## Step 8: Wiring Up Multiple LEDs in Parallel

Unlike LEDs that are wired in series, LEDs wired in parallel use one wire to connect all the positive electrodes of the LEDs your using to the positive wire of the power supply and use another wire to connect all the negative electrodes of the LEDs your using to the negative wire of the power supply. Wiring things in parallel has some distinct advantages over wiring things in series.

If you wire a whole bunch of LEDs in parallel rather than dividing the power supplied to them between them, they all share it. So, a 12V battery wired to four 3V LEDs in series would distribute 3V to each of the LEDs. But that same 12V battery wired to four 3V LEDs in parallel would deliver the full 12V to each LED - enough to burn out the LEDs for sure!

Wiring LEDs in parallel allows many LEDs to share just one low voltage power supply. We could take those same four 3V LEDs and wire them in parallel to a smaller power supply, say two AA batteries putting out a total of 3V and each of the LEDs would get the 3V they need.

In short, wiring in series divides the total power supply between the LEDs. Wiring them in parallel means that each LED will receive the total voltage that the power supply is outputting.

And finally, just some warnings...wiring in parallel drains your power supply faster than wiring things in series because they end up drawing more current from the power supply. It also only works if all the LEDs you are using have exactly the same power specifications. Do NOT mix and match different types/colors of LEDs when wiring in parallel.

OK, now onto to actually doing the thing.

I decided to do two different parallel setups.

The first one I tried was as simple as it could be - just two 1.7V LEDs wired in parallel to a single 1.5V AA battery. I connected the two positive electrodes on the LEDs to the positive wire coming from the battery and connected the two negative electrodes on the LEDs to the negative wire coming from the battery. The 1.7V LEDs didn't require a resistor because the 1.5V coming from the battery was enough to light the LED, but not more than the LEDs voltage - so there was no risk of burning it out. (This set up is not pictured)

Both of the 1.7V LEDs were lit by the 1.5V power supply, but remember, the were drawing more current from the battery and would thus make the battery drain faster. If there were more LEDs connected to the battery, they would draw even more current from the battery and drain it even faster.

For the second setup, I decided to put everything I had learned together and wire the two LEDs in parallel to my 9V power supply - certainly too much juice for the LEDs alone so I would have to use a resistor for sure.

To figure out what value I should use I went back to the trusty formula - but since they were wired in parallel there is a slight change to the formula when it comes to the current - I.

R = (V1 - V2) / I

where:
V1 = supply voltage
V2 = LED voltage
I = LED current (we had been using 20 mA in our other calculations but since wiring LEDs in parallel draws more current I had to multiply the current that one LED draws by the total number of LEDs I was using. 20 mA x 2 = 40 mA, or .04A.

And my values for the formula this time were:

R = (9V - 1.7V) / .04A
R = 182.5 Ohms

Again, since the variety pack didn't come with that exact value resistor I attempted to use the two 100 Ohm resistors bundled together in series to make 200 Ohms of resistance. I ended up just repeating the mistake that I made in the last step again though, and wired them together in parallel by mistake and so the two 100 Ohm resistors only ended up providing 50 Ohms of resistance. Again, these LEDs were particularly forgiving of my mistake - and now I have learned a valuable lesson about wiring resistors in series and in parallel.

One last note about wiring LEDs in parallel - while I put my resistor in front of both LEDs it is recommended that you put a resistor in front of each LED. This is the safer better way to wire LEDs in parallel with resistors - and also ensures that you don't make the mistake that I did accidentally.

The 1.7V LEDs connected to the 9V battery lit up - and my small adventure into LED land was completed.
<p>I learnt this the hard way when a year or so ago, I bought some flashy new UV LEDs and put them in parallel to make a lamp, then watched them all fr</p>
<p>hi</p><p>thanks for this instructable but i have a related question....i have numerous LEDs powered by a 9 volt battery....trouble is it's draining the battery way too quickly....now part of the circuit i just added is a separate circuit board that makes the LEDs breath/ fade up/ fade down....now those LEDs did not light at all, although the others in the circuit did....until i changed the battery then all worked as it should.....so i can only conclude that the battery is draining too quickly on this circuit...having to change every couple of days on ittermittent use. i suspect it's a Q of soemthing i need to do with the switch...but only way to wire a switch is as a circuit breaker on the +ve wire....? i know i have had this problem on 3volt circuit before....just can't remember what i did to fix it! :-(</p>
<p>Thank you very much for these instructions.</p><p>I'm using LEDs with a forward current of 20mA and a peak forward current of 120 mA (Conditions: ulse width less than 10 msec and duty less than 1/10). How much more light will the LEDs emit with increasing current (so long as the conditions aren't exceeded)?</p>
I didn't have a breadboard or whatever it's called. My Soldering iron came in the mail so I wanted to make something to practice with until the rest of my parts for my real project came. I stopped by 2 different radio shacks looking for the regular small leds in white but none of them had them so I used a big white one and 3 small red ones. Got 2 3x AAA battery holders some resistors, some button's, and some wire. Through trial and error burning my fingers a few times, accidentally almost blowing up the batteries, I got the lights to turn on. Figured it was time to solder. Learned on the tiny buttons, I had wired the 3 leds in series, skipped the resistors as I figured splitting it 3 ways shouldn't fry the leds. The jumbo led I had gotten some info about while at radio shack and he told me I wouldn't need resistors. Soldered them, and they worked! The buttons only turn the light off if I hold them so those are probably the wrong buttons. But they were so tiny and cute for my little altoids tin light.
<p>Thanks! I'm tryinig to look for a tutorial using an LED with a 3 position slide switch, OFF-ON-ON. So the first ON would be a bright light and the second a dim light. But I've been searching google for 2 hours and can't find anything with a 3 position slide switch. Can you make a tut about that? Thanks again! :)</p>
<p>If you have your three positions, I imagine there would be two pins for the two &quot;on&quot; positions, in which case you could wire the led to both pins, but use a different value resistor for each, and the negative can come in on the common &quot;input&quot; pin.<br>I imagine the switch would break the circuit when it's off, join the circuit to pin1 at one position, then cut that and join the circuit to pin2. If it's a sliding thing and there is a chance that pin1 and pin2 can be bridged while the switch is sliding, I guess you'd have two inputs with two resistors in parallel which reduces the resistance and could overdrive the LED. There is probably a way around that or you could make sure the sliding switch (brass spring contact type) has a large enough separation that it never contacts pin1 and pin2 simultaneously.</p>
<p>Hey, just a quick question, your &quot;stand&quot; or holder for the LEDs where can you get these?</p>
<p>I have a small string of &quot;fairy lights&quot; LED's that are powered by two CR2032 batteries wired in series producing 6.2 volts (measured). I wanted more capacity so I connected the LED light string to 4 AA batteries wired in series producing 6.25 volts (measured). The 4 AA batteries burn out the LED string. Why?</p>
<p>Any additional voltage will change the formula, so it's possible that you are letting the LEDs draw too much current, which burns them out. It's also possible that for cheapness' sake, there is no resistor on your fairy lights as lithium batteries tend to have a very low discharge rate which is probably just right for the LED's, but now you have AA's which can dump way more current, that this is what has burnt them out.</p>
You can mix colors in parallel if you calculate the correct resistor for the gnd side. E.g. A red with a small resistor and a green with a larger resistor. If it's dead on they'll have the same brightness as well.
<p>Hi, I currently have 10 solar powered decking lights with each <br>consisting of 3 leds running off a 1.2v AAA rechargable 600mah <br>batteries. I would like to replace the solar power as its unpredictable <br>with a power supply. Could i simply connected all 10 lights in series <br>which would require a 12v power supply? What current/amps should is use as of course i have no spec sheet for the leds? TIA</p>
<p>Would i need a 12v led driver power supply at 600mA = 7.2w - i have found a 12v - 500mA would this be sufficient? if used the a 12v - 833mA led driver power supply would this be too much as i read the LED Driver power supply is more efficient? TIA</p>
<p>Hi. I have 2 LED lamps that they are 6V each. How can I connect them to a 12V source? Just connect them in series?</p><p>Thanks</p>
<p>I have a question, I am setting up 83 LEDs in series in my car, I want to put this in my tail light and use the same lights for running lights and brake lights. The orginal bulb has a three prong plug, two 12v sources and a ground in the middle. I want to run the LED's at 9v until it gets the second signal from the brake line then to run it at 12v. I would love any help I can get, and if I should post this elsewhere????</p>
<p>Use a Relay to switch from 9V to 12V when applying brakes. (You'll have IN-A 9V, IN-B 12V and the IN-B will also activate the Relay).</p><p>On the &quot;daylight&quot; wire, just add an LM7809 to get 9V Output.</p><p>You might also want to add a capacitor the the output of the relay to prevent the light from closing / opening everything you apply the brakes.</p>
<p>thanks</p>
<p>ur lights are cool</p>
<p>wow I can't believe you did this ! It seemed so hard to me!</p>
<p>I hope I may ask a question, I have a led light that I recycled from a spot lamp that packed it in. the original unit was 6volt but I do not have a 6 volt power source I do however have a 9 volt one. is there a way to use this power source for that led pack? or would I burn them out. I have basic knowledge of electronics but Im unclear on how much these lights would draw if they would take what they need and disregard the rest </p>
<p>There's a circuit called a voltage divider. It's made by placing two resistors in series, then connecting a wire to the center of this series. Look it up, perform calculations to figure out which resistors will split the voltage into 6V, and use it.</p>
<p>awesome thank you, since I posted this I ordered some LEDs and protptype boards from china the LEDs came with a bunch of resisters and thanks to your reply I now know what they are for lol :) </p>
<p>Nothing wrong with a voltage divider circuit. There is also a thing called a voltage regulator that will do an astounding job at maintaining the voltage at whatever it is designed to produce. Some have built-in adjustability. Others are set voltage like LM7805 which is a 5 vdc version (and inexpensive), so you would need to find a 6 vdc version. In fact there might even be one built into the device, but I doubt it. If so, it might handle 9 vdc.</p><p>And, no, they won't take what they need and disregard the rest. It depends on what is in the package that runs the device at 6 vdc. If it has a regulator or a voltage divider that somehow handles the 9 vdc, you will have lucked out. Otherwise, you have toasted it most likely.</p>
<p>I preface this with, my dad was an electrician and I wanted nothing to do with his long drawn-out lectures on the theory of electricity when i was a kid....</p><p>I'm helping my son with a project. His assignment was to build a 3-story building with 7 rooms. Each room has to have 2 lights and an independent switch but all seven rooms have to be powered by one power source. We bought these battery powered tea lights that come with a CR2032 battery and removed the battery (http://www.hobbylobby.com/Home-Decor-Frames/Candle... and then soldered pairs of two in series (7 pairs in total). I can't find any specifications for the lights as far as voltage so the only information I have is what a CR2032 battery provides. These only have to stay powered for a few minutes while the teacher gives the structure an &quot;earthquake resistance test&quot;. Any help or ideas on an appropriate power source?</p>
<p>I suspect that your situation is long since passed, but I think a single 3 vdc 2032 would run all of these lights if hooked in parallel. In other words, the two lights should have been wired in parallel, not series. I am assuming that these are LED lights. </p><p>Place all of the switches where they will be located in your model. Once the switches are in place, wire the battery plus (+) to each switch. I am assuming that your switches are only going to 'break' the plus (+) line.</p><p>Start with a wire that runs to the nearest room from where you want to place the battery. Run the wire from the battery plus (+) to the switch and then from there to the next nearest room's switch and so on until all seven switches are connected on their input side. (It might be easier to connect all of the first floor on one line and then all of the second floor on another line, but that part doesn't matter.) </p><p>When all of the switches are connected on the plus side, connect the lights in each room to the plus (+) side of each light and then to the other side of the switch it will be operated by. None of these light connections should go to any other switch - just the switch in that room and NOT to the battery minus (-). </p><p>Now all of the minus (-) side connections of the parallel lights should be run to all of the other minus (-) connections so that all of the minus (-) side connections are together and attached to the battery (-). </p><p>One CR2032 should run all of it and the switches should control each independently.</p><p>Now I suspect that there is a reason for wanting to wire the house and then show a resulting fault that an earth quake or something like that would have on the wiring as the house falls. You might want to wire each switch directly from the battery plus (+) to avoid a portion of the house falling from taking the power away from the rest of the house. You could run the plus(+) and the minus(-) together to each switch, etc. Perhaps the young man knows what the teacher is trying to achieve so that you can route your wires accordingly.</p>
<p>&quot;Once I knew that I needed a resistor of 140 ohms to get the correct amount of voltage to the LED&quot;</p><p>I think you mean to say that the resistor ensures that the <strong>CURRENT</strong> flowing through the circuit does not exceed LED's <strong>CURRENT</strong> rating of 20 mA. In your previous step...if you had an LED rated at 1.5 volts and you used a 1.5 volt battery WITHOUT a resistor, your applied voltage would be perfect but you would burn out the LED since the current flowing through it would be extremely large.</p>
<p>I understood your post and like to ask you something:</p><p>If ideally we had a 1.7v input, a 1.7v led that consumes 20ma, which resistor you would need to securely feed the led?</p>
<p>The situation here is what resistor do you use when the LED and the voltage source are the same? What should be done to overcome the 'no resistor needed' situation? You can approach it by increasing the supply to a higher value or look at the spec sheet to find out just where 1.7 v is and where 20 ma is with regard to the device specs. If the 20 ma is the optimum operational current for the diode, then design to that current. If the light output doesn't change appreciably, try using a lower voltage. That is, suppose the diode works ok at 1.5 vdc at 20 ma. 1.7 - 1.5 = 0.2 v. At 20 ma the resistor in series with the diode would have to be 0.2/0.02 = 10 ohms. That would work, but the resistor really should be larger, so it would be better to up the source voltage if you can. The 10 ohm resistor is better than none.</p>
<p>So you need to ensure that only 20mA flows through your LED correct? What size of resistor would permit only 20mA to flow when a voltage of 1.7V is connected across its leads? R = V/I, so R = 1.7 V/ 20 mA = 85 Ohm.</p><p>So 85 Ohms is the resistor you need, choose the closest available resistor higher or equal to this value.</p>
<p>This isn't quite correct... 1.7V(in)-1.7V(f)=0V, 0V/0.02A=0 Ohms. It should be fine without a resistor, though a resistor won't hurt if your voltage source has potential to spike.</p>
<p>I thought that the LED itself can control the current passed through it when the applied voltage doesn't exceed its forward voltage? My teacher told me resistor only needed when the applied voltage exceed the forward voltage of the LED.</p>
<p>But isn't that always the case? If not, the LED would not work.</p>
<p>I disagree. He stated it correctly. He was designing so that at the current spec'ed the voltage would be at the right amount as seen by the diode. </p><p>And I think you are incorrect in the second one, too. If the diode is rated at its forward voltage to accept 20 ma when 1.5 vdc is applied, then the current that should be flowing with that battery hooked without a resistor is 20 ma, not extremely large. Unless, of course, the real 20 ma current spec is for a 1.2 vdc supply, then the current might get higher, but not an awful lot - though probably enough to destroy the LED</p><p>The only thing about not having a resistor is that the power might fluctuate (usually in battery-operated situations that is not a problem) such as arcing in the line caused by poor contact, when a power supply gets hit by a transient it can't handle or an automobile starts up charging its battery at 14.3 rather than 12 vdc. These are the reasons to not leave out the resistor.</p>
<p>I understood your post and like to ask you something:</p><p>If ideally we had a 1.7v input, a 1.7v led that consumes 20ma, which resistor you would need to securely feed the led?</p>
<p>I'm a bit confused. I decided to find out why cheap LED torches pop their LED's. Not a resistor in sight. 7 white LED's on 3 rechargeable 1.2 V batteries still pop. They are all wired in parallel. The article states how to connect LED's in parallel with one resistor and further on someone states each LED in parallel must have it's own LED. It does seem that should an LED pop the load on the others will go up if only one resistor is used therefore it would be a good idea to run them at a slightly lower current than maximum. Should an LED pop then the current would still be within tolerance and the batteries / LED's would last longer. The light would degrade slightly though. </p>
<p>It is not that each LED 'must' have its own resistor, it is just less problems if they do. LEDs will still 'pop' because they fail. Not knowing what the spec sheet says, one is running a fairly large risk of failure if the current gets too high. These diodes are only tested for operation not longevity. </p><p>If you put one resistance (combo of resistors or just one) in series with each diode, the effects of one 'popping' is not noticed by the others. And the problem goes up linearly with the number of LEDs sharing a resistor.</p><p>If you look at the spec sheets, there is a range of values where the light output is relatively constant. The design point should be near the downward curve of the output and further from the max where it will 'pop'.</p>
<p>I need to make lamp battery 3,7V it should be commanded by a momentary switch </p><p>1click 1 led on, 2nd click 2 led on, 3rd click off.<br>Anyone can help me?</p>
<p>I don't understand why people recommend microcontrollers. It's stupid and a waste of money, just get a small electrical switch and solder the power cable to it. Doesn't get much simpler than that.<br>Not sure if these switches are toggle or hold:<br></p>
<p>I think the main reason that they recommend them is that they understand them. It takes more than hooking wires together to make it work. Would you guys please explain how you get this microcontroller to do anything at all? I think that you have to write a program, don't you? And don't you have to understand how to write that program into the controller as well? Can you modify the program when you find it doesn't work? </p><p>Don't get me wrong because I think you are right. But we tend to go with what we know. This might all be fun for you, but not so much for someone who doesn't have the knowledge or inclination to attempt it.</p><p>And another thing, I worked on a tractor that had an instrument panel operating off ordinary electric gauges. It had broken and for \$200 the people who owned the tractor replaced it with one that uses a microcontroller. It worked for awhile and then broke again. I looked at both the old one and the newer one and found that the old one was still operable, but I could get nowhere with the newer one because I couldn't find any specs on the microcontroller. I fixed the problem with the original analog panel and it works fine. (BTW, the problem was that the ammeter in both units overheated and burned things up. (There is no connection between the ammeter and the other gauges except proximity.) I could find no reason for the overheating so I took the ammeter out, tied its input and output wires together outside of the instrument panel and replaced it with a cute little digital voltmeter. No problem since.)</p>
The microcontroller provides a reliable/stable current/voltage control than a simple resistor, and also provides more possibilities such as: pattern flashers, and transmission of data. Led's connected to microcontrollers can also be used as testing equipment.
<p>microcontrollers can easily do what weeks of analog or IC design would struggle with. at &pound;3 a pop for a arduino nano with 13 io ports, ic2, 8 analog ports, serial connections and 6 PWM lines, which will also run many other functions, why not? otherwise, you end up with a huge horrible mess of ANDs, 555s, 4017s, ORs, capacitors, NANDs, resistors, diodes and wires ect. whilst trying in vain to find that ONE dry joint in a million connections- I've been there, and digital is easier trust me.</p>
<p>A microcontroller with 4 outputs can go for as little as 50 cents (if you have the software/hardware to upload). Here's an example[http://electronics.stackexchange.com/questions/74201/is-there-an-ic-chip-to-toggle-through-three-outputs] without microcontrollers. You can also build toggles using simple circuits like JK Flipflops, or buy toggle switches.</p>
<p>so, the most easy way would likely be a microcontroller, but you could (if you're willing to settle for a toggle/slider) just use a single pole triple throw switch:</p><p>one lead unconnected, the other two linked by a diode and each two a LED. both other ends of the LED to ground via an appropriate diode each, and the switch's communicator to +3.7v. </p><p>or if you like a challege and are rich, you could use latching AND gates, a NOT gate and a triple pole momentary switch <strong>but</strong> unless you have custom pcb tools and surface mount soldering gear, it will end up massive. also, trying to describe it with ascii charactors will be near impossible</p><p>a slider may also be more easy to use tho</p>
<p>You're probably better off going to an electrical shop (a decent one) and talking to a member of staff </p>
Youre gonna have to use a microcontroller for that, an arduino should do
<p>Ok, let me see if I've got this straight. </p><p>I want to do 50 LEDs in parallel. They're all 3v, 20mA. Assume they're all pretty well matched.</p><p>If I wanted to use a 5V power supply, and just use 1 resistor, my math would look like this:</p><p>50 * .02A = 1A (total current)</p><p>R = (5Vpower supply - 3Vload) / 1A</p><p>R = 2V / 1A</p><p>R = 2&Omega;</p><p>Also, 3V * 1A = 3Watts, </p><p>so I would need a 3W 2&Omega; resistor, correct? </p>
<p>Yes, it looks right, but I would say 'don't do it', first because they are all not 'well matched' and second because the large resistors are more than 50 times the cost of little 1/4 W resistors. For a little more cost in time, you can put 100 ohm resistors in serial and avoid all of the problems the heat on the 3W resistor (do you have one of those?) will cause. You can purchase 50 qty 100 ohm resistors for about \$1, put 100 ohms in series with each diode and avoid the problems your one resistor solution will create. This method will give some safety margin and will prevent individual failures from burning up your string (not the usual failure, but consider what a short would do). A 100 ohm resistor in the line drawing 0.02 A is 100 x .02 = 2 v drop on each leg. I would consider increasing that to 120 ohm if you can find them, but I know the 100 ohm resistors are available cheap .</p>
You may need more than 2ohms but 3 watts should be sufficient, although I would personally prefer slightly larger, like say, 5 watt. (I look at it this way: better to be safe than sorry.)
<p>Nooo. First rule of leds is that they each need a resistor in parallel.</p><p>In parallel, the variance in forward voltages (differences from specification) becomes very exaggerated, so they will all draw different currents. In fact, the majority of the current will travel through the led with the lowest voltage drop, and burn it out. This will occur for every led until eventually none of them work. In series, this problem is more or less solved because there's just one voltage drop (across multiple leds) for one resistor.</p><p>For each parallel circuit, the voltage will be 5V. The voltage after travelling through the led will be approximately 2V(+-variance). </p><p>V=IR</p><p>V=5V input - 2V consumed by loads</p><p>R=V/I</p><p>R = 2Volts/0.02Amps = 100 Ohms</p><p>Each led in parallel will require at least 100 Ohms of resistance.</p><p>You could also do 25 parallel circuits of 2 leds in series. Each pair of leds would then require only a 50 Ohm resistor.</p><p>5V - 4V consumed = 1V, 1V/0.02A = 50 Ohms</p>
<p>I learnt this the hard way when a year or so ago, I bought some flashy new UV LEDs and put them in parallel to make a lamp, then watched them all fry</p>