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Circuit Problem

  1. Dec 3, 2003 #1
    Well, i finished my studies but have continued along with my love of electronics.

    But, i have come across a problem.
    I have four 1.5V batteries in series, thus producing 6 volts.

    this is to power 4 blue LED's in paralell. The problem being the max volatage of the LED's is 4 volts (3.5 is best).

    So, first i need to drop the voltage given to the LED's from 6v to 3.5v. A resiter in series would do this right?

    each led is 130mW at 4 volts, so thats 30mA

    So my question is how do i supply the 4 leds in paralell with 3.5?
    please show your working so i can learn how to do it.
     
  2. jcsd
  3. Dec 3, 2003 #2
    yes, theoretically. Do you want to physically do the circuit? If so you have to check also if the battery supports the current for the 4 leds (that is 4*30 = 120 mA) and if the resistor can disipate te power of 120mA*2.5V = 300mW
     
  4. Dec 3, 2003 #3

    ShawnD

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    Science Advisor

    First you have to know the resistance in each of the LEDs. From there, you find the current that you want through each of the LEDs (based on their resistance and the voltage you want). Then you calculate the resistance you need to have to get that current for the voltage you have.

    Lets say you have 10V supply, you wanted 6V and the load was 3ohms. If I want 6V on the 3ohm load, the current would be like this:
    I = v/r
    I = 6/3
    I = 2A

    Since I want 2A for a 10V suply, we find the resistance:
    R = V/I
    R = 10/2
    R = 5ohm

    OK so now I know that I want 5 ohms but I only have 3. If I add a 2ohm resistor, I will get the voltage I want.
     
  5. Dec 3, 2003 #4
    When your dealing with LED's you don't need to know the resistance of the LED. This is because LED's are factory rated with a current/voltage spec for optimal brightness and longevity. If you exceed the factor values of an LED you shorten its life while only getting a minimul increase in lux. There are two routes you can take to do this: one assume the battery has no internal resistance. I did it this was and this is what I came up with:
    [​IMG]

    The other method is to do this using a single resistor, or group of resistors, in series with the battery as Guybrush Threepwood mentioned.


    Also, you can assume the battery has some nominal resistance (say .1 ohms) and then find the voltage drop associated with that internal resistance using the expected load. The voltage availabe to each leg of the circuit would then be reduced by the voltage dropped by the internal resistance.

    Or as I did just assume the battery has no internal resistance.

    In both cases, the LED legs are all in parallel with the battery thus the voltage drop of each leg will be equal (the actual voltage depends how you decide to analyze the battery resistance).

    What you should do is set up a series of voltage/current loops to find the current/voltage through each leg. But since LEDs are factory rated, you can skip that step and go right to meat and potatoes by saying the resistor has to drop the voltage that the LED is unable to drop. Use the LED's rated current and viola, you have a resistance.


    Good luck in your endevors.
     
    Last edited: Dec 3, 2003
  6. Dec 3, 2003 #5
    LED’s have a fixed voltage drop across their terminals as do silicone or germanium diodes, except the drop across an LED will be higher. Resistance through the LED is negligible (it will burn itself out if you don't take precautions), and the remainder of the supply voltage will be dropped across the current limiting resistor. Red LED’s typically drop about 1.6 volts, and they will do this whether you are using a 6-volt or a 60-volt supply. The difference will, of course, be dropped across the current limiting resistor, which as the name implies, limits the current passing through the LED to an acceptable level (or at least it should, hehe). The method I would use involves knowing what the forward voltage drop of a particular LED is, as well as knowing what the desired current level passing through it should be (all of which can be found in the manufacturer’s literature). Then;

    R (current limiting resistor) = (supply Voltage – forward voltage drop of the LED) / Amperes

    For the figures I’m taking from your post I’m assuming 3.5V is the nominal forward voltage drop across a blue LED (typical) and 30 milliamps is the desired current to have passing through the device. If this is true, then;

    R = (6V-3.5V)/0.03A
    R = (2.5V)/0.03A
    R = 83.3 Ohms

    Personally, never having experimented with blue LED’s, I think 30ma might make them burn too bright for my taste. I like LED’s to be bright enough to see that they’re on, and not much more.

    What you will likely find too, if you deal with these devices a lot, is that some LED’s will be naturally brighter than others. Therefore, when building a circuit where multiple LED’s are used it is sometimes of benefit to string a number of them in series (so that the current passed by one is passed by all the others) then pick out the odd-ball LED’s that do not match in brightness with the majority.
     
    Last edited by a moderator: Dec 3, 2003
  7. Feb 3, 2004 #6

    mac

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