Need some help designing and building LED light strips for an interactive table
March 3, 2007 8:31 PM Subscribe
Need some help designing and building LED light strips for an interactive table -- help me select 880nm IR LEDs, understand the circuitry, and figure out how to deal with power.
As part of the table we need to build a couple of LED light strips making use of high output 880nm IR LEDs (yes, it's an FTIR table). One strip is 4' long, the other is 3' long and the LEDs are probably going to be 1" on center. Ideally, I'd like to power the strip from regular wall current.
So, the questions:
- Are there LED strips that I can readily adapt to this task?
- I love / hate digikey. For a novice, there's too much information. Can you help me find what I mean when I say "high output 880nm IR LED"? (I'm just thinking little bulb shaped guy with the 2 leads sticking out the bottom...)
- What do I need to do resistor-wise? The LEDs are going to be in series... Should I just be doing 1 stronger resistor at the end of the series or little ones in-between?
- How do I deal with the power? Is there a power brick that'll get me down into the range that I need or is will it end up being something custom?
As part of the table we need to build a couple of LED light strips making use of high output 880nm IR LEDs (yes, it's an FTIR table). One strip is 4' long, the other is 3' long and the LEDs are probably going to be 1" on center. Ideally, I'd like to power the strip from regular wall current.
So, the questions:
- Are there LED strips that I can readily adapt to this task?
- I love / hate digikey. For a novice, there's too much information. Can you help me find what I mean when I say "high output 880nm IR LED"? (I'm just thinking little bulb shaped guy with the 2 leads sticking out the bottom...)
- What do I need to do resistor-wise? The LEDs are going to be in series... Should I just be doing 1 stronger resistor at the end of the series or little ones in-between?
- How do I deal with the power? Is there a power brick that'll get me down into the range that I need or is will it end up being something custom?
I don't really know what an FTIR table is. I looked it up so I have some idea but I really don't know what you have in mind.
The problem with wiring in series is going to be voltage. Let's say for simplicities sake that you have a string of N leds in series, and a resistor in series. Each LED is going to drop between 1.5V and 3V. Let's just say 1.5V. So the LEDs total will drop 1.5N volts. A 4' long strip with 1" spacing means 48 leds, so a total of 72V. If they each drop 3V then we're now talking 144V. That means your power supply needs to be at least 144V DC. This is not impossible from wall-level power supply. I have designed tube amps that use several hundred volts DC as the source supply. But now you're talking about a pretty beefy power transformer, filtering caps, etc, etc, not just a little wall wart.
So I'm thinking you have to wire in parallel. I am guessing any item with more than a few LEDs that runs off low voltage does this. Although you may see problems with uneven brightness I don't see why this would cause a problem unless one of the LEDs varied wildly from the others in terms of conducting current at the given voltage.
Now that I think about it, lots of LEDs are pretty much always wired in parallel.
The point of the resistor, by the way, is to set the current. LEDs will conduct infinite current up to the point where they destroy themselves, so you need to limit current for them. Consider the case where you have 1 LED, 1 resistor, and, say, a 5V power supply. Say the LED drops 1.5V across it. This means that the resistor will drop 3.5V (the total drop across all devices in a loop must equal the source voltage). If you want 1mA of current, then use the formula V=IR and solve for R (R=V/I, in this case you'd want a resistor of 3500 ohms (3.5V/0.001A). Generally you choose an amperage from examining the data sheet and then picking an amperage that gives you enough brightness, with minimal consumption. 1mA is just an example here, it really varies.
With a parallel setup, you can still use 1 resistor, but if each LED uses, say, 1mA and you have 100 LEDS in parallel, then you're now using 100mA. With a 3500 ohm resistor, as in my example, that resistor is dissipating 1/3w. If you need 10mA per LED then it's 3w and so forth. In general you wouldn't want to use a higher voltage source than needed: the lower voltage source, the less you drop across the resistor, the less power it needs to dissipate. Alternately you can use 1 resistor per LED and only have to dissipate a miniscule amount of voltage. Honestly this is how I usually see it.
I think you can use pretty much any wall-wart power supply you want. Use the lowest voltage that you'll need and make sure it can deliver enough amperage to power all the lights and anything else in the circuit.
posted by RustyBrooks at 9:57 PM on March 3, 2007
The problem with wiring in series is going to be voltage. Let's say for simplicities sake that you have a string of N leds in series, and a resistor in series. Each LED is going to drop between 1.5V and 3V. Let's just say 1.5V. So the LEDs total will drop 1.5N volts. A 4' long strip with 1" spacing means 48 leds, so a total of 72V. If they each drop 3V then we're now talking 144V. That means your power supply needs to be at least 144V DC. This is not impossible from wall-level power supply. I have designed tube amps that use several hundred volts DC as the source supply. But now you're talking about a pretty beefy power transformer, filtering caps, etc, etc, not just a little wall wart.
So I'm thinking you have to wire in parallel. I am guessing any item with more than a few LEDs that runs off low voltage does this. Although you may see problems with uneven brightness I don't see why this would cause a problem unless one of the LEDs varied wildly from the others in terms of conducting current at the given voltage.
Now that I think about it, lots of LEDs are pretty much always wired in parallel.
The point of the resistor, by the way, is to set the current. LEDs will conduct infinite current up to the point where they destroy themselves, so you need to limit current for them. Consider the case where you have 1 LED, 1 resistor, and, say, a 5V power supply. Say the LED drops 1.5V across it. This means that the resistor will drop 3.5V (the total drop across all devices in a loop must equal the source voltage). If you want 1mA of current, then use the formula V=IR and solve for R (R=V/I, in this case you'd want a resistor of 3500 ohms (3.5V/0.001A). Generally you choose an amperage from examining the data sheet and then picking an amperage that gives you enough brightness, with minimal consumption. 1mA is just an example here, it really varies.
With a parallel setup, you can still use 1 resistor, but if each LED uses, say, 1mA and you have 100 LEDS in parallel, then you're now using 100mA. With a 3500 ohm resistor, as in my example, that resistor is dissipating 1/3w. If you need 10mA per LED then it's 3w and so forth. In general you wouldn't want to use a higher voltage source than needed: the lower voltage source, the less you drop across the resistor, the less power it needs to dissipate. Alternately you can use 1 resistor per LED and only have to dissipate a miniscule amount of voltage. Honestly this is how I usually see it.
I think you can use pretty much any wall-wart power supply you want. Use the lowest voltage that you'll need and make sure it can deliver enough amperage to power all the lights and anything else in the circuit.
posted by RustyBrooks at 9:57 PM on March 3, 2007
Looking at Kadin's data sheet, 100mA per led, woof! You need like 84 LEDs right, for 7' of LEDs? That's about 8.5A, a LOT for any wall-wart supply. Oh, that's maximum though. You'll probably use less than that. With IR LEDs you're going to have to find some way to figure out how bright is bright enough. With visible LEDs you can just put a pot in series with an LED, attach to your power source, and make sure you don't turn the pot down to the point where you exceed the max current of the LED. Dial it around until you find a brightness you like, measure the resistance and current, see if you'd like the current to be lower, and if it's OK then use that resistance.
posted by RustyBrooks at 10:01 PM on March 3, 2007
posted by RustyBrooks at 10:01 PM on March 3, 2007
The more I think about it, the more I think it's a good idea to have each LED have it's own resistor. This seems to be a standard practice any way.
posted by RustyBrooks at 10:04 PM on March 3, 2007
posted by RustyBrooks at 10:04 PM on March 3, 2007
This isn't specific to what you're doing in particular, but you should look into reading the chapter on diodes in the Art of Electronics to get an idea of how to use and design circuits with diodes (e.g. LED's).
posted by scalespace at 10:45 PM on March 3, 2007
posted by scalespace at 10:45 PM on March 3, 2007
Also, if you need to optimize your device for power consumption and could tolerate the fact that your led's might not be on all the time (i.e. they are refreshed at a rate higher than 24 Hz, but are not literally on all the time), you could build a circuit that has an oscillator and phase shift network to turn on a transistor that switches on a subset of your string of LED's for one period of time, then another set, etc. at a rate high enough that does not affect your detection circuitry.
This allows you to pump more current to the LED's, thus increasing their brightness, but you can minimize your power consumption becuase not all your LED's are on at the same time.
Also, if you do use this method of refreshing the LED's, you can possibly use this to your advantage in designing your detection circuitry for the FTIR (I'm guessing this is frustrated total internal reflectance) interface.
(Note, what I said above is more-or-less back-of-the-envelope brain storming.)
By the way, if you haven't done a literature search yet, you should get your hands on this paper. I'm at home, so I can't access the ACM's papers but it seems like Jefferson Han at NYU did some pretty cool work that got published and was featured in the blue a year or two back.
Han, J. Y. 2005. Low-cost multi-touch sensing through frustrated total internal reflection. In Proceedings of the 18th Annual ACM Symposium on User interface Software and Technology (Seattle, WA, USA, October 23 - 26, 2005). UIST '05. ACM Press, New York, NY, 115-118.
posted by scalespace at 11:03 PM on March 3, 2007
This allows you to pump more current to the LED's, thus increasing their brightness, but you can minimize your power consumption becuase not all your LED's are on at the same time.
Also, if you do use this method of refreshing the LED's, you can possibly use this to your advantage in designing your detection circuitry for the FTIR (I'm guessing this is frustrated total internal reflectance) interface.
(Note, what I said above is more-or-less back-of-the-envelope brain storming.)
By the way, if you haven't done a literature search yet, you should get your hands on this paper. I'm at home, so I can't access the ACM's papers but it seems like Jefferson Han at NYU did some pretty cool work that got published and was featured in the blue a year or two back.
Han, J. Y. 2005. Low-cost multi-touch sensing through frustrated total internal reflection. In Proceedings of the 18th Annual ACM Symposium on User interface Software and Technology (Seattle, WA, USA, October 23 - 26, 2005). UIST '05. ACM Press, New York, NY, 115-118.
posted by scalespace at 11:03 PM on March 3, 2007
First, try registering at globalspec (free) instead of digikey.
Second: FTIR = "Frustrated Total Internal Reflection" or "Fourier Transform Infrared (Spectrometry)"?
In any case, I found this one at $0.59 each. And the forward current is also 100mA and the drop is 1.7V.
Pay attention to Rusty's post, as he has it nailed. If I'm guessing right, you're going to the 4x3 light bars at a right angle to one another, and an array of detectors opposite them? You are not turning on the LED's individually, right?
I don't think an array of LEDs is the smartest way to do this. Rusty mentioned you'd need over 8amps just for the LEDs. Consider that most likely the breaker on the junction box for the wire that powers that entire room is going to be 15 amps. In an ordinary house, each room gets 15 amps. You are using 8 just for LEDs. Throw a computer in the room and then turn on the lights and you just blew the circuit.
Consider a single very bright led and strands of fiber optic "wires" running down the 4' length, a few strands for each 1" location (notice the use of hte phrase "internal reflection" in the link, hint hint). This will cut the power requirements tremendously, and make wiring the circuit easier. Remember, if you wire them serial, and a connection breaks, the whole light guide goes out, and good luck finding the break.
FYI, and and LCD display uses an optical waveguide to evenly spread the light from a single bulb across the entire screen.
posted by Pastabagel at 11:30 PM on March 3, 2007
Second: FTIR = "Frustrated Total Internal Reflection" or "Fourier Transform Infrared (Spectrometry)"?
In any case, I found this one at $0.59 each. And the forward current is also 100mA and the drop is 1.7V.
Pay attention to Rusty's post, as he has it nailed. If I'm guessing right, you're going to the 4x3 light bars at a right angle to one another, and an array of detectors opposite them? You are not turning on the LED's individually, right?
I don't think an array of LEDs is the smartest way to do this. Rusty mentioned you'd need over 8amps just for the LEDs. Consider that most likely the breaker on the junction box for the wire that powers that entire room is going to be 15 amps. In an ordinary house, each room gets 15 amps. You are using 8 just for LEDs. Throw a computer in the room and then turn on the lights and you just blew the circuit.
Consider a single very bright led and strands of fiber optic "wires" running down the 4' length, a few strands for each 1" location (notice the use of hte phrase "internal reflection" in the link, hint hint). This will cut the power requirements tremendously, and make wiring the circuit easier. Remember, if you wire them serial, and a connection breaks, the whole light guide goes out, and good luck finding the break.
FYI, and and LCD display uses an optical waveguide to evenly spread the light from a single bulb across the entire screen.
posted by Pastabagel at 11:30 PM on March 3, 2007
Warhol, email me at spamtrap2000 at the popular google email service, and I'll send that paper over. In return, of course, I want you to make me one too.
posted by metaculpa at 11:33 PM on March 3, 2007
posted by metaculpa at 11:33 PM on March 3, 2007
Best answer: Here's what looks to be a very useful wiki on the subject of FTIR. The same group has a message board here. They say:
Step 3: Circuit
You don’t need a series resistor for each IR led. You can put several led’s in parallel. Putting 100’s of them in parallel is not a good solution either, since the current becomes very high for the series resistor (hot). A combination of series/parallel seems the most convenient.
I used 2 strings of both 18 leds (=36 in total)
Every string has 3 strings of 6 leds (=18 total)
For every 6 leds i used 1 resistor, so for 18 leds i used 3 resistors, each 68 ohm
You’ll probably work with different numbers since you might use less or more leds or less or more powerful onces.
Also, thanks for mentioning this kickass project! I've seen the demo video floating around but had no idea the materials required were so simple. Might have to get started on one myself.
posted by contraption at 11:59 PM on March 3, 2007
Step 3: Circuit
You don’t need a series resistor for each IR led. You can put several led’s in parallel. Putting 100’s of them in parallel is not a good solution either, since the current becomes very high for the series resistor (hot). A combination of series/parallel seems the most convenient.
I used 2 strings of both 18 leds (=36 in total)
Every string has 3 strings of 6 leds (=18 total)
For every 6 leds i used 1 resistor, so for 18 leds i used 3 resistors, each 68 ohm
You’ll probably work with different numbers since you might use less or more leds or less or more powerful onces.
Also, thanks for mentioning this kickass project! I've seen the demo video floating around but had no idea the materials required were so simple. Might have to get started on one myself.
posted by contraption at 11:59 PM on March 3, 2007
Best answer: Going by that LN75X datasheet, you want about 50mA forward current, 100mA is where the diodes burn down. 1.5V forward voltage means you can get 6 of them in a chain supplied by 12V and have 3V for the resistor to drop. Yes, you could theoretically fit 7 in the chain but the manufacturing tolerances in the diodes mean you have lots more uncertainty in the forward current, which means some strings run singificantly dimmer than others. And 7 is a sucky number to make multiples from.
I suggest strings of 6 LEDs in series with a 0.5W 56-ohm resistor for each string. For 84 LEDs, you need 14 of these series strings, which (connected in parallel) will pull 750mA from a 12V supply. Go buy a 12V, 1A switch-mode regulated supply (wall wart) and power it from that; they're very common.
If you decide you need more brightness and reduce the pitch to under 1", you can just add more strings-of-6, up to 18 strings for every amp of current your wall wart is rated for. You can also increase the brightness by using 43 ohm resistors instead of the 56 ohm but that means you're pulling 70mA per string and can use only 14 strings per 1A supply. Cranking the current per LED up like that won't give you much extra brightness though; they'll be very near max output at 50mA.
Putting all the LEDs in parallel with their own resistor is bad because you waste 87.5% of your power in the resistors so need a much larger supply. Putting them all in series is no good because you need high (and therefore dangerous) voltages to power it... hence the choice of a number of shorter strings that use the available voltage efficiently.
Your parts list: 84 of the LN75X, 14 of 0.5W 56-ohm resistors, 12V 1A plugpack (or old computer power supply), wire and solder.
posted by polyglot at 12:01 AM on March 4, 2007
I suggest strings of 6 LEDs in series with a 0.5W 56-ohm resistor for each string. For 84 LEDs, you need 14 of these series strings, which (connected in parallel) will pull 750mA from a 12V supply. Go buy a 12V, 1A switch-mode regulated supply (wall wart) and power it from that; they're very common.
If you decide you need more brightness and reduce the pitch to under 1", you can just add more strings-of-6, up to 18 strings for every amp of current your wall wart is rated for. You can also increase the brightness by using 43 ohm resistors instead of the 56 ohm but that means you're pulling 70mA per string and can use only 14 strings per 1A supply. Cranking the current per LED up like that won't give you much extra brightness though; they'll be very near max output at 50mA.
Putting all the LEDs in parallel with their own resistor is bad because you waste 87.5% of your power in the resistors so need a much larger supply. Putting them all in series is no good because you need high (and therefore dangerous) voltages to power it... hence the choice of a number of shorter strings that use the available voltage efficiently.
Your parts list: 84 of the LN75X, 14 of 0.5W 56-ohm resistors, 12V 1A plugpack (or old computer power supply), wire and solder.
posted by polyglot at 12:01 AM on March 4, 2007
Oh yeah... having LEDs directly in parallel without their own current-limit resistor is a bad idea. Even connecting a couple of series-strings in parallel is bad; the reason is that each LED has subtly different forward voltages. One string will take the vast majority of the current and the other will get very little, resulting in highly uneven illumination and perhaps destroyed LEDs.
Make sure you have one resistor per series string. It's not like they (resistors) cost more than about 10c each.
If I read Pastabagel correctly, he's assuming connection directly to the mains. DO NOT do this, it's highly dangerous.
posted by polyglot at 12:07 AM on March 4, 2007
Make sure you have one resistor per series string. It's not like they (resistors) cost more than about 10c each.
If I read Pastabagel correctly, he's assuming connection directly to the mains. DO NOT do this, it's highly dangerous.
posted by polyglot at 12:07 AM on March 4, 2007
Rusty mentioned you'd need over 8amps just for the LEDs. Consider that most likely the breaker on the junction box for the wire that powers that entire room is going to be 15 amps. In an ordinary house, each room gets 15 amps. You are using 8 just for LEDs. Throw a computer in the room and then turn on the lights and you just blew the circuit.
That's 8A at ~1.5V or just 12W, plus the draw of the resistors. 12W at the 120v being supplied by your wall outlet is only a 1/10th of an Amp. Even if the resistors are drawing 10X the power of the LEDs, unless you've got a PDP-11 or something you're safe to use your computer and light while the LEDs are on.
posted by Mitheral at 12:31 AM on March 4, 2007
That's 8A at ~1.5V or just 12W, plus the draw of the resistors. 12W at the 120v being supplied by your wall outlet is only a 1/10th of an Amp. Even if the resistors are drawing 10X the power of the LEDs, unless you've got a PDP-11 or something you're safe to use your computer and light while the LEDs are on.
posted by Mitheral at 12:31 AM on March 4, 2007
Wire the LEDs in parallel. That way, when you want to increase the number of them (if you do!), you just have to make sure you have a sufficient supply.
The math will ALWAYS be the same for each LED. (Resistor value = (Vps-Vled)/Iled. Resistor power rating = Iled^2 x Resistor value). Do it right, one time.
Lower voltages on the supply. In practice, for amateurs, low voltage is a good thing. (I try to design for less than 28V in about everything I do.)
Troubleshooting a dead LED is easier. Turn on the table, view it with a camcorder and you can see each LED.
Series units, if one blows, will leave you with a dead string of lights and no way to quickly determine which is at fault.
If you do this on a PCB, (which you should), wiring won't be any more complicated parallel versus series. Board size is practically limited to maybe 10-12" for most el-cheapo board prototypes houses. (Consult Nuts & Volts magazine or similar rags for low cost board houses. )
You can run the lights dimmer or brighter with a variable voltage power supply. A 10 amp supply under 5 V is not THAT expensive, and their are bricks out there that are small at those power levels (< 50 w). if you get to that point, email me and i'll help you find a good>
The slight drawback of having individual resistors is higher parts count, but it's traded off (IMO!) for other benefits.
posted by FauxScot at 5:39 AM on March 4, 2007
The math will ALWAYS be the same for each LED. (Resistor value = (Vps-Vled)/Iled. Resistor power rating = Iled^2 x Resistor value). Do it right, one time.
Lower voltages on the supply. In practice, for amateurs, low voltage is a good thing. (I try to design for less than 28V in about everything I do.)
Troubleshooting a dead LED is easier. Turn on the table, view it with a camcorder and you can see each LED.
Series units, if one blows, will leave you with a dead string of lights and no way to quickly determine which is at fault.
If you do this on a PCB, (which you should), wiring won't be any more complicated parallel versus series. Board size is practically limited to maybe 10-12" for most el-cheapo board prototypes houses. (Consult Nuts & Volts magazine or similar rags for low cost board houses. )
You can run the lights dimmer or brighter with a variable voltage power supply. A 10 amp supply under 5 V is not THAT expensive, and their are bricks out there that are small at those power levels (< 50 w). if you get to that point, email me and i'll help you find a good>
The slight drawback of having individual resistors is higher parts count, but it's traded off (IMO!) for other benefits.
posted by FauxScot at 5:39 AM on March 4, 2007
Best answer: Wiring LEDs in parallel without something to control the current will result in a disaster. If one LED fail, the current it would have drawn will be spread amongst the other LEDs. That will push some other LED closer to failure, when it goes, that current goes to the rest. Then you hit the magic point where you overdrive every remaining LED. Some milliseconds later, they're all dead.
When you do parallel LEDs, you need current control, or you need to seriously derate the LEDs. If you run them at 25mA, not 50mA, then a cascade failure is much less likely.
Given that LEDs aren't going to be identical, one LED is going to draw more current than the rest -- and it'll be the one.
Far better for driving strings and arrays of LEDs is a constant current controller. These put out a given amount of current, no matter the resistance of the downstream string. Here's an example optimized for the .3-3A range for power LEDs, but seven 50mA dies are the equivalent of one 350mA die, power wise.
Derate to 30mA per die, use a constant current controller, and it will last. Running them at the edge as a long parallel string, and it will fail.
Series units, if one blows, will leave you with a dead string of lights and no way to quickly determine which is at fault.
That's easy to test out. You'll know what part of the string is out. You bridge one of the LEDs with a resistor until the other ones come on, if not, move the resistor to the next LED.
posted by eriko at 9:02 AM on March 4, 2007
When you do parallel LEDs, you need current control, or you need to seriously derate the LEDs. If you run them at 25mA, not 50mA, then a cascade failure is much less likely.
Given that LEDs aren't going to be identical, one LED is going to draw more current than the rest -- and it'll be the one.
Far better for driving strings and arrays of LEDs is a constant current controller. These put out a given amount of current, no matter the resistance of the downstream string. Here's an example optimized for the .3-3A range for power LEDs, but seven 50mA dies are the equivalent of one 350mA die, power wise.
Derate to 30mA per die, use a constant current controller, and it will last. Running them at the edge as a long parallel string, and it will fail.
Series units, if one blows, will leave you with a dead string of lights and no way to quickly determine which is at fault.
That's easy to test out. You'll know what part of the string is out. You bridge one of the LEDs with a resistor until the other ones come on, if not, move the resistor to the next LED.
posted by eriko at 9:02 AM on March 4, 2007
Regarding waste wattage due to each LED having it's own resistor: don't make the supply too high, and you'll waste less. The series-parallel idea might work (3 or 4 led's in series, with those clumps wired parallel).
Regarding power consumption: the total consumption is the current times the total power supply voltage, not the voltage dumped across the LEDs. So a 5V supply, with 8 amps, is 40W. There is also considerable wastage at the voltage regulator. But also, just because your device "consumes" 8A does not mean it'll consume 8A AC at the wall... the wall wart contains a step-down transformer that makes the voltage you'll be using smaller and the current larger. The actual current through the wall circuit, for a 5V wall wart, would be 8 * 5 / 120 = 0.3A. This is actually just a back-of-the-envelope calculation, because actually the transformer is probably going to be 120:8 or something, so that it can be rectified/regulated down to 5V, and also, there is an AC to DC conversion in there, the ratio of which depends on the kind of rectification (half wave, full wave, bridged full wave). The actual amperace in AC at the wall might be something like .75A. You're still not blowing any fuses.
However, you'll probably be blowing the transformer in your wall wart, or some of the components in your voltage regulator. By the way, you can make your own voltage regulator. Here's an example of a simple design. Note that this design supplies only 200 mA. In the world of guitar pedals, which is what this is for (it's probably more filtered and cleaner than what you'd need for the LEDs) this is a lot: I can power 6 or 7 pedals off this, easy.
Provided the series-parallel thing works OK, and assuming you want 50mA for each IR led, and using a 5V source, you could put 3 LEDs in series with a resistor. The resistor value you'd want would be .5V/50mA = 10ohms. You'd have 28 of these "clumps" so 1.4A, not too bad. I'd probably suggest a slightly higher supply voltage or less LEDs per clump, I don't think you want to cut it that close (less than a 10% margin).
Using one resistor for a couple of parallel LEDs seems like a bad idea. If the voltage drop for any of the LEDs is higher than the others, then that parallel loop has to drop extra voltage across just the metal leads, which are close enough to 0 ohms that you'll get a lot of extra current in that loop. Which is bad.
posted by RustyBrooks at 9:06 AM on March 4, 2007
Regarding power consumption: the total consumption is the current times the total power supply voltage, not the voltage dumped across the LEDs. So a 5V supply, with 8 amps, is 40W. There is also considerable wastage at the voltage regulator. But also, just because your device "consumes" 8A does not mean it'll consume 8A AC at the wall... the wall wart contains a step-down transformer that makes the voltage you'll be using smaller and the current larger. The actual current through the wall circuit, for a 5V wall wart, would be 8 * 5 / 120 = 0.3A. This is actually just a back-of-the-envelope calculation, because actually the transformer is probably going to be 120:8 or something, so that it can be rectified/regulated down to 5V, and also, there is an AC to DC conversion in there, the ratio of which depends on the kind of rectification (half wave, full wave, bridged full wave). The actual amperace in AC at the wall might be something like .75A. You're still not blowing any fuses.
However, you'll probably be blowing the transformer in your wall wart, or some of the components in your voltage regulator. By the way, you can make your own voltage regulator. Here's an example of a simple design. Note that this design supplies only 200 mA. In the world of guitar pedals, which is what this is for (it's probably more filtered and cleaner than what you'd need for the LEDs) this is a lot: I can power 6 or 7 pedals off this, easy.
Provided the series-parallel thing works OK, and assuming you want 50mA for each IR led, and using a 5V source, you could put 3 LEDs in series with a resistor. The resistor value you'd want would be .5V/50mA = 10ohms. You'd have 28 of these "clumps" so 1.4A, not too bad. I'd probably suggest a slightly higher supply voltage or less LEDs per clump, I don't think you want to cut it that close (less than a 10% margin).
Using one resistor for a couple of parallel LEDs seems like a bad idea. If the voltage drop for any of the LEDs is higher than the others, then that parallel loop has to drop extra voltage across just the metal leads, which are close enough to 0 ohms that you'll get a lot of extra current in that loop. Which is bad.
posted by RustyBrooks at 9:06 AM on March 4, 2007
And eriko is right, a "constant current source" is not a bad idea either. I tend to think of resistors for limiting current because I work with audio equipment and you don't see constant current sources used too often for biasing (partly because the exact bias point is not crucial, but also because you're never talking about biasing dozens or hundreds of devices, usually just one or 2)
The Art of Electronics looks interesting. But depending on your level of sophistication (i.e. if you understand basically what resistors, capacitors, and inductors do) then a decent textbook on semiconductor devices/circuits would probably be helpful. I picked one up at half price books for a dollar or two a few years ago when I wanted to brush up on my amplifier design circuits. The first few chapters in the one I have are about diodes, leds, power supplies, etc, and would probably be helpful.
posted by RustyBrooks at 9:11 AM on March 4, 2007
The Art of Electronics looks interesting. But depending on your level of sophistication (i.e. if you understand basically what resistors, capacitors, and inductors do) then a decent textbook on semiconductor devices/circuits would probably be helpful. I picked one up at half price books for a dollar or two a few years ago when I wanted to brush up on my amplifier design circuits. The first few chapters in the one I have are about diodes, leds, power supplies, etc, and would probably be helpful.
posted by RustyBrooks at 9:11 AM on March 4, 2007
Wow, this thread really illustrates the rule "beware advice on the internets." Listen to polyglot and eriko.
For example, one resistor per LED is not the way LED arrays are normally wired. It is extremely power inefficient. Multiple strings of 3 to 6 LEDs, depending on voltage, is the way to go. While one resistor per LED is feasible, it results in over 8 amps and 40 watts. That requires some hefty wiring. The 6 LED string method results in less than 1.5 amps and 17 watts.
Wiring LEDs in parallel with a single resistor is very bad for a lot of reasons and should not even be considered.
Pulsing LEDs does not generally increase the perceived brightness unless you are severely current limited. The human eye integrates the light so that the brightness is the average power applied to the LED. You get about the same results whether you apply 20 mA continuously or 40 mA with a 50% duty cycle. The general reason for pulsing LEDs is to reduce the amount of circuitry required to individually control single LEDs in large arrays. You don't need to do that for this application.
posted by JackFlash at 12:39 PM on March 4, 2007
For example, one resistor per LED is not the way LED arrays are normally wired. It is extremely power inefficient. Multiple strings of 3 to 6 LEDs, depending on voltage, is the way to go. While one resistor per LED is feasible, it results in over 8 amps and 40 watts. That requires some hefty wiring. The 6 LED string method results in less than 1.5 amps and 17 watts.
Wiring LEDs in parallel with a single resistor is very bad for a lot of reasons and should not even be considered.
Pulsing LEDs does not generally increase the perceived brightness unless you are severely current limited. The human eye integrates the light so that the brightness is the average power applied to the LED. You get about the same results whether you apply 20 mA continuously or 40 mA with a 50% duty cycle. The general reason for pulsing LEDs is to reduce the amount of circuitry required to individually control single LEDs in large arrays. You don't need to do that for this application.
posted by JackFlash at 12:39 PM on March 4, 2007
Best answer: Designing your own PCBs will be nice if you're not on a tight budget. You'd have to do something like come up with a 12" by 1" unit and build strips out of 3 or 4 of those. It will probably cost about $150. Put nice mounting holes in each PCB, and find some connectors to use for the power (soldering wires directly into PCBs isn't horribly wrong, but is more likely to break in a pain-in-the-ass way.) I've used expresspcb.com, which provides free simple board design software that you can learn immediately. It doesn't have a lot of capabilities, but it's fine for a simple undemanding job like this.
posted by TheOnlyCoolTim at 12:55 PM on March 4, 2007
posted by TheOnlyCoolTim at 12:55 PM on March 4, 2007
Jackflash: I just opened up some equipment I had, notably an old big router. It has 3 leds per port, 16 ports (48 leds). Each one has it's own current limiting resister (they're wired in parallel of course).
Also, I checked out the plans for this midi control box. There are 200-ish resistors, each with it's own current limiting resistor.
For huge arrays the rules might be different but certainly I don't think it's terribly bizarre to use one resistor per LED. Also, the energy wastage/large amperage is only a problem if you are using a voltage that is considerably higher than the voltage drop for the LEDs. Certainly, semi-series wiring could be more efficient. But I'm just sayin.
posted by RustyBrooks at 1:49 PM on March 4, 2007
Also, I checked out the plans for this midi control box. There are 200-ish resistors, each with it's own current limiting resistor.
For huge arrays the rules might be different but certainly I don't think it's terribly bizarre to use one resistor per LED. Also, the energy wastage/large amperage is only a problem if you are using a voltage that is considerably higher than the voltage drop for the LEDs. Certainly, semi-series wiring could be more efficient. But I'm just sayin.
posted by RustyBrooks at 1:49 PM on March 4, 2007
Rusty, I don't know about your applications, but if you need to control each light individually, like the midi box, then sometimes individual resistors may be necessary. But if you are just trying to light up everything at once individual resistors are inefficient.
posted by JackFlash at 2:12 PM on March 4, 2007
posted by JackFlash at 2:12 PM on March 4, 2007
Actually that's a point I hadn't considered... in both of those applications you need to be able to turn each LED on and off independently, so they'd have to have individual resistors by necessity.
I still maintain that you could get decent efficiency by using a lower voltage power supply. You'd have a lower part count with the parallel-series arrangement but probably neither one is considerably more complex to wire.
posted by RustyBrooks at 3:15 PM on March 4, 2007
I still maintain that you could get decent efficiency by using a lower voltage power supply. You'd have a lower part count with the parallel-series arrangement but probably neither one is considerably more complex to wire.
posted by RustyBrooks at 3:15 PM on March 4, 2007
Response by poster: There is a ton of *briliant* advice in this thread. Contraption - man, that wiki is perfect with a couple of perfect links. Eriko and Polyglot especially, thanks for sound advice and thinking. TheOnlyCoolTim - a custom PCB, yeah, duh! And I probably wouldn't have thought of it until you mentioned it.
Scalespace and metaculpa, thanks for the leads and offer, but I already have the paper. We were at a point in our development where we're deciding to use tech that we had developed until I was reminded of Jeff Han's work which saves us a bunch of effort and research in materials.
So, for anyone wondering, this will be part of an Impressionism Exhibit that will be at the High Museum in Atlanta, Denver Art Museum, and Seattle Art Museum. (It's being developed by two of us at Denver). If anyone wants to know how things pan out or wants to keep in touch, send me email at bwyman@denverartmuseum.org.
Thanks for all the help so far!
posted by warhol at 6:35 PM on March 4, 2007
Scalespace and metaculpa, thanks for the leads and offer, but I already have the paper. We were at a point in our development where we're deciding to use tech that we had developed until I was reminded of Jeff Han's work which saves us a bunch of effort and research in materials.
So, for anyone wondering, this will be part of an Impressionism Exhibit that will be at the High Museum in Atlanta, Denver Art Museum, and Seattle Art Museum. (It's being developed by two of us at Denver). If anyone wants to know how things pan out or wants to keep in touch, send me email at bwyman@denverartmuseum.org.
Thanks for all the help so far!
posted by warhol at 6:35 PM on March 4, 2007
This thread is closed to new comments.
For power, you could use any sort of low voltage power supply. Not sure how many LEDs you're thinking of having, but a 5V wall wart might do it for just a few, or a 12V computer supply for a few more. LEDs are current devices, so they (most of them -- get and read a datasheet for the part you end up using, obviously, before you hook it up) can work with a fairly wide range of supply voltages, you just need to calculate the accompanying resistor value correctly, so that the proper amount of current is delivered. You might want to read this basic guide.
As for your DigiKey issues, not sure how to help you there. I just went to their site and typed in "880nm LED" into the search box, and exactly one item came up under "Parts, Discrete." It's part number "LN75X-ND" description "IR LED 880NM 25 DEG SIDE VIEW" which is I think what you want. $1.03 each if you buy 10.
The datasheet for that part (a quick Google later) is here (PDF). It has a current consumption of 100mA and a forward voltage drop of 1.5V, if I'm reading it correctly.
posted by Kadin2048 at 8:53 PM on March 3, 2007