing. I'm not sure how to short out the resistor without blowing the LED itself, unless that is what you mean. I've got one extra LED that I can use to test with but I'm worried about doing something wrong and having to get another set of 5 or more (I'm already using four of them in another simpler circuit) \$\endgroup\$ – Alister M Jan 19 '17 at 15:31
  • \$\begingroup\$ @AlisterM I'm assuming, from reading the page, that they include a \$100\:\Omega\$ resistor in series. Looking closely at the picture, I think I can see the SMT resistor there. I was suggesting soldering a wire across that resistor. But I'd say it's just fine NOT doing that and simply trying out two in series with your 9 V source. Might be a little less bright than you want. But it is safe and easy to test. Try it out that way and don't bother to short out a resistor, for now. \$\endgroup\$ – jonk Jan 19 '17 at 18:54

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