A useful website for LED array design

[[email protected]]

Hey folks!! I just created a website for those of you interested in
creating LED arrays. It will calculate the resistors values and design
the circuit for you.

Check it out here
http://ledcalculator.net

Your feedback is welcome

 

[IanM]

Hey folks!! I just created a website for those of you interested in
creating LED arrays. It will calculate the resistors values and design
the circuit for you.

Check it out here
http://ledcalculator.net

Your feedback is welcome

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.

 

[[email protected]]

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 = 10.5!!

Your LEDs wont be DEDs!! hehehe

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

Try this: http://ledcalculator.net/default.aspx?values=10.5,3.5,20,4,0

 

[Nicholas Sherlock]

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 = 10.5!!

Your LEDs wont be DEDs!! hehehe

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

Try this: http://ledcalculator.net/default.aspx?values=10.5,3.5,20,4,0

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

Cheers,
Nicholas Sherlock

 

[David L. Jones]

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 = 10.5!!

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.

 

[IanM]

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 = 10.5!!

Your LEDs wont be DEDs!! hehehe

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

Try this: http://ledcalculator.net/default.aspx?values=10.5,3.5,20,4,0

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 <http://www.duracell.com/oem/Pdf/others/ATB-5.pdf>

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.

 

[IanM]

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.

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:
<http://led.linear1.org/led.wiz>

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.

 

[[email protected]]

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:
<http://led.linear1.org/led.wiz>

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.- Hide quoted text -

- Show quoted text -

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.

 

[IanM]

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 (<POP> 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 <http://ledcalculator.net/default.aspx?values=24,3.42,350,8,0>
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.

 

[Wim Lewis]

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.

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.

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

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.

 

[Jasen Betts]

But what's wrong with the 1 ohm resistor?

plenty, it might a well not be there.

LEDs do not have internal resistance

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)

and that's why you should place at least 1 ohm resistor to
limit the current on that line.

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

 

[[email protected]]

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

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:
http://ledcalculator.net/default.aspx?values=10.5,3.5,30,3,0

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

 

[Jasen Betts]

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:
http://ledcalculator.net/default.aspx?values=10.5,3.5,30,3,0

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)

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

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

Bye.
Jasen

 

[IanM]

Thank you everyone for your feedback.

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

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:
http://ledcalculator.net/default.aspx?values=10.5,3.5,30,3,0

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

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:
http://ledcalculator.net/default.aspx?values=10.55,3.5,30,3,0
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.