LED Circuit Help - Page 1
Forum section: electronics / beginners

author: SRoh
author_rank: Newbie
timestamp: September 18, 2014, 02:14:40 pm
content: Hello Everyone,I need to make a circuit with 4 high intensity LEDs. I have following questions (circuit attached):1. Where can I buy the high intensity LEDs?2. In the circuit do I need to throw in resistor if I am using 3 R44 (button cells) to power up the LED?3. Is the set of 4 , 3 R44 cells (12 in total) going to be enough for running the LED?Thank you!

author: katzohki
author_rank: Frequent Contributor
timestamp: September 18, 2014, 04:20:23 pm
content: You can get high intensity LEDs at Digikey and Mouser.Yes, you should have a current limiting resistor with LEDs.12 x LR44 batteries will be enough voltage to run your LEDs, but it will be a matter of time as to how long they will last. That depends on what LED you choose.Your circuit arrangement is a little unusual. I would say that you could use 3 of these batteries alone and wire the LEDs in parallel...

author: mariush
author_rank: Super Contributor
timestamp: September 18, 2014, 04:20:23 pm
content: 1. You can buy high intensity leds from lots of places, it depends on where you live.Besides local stores, there's digikey.com in US and Canada and possibly other countries, there's Newark.com / Farnell.com / Element14 (same company, different sites for different regions of the world), Mouser.com , tme.eu .. all these are electronic part distributors selling genuine parts.Then, there's ebay, lots of leds of various quality levels and various prices.2. Resistors are needed to limit the current going into the LEDs, so that they won't burn up. LEDs aren't like incandescent lightbulbs, if you give them "juice", they'll consume as much as they can and get damaged.Now, you have to imagine any power source or in your case the battery as having a resistor connected to it ... any battery has a particular amount of resistance, and with some batteries due to high internal resistance the current can not come out big enough quantity to damage the LEDs.LR44 batteries have some resistance, but they're not really designed to power leds, they're designed for low current, for calculators, for watches, stuff that uses a very small amount of energy.For example, if you check a datasheet (here's a random one from Google:http://www.gpbatteries.it/allegati/datasheet/specialistiche/alcaline/102002_GP76A.pdf) you can see that they recommend loads of 3 kohm or 6 kohm, which amounts to something like 0.25 - 0.5 mA. Even with high intensity leds the leds would probably need about 3-5mA, maybe even more.So the internal resistance of a LR44 battery would probably be big enough that you won't need resistors, but at the same time the batteries will discharge very fast.I would recommend replacing the batteries with something that can provide more current.For example, as standard CR2032 battery (like the ones on motherboards for BIOS) can do about 30mA for a few seconds and 5-10mA for a long time, enough for one high intensity led. They're also 3v from the start, so if you buy a led with a forward voltage smaller than 3v, then you can run a led straight from a single battery, with an optional resistor to limit current if you wish.Now... question 3... it really depends on how you wire up the leds and what batteries you decide to use.You can wire the leds in series or in parallel. If you connect the LEDs in series, then the leds act as a single LED with a higher voltage required to light up. You may need to add one resistor to limit the current to the maximum you wish (or the maximum current the led supports).If you connect the leds in parallel, then each led would need its own resistor to limit the current but the voltage remains small, which may be good if you have small voltage batteries.Let's do a practical example... say you buy 4 leds like these:http://www.digikey.com/product-detail/en/HLMP-EH08-Y2000/516-2429-ND/1232866Color RedWavelength - Dominant 615nmWavelength - Peak 621nmMillicandela Rating 18150mcdVoltage - Forward (Vf) (Typ) 1.94VCurrent - Test 20mASo 1.94v and 20mA maximum current. Let's assume that some leds may be better than others, so just round up to 2v. At 20mA, the leds may be too bright, so let's say you want to limit the current to 10mA.Now you have to think about the batteries you're going to use.I'll go with a CR2032, which means the voltage of each battery will go down from about 3v to about 2.4v when they're fully discharged. Also, if you have a look on page 6 of this white paper, you'll see that CR2032 have an internal resistance of about 5 ohm when they're fresh and it slowly goes up to about 25-30 ohm when they're almost fully discharged. This will be important later.If we connect the four LEDs in series, then the leds will behave like a single LED that needs a voltage of 4x2v = 8v to work, and the current remains the one we chose, 15mA.We can stack 3 of those CR2032 batteries to obtain a battery that at best will act like 3x 3v = 9v.Now we need to limit the current going into the leds to that 10mA (or 0.01A)... we have a simple formulaVoltage power supply - Voltage led = current x Resistance... so we have 9v - 8v = 0.01A x R therefore R = 1v / 0.01A = 100 ohm.The internal resistance of the battery stack will be 3 x 5-10 ohm (when really new) or about 30 ohm, which is lower than this 100 ohm value we calculated. So it's obvious we'll need a resistor to limit the current going to the LEDs to 10mA, but instead of using a 100 ohm resistor, it would make sense to keep in mind the internal resistance and use a lower value.But when the batteries discharge, not only the internal resistance increases but the voltage also goes down. So let's do the math for example with 2.8v, in which case reusing the formula:3 x 2.8v - 8v = 0.01 x R therefore R = 0.4/0.01 = 40 ohmAt this voltage, the internal resistance of the battery probably increased to about 15 ohm, and since we have a stack of 3 x 3v batteries, the resistance will be about 3x 15ohm = 45 ohm so we actually don't even need a resistor to limit the current at 10mA, the resistance of the battery already increased enough to make the resistor optional... an additional resistor will only decrease the current going to the leds.So with this in mind, i would probably go with a 20-30 ohm resistor just to make sure the leds don't get damaged when the batteries are very new.

author: SRoh
author_rank: Newbie
timestamp: October 06, 2014, 11:03:52 pm
content: Quote from: mariush on September 18, 2014, 04:20:23 pmHello Maurish, Thank you for detailed answer. I can use the LR44 batteries only due to size limitations. Moreover, I am planning to have the LEDs connected in Parallel which is different from the illustration I attached. Does that change the calculations a lot? I have mechanical background, so I am sorry if I am asking very simple stuff here.