# LED array

Discussion in 'Electronic Basics' started by Jonathan Leppert, Nov 18, 2003.

1. ### Jonathan LeppertGuest

Hi,

I am creating a 144-element LED array, wired in parallel. My supply voltage
is 5.0 volts at around 5-6 amps. The LEDs take a forward current of 3.5
volts, being driven at around 30 ma. My question is should I assume the
array acts as one gigantic LED with its current consumption summed, with the
same forward voltage so hence I will need to limit voltage with a resistor,
right? I am thinking a 1.0 Ohm 10 W resistor.

Would this work or am I going to destroy 144 LEDs? Any explanations or
insight would be greatly appreciated.

Thanks,

Jon

2. ### Costas VlachosGuest

I think you will run into problems with this setup. The 144 LEDs will not
have identical forward voltages, due of manufacturing tolerances, etc. The
LED (or group of LEDs) with the lowest Vf will light up first, will draw
most of the current, and will probably burn shortly, followed by the others
in a chain-reaction, until all LEDs are gone. This occurs because the V-I
curve of a LED is steep around its Vf, so small changes in Vf cause large
changes in If. I don't know what the best configuration would be, but don't
connect them in parallel.

Costas
_________________________________________________
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3. ### Ben WeaverGuest

Hi Jon...

I think you have the right idea. But I get about 0.3 ohms. (I did
144x30mA = 4.32A, and then 1.5V/4.32A = 0.347 ohms. That'll dissipate
6.48W so 10W would be a good choice. It'll get really hot though.)

Anyway, I did a similar thing some time back and wired the LEDs up in
series to give a total voltage drop equal to the power supply. That way
you don't need the series resistor (which will effectively be wasting
power).

You might try this: Get two of your LEDs and wire them up in series. Put
them across the 5V power supply. If they're bright enough (remember this
time they're only getting 2.5V each) then you're in luck. Get 72 of your
LEDs and wire them in parallel to form a gigantic LED, and do the same
for the other 72. Now wire those two big "cells" in series. Ta da! No
resistor getting hot!

But of course, you may think that the LEDs aren't bright enough and so
you'll have to go back to your original idea.

HTH

Ben
~~~

4. ### Steve J. NollGuest

Costas is right. You need a series resistor for each LED.
At the day job we manufacture 180 element LED arrays (from a single
die.) Each has its own series resistor. And, if you want very good
uniformity you have to select the resistors.

Steve J. Noll | Ventura California |
| The Used High-Tech Equipment Dealer Directory
| http://www.big-list.com
| The Peltier Device Information Site:
| http://www.peltier-info.com

5. ### Ben WeaverGuest

I didn't find this too much of a problem. But worth bearing in mind.

Ben
~~~

6. ### John FieldsGuest

---
The LEDs will have forward voltages which can lie anywhere in the range
specified by the data sheet and a negative temperature coefficient of
forward voltage, so the LED with the lowest Vf will hog more current
than the rest, which will cause it to heat up more than the rest which
will cause it to allow more current to flow through itself which will
cause it to heat up more, which will eventually lead to its destruction.
If it fails open, then its fate awaits some of the other LEDs in the
array, so you need to bite the bullet, use a series current limiting
resistor for _each_ LED, and connect all of the series resistor-LED
circuits in parallel across the supply. Since your LEDs have a Vf of
3.5V at 30mA (?) you'll need to drop 1.5V at 30mA in each series
resistor, so since R = E/I, R = 1.5V/0.035A ~ 42.9 ohms. The closest
standard 5% value is 43 ohms, and each resistor will dissipate P = IE =
0.035A*1.5V ~ 53 milliwatts, so you could use 1/4 watters no problem.

^^^^^^ ^^
0.030A 45

Oops...

8. ### Bill BowdenGuest

Depends on how well the LEDs are matched. LEDs in parallel
tend to draw different currents due to differences in forward
voltage drop. It might work, you could try 3 or 4 in parallel
to see how well the brightness matches. Why not use a higher supply
voltage so you can arrange the LEDs in series strings, each
with their own resistor?

-Bill

9. ### Jonathan LeppertGuest

will cause it to allow more current to flow through itself which will
How much current are we talking about here? The LEDs I am using are "under
rated" according to the distributor, and can be run at higher currents (with
a decreasing life, of course). I'm not too concerned with making sure the
brightness is the same for each LED, and the array will only be on for 2
hours a day. Do you really think I'll need resistors (not worried about the
price, just the logistics of wiring up 144).

Thanks,

Jon

10. ### Jacobe HazzardGuest

I agree that LEDs in parallel is a bad idea. Try it with just a few, and you
will see how bad it can be.
If you want to use less series resistors, then increase the supply voltage
and string the LEDs in series, one resistor per string and as many strings
as you need.

11. ### John FieldsGuest

---
most of the "standard" white LEDs I've seen are rated for 20mA nominal
and 30mA absolute max continuous, so I'd be leery about what the
distributor says unless he can back it up with a data sheet. He may be
referring to the pulse rating which is probably around 100mA with a 10%
duty cycle.
---

12. ### JeffMGuest

Jonathan Leppert
I'm with Bill on this one. Less energy wasted in resistors.

13. ### Jonathan LeppertGuest

Ok. Assuming I up the voltage to 12 VDC, and use 36 series strings of 4
LEDs, what value/type of series resistor should I get for each string?

Vf = 3V (2.8-3.8V range)
Forward Current: 30 ma

I know this isn't the most efficient way, but its better than 144 resistors,
and I only have a choice of 5V, 9V, and 12V for power. I'm not familiar with
a regulator or DC-DC converter that would efficiently up 12V to something
like 24V or higher. I'm open to better configurations if anyone has any
ideas.

Jon

14. ### Robert MonsenGuest

You should measure the voltage across a set of the LEDs at 30mA. Average the
value, and call it Vf. Then your resistor value should be a standard value
close to

R = (12 - 4 * Vf)/0.030

Negative R means you won't get 30mA out of it.

Since you are running the LEDs at 30mA, I'm guessing they will get hot.
Since this is the case, and since Vf is dependent on temperature, you should
let them heat up before taking the measurements.

You can build a 30mA current source as follows:

VCC
+
+---+---+
| |
| .-.
| | |100
| | |
| '-'
| |
| |
.-. |<
1k | |<---| PNP
| | |\
'-' |
| |
| |
| LED
| |
| |
| |
+---+---+
|
===
GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

Hook the + lead of your multimeter to the collector of the PNP transistor,
and the - lead to ground, and put the multimeter into current measurement
mode. Adjust the pot to get 30mA. Then, hook up the LED between the
collector and ground, and measure the voltage at the collector (remember to
wait till it heats up). Thats Vf(30mA). It won't be exact, since the current
will vary depending somewhat depending on the voltage at the collector, but
it should be close enough for your measurement.

Regards,
Bob Monsen

15. ### Costas VlachosGuest

If you arrange them in strings of 4 LEDs in series, then you have 4 * 3 =
12V total Vf (i.e., you don't need a resistor). This might work, but the
resistor is generally a good thing to have, as it limits the current in
overvoltage situations, etc., and makes the overall V-I curve less steep
which is a good thing. So you could string two or three LEDs in series, and
work out the resistor. For 2 series LEDs at 30mA, R = 6V/30mA = 200 Ohm. For
3 LEDs, R = 3V/30mA = 100 Ohm. And 144 is divisible by both 2 and 3, which
is a good thing.

cheers,
Costas
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16. ### John FieldsGuest

---
If you wanted to run it off the 120V (?) mains and you could assure that
no one could come in contact with any of the wirng you could do this
twice, which would get you down to two resistors:

ACIN>---[R]--+--[36LED>]---+
| |
+--[<36LED]---+
|
ACIN>----------------------+

36LED is a string of 36 LEDs wired in series, with two strings wired in
anti-parallel so that each string lights alternately as the polarity of
the input voltage reverses. Determining R analytically is tricky
because the LED array isn't ohmic, so I just scaled the array and ACIN
down by a factor of 9, built it, and adjusted R until I got about 30mA
RMS through it:

-->| |<---1.7VRMS

13.3AC>---[47R]---+--[4LED>]---+
| |
+--[<4LED]---+
|
13.3AC>------------------------+
<--28.5mA RMS-->

Scaled back up it should look like this:

120VAC>---[430R]---+--[36LED>]--+
| |
+--[<36LED]--+
|
120VAC>-------------------------+

With approximately 28mA flowing through the 430 ohm resistor it should
dissipate about 337mW, so I'd use a standard 5% 430 ohm 1/2 watt
resistor. There is one other problem to consider, and that's line
transients. While they'll be attenuated by the 430 ohm resistor, a sure
way to get rid of them would be to split the series R and use a couple
of Zeners in series opposition to soak it up, like this:

R1 R2
120VAC>---[220R]-+-[220R]--+--[36LED>]--+
| | |
| +--[<36LED]--+
[TVS] |
| |
| |
120VAC>----------+----------------------+

Since you'll have 36 LED's in each string they'll be dropping 126V peak,
and since 120VRMS ~ 170VP, the peak voltage at the junction of the
resistors will be 170 - ((170-126)/2) = 148VP with nominal (120V) mains.

For high line (132V) the voltage at the junction will be 187 -
((187-126)/2) ~ 163V, so you'll want to get something with a breakdown
voltage higher than 163V at the low end of its breakdown voltage
tolerance.

If you get, say, a 1000V spike across the mains and the TVS is rated for
180V, R1 will be dissipating (1000V-180V)²/220R ~ 3000 watts (YOW!!!!)

That's not as bad as it sounds, though... Since 1 watt is 1 joule per
second, if the spike lasts for 1 millisecond the resistor will only be
as hot (after 1 second) as if was dissipating 3 watts, _but_ it needs to
be something that can handle the pulse. Carbon composition or some sort
of bulk resitor (like carborundum) would work. Also, for a 1000V spike
the current which would be flowing through the TVS would be
(1000V-180V)/220R ~ 3.7A and it would be dissipating 670W during the
time of the pulse, so you'd need to get something that could handle
that.

If you're interested in following this approach but you need isolation
from the mains, you could use a little isolation transformer to do the
job.

http://www.premmagnetics.com/pdf/Page9.pdf

17. ### Watson A.Name \Watt Sun - the Dark Remover\Guest

You must use a resistor for each individual LED. 1.5V / .03 = 50 ohms.
Use the next higher common value, 51 ohms. You can also put two 100
ohm resistors in parallel.

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18. ### Watson A.Name \Watt Sun - the Dark Remover\Guest

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19. ### Watson A.Name \Watt Sun - the Dark Remover\Guest

How'd it change from 30 mA to .035A?

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Don't be ripped off by the big book dealers. Go to the URL
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Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
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20. ### Costas VlachosGuest

Whoa! As if 144 resistors weren't enough!

Costas