Voltage step down - Page 1
Forum section: electronics / beginners

author: kirill578
author_rank: Newbie
timestamp: March 30, 2012, 01:07:21 pm
content: Hey, I'm working on lighting system, which is charged via a solar cell during the day, and works during the evening.I have a 6V@4Ah battery and 30 leds that draw about 750mA. there is a linear voltage regulator which takes the voltage down to 3.3V (LM1117T). the main problem is that the linear regulator dissipate huge amount of power, 2W when the battery is fully charged , I thought using a switching regulator. but the input voltage is not constant.I thought may be you can suggest something.

author: mariush
author_rank: Super Contributor
timestamp: March 30, 2012, 01:31:50 pm
content: Digikey's website is a bit screwed right now, it shows wrong pictures and datasheets so I can't give you straight links.Check these out...MIC4680-3.3YM -http://www.micrel.com/_PDF/mic4680.pdfworks with 4v-30v input , 3.3v @ 1A (up to 1.3A) :LM2576S-3.3 -http://www.ti.com/lit/ds/symlink/lm2576.pdf(about 1v voltage drop at 1A as far as I can see in datasheet)TL2575-33IKTTR -http://www.ti.com/lit/ds/slvs638b/slvs638b.pdf(this one needs at least 4.75v but it's cheaper than the others esp in volume)They're fairly easy to use, needing only a few extra components like an inductor, a diode and a capacitor.

author: IanB
author_rank: Super Contributor
timestamp: March 30, 2012, 05:29:09 pm
content: You have a few challenges with your system here in matching the power source with the load.Firstly LEDs need a current source, not a voltage source. So primarily you need a current regulator rather than a voltage regulator.Secondly, you have 30 LEDs which I presume you are arranging in parallel, each requiring about 25 mA? The trouble here is that if you arrange a 750 mA current source, there is no assurance the current will be evenly divided 30 ways between each of the LEDs. Some may get much more than 25 mA and some much less. This will lead to failures and an unreliable system.A better way is to put many LEDs in series and drive them with a higher supply voltage. This will ensure each LED gets exactly the right current.For example you could put 10 LEDs in series and feed them with a 25 mA constant current boost converter. Three of these converters would drive all of your LEDs. However, this boost converter would also have a certain efficiency (guess ~80%).There is going to be no economical and practical way to get close to 100% efficiency of power conversion with almost any arrangement.

author: T4P
author_rank: Super Contributor
timestamp: March 30, 2012, 07:23:45 pm
content: It would be easier to use a LED boost converter and use higher power LED's instead of 30 LED's or use the same 10 LED's in series in 3 pairs having it's own bosot converter .There's a reason for having SO many LED Boost converters out there .

author: Jeff1946
author_rank: Regular Contributor
timestamp: March 30, 2012, 07:33:23 pm
content: Seems to me the best choice might be to wire the LEDs in 5 strings of 6 LEDs with each string having its own series resistor. Something like a 120 ohms would work for 20 volts to give about 25 ma per string. Use a boost converter to convert battery voltage to the 20 volts. Go to the web sitehttp://dics.voicecontrol.ro/tutorials/mc34063/to get the right value components. The MC34063 is cheap and available in a DIP package. To max efficiency you need to use an inductor with low resistance and a Schottky diode.

author: bruce273
author_rank: Contributor
timestamp: March 30, 2012, 07:42:46 pm
content: How are the LEDs wired or have you not decided yet?

author: Psi
author_rank: Super Contributor
timestamp: March 30, 2012, 10:35:33 pm
content: Yeah, you look to have the leds in parallel if you're running them all from 3.3VThis power supply module seems to do what you want. It's a stepdown DC/DC with 4.5-36V input and a 700mA current limited output.4.5V is enough to discharge your 6V SLA to well below min volts.http://search.digikey.com/scripts/DkSearch/dksus.dll?vendor=0&keywords=945-1123-ND&cur=USDBut you will find it much more efficient if you put your leds in series (or multiple series runs in parallel) then power this module from a higher voltage. (It can only step down)

author: kirill578
author_rank: Newbie
timestamp: March 31, 2012, 10:11:22 am
content: Well, There is no electronics store in my place, and basically my only way of getting parts is ebay and old electronics. (digikey takes 50$ for the shipping).I've managed to find in my jankbox LM2576 which is a pretty straight forward solution, but as you suggested not that efficient.Unfortunately, I have found just one MC34063A. I'm using 2.2V leds, should I boost it up to 66V to connect it in a long series, or boost it up to 22V (with one chip) and have 3 parallel series?lets say I boost it up to 66V do I need a resistor, Or I should boost it up to 67V and add a 40ohm?

author: IanB
author_rank: Super Contributor
timestamp: March 31, 2012, 04:27:38 pm
content: Don't boost the voltage to more than 20-30 V. 60 V or so is too high.Remember that you must have current limiting in your circuit. Your LEDs are not 2.2 V LEDs, they are 20 mA LEDs.Three (or more) parallel series would probably be most appropriate to avoid the need to boost to higher voltages.

author: TerminalJack505
author_rank: Super Contributor
timestamp: March 31, 2012, 06:08:38 pm
content: Since you seem to be trying to do your project with what you have on-hand you might consider using a 555 timer as a boost converter.Either that or look into the so-called 'Joule thief' circuit. It is a minimal boost converter circuit. This circuit would require you to make your own transformer, however, so it would be a little trickier.I'd go with Ian's advice and keep the voltages low so a couple of these boosting up to 35V should work.

author: Zero999
author_rank: Super Contributor
timestamp: March 31, 2012, 06:27:31 pm
content: It would be best to boost the voltage to something like 24V and connect the LEDs as three strings of ten, each with a series resistor.

author: bruce273
author_rank: Contributor
timestamp: April 02, 2012, 03:57:18 pm
content: Putting them in parallel would work well if you want to lower the voltage of the power supply. However like Ian B said they will require a certain current to operate not voltage. This will meen that you will need a high current power supply as each parallel branch will require the 750mA or whatever the current rating of your LEDs are. So you will have to multiply the LED current by the number of parallel branches. As far as power supplies are concerened I have used a linear technology lt1073 chip which is designed to boost or reduce voltage from small batteries to a constant voltage. so the best thing is to go onto linear technologies website after figuring out what voltage and current you need to drive your circuit and use their online search tool to find the best one. You can also buy off the shelf DC-DC converters which will also work however they are more expensive

author: TerminalJack505
author_rank: Super Contributor
timestamp: April 02, 2012, 06:16:16 pm
content: If you plan on boosting with the Joule thief circuit or a 555 timer like I suggested then running the LEDs in parallel will be a little tricky. You will need to be sure to match the strings of LEDs so that the total voltage drop of each parallel branch is equal.This is due to the way the Joule thief circuit and most 555 boost circuits work. The output voltage isn't really regulated. This happens to work great for illuminating LEDs, though. During flyback when the inductor's voltage spikes, once the voltage reaches the point that the LEDs will forward bias, the voltage will peak and the power will be transferred to the LEDs. If one of the parallel branches has a higher total Vfthen it may get less current than the other branches. If you think about it, the LEDs are doing exactly the same thing that a flyback diode does when you put one across an inductive load.If you're not too concerned with higher voltages then you can run all the LEDs in series. This is really how the Joule thief circuit is designed to work. There are a ton of videos on You Tube of people using the circuit to run a string of 30 to 50 LEDs from just one or two AA batteries. Be careful with some of those--the Joule thief circuit has a bit of a following with the so-called 'free energy' folks. So you will see videos where A) the person that posted it make the circuit out to be more than it really is and/or B) the person that posted it doesn't truly understand how the circuit works.