A useful website for LED array design

P.O.S.

Ask it to design a circuit for 3 LEDs with Vf 3.5V @ 20 ma, from a 10.5 V supply and it will tell you to put them all in series with a 1 ohm resistor! One used to get *far* better design guidance on the back of a Tandy blister pack. LOL.

Should come with a health warning: This site teaches you how to convert your LEDs into DEDs with emmission of toxic fumes from the epoxy encapulation.

Common!! What's wrong with the 3 LEDs in series if the power suuply voltage is 10.5 and the LED drop voltage is 3.5? 3.5 x 3 =3D 10.5!!

Your LEDs wont be DEDs!! hehehe

Try adding another LED and it will give you a series/parallel configuration..

Try this:

What's the current through the LEDs in that situation? Hint: V=IR.

Cheers, Nicholas Sherlock

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There is plenty wrong with that. If you don't know why then I suggest you try following some of the "useful links" on your site for starters.

Dave.

Consider the consequences of one or more of the LEDs having a Vf that is lets say 5% low, or 5% too much supply voltage. Also consider that a typical LED has a Vf that decreases by about 2mV per deg C.

You are also actively encouraging the use of a non regulated battery supply.

A new 1.5V nominal Alkaline cell typically gives 1.55V. At 50% discharge they tend to be down to 1.2 to 1.3V. A practical end of life voltage is around 1V. If I use 8 alkaline cells for a nominal 12V battery, it will start its life at 12.4V and be pretty much dead at 8V. Hence my interest in what happens around 10.5V

If you are dropping less than about 1/3 of the supply across the series resistor then either you will be throwing away batteries with a lot of life left in them or you will be overdriving the LEDs with a new battery hence my comment about DEDs (old geek humour - see early 70's AFJ datasheet from TI). Also the brightness tracking in the 4 LED case you gave will be abysmal. If the supply drops a mere 10%, the 3 series LEDs will nearly extinguish while the single LED will be virtually the same brightness.

You should read

For a regulated supply I *MIGHT* push my luck as far a 20% drop across the series resistor, but only if I was operating at less than 50% of the rated LED current.

Rewrite your calculator to balance the series chains as well as possible for the number of LEDs and allow more headroom for the current limiting resistor to do its work and provide for input of a supply voltage range, give current in each branch for maximum and minimum supply voltage and you *MIGHT* have something. Perform a worst case analysis (e.g 10% over voltage, all LEDs with a Vf 10% low and resistors at minimum value for tolerance). Also add a check box for 10% tolerance resistors.

Expect geeks to pound on your code till it breaks :-) Its not personal, its just geek nature. If you register a domain rather than putting it on your personal home page, expect harsher criticism. I do appreciate the effort you've put in to making the site pretty, now make it useful . . .

Meanwhile, most of us will continue to solve simple stuff like this on the back of an envelope or in our heads. Usually the process is more driven by what resistors we've got handy than by what's ideal.

His calculator is functionally identical to the first (often daft) result from Rob Arnold's one (which gives the same sigle idiotic result for the 10.5V 3 LED test case). If he can't fix his code, one can only conclude he's nicked it and doesn't understand it. Its probably a bad case of Cargo Cult programming IMHO but lets give him a chance to prove he understands it and probably wrote it by fixing it ;-)

Anyone still interested in LED Calculators should try:

At least Rob's calculator gives some alternatives that are useable if you increase the number of LEDs and Luxeon think his site's good enough to link to.

n 5

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Yes, ledcalculator.net offers the same results as Rob Arnold's one but with a much user friendly UI. I didn't come up with something new.. It's Ohm's law.

But what's wrong with the 1 ohm resistor? LEDs do not have internal resistance and that's why you should place at least 1 ohm resistor to limit the current on that line.

Please correct me if I am wrong. I am just a hobbyist.

LEDs *do* have internal resistance, but not enough to limit the current if you try to power them from a 'stiff' voltage source. If you assume they don't, as required by the formula you are using, then in the case of the 10.50 V supply and three 3.50 V Vf LEDS, the current would be ZERO as there would be ZERO volts across the 1 ohm resistor. However if the supply increases by 1% there is now just over 0.10 V across the resistor which would give a current of just over 100 mA.

[***[***BANG**]***] + a nasty cloud of smoke

The 1 ohm resistor does *NOTHING* usefull to limit the current if the maximum permitted current is 30 mA. Its FAR too low value.

In practice its not quite that bad as the slope resistance of the LED will be rather more than 1 ohm for normal 5mm and 3mm LEDs but you can easily end up with the situation that they are running at say 30mA with a milliammeter (or DVM on a mA range) in circuit but go bang (or more accurately ( fizzle smoke) when you connect them without the milliameter.

When I got into this sort of thing, there was no internet and you could buy a round of drinks for the price of one or two LEDs. They also tended to melt their plastic when you soldered them. A *lot* of care went into choosing resistor values for the average hobby user, even to the point of checking the Vf of individual LEDs. One also derated them by about 50% if you wanted them to last. Now they are dirt cheap. It would be a real shame however if someone tried to power 8 white Luxeon Star LEDs (Vf 3.42 V @ 350 mA) to put on their truck or boat using your circuit and the result would be a *very* unhappy user.

My other detailed reply to you had some suggestions for improving your calculator which if you coded it yourself should be quite easy to implement. Fix your calculator and I'll apologise for accusing you of stealing Rob's design and code.

The Vf of the LEDs will vary with temperature and also from one device to the next (manufacturing variations). With only 1 ohm of "ballast" resistance, the current will vary widely when the Vf varies slightly. On the other side of the coin, the power supply is unlikely to be exactly

10.50000 volts; it'll vary a bit. The purpose of the resistor is not merely to *limit* current but also to provide a bit of "give" in a real-world circuit. In practice this means that you need a good fraction of the power supply voltage to be dropped across the resistor, which means that a good fraction of your total power is wasted as heat; this is why high-power or high-efficiency LED supplies are switching current regulators rather than big resistors.

It's a nice webpage! But, like most automated tools, there are some real-world considerations that it doesn't deal with.

It would be handy if the user could plug in the percentage tolerance of the Vf, of the power supply, and of the resistors (or fix them at 5%), and get a listing of the worst-case variations in current and the worst-case difference in current between LED strings.

--
   Wim Lewis , Seattle, WA, USA. PGP keyID 27F772C1
  "We learn from history that we do not learn from history." -Hegel

plenty, it might a well not be there.

yes they do, not enough, but some, but what's worse is the voltage drop they present is not constant it varies according to temperature (and probably other factors)

1 ohm is not nearly enough.

buit even assuming perfect 3.5 V leds and a perfect 10 V supply with a 1 ohm resistor there won't be any current flowing.

Bye. Jasen

hey

Thank you everyone for your feedback.

I will improve the calculator to take into account the Vf and supply voltage ranges (%).

One more question. How should the design be if you have a perfect 10.5 power supply voltage and perfect 3 x 3.5 fV LEDs like in this case:

Should the 1 ohm resistor be ommitted? or should we put the third LED on a new parallel line?

in the a perfect world you could use no resistor.

in the real world you should to have resistors on all parallel circuits that drop atleast 20% of the supply voltage. (more for unregultated supplies)

Go for the new parallel line, I expect that most visitors to your site will want real-world solutions,

Bye. Jasen

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You still don\'t get it, do you?

Because LEDs are diodes and have a sharply rising forward current for a
miniscule change in forward voltage at their knee, they should be driven
by a constant current source so that if their Vf or the supply voltage
wanders their If will remain constant.

Failing that, for safety you need to _limit_ the forward current by
choosing the series resistor considering that the supply voltage will be
at its maximum and the Vf of the LED(s) will be at its minimum, then
calculating the series resistance like this:


           Vs(max) - Vf(min)
     Rs = ------------------
                 If

That will assure If will never rise above the value plugged into the
formula.

It\'ll fall if the supply voltage decreases or if Vf increases, causing
the illumination to diminish, but at least you won\'t be hurting the LED.
JF

Good.

Offer the user a field to enter minimum percentage of the supply voltage to drop across any series resistor. Start another parallel chain of series LEDs when the voltage across the resistor would otherwise be less than the minimum value. Initialize this field to something reasonable like 30% as a default.

Otherwise we'll just give you a hard time for the resistor being too small for safety for cases like:

1.8 ohm!!

Fix the divide by zero error if zero current entered (eg. current less than 0.1 mA should prompt for a value in range like you do for other daft entries.

Its a matter of style, but it would be far more usual to put the resistor between the LEDs and the positive supply, not between the LEDs and ground. (It also reduces the risk of damage if the circuit is in a grounded metal case and one of the LEDs gets shorted to the case.)

Also it would be to force the number of LEDs in each chain to be as near the same as possible instead of putting a single left over LED in the last chain. (Integer divide total LED number by number of chains, then add 1 LED to each chain until you've used up the remainder.) That way the relative brightness of different chains does not change so much when the supply voltage varies. As the aim is to get the same current in each LED, the efficiency and total power dissipated in the LEDs and also in the resistors will remain the same (neglecting small variations in the current in each chain due to rounding the resistors to the next preferred value) as that given by the current version.

I'm keen to see a check box to choose between 5% and 10% resistors (always choose a higher value as the chosen current should never be exceeded).

That should keep you busy for a while . . . I'll try to remember to check back next week to see your progress if I've got any spare time.

Good user interface, looks good, but as others have pointed out, there is a problem with the use of such small resistor values. Fix that and you should have a nice calculator.

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