I'm confused with voltage/current in my first LED array

Discussion in 'The Projects Forum' started by Fixxxer, Feb 15, 2007.

  1. Fixxxer

    Thread Starter Member

    Feb 15, 2007
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    Hello, all!

    I've just started teaching myself about electronics (I'm currently in Volume 1, Chapter 12 on this site). As a project, I want to make an LED "sign" to put in my game room. The sign will consist of 185 10mm LEDs (105 green, 80 blue). The specs for the LEDs that I'm considering purchasing can be found here (green) and here (blue).

    Both colors have a MIN Vf of 3.00 V, a TYP Vf of 3.15 V and a MAX Vf of 3.30 V.

    Here's the If vs. Vf graph for both colors:
    [​IMG]

    Like I said, I'm a beginner, so please correct me if I'm wrong here...
    According to that graph, if I supply approximately 5 mA of current to each LED, the voltage dropped across each one will be the "typical" 3.15 V. Or, if I supply 20 mA of current to each LED, the voltage dropped across each one will be the "maximum" 3.30 V. Is this correct?

    To begin planning my project, I wanted to find a suitable DC power supply. I've got a spare DC "wall wart" that outputs a maximum of 300 mA at a range of different voltages (it's adjustable from 1.5 V to 12 V). When testing it with my voltmeter last night, I found that it's actually supplying approximately 3 V higher than the number shown on the adjustment. For example, when I set it to the 7.5 V setting, my voltmeter read 10.23 V directly across the leads. That's no big deal though...

    As for the layout of my circuit (basically just an LED array), I plan on connecting several LEDs in series with a resistor, then connecting these groups in parallel with each other.

    First off, I want the LEDs to be as bright as possible without overdriving them and shortening their lifespan. So, would it be okay for me to supply them with 20 mA, resulting in each one dropping 3.30 V (the "maximum")? Or should I go with the "typical" voltage drop of 3.15 V and only supply them with 5 mA? If I do that, will the LEDs be significantly dimmer than they would be at 20 mA and 3.30 V?

    Assuming I choose to use the DC "wall wart" at 10.23 V, and I supply each LED with 5 mA (3.15 V across each one), then I could construct my array from groups of three LEDs in series with a resistor, all connected in parallel with each other (I'd have only two LEDs in one of the series groups, since I'm using a total of 185 LEDs). Now I need to choose my resistor. This is where I get confused.

    Here's my math (using a 10.23 V source, 5 mA current and 3.15 V drop for each LED):

    [Total voltage for all LEDs in a branch]
    3.15 * 3 = 9.45 V

    [Desired voltage drop for resistor]
    10.23 - 9.45 = 0.78 V

    [Resistor voltage drop divided by desired branch current]
    0.78 / 0.005 = 156 Ω

    Since there is no 156 Ω resistor, I'll have to use the next closest value of 160 Ω. Recalculating with that figure...

    [Find the voltage dropped by the 160 Ω resistor]
    160 * 0.005 = 0.8 V

    [Find the total voltage drop of all three LEDs]
    10.23 - 0.8 = 9.43 V

    [Find the voltage drop of one LED]
    9.43 / 3 = 3.14 V

    These numbers are still close to my target of 3.15 V and 5 mA for each LED, but I'm getting really confused as to why the current in the circuit stays the same, even though the voltage dropped across each LED has changed with my new resistor value. If each LED's voltage drop has changed (3.14 V from 3.15 V), wouldn't logic suggest that the current drawn by the LEDs would also change? It's very possible that I'm interpreting the math wrong, but according to my Ohm's Law equations above, the current would remain at 5 mA with either resistor. What am I missing?

    Just for argument's sake, lets say that I used a 300 Ω resistor...

    [Find the voltage dropped by the 300 Ω resistor]
    300 * 0.005 = 1.5 V

    [Find the total voltage drop of all three LEDs]
    10.23 - 1.5 = 8.73 V

    [Find the voltage drop of one LED]
    8.73 / 3 = 2.91 V

    See what I mean? By using a 300 Ω resistor, each LED's voltage drop has changed to 2.91 V, but the current remains at 5 mA. I know this is not right, but I don't understand what I'm doing wrong. If anyone could help me out, I'd REALLY appreciate it!

    Back to my project...
    So, at this point I've figured that I need a 160 Ω resistor in series with three LEDs (using a 10.23 V source) to get my desired current of 5 mA. Assuming that's correct (PLEASE correct me if I'm wrong), then why does the "LED series parallel array wizard" (found here) tell me that I should be using 180 Ω resistors instead?
    If you go to the wizard, plug in the same numbers that I've been using here:
    Source voltage = 10.23
    Diode Vf = 3.15
    Diode If = 5
    Number of LEDs = 185

    As you can see, I've got the basic concepts down, but I'm getting confused in the math/theory somewhere... I suspect that it has to do with the fact that the relationship between the LED's Vf and If is not linear, and I'm just not grasping the concept correctly. So, if anyone can help straighten me out, I'd REALLY appreciate it. I'm trying to learn as much as I can here, so feel free to go into as much detail as you'd like! :)

    Thanks in advance!
    -Lee
     
  2. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    Well your idea is right, if you connect the 300ohm resistor, than the current will drop, probably to 0, becaus 2.9V per LED is too low for it to open.

    This is because the LED is a nonlinear device, so unless you know the expression which matches the LED´s V/I graph, you can´t exactly count the current and resistor at the same time.

    All you can do is to take som resistor value, count the current, find the voltage drop in the graph, count the current again, find the drop again and so on until the difference between current in each stage is low enough (an order or so) lower than the current (untill the current doesn´t chage too much).
     
  3. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    I suspect that your wall wart voltage will drop when loaded also. It may also not be well filtered. If you have a scope, you should load it with approximately the expected current and then measure the voltage. If you don't have a scope, just use your multimeter.
    Wall warts are generally rated at full load, and are not regulated.
     
  4. Fixxxer

    Thread Starter Member

    Feb 15, 2007
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    I apologize, but I'm not sure I follow what you're saying... Could you give an example?

    That's an excellent point about it being rated at full load. I don't have a scope, but I'll test it loaded with a voltmeter when I get home from work tonight. Thanks!
     
  5. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    Sorry if I was confusing you. I was talking about the case when you know the resistor value and you want to know what current it will take.

    So you have 160ohm resistor.

    You estimated the curent to be 5mA, taking the 10.23V as ideal source which doesn´t change.

    You counted the drop to be 3.14V on one LED. Now you look at the table, which says that at 3.14V the LED will take only 4.5mA. Now you count the drop on the resistor, which is 160*0.0045=0.72V. That means 3.17V on each LED.

    3.17V on the LED is 6mA. 160*0.006=0.96V->3.09V
    3.09V is 4mA. 160*0.004=0.64V->3.2V
    3.2 is 10mA. 160*0.01=1.6V->2.87V
    2.8 is 0mA. 160*0=0->3.41V
    3.41 is 30mA
    ...

    hmm, it looks like this method somehow doesn´t work, even though in real the current would be stable.
    Can someone explain or say how to get the current on 160ohm resistor?

    ..now I made myself confused too...
     
  6. Fixxxer

    Thread Starter Member

    Feb 15, 2007
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    Glad I'm not the only one confused! :)
     
  7. n9352527

    AAC Fanatic!

    Oct 14, 2005
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    Because your calculation re-uses the 5mA current value. Instead, it should be:

    Vres = 160*I
    Vleds = 10.23 - 160I
    Vled = (10.23 - 160I)/3

    Now, you have to find the point that satisfies the last equation above on the characteristic curve of the LED. The current wouldn't be 5mA, it would be slightly lower than that. Not so easy.

    What you could do, like kubeek tried to suggest, is to do iterations in your calculation, using the 5mA value, and find a new voltage. Use the graph to find the new current corresponding to the new voltage and do a calculation again to find a new resistor value, it shoud converge to the 160 ohm that you have. Once you are reasonably close to the 160 ohm, within an acceptable accuracy, then you can read the correct values of voltage and current.

    This iterations is necessary, as kubeek suggested, because the characteristic curve is not linear, and you don't have any equation that describes the curve satisfactorily.
     
  8. hgmjr

    Moderator

    Jan 28, 2005
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    I think a "LOAD-LINE" analysis using your plot would provide the answer.

    If you take the VF value of 2.6V and mulitply it by 3 you get 7.8V. Subtract 7.8V from 10.23V to get 2.43V. If your plan is to use a 160 ohm resistor then divide 2.43V by 160 ohms to get 0.015 Amps.

    Put a dot on your plot at the point (2.6V, 0.015A).

    Next take the VF value of 3.2V and multiply it by 3 to get 9.6V.
    Subtract 9.6V from 10.23V to get 0.63V. Once again divide this value by 160 ohms to get 0.004 Amps.

    Put a dot on your plot at the point (3.2V, 0.004A).

    Now draw a line between these two points on your graph. The point at which this line intersects with the diode plot predicts the operating point of your series set of 3 diodes. Mind you the plot is little fuzzy but I get an operating point of (3.15V,0.0055A).

    hgmjr
     
  9. Fixxxer

    Thread Starter Member

    Feb 15, 2007
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    Excellent! Thanks for the help! I've written the seller to see if they can send me a couple of LEDs to test with before I order all of them. I don't know that they will, but I guess it doesn't hurt to ask. :)

    Speaking of that, do any of you know of a good LED supplier that isn't too expensive? Ebay is the cheapest place that I've found to order these 10mm LEDs (it's going to be about $37 for 250 LEDs shipped to my door), but some of these Ebay sellers are shady, and I'd prefer to order from a reputable source...
     
  10. hgmjr

    Moderator

    Jan 28, 2005
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    Greetings Fixxxer,

    I thought you might be interested in possibly using a constant current driver for LED array. Here is a link to a brief writeup that I prepared to illustrate the technique.

    This approach would add a potentiometer and a 3-terminal voltage regulator for each of the three LED arrays you plan to use. For the 62 banks of three LED arrays that would add another $60 to the overall cost.

    That may be a little pricey but these component have the potential for providing the design with features such as greater tolerance to variation on the power supply source and adjustablility for closer brightness matching among the 185 LEDs being used.

    hgmjr
     
  11. Fixxxer

    Thread Starter Member

    Feb 15, 2007
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    I really like the idea of using a constant current driver like that, but I can't justify spending that much extra money on it (my wife is already mad that I'm spending $40 on LEDs, and I still need to make the PCB!). :)

    I really appreciate you putting so much thought into this though... That constant current driver is definitely something that I'll save for use later! :)
     
  12. hgmjr

    Moderator

    Jan 28, 2005
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    OK, I understand.

    One thing before you kick the constant current source approach temporarily to the curb, if you are willing to use a 24 volt power supply that will allow you to use a 7 LED array per regulator and you can cut the number of voltage regulators down to 26 or 27. That is better than a 50% reduction.

    hgmjr
     
  13. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    I used constant voltage regulator for my displays, and it works quite good.

    It is a 3x 7segment display, so I adjusted the common voltage for all the segments by LM317.

    You could use few of these to feed the LEDs with constant voltage, which will make them have uniform brightness like with constant current sources, but the number of the regulators is only limited by power dissipation on them.
     
  14. Fixxxer

    Thread Starter Member

    Feb 15, 2007
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    Would something like this work? It looks like it only requires the use of one LM317L for the array, and then one (optional) MPS2222 transistor per LED series branch (for safety). Keep in mind that I'm still a beginner, so if this question seems blatantly obvious, please forgive me! :)

    (Warning: PDF link)
    http://www.onsemi.com/pub/Collateral/AND8109-D.PDF
     
  15. hgmjr

    Moderator

    Jan 28, 2005
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    Fixxxer,

    The circuit involving the transistor and the LM317 should work fine.

    I would stay away from the circuit in Figure 3. Parallel LED strings will almost certainly end up with brightness differences that you might find undesireable.

    A few words of caution.

    The LM317 requires around 2 volts of headroom between the Vin pin and the Vout pin to operate. That means that the power dissipation for the LM317 will increase at a rate of 2V*0.005A or 10 milliwatts per string of LEDs (assuming that the LED strings are set to 5 milliamps of current each).

    At 10 milliwatts per string, you will be looking at around 616 milliwatts if you want to stick with 3 LEDs per string. That is not too bad but it may call for a heatsink to keep things cool.

    This 2V dropped by the LM317 also will require you to make sure that the input power supply has the 2V plus 1.25 volts or at least 3.25V on top of the 3.3V per LED times the number of LEDs in the string.

    With a 3 LED string you will need to use a power supply whose output voltage is at least 9.9V + 3.25V or 13.15V. A 15 Volt power supply would be a reasonable choice.

    There are LDO (Low drop-out) voltage regulators which would overcome this impediment.

    The voltage regulator in the circuit I suggested is an LDO but it would not work in this application as it cannot alone supply the current for all of the LEDs in your display.

    hgmjr
     
  16. Fixxxer

    Thread Starter Member

    Feb 15, 2007
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    Thanks for the info! You guys have been a BIG help! I'm still undecided on what I'm going to end up doing... I want to keep it as cheap as possible, and the power supply I want to use only has a range of 1.5 V to 12 V. I want to stay away from batteries, since the final product will be hanging on a wall or sitting on an end table in my game room. We'll see...

    On a good note, I received the free sample 10mm LEDs yesterday, so I can test them out tonight to get some more accurate figures (since that graph in the specs is a little blurry).
     
  17. hgmjr

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    Jan 28, 2005
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    We are all looking forward to hearing back from you.

    hgmjr
     
  18. Fixxxer

    Thread Starter Member

    Feb 15, 2007
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    Okay, here are the results of my testing with the free sample LEDs... All tests were done on a breadboard with a battery as a voltage source:

    FIRST TEST:
    One (1) blue 10mm LED in series with four (4) resistors

    Battery voltage (unloaded) = 6.29 V
    Battery voltage (loaded) = 6.28 V

    R1 = 10 Ω
    R2 = 386 Ω
    R3 = 118 Ω
    R4 = 46 Ω
    R total = 560 Ω

    I = 5.77 mA

    Voltage dropped across components:
    LED = 3.004 V
    R1 = 0.0572 V
    R2 = 2.254 V
    R3 = 0.692 V
    R4 = 0.271 V
    R total = 3.274 V


    SECOND TEST:
    One (1) green 10mm LED in series with four (4) resistors

    Battery voltage (unloaded) = 6.29 V
    Battery voltage (loaded) = 6.28 V

    R1 = 10 Ω
    R2 = 386 Ω
    R3 = 118 Ω
    R4 = 46 Ω
    R total = 560 Ω

    I = 5.92 mA

    Voltage dropped across components:
    LED = 2.917 V
    R1 = 0.0587 V
    R2 = 2.312 V
    R3 = 0.710 V
    R4 = 0.278 V
    R total = 3.358 V


    THIRD TEST:
    Three (3) blue 10mm LEDs in series with one (1) resistor

    Battery voltage (unloaded) = 9.07 V
    Battery voltage (loaded) = 9.00 V

    R1 = 1 Ω

    I = 6.13 mA

    Voltage dropped across components:
    LED1 = 3.016 V
    LED2 = 2.983 V
    LED3 = 2.983 V
    R1 = 0.0062 V


    FOURTH TEST:
    Three (3) green 10mm LEDs in series with one (1) resistor

    Battery voltage (unloaded) = 9.02 V
    Battery voltage (loaded) = 8.97 V

    R1 = 1 Ω

    I = 8.45 mA

    Voltage dropped across components:
    LED1 = 2.981 V
    LED2 = 3.038 V
    LED3 = 2.930 V
    R1 = 0.009 V


    FIFTH TEST:
    Three (3) blue 10mm LEDs in series with one (1) resistor
    ...in parallel with...
    Three (3) green 10mm LEDs in series with one (1) resistor

    Battery voltage (unloaded) = 9.04 V
    Battery voltage (loaded) = 8.95 V

    R1blue = 1 Ω
    R2green = 1 Ω

    I1blue = 5.32 mA
    I2green = 7.74 mA

    So, it looks like I could use a little higher resistor value on the green LED branch(es), but everything looks pretty good for the most part. When I had the array of 6 LEDs connected, they all appeared to be the same brightness. Next, I need to test them with my wall wart, since it supplies at a little more than 9 V on the 9 V setting.
     
  19. hgmjr

    Moderator

    Jan 28, 2005
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    Good work. Very methodical empirical measurements.

    hgmjr
     
  20. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Just be aware that 1 ohm is too low to be a good current limiter. If the voltage across it changes by 100mV, the current changes by 100mA!
     
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