Measuring LED forward voltage

Discussion in 'General Electronics Chat' started by zazas321, Dec 9, 2015.

  1. zazas321

    Thread Starter Member

    Nov 29, 2015
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    I need to find the right voltage drop of the LED, but came up with problem.. Can i measure the voltage drop if i connect 4,2v batteries straight to the LED and measure the voltage across it. For some reason for the multimeter showed the same voltage across both, red and green LEDs i measured. it showed 2,1 voltage. even thought on the package label, where i bought the led is written, that red led has 2,5V drop and the green- 2.2V. Is there any easy way to find out the actual voltage drop?
     
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  2. nigelwright7557

    Senior Member

    May 10, 2008
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    You need a resistor in series with the LED to measure its forward voltage drop.
    If you don't the LED will blow.
     
  3. #12

    Expert

    Nov 30, 2010
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    A 9V battery with something between 470 ohms and 1 k will get you in the ball park.
     
  4. zazas321

    Thread Starter Member

    Nov 29, 2015
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    Yeah. thanks
     
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  5. dl324

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    As I mentioned in your other thread. If you're going to measure the voltage on a lot of LEDs, it will be more convenient if you build a 20mA current source.
     
  6. #12

    Expert

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    One way that is fairly close might be to use a 9 V battery with a 330 ohm resistor in series with the LED. There are several ways to do this, depending on how much accuracy is required. For instance, this project I did with tracecom could be adjusted to 0.020 amps.

    http://forum.allaboutcircuits.com/threads/10-ma-to-01-ohm-resolution-dvm-adapter.89321/
     
  7. dl324

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    OP said he was measuring several different colors of LEDs and wasn't getting the voltage given on the bags. Setting up a 20mA current source would let him measure LED forward voltage at the current Vf is typically specified at without needing to worry about current variations caused by forward voltage or resistor.

    Could use a fixed series resistor, but I don't think that was what he was asking for. He wanted to know why his measured voltages were't matching the label on the bag.
     
  8. #12

    Expert

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    I am not aware of the, "other Thread" so this clears up a lot for me.

    There are several answers as to why the label wouldn't match the device, especially including the idea that, without a resistor, he was mostly measuring the battery voltage.:D Then there are batch variations, temperature effects, whether he bought Chinese discards, and how bad the translation from one language to another. The finished project which I mentioned in post #6 could clear up any lingering doubts, or one could fake it pretty closely (within 1%) with an LM317 and a trim pot.

    If you need a schematic, I can do that.;)
     
  9. #12

    Expert

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    This circuit is expecting to use a 9V battery.
    Finding a 10 ohm pot could be a problem, but there are lots of ways to use a larger pot and put a resistor in parallel with that. I designed this for deadly accuracy, but a low ohm, 10 turn or 20 turn trim pot could make it a lot easier to get the necessary parts and even leave out the 56 ohm resistor.
     
  10. dl324

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    Here's my go to current source (sink). It will work with 4.2V too, but a 4.5V adapter was being used in the other thread:
    upload_2015-12-9_17-18-33.png
    To measure voltage drop, voltmeter leads can be placed on supply and collector of Q2. LED current will be 18-22mA and can be trimmed by replacing R2 with a pot.
     
  11. lukasxx1

    New Member

    Jan 9, 2015
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    Okay i will try some of theese circuits. But what is not clear for me, if i connect the LED straight to the 4.2Volt battery and i am measurin voltage across both LED legs, shouldnt the multimeter show those 4.2Volts? If it shows 2,5V so where 1.7Volt goes?It transfers to the heat or what?
     
  12. #12

    Expert

    Nov 30, 2010
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    Either the LED blows up or your battery has so much internal resistance that it doesn't have enough power to destroy the LED. Some button cells are that weak, or you might just have some mostly discharged batteries, or they might be in pretty bad condition due to age.
     
  13. dl324

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    This is the IV curve for a high efficiency red LED I posted in your LDR thread
    [​IMG]
    Fortunately for you, you were using a weak battery and not your 4.5V 1A adapter.
     
  14. lukasxx1

    New Member

    Jan 9, 2015
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    I dont get what this chart has to with my question. I mean if i measure the voltage across the led its showing 2.5V, and the my power source is measured 4.2V.

    And about the 4.5V 1A adaptor. I mean if the LED requires 2.5V and 20mA, the power source will give the required ammount of current ? if i connect my 4.5V and 1A adaptor without a resistor to a LED, what current will i get?
     
  15. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    No, that is the wrong way to measure Vf (forward Voltage).

    First, you have to ask at what current should you be measuring the Vf? If the LEDs are rated for say, 20mA, then measure the Vf when 20mA is flowing in the LEDs.

    Second, how do you get 20mA flowing in the LED? I would connect a resistor in-series with the LED, and then use a variable bench power supply to adjust the applied voltage until the current is 20mA. Then use your mulimeter to measure the voltage across just the LED.


    Third, what resistor? My bench supply is adjustable over the range 0V to 30V. Suppose as a first cut, I would set the supply to ~10V. If the expected Vf is about 2.5V, then we know that if 10V is applied to series connected resistor and LED, then about 7.5V will appear across the resistor, and the remaining 2.5V will appear across the LED.

    Fourth, how to calculate a resistor value? By Ohm's Law, we want the resistor to drop 7.5V with 20mA flowing in it. R =E/I = 7.5/0.02 = 375Ω. That is not a standard value, but 330Ω or 390Ω is. So I would go to my parts box, and grab a 330Ω resistor.

    Fifth, connect it together, and twiddle the power supply voltage to get a current of 20mA (o.020A) on the supply's current meter...

    Sixth, now connect the multimeter across JUst the LED, and read Vf (at 20mA) !

    Seventh, substitute each different type of LED, but before reading Vf, tweak the power supply voltage to get 20mA before reading Vf.
     
    Last edited: Dec 10, 2015
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  16. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    Enough to destroy the LED and/or the adaptor (if it doesn't have inbuilt current-limiting). 20mA is the recommended current; not the maximum uncontrolled current which will result.
     
  17. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    But why you want to know the exact value of a diode forward voltage?
     
  18. zazas321

    Thread Starter Member

    Nov 29, 2015
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    Okay thankss alot! As i understand good, the voltage forward and the current of the LED for example 2.5V and 20mA just shows, how much voltage will approximately LED uses when there are 20mA flowing through it.. And if wire the circuit so the Led will get 30mA for example, the voltage droup should be higher? do i get it right?
     
  19. dl324

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    It illustrates how lucky you were that your LED wasn't destroyed. LEDs are current controlled devices and require something to limit current. In simple circuits, that's a resistor. In more exacting applications, it's a current source.

    You used a voltage source with no current limit other than it's series resistance. Fortunately for you, the series resistance of your battery was sufficiently high to limit the current to a value low enough to protect the LED.
    As several have already mentioned, if you had used the adapter without a current limiting, the LED would have been destroyed. If the supply did it's job, you would have had 1A in the LED; or something approaching that value, very briefly.

    As I suggested in your LDR thread; you need to take a break from building and do some studying so you can understand what you're doing.
     
    Last edited: Dec 10, 2015
  20. dl324

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    Mar 30, 2015
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    That's it. LEDs of different colors will have different IV curves. Those curves dictate the current required to get a specific forward voltage. Forward voltage for modern LEDs are typically specified at 20mA; hence my suggestion to build a 20mA current source. I posted a current sink because I knew you already had BC547 transistors.

    LEDs from the same "batch" are typically binned (sorted) by brightness. So operating LEDs from the same bin at the same current should have similar brightness.

    If you use 5% resistors in your 50 LED circuit, the highest current variation would be 10% and that would cause a brightness variation that would be too small to be noticeable.
     
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