Help with voltage too low for LED's

Discussion in 'The Projects Forum' started by N00bish-UltraLowVolts, Dec 18, 2007.

  1. N00bish-UltraLowVolts

    Thread Starter New Member

    Dec 18, 2007

    I have a very basic question. I appreciate any help. I know a *little* about DC circuits, and I have done some research online, and I will admit that I do not completely understand the calculations required.

    I have a device that I am adding several LED lights to. The device runs on 2 AA batteries, but the 3v is stepped down to 2 volts, so effectively the circuit is running at 2v. I am not sure of the current, because my volt meter doesn't gave an ma ratting for it. I assume it is above 200ma, though that seems high to me.

    Now my LED's run between 2.2 and 3.7 volts. Yes, blue is real power hungry. current for the LED's range from 10 to 36ma. Total current ratings for all LED's is roughly 190 ma.

    I tried this...
    Bought a transistor. TIP31 seemed the best fit from Radio Shack's collection (Which is limited).
    I ran a connection straight from the batteries into the collector (3v), put an LED on the emitter, and ran it to the negative lead. I ran a connection from the 2v output into the base, as this 2v is only available when the device is turned on. Seems to me, when the device turns on, the 2v current should close the transistor, and I should get right around 3volts out, but instead I get 1.6 volts out, at right around 200ma. ??????? why?

    So i am looking into an IC chip to step up the voltage. Found a number of diagrams, no clue how to do the math, but I'll work on that later. Right now i jsut want a prototype to prove the theory that what I am doing is possible. I know that using an IC is the proper way to handel this, and to me, will be much more fun to play with.

    I also know that I can use a capacitor and a diode (i think) to store the charge and release the current all at once, cycling very quickly, to power the LED's appropriately. I researched capacitors, farads, and coulombs, and all that good stuff, and I really don't understand how this relates to what current in will result in what current out the other side.

    What capacitor should I use for this?
    Should I use one to put out enough power for the 3.7v LED, and use resistors for the rest?
    would it be better to use a capacitor that is closer to the current rating for each color LED, and run the capacitors in parallel?

    I am running the LED's in parallel.

    Please help. My brain hurts. :cool:
  2. steelsiv

    New Member

    Dec 18, 2007
    If all you want to do is turn on LEDs, no blinking, you do not need a transistor. With only two volts to work with you will never drive a transistor into stauration, completely on. As you found out, this will only decrease your working voltage. You will need to know how many mA is needed for all of your LEDs. Say 100mA. Use Ohms Law to calculate the resistor needed to get 100mA from 2 volts. 2/.1=20 ohms.......E=I*R or R=E/I in your case. Simple enough if that's all you need.

    You can also purchase LEDs with lower power requirements, to decrease the amount of current needed to turn on an LED. It depends on your requirments. Lower current will make the two batteries last longer. A spec sheet for the LEDs will give the current needed.
  3. N00bish-UltraLowVolts

    Thread Starter New Member

    Dec 18, 2007
    Is I only need one 20ohm resistor, and i should run all LED's in parallel off that one resistor? That seems to simple... then again, sometimes.... ok.. most times, the simplest answer is the correct one.
  4. SgtWookie


    Jul 17, 2007
    OK, wait a minute ;)

    LEDs are rated as: current @ voltage
    For example, your typical standard red LED might be rated for 15mA @ 2.4V (or something like that)

    To get the correct current limiting resistor for that single LED:
    1) Subtract the LED's voltage rating from the supply voltage; in this case it's 3V - 2.4V = 0.6V
    2) Use Ohm's Law to calculate the resistance required to limit the current:
    R = E / I (Resistance = Voltage / Current)
    R = 0.6V / 15mA
    R = 0.6V / 0.015A
    R = 40 Ohms
    You can use the next higher standard value available to you; for example, 47 Ohms
    To then determine the new current flow through the resistor, therefore the LED:
    I = E / R
    I = 0.6 / 47
    I = 12.7mA (approximately)

    It is best to not run several LEDs in parallel on one resistor, because if one fails (burns open) the remaining LEDs will rapidly be burned up due to the remaining current flowing through them. You CAN do it, if the LED's have the same voltage @ current rating; I just don't recommend it. Remember, if you run several LED's in parallel from the same resistor, your current through the resistor will increase, and so will the power consumed in it.

    Let's say three 2.4v @ 15mA LEDs in parallel, the voltage drop and supply same from last example - but now our current is additive:
    3 x 15mA = 45mA
    R = 0.6V / 45mA
    R = 13.33... Ohms
    Let's figure the power consumed in the resistor:
    P = E x I
    P = 0.027 Watts
    You could still do it using a single 1/10 Watt resistor of 15 Ohms, but.. as before, I don't recommend it.

    As for the others... there is, or was, a LM3909 IC; designed for things like driving LEDs from a low voltage source. Don't have a datasheet for it; National Semiconductor discontinued it. However, there ARE charge pump devices around that can exchange your batteries' low voltage for higher voltage at lower current; I suggest you invest in a battery charger because you're going to go through a lot of batteries otherwise.

    I don't have any suggestions offhand except to look up boost/buck converters.
  5. hgmjr

    Retired Moderator

    Jan 28, 2005
    If you have already selected an led for your project, perhaps you can provide us with a part number. With a part number we can access a datasheet that will allow us to better advise you on the drive options available to you.

  6. N00bish-UltraLowVolts

    Thread Starter New Member

    Dec 18, 2007
    for SteelSiv, Well.. seeing as the unit is running 2 volts, and the biggest LED requires 3.7 at 20ma, i dont see how a resistor will help. in fact, i did the calculation, and no go with the correct resistor. i tried 1k, i tried 100k, nothing.

    for sgtWookie, I am actually familiar with those calculations, but unfortunately the input current is too low, rather than too high.

    for hgmjr... I don't have the part numbers handy, but they are 5mm LED's from radio shack. Green, Blue, Red and Yellow.

    I'm going to radio shack right now to get an assortment of timers capacitors, breadboard, etc etc. I am going to futz around with it and see if i can make it work. I am a network engineer, so logical testing is something I am good at. I'll try every combo to the point of insanity.

    For the record, if i respond later tonight with gibberish, i have lost my mind.
  7. gootee

    Senior Member

    Apr 24, 2007

    Disregarding, for the moment, that you think the voltage isn't high-enough: SteelSiv suggested 20 Ohms. You said "no go" with 20 Ohms and then you tried 1K Ohms and 100K Ohms and got nothing. 1K Ohms is 50x as much resistance as 20 Ohms, which means that you should expect something like 1/50th as much current when using 1K Ohms as you might get when using 20 Ohms. i.e. You were going in the wrong direction, by trying LARGER resistances. (However, randomly selecting widely-varying component values, when testing real circuits, is often a spectacularly-bad idea. You could very-easily "let the magic smoke out" of some of your components, when using that method. You really should just do the arithmetic, first. Alternatively, download the excellent, free LTspice circuit simulator, from

    Are you going to try to run the LED circuits from the 2V supply, or directly from the batteries' 3V? Do you know the voltage-drop vs current specs for each type of LED you are planning to use? You should probably try to set up some simple battery(+)-resistor-LED-battery(-) test circuits, to measure each type's voltage drops for a few different currents.

    If RS doesn't have datasheets on their website, maybe you could find some markings on them (with a magnifying glass, maybe), and determine who manufactured them, and then find the datasheets at the manufacturers' websites, possibly trying to verify exactly which LEDs you have by a simple measurement or two that you could then compare to possible matches' datasheets. Otherwise, try to find some datasheets for LEDs that look like they're reasonably similar, electrically.

    You should be able to fairly-easily find some relatively-simple charge-pump or boost-mode SMPS circuit that will give you the higher voltage you need, with sufficient current. BUT, your batteries may not last long. Have you considered just adding another battery, perhaps of a different type and/or voltage, instead of [or maybe: along with] more electronics?

    At , their online Webench power supply designer should be able to automatically design a simple boost-mode SMPS DC-to-DC converter, for you. Parts count might only be five to seven components, which are smaller for higher switching frequencies. Similarly, if you download LTspice from and select File--> Switch Selector Guide , it will automatically design a boost-mode SMPS for you.

    Both of those sites ( and, and many others, also have lots of very good Application Notes that you can download, with many schematics in them. And has some PDF files with large collections of circuits (search for "circuit collection", there, IIRC), with one or two specifically devoted to power supplies and DC-DC converters, e.g. AN66. Also, the SMPS ICs' datasheets almost always have example application circuits in them.

    You can also use those websites' parametric-search types of utilities. At , for eample, clicking on "Power Management" and then "Step-Up (Boost) Converters" allows you to enter min and max input voltages and the desired output voltage and max current. I just tried it with 2v min and 3v max input, and 5v at up to 200 mA out, and got a list of about eighteen different LTC chips that can be used to boost anything from 0.5v to 1.8v minimum input to 5v or more, at currents up to more than 1 Amp. Their free LTspice simulator software will do basically the same thing, except that selecting a device will cause it to also design the circuit and run the simulation. (Also: Note that the chips' datasheets usually have internal block diagrams of the chips, which might also give you some ideas about how you could do the same thing with simpler components.)

    Note, too, that the website appears to have a lot of good appnotes about driving LEDs.

    I just did a search at , and found this interesting webpage:

    However, also note that the typical 555 timer ICs, which some of those circuits use, cannot run on less than about 4.5 volts. :-(
    But you might get some ideas about defeating that limitation from AN29 at, page AN29-15, which also has a complete design for a 1.5V to 5V 200 mA converter.

    This design looks exactly applicable, but uses a special charge-pump IC:

    The possibly-applicable circuit in this one uses discrete transistors:

    Or you could build a circuit similar to this tiny SMPS, which requires something like's LT1111 (and four external components), or one of the many others that are similar:

    Or, maybe you could even try to do it "longhand": If you had a small toroidal choke or transformer core, maybe from an old computer power supply or monitor, perhaps you could re-wind it as a 2:1 or 3:1 (or whatever) step-up transformer. Then maybe you could take your 2V or 3V DC and make an oscillator circuit that runs on it, and put the oscillator's (now AC) output into the lower-turns side of the transformer, and take the output from the higher-turns side and rectify and filter it, to give you a higher-voltage DC.

    There are undoubtedly MANY, many more ways to do it, available on line. Google is your friend. You could do searches like: voltage boost OR "charge pump" circuit OR schematic .

    Good luck with that. :)

    - Tom Gootee
  8. Audioguru


    Dec 20, 2007
    Go to a battery manufacturer's website like and look at the datasheet for their AA alkaline cell.
    At 250ma the voltage of two in series is 3V for only a couple of minutes, it drops to 2.6V in a little over half an hour and is 2.2V in 6 hours. You would see red LEDs dim.

    If you use a circuit to stepup the voltage then the current in the battery will be much higher and its life will be shorter.

    This project has too many LEDs or its battery is too small.
  9. hgmjr

    Retired Moderator

    Jan 28, 2005
    Greetings N00bish-UltraLowVolts,

    You mentioned early on that you were dealing with "several" LEDs but I haven't seen any mention of the exact count you have in mind.

  10. iamspook


    Aug 6, 2008
    I've skimmed the thread only so forgive me if I missed whether this point was made correctly.

    LEDs are CURRENT driven devices. What this means is that when they are ON, they drop a certain voltage across them more or less independently of how much current flow through it. That's the spec of the LED and it varies with make and colour. You absolutely need to control the current and the purpose of a resistor is a poor but often adequate way to create a crude current limiter. Once conducting, the LED will happily suck on that current source until it blows up which is why you need to control it. The transfer-characteristics of a transistor show that it too is a current-controlled device not a voltage controlled device. So a properly designed driver for a LED will provide a fixed current through the LED somewhat independent of the supply voltage and loads on the rest of the circuit. In order to make the LED conduct and therefore shine, you need to supply enough potential to only forward-bias the LED, after that, your source needs to be able to supply a minimum current. So if your 2v source is only capable of 20mA, then the LED won't work. You will need to use that source to switch in a higher power source like an external battery.
  11. Tahmid

    Active Member

    Jul 2, 2008
    Since getting 3.7v from 2v is a problem in lighting the leds, you can use lt1073, set output at 12v/5v, and run the leds.

  12. SgtWookie


    Jul 17, 2007
  13. Tahmid

    Active Member

    Jul 2, 2008