Newbie LED question

J

Jill

Guest
I am trying to put together a decorative project that will have four LEDs in
it
(http://www.mouser.com/ProductDetail/Seoul-Semiconductor/LW514/?qs=gLyyx31KZaA5z9f8mONvwA%3d%3d).
The LEDs are 3.2 volts each. I have a "wall wart" transformerless AC adapter
that is 12V.

Can I just connect up the LEDs in parallel and attach them to the PS? Do I
need a dropping resistor (what value)?

Thanks all
 
On Fri, 1 Jul 2011 16:42:07 -0400, "Jill" <N@NE.nothing> wrote:

I am trying to put together a decorative project that will have four LEDs in
it
(http://www.mouser.com/ProductDetail/Seoul-Semiconductor/LW514/?qs=gLyyx31KZaA5z9f8mONvwA%3d%3d).
The LEDs are 3.2 volts each. I have a "wall wart" transformerless AC adapter
that is 12V.

Can I just connect up the LEDs in parallel and attach them to the PS? Do I
need a dropping resistor (what value)?
Click to expand...
The LEDs will want a nominal 20 mA of current and will drop a nominal
3.2V at that drive. You could put two LEDs in series with a 330 ohm
resistor in series with each pair. That would give you about 17 mA
through the LEDs (should be plenty) and you'll drop about 100 mW in the
resistor, so a common 1/4 W 5% 330 ohm resistor (available at Radio
Shack) will be fine.


.------o------------.
| | |
| .-. .-.
o | | 330 | | 330
.-----. | | | |
| | '-' '-'
| + | | |
| | | |
|12 V | | |
| | V -> V ->
| - | - -
| | | |
| | | |
'-----' | |
o V -> V ->
| - -
| | |
'------o------------'
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)


If the LEDs are separated enough that you'll want them wired
individually, then use each LED in series with a 470 ohm resistor and
drive the four sets in parallel from the battery.

--
Rich Webb Norfolk, VA
 
On Fri, 1 Jul 2011 16:42:07 -0400, "Jill" <N@NE.nothing> wrote:

I am trying to put together a decorative project that will have four LEDs in
it
(http://www.mouser.com/ProductDetail/Seoul-Semiconductor/LW514/?qs=gLyyx31KZaA5z9f8mONvwA%3d%3d).
The LEDs are 3.2 volts each. I have a "wall wart" transformerless AC adapter
that is 12V.

Can I just connect up the LEDs in parallel and attach them to the PS?
Click to expand...
---
No.
---

Do I need a dropping resistor (what value)?
Click to expand...
---
Yes; 300 ohms +/- 5%, 1/4 watt.
---

WALL-WART
+-------+
MAINS>--|~ +12|---+-[300]--[LED1>]--[LED2>]--+
| | | |
| | +-[300]--[LED3>]--[LED4>]--+
| | |
MAINS>--|~ 0V|------------------------------+ +-------+

Do you want to know why?

--
JF
 
"John Fields" <jfields@austininstruments.com> wrote in message
news:qocs07d2m2818vjtj0i0l52pgqgr68e4eq@4ax.com...
On Fri, 1 Jul 2011 16:42:07 -0400, "Jill" <N@NE.nothing> wrote:

I am trying to put together a decorative project that will have four LEDs
in
it
(http://www.mouser.com/ProductDetail/Seoul-Semiconductor/LW514/?qs=gLyyx31KZaA5z9f8mONvwA%3d%3d).
The LEDs are 3.2 volts each. I have a "wall wart" transformerless AC
adapter
that is 12V.

Can I just connect up the LEDs in parallel and attach them to the PS?

---
No.
---

Do I need a dropping resistor (what value)?

---
Yes; 300 ohms +/- 5%, 1/4 watt.
---

WALL-WART
+-------+
MAINS>--|~ +12|---+-[300]--[LED1>]--[LED2>]--+
| | | |
| | +-[300]--[LED3>]--[LED4>]--+
| | |
MAINS>--|~ 0V|------------------------------+ +-------+

Do you want to know why?

--
JF
Click to expand...
Thanks for your reply.

Not really sure how to read your diagram. I think it is saying:
+12 to 300ohm to LED1 to LED2 to gnd
+12 to 300ohm to LED3 to LED4 to gnd
(basically group1 in series through resistor to, group2 in parallel through
second resistor)
the vertical lines (2nd, 4th row) are confusing me.

And yes, I would love to know why?

Thanks again.
 
"John Fields" <jfields@austininstruments.com> wrote in message
news:qocs07d2m2818vjtj0i0l52pgqgr68e4eq@4ax.com...
On Fri, 1 Jul 2011 16:42:07 -0400, "Jill" <N@NE.nothing> wrote:

I am trying to put together a decorative project that will have four LEDs
in
it
(http://www.mouser.com/ProductDetail/Seoul-Semiconductor/LW514/?qs=gLyyx31KZaA5z9f8mONvwA%3d%3d).
The LEDs are 3.2 volts each. I have a "wall wart" transformerless AC
adapter
that is 12V.

Can I just connect up the LEDs in parallel and attach them to the PS?

---
No.
---

Do I need a dropping resistor (what value)?

---
Yes; 300 ohms +/- 5%, 1/4 watt.
---

WALL-WART
+-------+
MAINS>--|~ +12|---+-[300]--[LED1>]--[LED2>]--+
| | | |
| | +-[300]--[LED3>]--[LED4>]--+
| | |
MAINS>--|~ 0V|------------------------------+ +-------+

Do you want to know why?

--
JF
Click to expand...
Assuming the transformerless 12 volt adapter is actually 12 volts, then your
solution is good. Common 12 volt transformer types will almost certainly be
under loaded, in that case the actual voltage will be 14 volts plus. That
permits all 4 LEDs in series and a single 100 ohm resistor will likely work
fine.
For the OP, each parallel LED needs one resistor to limit the current. 12
volts would cause too much current for the LED, and it would burn out. So
you add the dropping resistor. Since the LED drops 3.2 volts, each resistor
will drop 8.8 volts. (12 minus 3.2)
Since voltage divided by current equals the desired resistance, then 8.8
divided by 10 milliamps equals 880 ohms. 10 milliamps will give good
brightness without risking the LED. Each LED can be made a different
brightness by varying the resistance to each LED. You could try 1000 ohm
resistors (or higher) to see how changes in resistance change the
brightness. Don't go below 680 ohms if you use one resistor for each LED.
 
"bw" <bwegher@hotmail.com> wrote in message
news:iulgfd$nfu$1@dont-email.me...
"John Fields" <jfields@austininstruments.com> wrote in message
news:qocs07d2m2818vjtj0i0l52pgqgr68e4eq@4ax.com...
On Fri, 1 Jul 2011 16:42:07 -0400, "Jill" <N@NE.nothing> wrote:

I am trying to put together a decorative project that will have four LEDs
in
it
(http://www.mouser.com/ProductDetail/Seoul-Semiconductor/LW514/?qs=gLyyx31KZaA5z9f8mONvwA%3d%3d).
The LEDs are 3.2 volts each. I have a "wall wart" transformerless AC
adapter
that is 12V.

Can I just connect up the LEDs in parallel and attach them to the PS?

---
No.
---

Do I need a dropping resistor (what value)?

---
Yes; 300 ohms +/- 5%, 1/4 watt.
---

WALL-WART
+-------+
MAINS>--|~ +12|---+-[300]--[LED1>]--[LED2>]--+
| | | |
| | +-[300]--[LED3>]--[LED4>]--+
| | |
MAINS>--|~ 0V|------------------------------+ +-------+

Do you want to know why?

--
JF

Assuming the transformerless 12 volt adapter is actually 12 volts, then
your solution is good. Common 12 volt transformer types will almost
certainly be under loaded, in that case the actual voltage will be 14
volts plus. That permits all 4 LEDs in series and a single 100 ohm
resistor will likely work fine.
For the OP, each parallel LED needs one resistor to limit the current. 12
volts would cause too much current for the LED, and it would burn out. So
you add the dropping resistor. Since the LED drops 3.2 volts, each
resistor will drop 8.8 volts. (12 minus 3.2)
Since voltage divided by current equals the desired resistance, then 8.8
divided by 10 milliamps equals 880 ohms. 10 milliamps will give good
brightness without risking the LED. Each LED can be made a different
brightness by varying the resistance to each LED. You could try 1000 ohm
resistors (or higher) to see how changes in resistance change the
brightness. Don't go below 680 ohms if you use one resistor for each LED.
Click to expand...
Thank you
 
Jill wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
news:qocs07d2m2818vjtj0i0l52pgqgr68e4eq@4ax.com...
On Fri, 1 Jul 2011 16:42:07 -0400, "Jill" <N@NE.nothing> wrote:

I am trying to put together a decorative project that will have four LEDs
in
it
(http://www.mouser.com/ProductDetail/Seoul-Semiconductor/LW514/?qs=gLyyx31KZaA5z9f8mONvwA%3d%3d).
The LEDs are 3.2 volts each. I have a "wall wart" transformerless AC
adapter
that is 12V.

Can I just connect up the LEDs in parallel and attach them to the PS?

---
No.
---

Do I need a dropping resistor (what value)?

---
Yes; 300 ohms +/- 5%, 1/4 watt.
---

WALL-WART
+-------+
MAINS>--|~ +12|---+-[300]--[LED1>]--[LED2>]--+
| | | |
| | +-[300]--[LED3>]--[LED4>]--+
| | |
MAINS>--|~ 0V|------------------------------+
+-------+

the vertical lines (2nd, 4th row) are confusing me.
Click to expand...
If you mean the ones directly under the W and the T in "wall wart,"
they're just the sides of the box that the 12V supply is in. All the
rest are wires.
And yes, I would love to know why?

LEDs need current limiting, because their voltage/current curve
Click to expand...
is ridiculously exponential - a very small change in voltage can
cause a dramatic change in current. If the current is regulated
(by, in this case, the 300 ohm resistors), then the voltage will
be whatever it needs to be at that forward current.

The reason he's got two LEDs in series is so that there's less
power dissipated in the series resistors.

Hope This Helps!
Rich
 
On Fri, 1 Jul 2011 17:59:16 -0400, "Jill" <N@NE.nothing> wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
news:qocs07d2m2818vjtj0i0l52pgqgr68e4eq@4ax.com...
On Fri, 1 Jul 2011 16:42:07 -0400, "Jill" <N@NE.nothing> wrote:

I am trying to put together a decorative project that will have four LEDs
in
it
(http://www.mouser.com/ProductDetail/Seoul-Semiconductor/LW514/?qs=gLyyx31KZaA5z9f8mONvwA%3d%3d).
The LEDs are 3.2 volts each. I have a "wall wart" transformerless AC
adapter
that is 12V.

Can I just connect up the LEDs in parallel and attach them to the PS?

---
No.
---

Do I need a dropping resistor (what value)?

---
Yes; 300 ohms +/- 5%, 1/4 watt.
---

WALL-WART
+-------+
MAINS>--|~ +12|---+-[300]--[LED1>]--[LED2>]--+
| | | |
| | +-[300]--[LED3>]--[LED4>]--+
| | |
MAINS>--|~ 0V|------------------------------+ +-------+

Do you want to know why?

--
JF

Thanks for your reply.

Not really sure how to read your diagram. I think it is saying:
+12 to 300ohm to LED1 to LED2 to gnd
+12 to 300ohm to LED3 to LED4 to gnd
(basically group1 in series through resistor to, group2 in parallel through
second resistor)
the vertical lines (2nd, 4th row) are confusing me.

And yes, I would love to know why?

Thanks again.
Click to expand...
---
My pleasure. :)

sorry about the garbled diagram; here it is corrected:

.. WALL-WART
.. +-------+
..MAINS>--|~ +12|---+-[300R]--[LED1>]--[LED2>]--+
.. | | | |
.. | | +-[300R]--[LED3>]--[LED4>]--+
.. | | |
..MAINS>--|~ 0V|-------------------------------+
.. +-------+

The "vertical" lines in the second and fourth row are braces/brackets
and I like to use them to bound components.

The circuit is connected the way you read it, and the cathode ends of
the LEDs are identified by the ">" inside the brackets.

The tech stuff:

LEDs are basically just diodes and, once they're forced into forward
conduction in the region of the knee, a very small change in voltage
across the diode can/will result in a very large change in current
through it.

All LEDs have negative temperature coefficients of resistance, so if
they're driven with a stiff enough voltage source, as they warm up
their resistance will drop, allowing more more current through them,
which will heat them up more and more and allow more and more current
through them, and on and on, until the LED gives up the ghost and
releases the magic smoke.

In order to keep that from happening, what's been done is that the
voltage band across an LED is specified with a certain current through
it, as specified on the LED manufacturer's data sheet.

Knowing the width of that band, we can then approximate a
constant-current source which will keep the LED from committing
suicide.

From the data sheet you provided, the manufacturer hasn't specified a
band but, we do have that, with 20mA through the LED, the voltage
dropped across it will be 3.2 volts so, two in series will drop 6.4
volts.

Hence, if we have a stiff 12V supply and we want it to supply 20mA to
a load which will drop 6.2V with 20mA through it, we must supply the
means to limit the current into the load to 20mA while dropping the
unwanted 6.4V from the supply.

We can do that by subtracting the load's forward voltage from the
supply voltage and then dividing by the current into the load:

Vs - Vl 12V - 6.4V
R = --------- = ------------ = 280 ohms
Il 0.02A

and interposing that resistance between the 12V source and the 2 LEDs
in series load.


300 ohms is a standard 5% value, and will be fine in this application.

--
JF
 
"John Fields" <jfields@austininstruments.com> wrote in message
news:dkhs07tudcakhaf41r7fkgbg1dqt09btec@4ax.com...
On Fri, 1 Jul 2011 17:59:16 -0400, "Jill" <N@NE.nothing> wrote:


"John Fields" <jfields@austininstruments.com> wrote in message
news:qocs07d2m2818vjtj0i0l52pgqgr68e4eq@4ax.com...
On Fri, 1 Jul 2011 16:42:07 -0400, "Jill" <N@NE.nothing> wrote:

I am trying to put together a decorative project that will have four
LEDs
in
it
(http://www.mouser.com/ProductDetail/Seoul-Semiconductor/LW514/?qs=gLyyx31KZaA5z9f8mONvwA%3d%3d).
The LEDs are 3.2 volts each. I have a "wall wart" transformerless AC
adapter
that is 12V.

Can I just connect up the LEDs in parallel and attach them to the PS?

---
No.
---

Do I need a dropping resistor (what value)?

---
Yes; 300 ohms +/- 5%, 1/4 watt.
---

WALL-WART
+-------+
MAINS>--|~ +12|---+-[300]--[LED1>]--[LED2>]--+
| | | |
| | +-[300]--[LED3>]--[LED4>]--+
| | |
MAINS>--|~ 0V|------------------------------+ +-------+

Do you want to know why?

--
JF

Thanks for your reply.

Not really sure how to read your diagram. I think it is saying:
+12 to 300ohm to LED1 to LED2 to gnd
+12 to 300ohm to LED3 to LED4 to gnd
(basically group1 in series through resistor to, group2 in parallel
through
second resistor)
the vertical lines (2nd, 4th row) are confusing me.

And yes, I would love to know why?

Thanks again.

---
My pleasure. :)

sorry about the garbled diagram; here it is corrected:

. WALL-WART
. +-------+
.MAINS>--|~ +12|---+-[300R]--[LED1>]--[LED2>]--+
. | | | |
. | | +-[300R]--[LED3>]--[LED4>]--+
. | | |
.MAINS>--|~ 0V|-------------------------------+
. +-------+

The "vertical" lines in the second and fourth row are braces/brackets
and I like to use them to bound components.

The circuit is connected the way you read it, and the cathode ends of
the LEDs are identified by the ">" inside the brackets.

The tech stuff:

LEDs are basically just diodes and, once they're forced into forward
conduction in the region of the knee, a very small change in voltage
across the diode can/will result in a very large change in current
through it.

All LEDs have negative temperature coefficients of resistance, so if
they're driven with a stiff enough voltage source, as they warm up
their resistance will drop, allowing more more current through them,
which will heat them up more and more and allow more and more current
through them, and on and on, until the LED gives up the ghost and
releases the magic smoke.

In order to keep that from happening, what's been done is that the
voltage band across an LED is specified with a certain current through
it, as specified on the LED manufacturer's data sheet.

Knowing the width of that band, we can then approximate a
constant-current source which will keep the LED from committing
suicide.

From the data sheet you provided, the manufacturer hasn't specified a
band but, we do have that, with 20mA through the LED, the voltage
dropped across it will be 3.2 volts so, two in series will drop 6.4
volts.

Hence, if we have a stiff 12V supply and we want it to supply 20mA to
a load which will drop 6.2V with 20mA through it, we must supply the
means to limit the current into the load to 20mA while dropping the
unwanted 6.4V from the supply.

We can do that by subtracting the load's forward voltage from the
supply voltage and then dividing by the current into the load:

Vs - Vl 12V - 6.4V
R = --------- = ------------ = 280 ohms
Il 0.02A

and interposing that resistance between the 12V source and the 2 LEDs
in series load.


300 ohms is a standard 5% value, and will be fine in this application.

--
JF
Click to expand...
Got it (some anyway)
 
On Sat, 2 Jul 2011 13:30:53 -0400, "Jill" <N@NE.nothing> wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
news:dkhs07tudcakhaf41r7fkgbg1dqt09btec@4ax.com...
On Fri, 1 Jul 2011 17:59:16 -0400, "Jill" <N@NE.nothing> wrote:


"John Fields" <jfields@austininstruments.com> wrote in message
news:qocs07d2m2818vjtj0i0l52pgqgr68e4eq@4ax.com...
On Fri, 1 Jul 2011 16:42:07 -0400, "Jill" <N@NE.nothing> wrote:

I am trying to put together a decorative project that will have four
LEDs
in
it
(http://www.mouser.com/ProductDetail/Seoul-Semiconductor/LW514/?qs=gLyyx31KZaA5z9f8mONvwA%3d%3d).
The LEDs are 3.2 volts each. I have a "wall wart" transformerless AC
adapter
that is 12V.

Can I just connect up the LEDs in parallel and attach them to the PS?

---
No.
---

Do I need a dropping resistor (what value)?

---
Yes; 300 ohms +/- 5%, 1/4 watt.
---

WALL-WART
+-------+
MAINS>--|~ +12|---+-[300]--[LED1>]--[LED2>]--+
| | | |
| | +-[300]--[LED3>]--[LED4>]--+
| | |
MAINS>--|~ 0V|------------------------------+ +-------+

Do you want to know why?

--
JF

Thanks for your reply.

Not really sure how to read your diagram. I think it is saying:
+12 to 300ohm to LED1 to LED2 to gnd
+12 to 300ohm to LED3 to LED4 to gnd
(basically group1 in series through resistor to, group2 in parallel
through
second resistor)
the vertical lines (2nd, 4th row) are confusing me.

And yes, I would love to know why?

Thanks again.

---
My pleasure. :)

sorry about the garbled diagram; here it is corrected:

. WALL-WART
. +-------+
.MAINS>--|~ +12|---+-[300R]--[LED1>]--[LED2>]--+
. | | | |
. | | +-[300R]--[LED3>]--[LED4>]--+
. | | |
.MAINS>--|~ 0V|-------------------------------+
. +-------+

The "vertical" lines in the second and fourth row are braces/brackets
and I like to use them to bound components.

The circuit is connected the way you read it, and the cathode ends of
the LEDs are identified by the ">" inside the brackets.

The tech stuff:

LEDs are basically just diodes and, once they're forced into forward
conduction in the region of the knee, a very small change in voltage
across the diode can/will result in a very large change in current
through it.

All LEDs have negative temperature coefficients of resistance, so if
they're driven with a stiff enough voltage source, as they warm up
their resistance will drop, allowing more more current through them,
which will heat them up more and more and allow more and more current
through them, and on and on, until the LED gives up the ghost and
releases the magic smoke.

In order to keep that from happening, what's been done is that the
voltage band across an LED is specified with a certain current through
it, as specified on the LED manufacturer's data sheet.

Knowing the width of that band, we can then approximate a
constant-current source which will keep the LED from committing
suicide.

From the data sheet you provided, the manufacturer hasn't specified a
band but, we do have that, with 20mA through the LED, the voltage
dropped across it will be 3.2 volts so, two in series will drop 6.4
volts.

Hence, if we have a stiff 12V supply and we want it to supply 20mA to
a load which will drop 6.2V with 20mA through it, we must supply the
means to limit the current into the load to 20mA while dropping the
unwanted 6.4V from the supply.

We can do that by subtracting the load's forward voltage from the
supply voltage and then dividing by the current into the load:

Vs - Vl 12V - 6.4V
R = --------- = ------------ = 280 ohms
Il 0.02A

and interposing that resistance between the 12V source and the 2 LEDs
in series load.


300 ohms is a standard 5% value, and will be fine in this application.

--
JF

Got it (some anyway)
Click to expand...
---
I probably clouded the issue with too much irrelevant data; sorry
about that.

What is it that's not clear to you?

--
JF
 
(created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)
Click to expand...
AACircuit... interesting... Thanks!

--
@~@ You have the right to remain silent.
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