LED Resistor - Page 1
Forum section: electronics / beginners

author: dark_hawk
author_rank: Regular Contributor
timestamp: July 02, 2013, 06:38:54 am
content: Hello,This is a very newbie question so please be nice. (It's in the beginners section).I'm trying to connect 18 LEDs to a 6V power source. LEDs Voltage drop is 3.5V @ 20ma. Question is: Which is better and why, Connect a resistor (130ohm 1/8watt) to each LED or Can I connect a resistor (13ohm 1watt) to 9 of them?Thanks.

author: Unknown
author_rank: Unknown
timestamp: July 02, 2013, 07:17:47 am
content: EDIT: I think I answered the wrong question here. In this case, I think it would be better to have one resistor per LED. The reason being that since E = IR, as the current through the resistor changes, so will the voltage drop, so you might end up with a toasty LED if you have a resistor rated for all 18 and only one is on. With one resistor per LED, the voltage drop and current are both known, so you can be reasonably comfortable with the value of the resistor.And someone pleeeeze correct me if I'm wrong, I want to know.

author: casper.bang
author_rank: Frequent Contributor
timestamp: July 02, 2013, 07:30:19 am
content: Quote from: dark_hawk on July 02, 2013, 06:38:54 am18 small-signal resistors are going to be cheaper than 2 power resistors, but then again, you could just put small-signal resistors in parallel. Parallel coupling always opens up the chance of creeping single-point-failures, otoh it will also allow you to hit an optimal target resistance more precisely. Regardless, either solution is going to dissipate about 1W of energy into heat, which may or may not warrant a look at more efficient alternatives like a cheap step-down switching regulator.

author: Unknown
author_rank: Unknown
timestamp: July 02, 2013, 07:54:39 am
content: Quote from: dark_hawk on July 02, 2013, 06:38:54 amHello,In my experience, it is customary to typically use one(1) resistor per LED. Mainly because it is cheaper like mentioned before.

author: dark_hawk
author_rank: Regular Contributor
timestamp: July 02, 2013, 08:04:16 am
content: All of the LEDs will be lit together, so there is no fear of a toasty LED taking the full voltage.If the expense is not an Issue, Will I able to get away with 2 resistors instead of 18?The space is really small and putting 18 resistors in there will be a pain.

author: Unknown
author_rank: Unknown
timestamp: July 02, 2013, 08:27:09 am
content: Quote from: dark_hawk on July 02, 2013, 08:04:16 amthey both pretty much weigh the same. (9) 1/8th watt resistors is 1 and 1/8 wattsin other words I do not see a major problem with it, so long as your resistances are correct for your load(s)

author: sleemanj
author_rank: Super Contributor
timestamp: July 02, 2013, 09:14:52 am
content: Since you can't run more than one in series off that 6v supply (unboosted), then ideally go with a resistor each, as you correctly conclude if you used one resistor for all the LEDs in parallel, every LED that goes out, increases the current to the remaining ones, causing a cascade failure, eventually - not likely instantly.If you can use say 12v, then six series strings of three with 1 resistor per string would be a good compromise.

author: IanB
author_rank: Super Contributor
timestamp: July 02, 2013, 09:24:18 am
content: Quote from: dark_hawk on July 02, 2013, 06:38:54 amIt doesn't look like anyone has given you the full answer to your question. You should normally use one resistor per LED to ensure proper current distribution between the devices (each LED should get 20 mA).If you use a 13 ohm resistor you will have 200 mA available, but how will you ensure that is split evenly between the LEDs so that each one only gets 20 mA? Instead of, for example, one LED getting 40 mA and the others getting 18 mA? Using one resistor per LED ensures this cannot happen. The individual resistors act as current dividers.If space is at a premium, have you considered using surface mount resistors? They would take up much less room than through hole resistors.

author: lapm
author_rank: Frequent Contributor
timestamp: July 02, 2013, 09:24:47 am
content: I give one more reason to use resistor per led....There is manufacturing variations in leds. At same voltage some take just little more current then others. Putting several of them in parallel would result usually one led burning itself out by hoarding massive current, then next... I have seen this happen myself.Resistor per led, avoids this by adjusting single led's current with single resistor. This way one led cant rob all the current from its neighbors.

author: mariush
author_rank: Super Contributor
timestamp: July 02, 2013, 09:34:16 am
content: Well, 6v input , 3.5v on led ... that means you drop 2.5v on resistor. V = I x R = > I = 2.5 / 130 = 19mA.I guess the value is right for 20mA but you should pay attention to the actual rating of the leds. You don't want to run the leds close to the maximum power rating.20mA might be super bright leds which may shorten their life and if the leds are close together this brightness may be annoying or cause ugly effects.If you connect multiple leds in parallel and somehow one fails, the other leds get more current so their color may be distorted, for example a red led might turn to slight orange tint.You might want to investigate using led drivers instead of resistors. They'll be slightly more expensive than simple resistors but they'll work better with varying voltage, since they do constant current on a wide range of input voltages.For example, this BCR405U can do 50mA constant current by default, up to 65mA with an external resistor, with input voltages up to 40v :http://uk.farnell.com/infineon/bcr405u/led-driver-low-power-leds-6sc74/dp/1791064datasheet:http://www.infineon.com/dgdl/bcr405u.pdf?folderId=db3a30431400ef68011407a9cfc70181&fileId=db3a304333b8a7ca0133f9d1e2ca4349You can put 4-6 leds on one of these, 10-15mA per led should still get the leds quite bright.There's also other led drivers in similar footprints, and even smaller ones that are like smd resistors, such as these:http://uk.farnell.com/on-semiconductor/nsi50010yt1g/ic-led-driver-50v-0-01a-sod123/dp/1794979http://uk.farnell.com/on-semiconductor/nsi45020at1g/ic-led-driver-45v-0-02a-sod123/dp/1794973http://uk.farnell.com/on-semiconductor/nsi45030at1g/ic-led-driver-45v-0-03a-sod123/dp/1794976RL

author: casper.bang
author_rank: Frequent Contributor
timestamp: July 02, 2013, 09:50:07 am
content: Just thought it should be mentioned, that you can also avoid the discrete resistors entirely if space is really a concern, by using the appropriate (length and thickness) constantan wire. It's very cheap and will last you many projects, just don't try to dissipate more than 1W/30cm as it starts to feel warm to the touch.

author: dark_hawk
author_rank: Regular Contributor
timestamp: July 02, 2013, 11:24:22 am
content: Got to love this community.I asked a simple question and in the process learned of the existence of an LED driver and constantan wire.Will go with one resistor per led for this project and will get some LED drivers, smd leds/resistors and contantan wire next time I'm making an order.Thank you all for your answers.

author: bingo600
author_rank: Super Contributor
timestamp: July 02, 2013, 05:48:39 pm
content: Another solution since you will light all 18 up at once , to use a LM317 to generate the voltage the LED's require from the 6v psu.Voila ... No resistors needed.But i'd prob go for the one resistor per led also/Bingo

author: Unknown
author_rank: Unknown
timestamp: July 02, 2013, 06:03:04 pm
content: There are SIP resistor networks that might take less space on the board.

author: IanB
author_rank: Super Contributor
timestamp: July 02, 2013, 06:32:14 pm
content: Quote from: bingo600 on July 02, 2013, 05:48:39 pm1. LEDs do not require a regulated voltage, they require a regulated current2. Even if you configured the LM317 as a current regulator, it would not provide for current balancing between the parallel LEDs3. You could put several LEDs in series to ensure equal current, but then you would need a boost converter which the LM317 is not4. An LM317 still needs resistors to set the output voltage or currentQuoteWhy would you ignore your own advice? Is it perhaps because you know your advice is bad?

author: bingo600
author_rank: Super Contributor
timestamp: July 02, 2013, 07:13:09 pm
content: Quote from: IanB on July 02, 2013, 06:32:14 pmWhoaaa ... having a bad day ?As i see it ,the LM317 wouldn't be worse then the "connect a resistor (13ohm 1watt) to 9 of them"But you might be right in that it wasn't any better either.I would chose 1 resistor per LED for getting an even light distribution on all 18.@OP sorry about the advice.@Ian B- Hope your day gets better/Bingo

author: casper.bang
author_rank: Frequent Contributor
timestamp: July 02, 2013, 07:49:22 pm
content: Quote from: IanB on July 02, 2013, 06:32:14 pmSemantics; matching the appropriate forward voltage, the current is predictable and in accordance with the datasheet.QuoteMore importantly, it would do absolutely nothing to solve the OP's problem of adapting 6V to 3.5V!QuoteA boost converter would need several resistors, caps and an inductor - now we're needing a PCB and lots of space.QuoteSure but there *are* fixed voltage regulators that can do 3.5V like LM78D35 or AN77035SP.

author: IanB
author_rank: Super Contributor
timestamp: July 02, 2013, 08:02:12 pm
content: Quote from: casper.bang on July 02, 2013, 07:49:22 pmThis statement is not correct. The forward voltage is a range that varies from batch to batch, varies between samples, and varies with operating temperature. Given an accurately regulated voltage across an LED you cannot predict the current passed with any degree of confidence. Any attempt to voltage regulate an LED without current limiting in the circuit will dispose the LED to premature failure.The inverse statement is what is correct. Given a regulated current through the LED, you can use the datasheet toestimatethe forward voltage with reasonable confidence. But datasheet numbers are typical. Individual samples may vary.

author: Jon Chandler
author_rank: Frequent Contributor
timestamp: July 02, 2013, 08:04:56 pm
content: Quote from: casper.bang on July 02, 2013, 07:49:22 pmNo, Ian is exactly right. The currentmustbe balanced between the LEDs with individual resistors.IFthe LEDs wereabsolutely perfectly matchedthis wouldn't be true and the current would be split equally but this is never the case.LEDs in parallel must have individual resistors. Always.LEDs in series only need one resistor for the group as the current is the same through each but the supply voltage must be greater than the sum of the LED forward voltages.

author: IanB
author_rank: Super Contributor
timestamp: July 02, 2013, 08:19:37 pm
content: Also, for others reading, don't fall into the trap of thinking the series resistor is a voltage dropping resistor. It is not. It is a current limiting resistor. There's a big difference.If you do the algebra you will see that a high voltage in series with a large resistance approximates a constant current source. Since a constant current supply is what an LED needs, the high voltage plus resistor combination is a cheap and easy way to drive small LEDs.

author: Unknown
author_rank: Unknown
timestamp: July 02, 2013, 08:25:14 pm
content: So what makes the difference between a voltage dropping resistor and a current limiting resistor? After all, they're the same resistor, and the current limiting resistor will also drop voltage. Is it because ohm's law is a linear equation?

author: IanB
author_rank: Super Contributor
timestamp: July 02, 2013, 08:44:45 pm
content: Quote from: duskglow on July 02, 2013, 08:25:14 pmTypically the difference is the situation in which the resistor is used and the purpose it is assigned.Voltage dropping resistors are typically used with other resistors in a potential divider situation. Suppose I put two resistors in series across a 6 V supply, say 1 k and 2 k. Then we know the 1 k resistor will be dropping 2 V and the 2 k will be dropping 4 V giving a total of 6 V to match the supply. The 4 V value will be stable and precisely known assuming the resistors are also stable and accurate. This voltage may now be used in other parts of the circuit.In contrast, when we put a resistor in series with an LED or other current driven device we don't intend to regulate the voltage, we intend to regulate the current. We don't much care what the actual LED operating voltage is just as long as we don't exceed the maximum current through the LED.

author: Thor-Arne
author_rank: Unknown
timestamp: July 02, 2013, 10:03:27 pm
content: Since no one suggested it, I'll just dropthishere.

author: MacAttak
author_rank: Unknown
timestamp: July 02, 2013, 10:11:27 pm
content: For 18 LEDs just use some bussed resistor networks. They are not very large, and aren't expensive either. You should have no trouble finding 7 (or 10) pin bussed resistors, which will be plenty for your 18 LEDs.A bussed resistor network is essentially a handful of resistors with one common lead tied together.150 ohm is probably a little safer to use than 130 ohm, just in case your power source is slightly more than 6v or the forward voltage is a little less than 3.5v. 150 is also a more common value ( = cheaper). You probably wouldn't even notice the difference in brightness from the slightly lower current.For example:http://www.digikey.com/product-detail/en/770101151P/770-101-R150P-ND/1000508You would only need two of those since it packs 9 resistors per package.

author: TerminalJack505
author_rank: Super Contributor
timestamp: July 02, 2013, 10:36:48 pm
content: Just one more option that the OP should be aware of is theJoule Thiefcircuit. These can be made fairly small. The transformer will be the biggest component but it still wouldn't be terribly bulky.They're a lot of fun just to play around with. They do require that you wind your own transformer and do some tweaking to get things optimized. (You don't want the circuit oscillating at audible frequencies, for example.)I've seen people run strings of 50 or more LEDs from a single 1.5V AA battery. The circuit is also very efficient. So if you're running your circuit from batteries and want to get the most juice out of the batteries then you should keep this circuit in mind.

author: dark_hawk
author_rank: Regular Contributor
timestamp: July 03, 2013, 04:37:12 am
content: Didn't know that asking a simple question will result in such an informative discussion, will make sure to ask so often.I had the voltage divider and current limiting resistor concept fuzzy in my head, Thank you Ian for the clarification.This Joule Thief circuit sounds like fun.The power source are 4 1.5V Type C batteries so there is no fear of an over-voltage running through circuit.The LEDs I got are generic with no datasheet, burned some to see their limit and they seem to pop at around 50-60ma with decent brightness at 3.5V at 20ma. Applying the full LED loads (18 5mm white + 5 x 10.5mm white) results in approximately 0.5A and voltage drop across a new set of batteries from 6.4V to 5.7V and an old set of batteries from 5.5V to around 4.4V. So correct me If I'm wrong the 130ohm resistor seems adequate.

author: cthree
author_rank: Frequent Contributor
timestamp: July 03, 2013, 05:38:14 am
content: An lm317 is going to require a heat sink sourcing 360mA and I think the op said space was an issue.

author: MacAttak
author_rank: Unknown
timestamp: July 03, 2013, 11:21:47 pm
content: Most typical LEDs are rated for 20 mA max, that doesn't mean they will pop at 21mA. It just means that once you exceed that rating the lifespan of the component is reduced. It won't "pop" until you greatly exceed that rating, but it will certainly be damaged.I'm not sure I understand your method for measuring voltage drop. A typical white LED should be something in the range of 3.5v. When lit, you should be able to read roughly a 3.5v difference between the anode and cathode. If your fully-charged battery source is at 6.4v then that means you need to drop the remaining 2.9v with the resistor. R = V / I, so the desired resistance is 2.9v/0.02A = 145 ohms. Anything less than that and you risk overdriving the LED. And even though the current rating is usually 20mA, most are efficient enough to where you almost can't tell the difference between ~15mA and ~20mA.