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Using a single resistor with several LEDs

W

wylbur37

Jan 1, 1970
0
When using an LED from a power source whose voltage is higher than the
rating of the LED, a resistor is typically used.
If more than one LED is to be used (let's say three),
each one would have its own respective resistor,
and the schematic would look something like this ...


+----------+-----------+-----------+--------------
| | | |
| | | |
| LED-1 LED-2 LED-3
power | | |
source | | |
| resistor-1 resistor-2 resistor-3
| | | |
| | | |
+----------+-----------+-----------+--------------


But if all the LEDs were of the same voltage and amperage
(e.g., if they were all of the typical 3.6V 20mA),
isn't there a simpler way to do this by using only ONE resistor?
In other words, can't it be done like the following ?


+--------------+-----------+-----------+---
| | | |
| | | |
power LED-1 LED-2 LED-3
source | | |
| | | |
| | | |
+---resistor---+-----------+-----------+---


And what would be the proper value of the resistor?

(I'm guessing it would be one third of what it was in the first diagram
because the required voltage reduction would be the same but the
effective current draw of the 3-LED assembly would be three times what it
was before.
So if the power source were 6V and the LEDs were 3.6V and 20mA each,
then the required resistance would be 120 ohms in the first case
and 40 ohms in the second case. Correct?)
 
M

Mark (UK)

Jan 1, 1970
0
Hi!

Or you could use LEDs with built in resistors?

Yours, Mark.
 
J

Jerry G.

Jan 1, 1970
0
The LED's will not normally be accurate enough to each other. You will most
likely have a situation where one would light normally, and the rest would
be dim, or not lit at all.

You need a separate resistor for each one, or you should look for LED's with
built in resistors. If they have built in ones, then you will be restricted
to the voltage range indicated for the LED's.

--

Jerry G.
==========================


When using an LED from a power source whose voltage is higher than the
rating of the LED, a resistor is typically used.
If more than one LED is to be used (let's say three),
each one would have its own respective resistor,
and the schematic would look something like this ...


+----------+-----------+-----------+--------------
| | | |
| | | |
| LED-1 LED-2 LED-3
power | | |
source | | |
| resistor-1 resistor-2 resistor-3
| | | |
| | | |
+----------+-----------+-----------+--------------


But if all the LEDs were of the same voltage and amperage
(e.g., if they were all of the typical 3.6V 20mA),
isn't there a simpler way to do this by using only ONE resistor?
In other words, can't it be done like the following ?


+--------------+-----------+-----------+---
| | | |
| | | |
power LED-1 LED-2 LED-3
source | | |
| | | |
| | | |
+---resistor---+-----------+-----------+---


And what would be the proper value of the resistor?

(I'm guessing it would be one third of what it was in the first diagram
because the required voltage reduction would be the same but the
effective current draw of the 3-LED assembly would be three times what it
was before.
So if the power source were 6V and the LEDs were 3.6V and 20mA each,
then the required resistance would be 120 ohms in the first case
and 40 ohms in the second case. Correct?)
 
N

Nigel Heather

Jan 1, 1970
0
I agree that there is no guarantee about the internal characteristics of
similar diodes. But then this argument equally holds true for the first
diagram - if they all used 300 ohm resistors you couldn't gurantee that they
all glowed identically, just the same as if you powered all three through a
single 100 ohm resistor.

My answer to the original question is -- Yes that should work and yes the
resitor will need to be a third the value of the individual ones in the
first diagram. Yes the diodes might differ internally but in my opinion the
differences will be small and I doubt that you will notice them. Give it a
try.

Cheers,

Nigel
 
S

Spehro Pefhany

Jan 1, 1970
0
When using an LED from a power source whose voltage is higher than the
rating of the LED, a resistor is typically used.
If more than one LED is to be used (let's say three),
each one would have its own respective resistor,
and the schematic would look something like this ...


+----------+-----------+-----------+--------------
| | | |
| | | |
| LED-1 LED-2 LED-3
power | | |
source | | |
| resistor-1 resistor-2 resistor-3
| | | |
| | | |
+----------+-----------+-----------+--------------


But if all the LEDs were of the same voltage and amperage
(e.g., if they were all of the typical 3.6V 20mA),
isn't there a simpler way to do this by using only ONE resistor?
In other words, can't it be done like the following ?


+--------------+-----------+-----------+---
| | | |
| | | |
power LED-1 LED-2 LED-3
source | | |
| | | |
| | | |
+---resistor---+-----------+-----------+---

Yes, generally it works okay for visual brightness and if you are not
running very close to the maximum current for the LED type and maximum
operating temperature. There's enough internal resistance in most LEDs
and the eye's log response is such that the current matching is "close
enough" for many purposes.
And what would be the proper value of the resistor?

(I'm guessing it would be one third of what it was in the first diagram
because the required voltage reduction would be the same but the
effective current draw of the 3-LED assembly would be three times what it
was before.

You guess correctly.
So if the power source were 6V and the LEDs were 3.6V and 20mA each,
then the required resistance would be 120 ohms in the first case
and 40 ohms in the second case. Correct?)

Best regards,
Spehro Pefhany
 
M

Michael A. Covington

Jan 1, 1970
0
Put the LEDs in series, not parallel. That is:

+ -----/\/\/\/-----LED-----LED----LED----- -

They will all carry the same current but need not have the same voltage
(e.g., you can have a 3.2-volt blue LED, a 2.1-volt green, and a 1.8-volt
red, all in series).

The power supply must be higher than the sum of the LED voltages, with the
resistor calculated to make up the difference.

Example: With the 3 LEDs I just mentioned, 3.2 + 2.1 + 1.8 = 7.1 volts.

Suppose the supply is 12 volts and you want 20 mA.

Then the resistor is (12 - 7.1)/0.020 = 245 ohms (use 270, which is a
standard value).

As someone pointed out, if you put the LEDs in parallel, they will not share
current equally unless their voltages are *exactly* equal, which depends on
temperature, manufacturing lot, etc., even if they're all the same color.
 
M

Michael A. Covington

Jan 1, 1970
0
Nigel Heather said:
I agree that there is no guarantee about the internal characteristics of
similar diodes. But then this argument equally holds true for the first
diagram - if they all used 300 ohm resistors you couldn't gurantee that they
all glowed identically, just the same as if you powered all three through a
single 100 ohm resistor.

My answer to the original question is -- Yes that should work and yes the
resitor will need to be a third the value of the individual ones in the
first diagram. Yes the diodes might differ internally but in my opinion the
differences will be small and I doubt that you will notice them. Give it a
try.

Yes; contrary to what I just said, they probably *do* have enough internal
resistance to even things out if you make a good faith attempt to use the
same kind. Obviously one red and one green will not do... even one bright
red and one dark red might not do.
 
S

si

Jan 1, 1970
0
There's no guarantee the LEDs will share the current equally due to internal
differences (even between LEDs of a similar batch). You will at best get one
or two at different brightness, at worse one or more LEDs will be totally
off. Hence the preference to use the seperate resistors.

Si.
 
W

Watson A.Name - \Watt Sun, the Dark Remover\

Jan 1, 1970
0
Mark (UK) said:
Hi!

Or you could use LEDs with built in resistors?

Yours, Mark.

si wrote:

Most LEDs don't come with a built-in resistor. Ones that do are more
expensive.

The LED driver chip makers get around this problem by connecting up to
maybe 6 LEDs in series. All the LEDs then get exactly the same current.

Also, if you are parelleling 3 or more LEDs to get more light, then
instead, use a single more powerful LED such as the Luxeon Star. There
are also other brighter LEDs such as the 'spider' LEDs.

With blue or white LEDs, if you do parallel LEDs without a separate
resistor for each, be prepared to have one or more LEDs that overheat
with a greatly shortened lifetime and/or much dimmer light output.
Also, one or more LEDs may overheat and its voltage drop will go up,
which then causes that LED to get dim, and the other LEDs will then end
up with more current. They then may overheat,, etc.

Murphy's Law always applies. :p
 
W

Watson A.Name - \Watt Sun, the Dark Remover\

Jan 1, 1970
0
Michael A. Covington said:
that through opinion Give it
a

Yes; contrary to what I just said, they probably *do* have enough internal
resistance to even things out if you make a good faith attempt to use the
same kind. Obviously one red and one green will not do... even one bright
red and one dark red might not do.

One factor needs further explanation. LEDs have a negative temperature
coefficient. So if one LED gets a bit more current, it gets hotter,
which then causes its voltage drop to be lower, which then causes it to
hog even more current. This vicious circle of current hogging usually
causes the LED to overheat and have a much shorter lifetime. This may
take hundreds or thousands of hours, but the closer the LEDs are run to
their maximum, the less time it will take.

One way to reduce this tendency is to have all the LEDs closely coupled
thermally. But then this usually means they will get hotter, because
there's a lot of heat in a small space. I put a dozen white LEDs on a
predrilled 0.1" hole spacing PCB spaced every other hole, and they got
too hot. So I spaced them out every third hole and they stayed cooler.
(View with Courier font)

Originally: 0 . 0 . 0 . 0 . 0 . 0
0 . 0 . 0 . 0 . 0 . 0

Much Better: 0 .. 0 .. 0 .. 0 .. 0 .. 0
0 .. 0 .. 0 .. 0 .. 0 .. 0

I've been leaving these cheap white LEDs from Hong Kong on 24/7 at their
maximum current, and they just get real dim after a few months and a
thousand or so hours. Just remember that whatever you do, heat is the
enemy of LEDs, and the hotter they get, the shorter they will live.
It's just a law of physics.
 
W

Watson A.Name - \Watt Sun, the Dark Remover\

Jan 1, 1970
0
Nigel Heather said:
I agree that there is no guarantee about the internal characteristics of
similar diodes. But then this argument equally holds true for the first
diagram - if they all used 300 ohm resistors you couldn't gurantee that they
all glowed identically, just the same as if you powered all three through a
single 100 ohm resistor.

My answer to the original question is -- Yes that should work and yes the
resitor will need to be a third the value of the individual ones in the
first diagram. Yes the diodes might differ internally but in my opinion the
differences will be small and I doubt that you will notice them. Give it a
try.

But see my other followup about the negative temperature coefficient
problem.
 
S

si

Jan 1, 1970
0
With different resistors LED characteristic differences are vastly swamped
by the series resistance. Under those circumstances the differences in
brightness will be minute and most probably not noticeable.
I only suggested NO GUARANTEE that the LEDs would share equally in direct
parallel it is not impossible, but I have found this so unlikely as to be
not worth what is saved in using the extra resistors! As you quite rightly
suggest, a R or 1/3rd original value will suffice.

Hey, but what about connecting them in series if you have sufficient supply
voltage? Current through the lot is the same and that only takes ONE
resistor too!

Si.
 
S

Sam Goldwasser

Jan 1, 1970
0
Or if the supply voltage is higher than the sum of the LED voltage drops AND
they have similar current and light output ratings, put the LEDs in series
with a single resistor.

With the example you gave though, there is no choice but to have individual
resistors for each LED.

--- sam | Sci.Electronics.Repair FAQ Home Page: http://www.repairfaq.org/
Repair | Main Table of Contents: http://www.repairfaq.org/REPAIR/
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Important: Anything sent to the email address in the message header is ignored.
To contact me, please use the feedback form on the S.E.R FAQ Web sites.
 
C

CWatters

Jan 1, 1970
0
wylbur37 said:
When using an LED from a power source whose voltage is higher than the
rating of the LED, a resistor is typically used.

.... and if the voltage is the same as the "rating of the LED" you have a
slight problem (eg you can't connect them without a resistor).
If more than one LED is to be used (let's say three),
each one would have its own respective resistor,
and the schematic would look something like this ...


+----------+-----------+-----------+--------------
| | | |
| | | |
| LED-1 LED-2 LED-3
power | | |
source | | |
| resistor-1 resistor-2 resistor-3
| | | |
| | | |
+----------+-----------+-----------+--------------


But if all the LEDs were of the same voltage and amperage
(e.g., if they were all of the typical 3.6V 20mA),
isn't there a simpler way to do this by using only ONE resistor?

Well if the power supply voltage is high enough you can connect the LED in
series and use one resistor in series with the lot.

In other words, can't it be done like the following ?


+--------------+-----------+-----------+---
| | | |
| | | |
power LED-1 LED-2 LED-3
source | | |
| | | |
| | | |
+---resistor---+-----------+-----------+---

Not a good idea because if the LED are not matched exactly you get a big
difference in brightness. They may also age differently - so it may start
out ok but in 3 years time one might be brighter than another.

Think about what happens if one LED is 3.59 and the other 3.61V
So if the power source were 6V and the LEDs were 3.6V and 20mA each,
then the required resistance would be 120 ohms in the first case
and 40 ohms in the second case. Correct?)


Colin
 
J

Jim Adney

Jan 1, 1970
0
In other words, can't it be done like the following ?


+--------------+-----------+-----------+---
| | | |
| | | |
power LED-1 LED-2 LED-3
source | | |
| | | |
| | | |
+---resistor---+-----------+-----------+---


And what would be the proper value of the resistor?

(I'm guessing it would be one third of what it was in the first diagram
because the required voltage reduction would be the same but the
effective current draw of the 3-LED assembly would be three times what it
was before.

If this were a perfect world and the 3 LEDs were exactly identical,
then this would work, but LEDs are never exactly alike and one of them
would hog the current, leaving it too brite and the others too dim.

-
 
D

Dan Fraser

Jan 1, 1970
0
To use a single resistor with several LEDS, put the LEDs in series. However,
unless the LEDs are the same color and from the same batch, the brightness
will vary. Also, the supply voltage has to be high enough to actually light
the series string.
 
J

Jan-Erik Söderholm

Jan 1, 1970
0
First, no LED has a "voltage rating". The have a "current rating" and
a *typical* forward voltage drop (at the rated current).

Second, *some* current limiting method is more or less always used.
Either passive (resistor) or active (like in the white LED
drivers) with current sensing feedback into the IC.

It's very hard to find a *voltage source* that has exactly
the correct voltage. And if the LED heats a little, the
neg tempco will lower the forward voltage drop of the LED,
and you'll end up with a current rush and a dead LED...

Jan-Erik.
 
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