Vin versus Vforward in LEDs

Discussion in 'General Electronics Chat' started by bkloehn, Jul 24, 2008.

  1. bkloehn

    Thread Starter New Member

    Jul 24, 2008
    I have a question about supplying a LED circuit (2 parallel branches of LEDs, each branch with 2 LEDs and a ballast R) with a voltage lower than the Vforward of the LEDs.

    LED Vforward range of 1.90 - 2.85 through all temperature ranges.

    In this case the circuit has one branch with LED Vf on the lower end of the range (1.90*2 = 3.8) and one branch with LED Vf on the upper end of the range (2.85*2 = 5.7). Both branches have a 13Ohm ballast to evenly distribute the current regardless of Vf. The target current for the circuit is 50mA so each branch would have 25mA. The objective is to test functionality of the strings by applying a voltage and checking current.

    IR + Vf = Vin gives me 2 Vin values for the range of Vf.
    for Vf = 1.90, Vin = 4.125
    for Vf = 2.85, Vin = 6.025

    The problem is that I can only use 1 Vin. If I use 6.025 then any circuit with Vf = 1.90, the LEDs will be driven past their current rating and risk damage to the LED.

    If I use 4.125 then any circuit with Vf = 2.85 will not have enough voltage to fully forward bias the LEDs.

    All that being said, if I under drive the circuit with say 3VDC, will the current be the same throughout the Vf range. I understand that the luminance of the LEDs will be lower but that is not the target of the test.
    Thanks for reading through and I look forward to any responses.
  2. SgtWookie


    Jul 17, 2007
    Sorry, this isn't the way it works. If your LED's Vf is different in the two branches, you must use a different resistance for the current limiting resistor.
    However, if you can match the sum of the Vf's, then you can use the same size current limiting resistor.

    For each series string of LEDs, you calculate Rlimit by:
    Rlimit = (Supply Voltage - Total(VfLED)) / LED Current

    So, let's look at your higher Vf LED string first. 5.7v is the total Vf. You really want at least a 0.5v "headroom" over the sum of the Vf, or any minor power supply fluctuations will cause the LEDs to flicker noticeably. So, you'll need a 6.3v supply.
    Rlimit = (6.3v - 5.7v) / 25mA
    Rlimit = 0.5/0.025 = 20 Ohms

    So, for the other string you'll need to calculate for a different Vf.
    Rlimit = (6.3 - 3.8) / 25mA
    Rlimit = 2.5 / .025 = 100 Ohms

    Let's see how many Watts we'll consume in the worst-case resistor (100 Ohms)
    P = EI (or, Power in Watts = Voltage x Current)
    P = 2.5 x .025
    P = .0625

    Good practice says to double the actual current in the circuit before determining the resistor wattage.
    .0625 x 2 = 0.125. A 1/8 Watt resistor or larger would work just fine for both.

    Now let's look at an improved scenario - balance the Vf on both sides by exchanging one low Vf and one high Vf.
    1.9+2.85 = 4.75 V

    Now if you had a really stable 5V supply, you could calculate Rlimit for it.
    Rlimit = (5v - 4.75) / 25mA
    Rlimit = .25 / 0.025 = 10 Ohms

    However, even a small change in the supply's voltage would cause a large change in the current flow through the LEDs.
    Let's see what happens if the supply voltage decreases by 1/10 of a volt (100mV)
    I = 4.9-4.75 / 10 = 15mA - the LED current has dropped by nearly 1/2!

    You would be much better off to use a 6v supply, as you would have plenty of "headroom".
    Rlimit = (6v - 4.75) / 25mA
    Rlimit = 1.25 / .025 = 50 Ohms
    The closest standard value resistor is 51 Ohms. Let's see what that does to the current.
    I = E/R
    I = 1.25 / 51 = 24.51mA - very close.
    Let's check the wattage needed.
    P = EI = 1.25 x .02451 = 0.0306
    Multiply by 2 = 0.0612 - we can use a 1/10 Watt resistor.

    Note that a "wall wart" is not a regulated power supply. It's output voltage will fluctuate wildly if it's load varies.
  3. bertus


    Apr 5, 2008
  4. bkloehn

    Thread Starter New Member

    Jul 24, 2008
    SgtWookie - I understand what you are saying.

    The LEDs are (in this case) binned by luminance and not by Vf. The range of Vf in the bins is 1.90 - 2.85. It is not feasible to test Vf on incoming LEDs so it is possible to have LEDs with Vf on both ends of the range on the same reel. These are things I can not change so it becomes a case of overcome, adjust, and adapt. As a Marine, you can appreciate that.

    As bertus suggested, a constant current source may solve the Vf differences but how would one know if one of the branches is in-op. You would not be able to check current as Iin = Iout regardless of the whether one or both branches is in-op. One branch would just have total current through it and would run "hotter" than the open branch with no current through it.

    If I apply a known voltage to the circuit, I would be able to monitor current to know if the current I am reading is the result of one branch or both branches. The problem is that if I choose the Vin (6.3V) that corresponds to the upper Vf and use that to power a circuit that has one/both branch Vf of 3.8V then the current would go well above the Ifmax and risk permanent damage to the LEDs. The problem is also vice versa, if I chose the lower Vin (4V) then if one/both branch happened to have Vf in the upper range, they would not be fully biased while one/both branch may have the lower Vf and be fully biased.

    That brings me to my question about using (3V) as Vin. The circuit will light (dimly) both branches at 3V. No combination of branches would be fully biased with the lower Vin, but does a difference in Vf effect current through the branches at such a low Vin.

    In other words

    If Vin = 3V and branch1 Vf = 3.8, does Ibranch1 = Ibranch2 when branch2 Vf = 5.7?

    Once again I appreciate the responses.
  5. SgtWookie


    Jul 17, 2007
    It's Improvise, Adapt, Overcome. ;)

    You have few options here.

    1) Regulate the voltage across both strings. Rlimit are SOT. (Select On Test). Very labor intensive.

    2) Design the board so that each string has an active current regulator in series, such as an LM317L (available in TO92 and 8-pin SOIC). You will experience about a 3v dropout across the regulator when used for a current regulator. An LM317L with a 50 Ohm resistor from the output terminal to the adjust terminal will output 25mA from the adjust terminal when supplied with current at the input terminal.
    Component costs; about $0.25/0.40 per board added to existing BOM. Increased cost of power supply due to power lost in LM317T regulator.

    3) Design some other type of constant current circuit. Complexity skyrockets, component inventory grows.

    4) Have your engineers use the pick & place machine(s) to pre-screen LED's, sorting them by Vf ranges into pick & place sorter boards. Then during PCB pick & place, select LEDs of complementary Vf ranges so that a single value of Rlimit may be used by both sides. Minimal human interaction once set up, minimal impact aside from pick & place time, no changes required to existing board, resulting product reliable and consistent quality.

    5) Say the heck with it, and produce an inferior product that will initially be cheap, but will put you out of business in the long run.

    Yes. One would be very noticeably brighter than the other, but they would both be far dimmer than with their specified forward current.

  6. bkloehn

    Thread Starter New Member

    Jul 24, 2008
    You must be a Parris Island Marine ;). I should have just said Semper Gumby

    Could you explain this a bit? I am not familiar.

    Or you make a product that blows every other product out of the water but since the cost is pennies more than your offshore competition, you can't even get a sniff because all OEMs want is cheap and functional. Quality takes a back seat to cost everytime. That is the world we are living in.

    Thank you for the prompt repsonses. It is certainly appreciated.
  7. SgtWookie


    Jul 17, 2007
    Negatory. Check the shades ;)

    For each board, measure the Vf of the diodes that were pick & placed and reflowed in the oven. Select the proper current limiting resistor for the measured Vf, and solder in place. If the Vf is too high or too low for the selected supply vs the resistors being used, scrap or send to rework for LED substitution. Very labor intensive, unless automatic test equipment and component insertion is used.

    You aren't telling me anything that I don't already know. :(

    That's why I'm suggesting that you use a pick & place machine that has component test capability to presort your LED's into Vf ranges - or SOME sort of ATE (automated test equipment) to pre-qualify the components. If they're not sorted on the front end, you're going to have rework, scrap, or have to design an active circuit to control the current. Everything will add cost.

    85% to 90% or so of your LEDs will fall into a fairly narrow Vf range. It's that 10%-15% at either end that are the killers, and will either require pre-sorting, scrap or rework. The more times a human has to touch an assembly, the more likely that it will become scrap. When an assembly has been reworked a couple of times and is finally scrapped, far more than the sum cost of components has been lost. You would save time by just throwing your profit dollars in the trash can.

    The only way you're going to be able to compete is to produce high quality products at high volume, which requires a high degree of automation in assembly and test, along with a well-engineered product.
  8. bkloehn

    Thread Starter New Member

    Jul 24, 2008
    Sgt Wookie,

    Thanks for the explanation, I can certainly see the labor involved in something like that.

    The pick n' place with test capability is something worth exploring but I can also see the overhead as something the bean counters would frown upon. Very helpful site here. Thanks again devil dog. (Cpl myself, 97 - 01).
  9. gerty

    AAC Fanatic!

    Aug 30, 2007
    1.   You must be a Parris Island Marine    
    Most people I've run into spell it wrong (Paris) It sure ain't no vacation resort..
  10. SgtWookie


    Jul 17, 2007
    Yep, there unfortunately is a good bit of labor involved unless you can automate the pre-selection portion.

    Using the pick & place to pre-sort is of course going to at least double the wear & tear on it, and complicate your storage binning somewhat if you're using it to pre-screen/pre-sort parts. If your LEDs are SMT/SMD, it won't be as bad as leaded LEDs on tape would.

    If they're on tape, probably the best thing to do would be to screen them while the boards are being pick & placed; just discard the LEDs that are too far off tolerance. Those discarded LEDs could possibly be re-sold as surplus. At least you'd minimize waste after assembly, and the pick & place machine time would not be dramatically increased.

    If your pick & place doesn't have some kind of testing capability, this would be a good time to justify purchase of it, particularly if you're going to be producing these in large volume. Either that, or hire a big team of technicians and solderer/rework folks to take care of your unacceptable reject rate.

    I strongly suggest that using ATE (automated test equipment) is a fundamental requirement, a cornerstone if you will, to producing a consistent, reliable and high-quality product with an absolute minimum of human intervention.

    Even if only 5% of the diodes are significantly out of tolerance, a normal distribution would give you a 20% rework rate (5 bad diodes spread across 25 assemblies, making 5 of them needing rework), which is clearly unacceptable. Your labor costs skyrocket, making your product non-competitive in the global marketplace.

    You can't work harder. Work smarter instead.

    One more option is to use a significantly higher-voltage supply, and higher-value (both in Ohms and wattage) resistors.
    3.8v to 5.7v @25mA being the goal, if you could use a 12v supply, a 330 Ohm resistor in series with two 1.9v LED's, that would give you 24.85mA. The two LEDs measuring 5.7Vf total would have 19.1mA current with a 330 Ohm resistor. That's still quite a bit of difference, but certainly much better than a lower voltage. The trouble here is that you would consume much more power in the resistors; about .2 W; you'd need to use 1/2W resistors. In other words, efficiency goes out the window.
    We're not always successful, but we try ;)
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    Last edited: Jul 28, 2008