Step 2: The LED

LEDs come in different sizes, brightnesses, voltages, colors and beam patterns, but the selection at Radioshack is pretty small and so I just picked up a couple different LEDs from what they had in a few different brightnesses and voltages. I kept close track of what LED was what voltage because I didn't want to accidentally send too much current through one of the low voltage LEDs.

The first thing I did with the LEDs was figure out which wire (its called an electrode) was positive and which was negative. Generally speaking the longer wire is the positive electrode and the shorter wire is the negative electrode.

You can also take a look inside the LED itself and see whats going on. The smaller of the metal pieces inside the LED connects to the positive electrode and the bigger one is the negative electrode (see picture below). But be warned - in the LEDs I picked up I didn't always find this to be true and some of the LEDs had the longer electrode on the negative when it should be on the positive. Go figure - its OK though, if it didn't light up I just flipped it around.

Once I knew what was positive and what was negative I just had to remember what the voltage of each LED was.

All my LEDs recommended 20mA of current. 20mA is standard for most LEDs.

<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>this might help you a bit. its 48 white led's powered by 9-12volts and its superbright with no resistors.</p>
<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>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>
<p>Hi guys. please be aware you are dealing with a total novice, with no knowledge of &quot;Electrics&quot; in any way, I build &quot;Card&quot; model Hangars and want to put led lights in them, My question to you is this, What do I need to install up to 20 3mm led lights, I have a 2xAA battery power unit and also the 3mm led lights ??? please help in the simplest terms, Thanks</p>
<p>You need to know the working current for the LED. If it is 20 ma at 1.5 vdc, then you can hook 20 LEDs in parallel using a resistor value of R = (3 - 1.5) / 0.02 = 75 ohms. I would use a resistor between 80 and 100 ohms for each of the LEDs. The maximum current would be I = .02 x 20 = 400 ma and each leg would consist of one LED which handles 1.5 vdc and one resistor of 100 ohms. The voltage drop to reduce the voltage by 3 - 1.5 = 1.5 vdc. </p><p>To get this drop, the current using a 100 ohm resistor is 1.5 / 100 = 15 ma. That should light the LED. If the current needed is 20 ma minimum then 15 ma is too little current and the resistor might have to be reduced to increase the current. Hook the resistor to the battery plus then the diode then the other side of the diode to battery minus. The LED should light. </p><p>When the combination to make one LED work is determined, then the others can be constructed using parallel methods. Wire the resistors to the diodes first, connect all of the other ends of the resistors together to battery plus and all of the other ends of the diodes together to the battery minus. </p><p>Note: It does not matter whether the diode is first in the circuit followed by the resistor, but it is best to do it the other way because a shorted diode will have devastating effects on the wiring that way, but not so if the resistor is first.</p><p>If you have switches, they can be inserted in each leg to allow controlling the on or off of each LED individually (20 switches), or in the line from battery plus between it and the first LED to control power to all of the LEDs at once. (Other combinations are possible.)</p><p>Note: if the LED does not light and you are sure the wiring is right, then make sure that you have the diode connected right. It does not hurt the diode to put it in backwards, so just switch the leads to see if it will light. If it won't then you have either a bad LED or you need a spec sheet on the LED to find out what the min and max values have to be. These were a guess based on other LED's specs.</p>
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