(One post up, wow) With no offense to frojoe or Sushi Warrior, a buck, buck-boost or constant current source is not a realistic project for someone who needs help calculating the value of a resistor for a string of LEDs. EDIT: Saw your edit. Again, incase you didn't read the original question (why post?) the design requirement was three strings of 20 LEDs at 20mA each. I don't know where 5 volts will get you, but supposing you rigged up a 3.5v boost circuit with 100% efficiency one AA battery will not supply the 2.8 amps at 1.5 volts required to run such a string.
complicated? i just said maxim makes chips that do it all. all in the datasheet: http://www.maxim-ic.com/quick_view2.cfm/qv_pk/5770 hell, its not even in the datasheet, its on that page. and a constant current source using a lm117 regulator and a current setting resistor is also spelled out in the datasheet of the lm117: http://www.maxim-ic.com/quick_view2.cfm/qv_pk/5770 page 18 so we are talking 2 IC's, an inductor, and 2 set resistors. All of which are spelled out in datasheets. Pretty easy, I think. They aren't interested in the theory, they just want the damn thing to work. edit1: a single AA battery, assuming 1000mA hours, will run 2.8A string for .35 hours. 2 in parallel will run for twice that. was there a time requirement? edit2: forgot to factor in the 97% efficiency of the boost regulator. so take 97% of the .35 hours. or something like that...
I completely agree. The theory was for the benefit of cpemma, and thus off topic. I didn't start it though Since they "Just want it to work" what is "less complicated"? A) * Go to radioshack, spend $2 on diodes/resistors * Solder the + to the - * Enjoy B) * Order a handful of MAX 8815's for a buck each. * Buy a SMD soldering iron, .015 solder, and tweezers because they're only available SOIC * Order the three capacitors, two current sense resistors, and the correct inductor * Wait the three days to get all the parts in * Lay out a PCB of some sort, and either pay $60 to have one cut, get the $30 etching kit, or mess with a Dremel for a few hours getting a board together * Solder everything to the board without overheating the chip, which is why you bought extras. * Spend many hours diagnosing why it is not working without propper tools * Finally get to test it, and enjoy. Again, I'm not sure if you've ever played with constant current sources, but from my experience with the 317 (the actual regulator, 117 is aviation/military grade) sourcing over 500mA requires very high tolerance current sense resistors* which are, again, not available at radioshack. The values are also nowhere close to what the data sheet indicates, so in reality you need to order a handful of values and hope one of them gives the current you're looking for. All of this also assumes that they have a multimeter, which is not a given. EDIT: * .15 ohm 0.01%, etc.
I already have 4.5 v, the whole idea is to get down to 3.5 abouts. If you got to 5 volts, then you would just have to drop the voltage again, adding more components and adding inefficiency. I want to use multiple D batteries, so that there is an acceptable run time, not one AA I have to change every 20 minutes. This is actually not an issue as the battery's and whatever power regulation I end up using will be in a separate, much less confined space. This is the whole reason I wanted to use one resistor(or combination of a few), as opposed to having a resistor on each led. The source will have a wire running from it to the leds, in a confined space, and it is in that area I do not want extra components or heat. I could put all the resistors near the source, but 60+ and 60- wires running in parallel isn't my idea of a good time. Didn't think my little question would start all this, but it has been interesting. Continue... -Joe-
you can solder flying leads to an soic. plus, when you go to order the current set resistor from digikey, you can can order the rest of your parts, and it will still be cheaper than all your parts bought at radioshack still don't think i'd call mine "complicated" though. but i'm stubborn. edit: you WANT to use D batteries? how do you have 4.5? 3x 1.5? you could that boost regulator with a single D and get about 7 hours out of it. whats the time requirement? we can make it fit, i promise! lol.
So far I spent a grand total of $1.04 on a two pack of diodes...I guess I need two for each array, so $3.12, is your solution cheaper than that? Yes, 3 D's in series for 4.5 volts. Is that 7 hours for one array or all 3? I want about 8 hours out of this, with more than one string running at once. It won't be running 8 hours in one shot. But part of the day over the 3 days I need it. Why not use 3? With 3 I am closer to my goal voltage, don't need to step up the voltage, and have a much longer run time.
Soldering fly leads to an SMD still requires a small tip temp-reg'ed iron and a lot of experience. If it is an integrated MOSFET (like the one you linked) fly leads will provide insufficient heat-sinking. I've been down this road before a few years back... To drive 20 white Luxeon III's at 2 amp 37.5v, using the supertex HV9930 in a Cuk configuration. EDIT: Incase someone noticed, these are not the same boards. The top one is actually for an 18s 1p setup for a 3rd brake light. The bottom one is the Cuk for the array sitting next to it. Doing this with a resistor would have been bad. However, on some of my other projects where having 2300 lumens isn't the goal, like this 240 LED 12v work light: They are strings of 3 LEDs all connected in parallel, with a single, 5W 1.3 ohm ceramic power resistor. It needed significantly less design time (If anyone is interested in the XLS with the parameters of the supertex chip let me know), and required no calibration or testing. It works with everything from 11v - 15v without issues, which lets me run it off cars batteries and just about any 12v switcher. If you just want it to work, KISS!
Using diodes to drop some volts from the battery - works fine until the battery starts to run down. As soon as the battery voltage dips below the total forward voltages of [LED + 2 diodes], no light. A resistor gives a buffer, the LEDs may dim but they'll stay lit until the battery has flattened to below the LED forward voltage. So you waste considerably more battery life with the diodes. You can run any number of parallel arrays off one source providing the source is capable of supplying the required current. Perhaps TheShadow27 can tell us what the current will be in his post#4 circuit?
Correct. It also provides an opportunity to have a 3 level selectable brightness setting. Do you have reason to believe it would be anything other than 400mA per string for a total of 1.2 A at full crank? I'm open to other ideas. . . EDIT: Just mocked up frojoe's configuration, 4.5v supply -> 2 1n4001 diodes -> 3.5 v white LED. The LED was illuminated down to 3.4v. Using one 1n4001 diode, the LED was pulling 18.7mA @ 4.25v in, so using both may be unnecessary; it depends on the LEDs you end up getting. Don't believe specs!
I don't believe your calculations are valid. A single D battery is about 12,000mAh or 12 amp hours at 1.5 volts. An LED runs at 3.5v, 20mAh; 60 LEDs are 1.2A at 3.5V. Conservation of energy, nothing's free, so 1.2A at 3.5V = 2.8A at 1.5v, assuming 100% efficiency, which is 4.2 hours on a D battery. In reality, a 1.5V D battery sourcing 2.8A will not stay at 1.5V very long, so the draw increases as the voltage decreases. I doubt it'd last 3 hours, but you're welcome to build it and prove me wrong. Take pictures Exactly. That's less then shipping for any of the above mentioned parts.
wikipedia says 20.5 amp hours. i was using that number. edit: interesting, one page says 20.5, another says 12.0 i'm not trying to say the way your doing it is wrong. its probably the simplest way to go. i'm just throwing up a different way to do it. i just wouldn't call my solution complicated.
Using only one won't provide the leds with too high a voltage? I thought that earlier you said they would each drop the voltage by .5. I know that 4v would be too much.
I found a sort of godsend - a 3V wallbug. I know it isn't very uncommon but now I don't have top buy squat. The only possible problem is that it only outputs 120mA. Does this mean I'll only be able to power 6 20mA LEDs? I assume so but I would be happy to be proven wrong.
Method 1: Boost switching supply. BOM: 28 components and 1 PCB, $34.72 with S&H Time: 7 hours calculations, parts selection, PCB layout, SPICE simulation. 3 Hours assembly (with PCB etch) and testing. Total = 10 Hours. Method 2: Resistor BOM: 1 component, 1.3 ohm power resistor, $0.98, $4.32 with S&H Time: 0.15 hours calculations, parts selection, PCB layout, simulations. 0.5 hours assembly time and testing. Total = 0.2 hours (12 minutes) You can call it whatever you want, but in engineering terms, in the best case method 1 is 40,000% more complex than method two. Yes but three alkaline batteries do not source 4.5 volts for very long. If you have a multimeter you can test it yourself. You might be able to get more since the LEDs won't pull 20mA at 3V. Is it a switching adapter or a transformer (Lightweight, 120/240v VS heavy, one voltage only)? The transformer may run well past it's rating, the switching adapter will not.
Case in point. This work light I built before I started messing with switching supplies runs off of a 32 volt PSU from a printer. It has strings of 9s x 12p (108 total) LEDs, with a total FV of 31.5v and a 2 ohm resistor to drop the half volt. For whatever reason, one burned out, taking out the whole series string. You can see the bright/discolored LEDs in random places elsewhere that the increase in current will burn out another row, and the whole array will go soon after.
