Rod
One and only one rule*: design the circuit to limit the current
through the LED to some particular value. The simplest way is
with a 1/2 watt (or larger) series resistor:
+Vcc-----[R]---[LED]---Gnd
Use the formula Resistance = (Vsupply - Vled)/Current
If LEDS are used in series: Vled = Vled(1) + Vled(2) + ... Vled(n)
* = until and unless you have a specific reason or reasons not to.
If you don't have specs on your LEDs, use these numbers for Vled:
Red LED ~1.8 volts; white (or blue, blueish white) LED ~ 3.4 volts.
and limit the current to about 20 mA (or less). (Other color
LEDS will range between these Vled values)
If you have the specs for the LED, you can use them to chose
the current limit, and you'll know Vled for each LED. Otherwise,
20 mA is generally a good value for the limit, and LEDS will light
*well* below 20 mA.
In a car, a nominal voltage of 14 for Vsupply can be used to
compute the size resistor you need. With a red LED, that computes
to 610 ohms, and I would recommend using 680 ohms (a standard value)
for a little extra safety margin. With a white LED, it computes
to 530 ohms. I'd recommend going up to at least 560 ohms, but you
could use the 680 ohm resistor and still get over 15 mA through
the LED. Probably won't see much change in brightness, either.
Do NOT wire LEDS in parallel. For multiple LEDS,
use series like this, and re-compute R:
Vcc---[R]---[LED1]---[LED2]---[LED3]---Ground
In the case above, for red LEDs, it computes to 430 ohms.
A 470 ohm standard resistor would be fine. With white
LEDs, R computes to 190 ohms, and a standard 220 ohm
resistor would be good.
If you want to use a parallel circuit, do this, and
use the 680 ohm resistor mentioned earlier:
Vcc---+---[R]---[LED1]---+---Ground
| |
+---[R]---[LED2]---+
| |
+---[R]---[LED3]---+
~ ~
+---[R]---[LEDn]---+
That's not the end of the story. The electrical "environment"
in a car is hostile. There's all kinds of electrical transients,
and it may be prudent to protect the LED's with a 15 volt TVS
diode from Vcc to ground. In addition, the ambient temperature
may be high, and you may want to limit the current to less
than the ~20 mA (or the specs, if you have them) to
compensate for high ambient temperature.
Ed
Rodney said:
"Connect the LED's in parallel with one another."
ehsjr said:
"Do NOT wire LEDS in parallel."
Right now you might be a little confused, and who can blame you!
Actually, what you want for automotive lighting is a combination,
series-parallel circuit. Take ehsjr's example:
"Vcc---[R]---[LED1]---[LED2]---[LED3]---Ground"
Shake it up a bit:
+12VDC---[R1]---[LED1.1]---[LED1.2]---[LED1.3]---Ground
|----[R2]---[LED2.1]---[LED2.2]---[LED2.3]-----|
|----[R3]---[LED3.1]---[LED3.2]---[LED3.3]-----|
|----[R4]---[LED4.1]---[LED4.2]---[LED4.3]-----|
Realize that the +12VDC is variable between about +11.5 and +14.5,
calculate your resistance based on optimal diode current at max auto
supply voltage (don't forget to use the current *and* voltage specs of
the LED that you choose in your calculations). You can now build the
series-parallel circuit, adding legs to your heart's content. For
this application, the auto's supply current is practically unlimited,
unless you want to try to emulate aircraft landing lights or summat.
Here's a few links that will help.
The first link is nice, as it will show you the circuit schematic.
http://led.linear1.org/led.wiz
Great links, thanks!
JazzMan
--
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The only thing I would add is to make sure that you calculate the power the
resistor will have to dissipate (Current * Current * Resistance) and select a
resistor with a power rating at least that large. Good practice is to use a
resistor with 2X the required rating.
Good luck.
--
James T. White
>It does happen to be in parallel with the other lights. Just had to
>solder a 220 ohm resistor to it.
Right, I think that's the source of the confusion in the discussion.
You are wiring resistor-LED-pair series circuits in parallel with
other devices using the same source, and I think the ehj-someone is
implying that non-current limited LED's should not be paralleled.
In pictures:
+V--[R]--[LED]--Rtn
|---[circuit]---|
and not:
+V--[R]---[LED]--Rtn
|-[LED]---|
|-[LED]---|
which would probably be ok *if* R is chosen for the current used by
the parallel group, and the LED's are closely matched for voltage
drop.
>On Fri, 19 Aug 2005 03:35:28 -0500,
>dako...@aol-dot-com.no-spam.invalid (redls1bird) wrote:
>
>>hello all. im interested in using some leds for automotive lighting
>>and such. i have a BASIC understanding of electronics and
>>electricity. i am familiar with diodes, but not leds. The very
>>small information i have found for auto use talks about using
>>resistors in parallel, but noone really goes into why. i believe its
>>to drop the voltage? but i wouldnt bet my next paycheck on it. can
>>anyone give me some basics on building led circuits? especially how
>>to connect multipes and determine the proper resistors to use?
>>Thanks for the help in advance.
>
>Here's a few links that will help.
>The first link is nice, as it will show you the circuit schematic.
>
>http://led.linear1.org/led.wiz
That one is really quite nifty. Thanks.
>http://www.superbrightleds.com/led_info.htm
>
>http://www.dannyg.com/examples/res2/resistor.htm
---
Since LEDs have a negative tempco, the common resistor is _never_
preferred because even small changes in diode characteristics can
cause one LED to hog current, overheat and, if it fails open, cause
the others to go, one by one.
--
John Fields
Professional Circuit Designer
Rod
No, they are not all in parallel, because LEDs don't play well
together in that configuration, and it would waste a lot of current.
--
Link to my "Computers for disabled Veterans" project website deleted
after threats were telephoned to my church.
Michael A. Terrell
Central Florida
That makes sense only if each LED has it's own separate resistor to set
the current:
| |
|--/\/\/\/--|>|--|
| |
|--/\/\/\/--|>|--|
| |
|--/\/\/\/--|>|--|
| |
|--/\/\/\/--|>|--|
| |
where -/\/\/\/- is a resistor and -|>|- is an LED.
Mark
---
Frayed knot. ;)
1. Each resistor will have to dissipate about 2.4 watts
2. What'll happen when the mains polarity reverses and puts the
cathodes of the LEDs at 170V positive WRT the anodes?
>>and not:
>>
>>+V--[R]---[LED]--Rtn
>> |-[LED]---|
>> |-[LED]---|
>>
>>which would probably be ok *if* R is chosen for the current used by
>>the parallel group, and the LED's are closely matched for voltage
>>drop.
>
>---
>Since LEDs have a negative tempco, the common resistor is _never_
>preferred because even small changes in diode characteristics can
>cause one LED to hog current, overheat and, if it fails open, cause
>the others to go, one by one.
Thanks for the clarification.
> ---
> Frayed knot. ;)
>
> 1. Each resistor will have to dissipate about 2.4 watts
>
> 2. What'll happen when the mains polarity reverses and puts the
> cathodes of the LEDs at 170V positive WRT the anodes?
>
>
I guess I could have added that low-voltage DC is being used, but I
thought that was obvious. :-)
Mark
---
I tend to think of a string of Christmas tree lights as working off
the mains, so to get that low-voltage DC they'd have to have a
power supply of some sort. Maybe a wall-wart? I don't know, I
haven't bought a string of lights in over ten years. Is that how
they do it?
> ---
> I tend to think of a string of Christmas tree lights as working off
> the mains, so to get that low-voltage DC they'd have to have a
> power supply of some sort. Maybe a wall-wart? I don't know, I
> haven't bought a string of lights in over ten years. Is that how
> they do it?
>
> --
> John Fields
> Professional Circuit Designer
I don't know either. I was originally responding to somebody who
said they THINK that Christmas lights are available as LED's wired
in parallel. My point was that IF that is the case, they would
each need a separate series resistor. And of course run off a
low voltage DC source.
When you think about it, the possible savings in efficiency are
likely offset by the inconvenience of using a wall-wart, the
power drain of the resistors, and the fact that they would only
be used for 3 or 4 weeks out of the year. But I honestly don't
know what is available.
Regards,
Mark
>
>John Fields wrote:
>
>> ---
>> I tend to think of a string of Christmas tree lights as working off
>> the mains, so to get that low-voltage DC they'd have to have a
>> power supply of some sort. Maybe a wall-wart? I don't know, I
>> haven't bought a string of lights in over ten years. Is that how
>> they do it?
>>
>> --
>> John Fields
>> Professional Circuit Designer
>
>I don't know either. I was originally responding to somebody who
>said they THINK that Christmas lights are available as LED's wired
>in parallel. My point was that IF that is the case, they would
>each need a separate series resistor. And of course run off a
>low voltage DC source.
I'm pretty sure that is the case. The new ones can have a few bulbs
fail and the string still works.
Tom
What is confusing about "Do NOT wire LEDS in parallel." ?
I am not implying that non-current limited LED's should not
be paralleled, I am stating it, forcefully:
Do NOT wire LEDS in parallel.
>
> In pictures:
>
> +V--[R]--[LED]--Rtn
> |---[circuit]---|
>
> and not:
>
> +V--[R]---[LED]--Rtn
> |-[LED]---|
> |-[LED]---|
>
> which would probably be ok *if* R is chosen for the current used by
> the parallel group, and the LED's are closely matched for voltage
> drop.
>
Wrong. Do NOT wire LEDS in parallel.
Ed
>>I don't know either. I was originally responding to somebody who
>>said they THINK that Christmas lights are available as LED's wired
>>in parallel. My point was that IF that is the case, they would
>>each need a separate series resistor. And of course run off a
>>low voltage DC source.
>
> I'm pretty sure that is the case. The new ones can have a few bulbs
> fail and the string still works.
They've invented bulbs that fail with a short circuit for that purpose,
you get a string of series wired christmas lights with 39 fail-short bulbs
and one (regular) fail open "fuse" bulb to stop a cascade failure from
starting a fire.
the fail short bulbs seem to have somethin wrapped around the terminal wires
inside the envelope, I'm guessing it's some sort of wire with poor
insulation that the mains voltage can punch through if the filament fails.
Bye.
Jasen
Find the + and - wires wire that connect to the existing
turn signal bulb, and the + and 1 wires that connect to the
running light bulb. You can connect them directly to the
two identical circuits below that you will build that feed
your led array(s).
From Turn Bulb + ---[R1]---+-----+--- To LED Array TURN+
| |
[TVS] [C1]
| |
From Turn Bulb - ----------+-----+--- To LED Array TURN-
From Running Bulb + ---[R2]---+-----+--- To LED Array DIM+
| |
[TVS] [C2]
| |
From Running Bulb - ----------+-----+--- To LED Array DIM-
LED Array:
D1 D2
From TURN+ ---[Rled]--->|----+---|<---[Rdim]--- + To DIM +
|
[LED1]
|
[LED2]
|
[LED3]
|
[LED4]
|
From TURN- -------------|<---+--->|------------ - To DIM -
D3 D4
C1,C2 = .1 uF 100 V
D1-4 = 1N400x diodes
TVS = 18 volt TVS diode such as P6KE18CALFCT-ND from Digikey
R1 = 10 ohm, 5 watt
Rled = 220 (assuming 2V drop for each LED, and 4 leds in the array)
or compute as Rled = (12.6 - Vledtotal)/.02 where
Vledtotal = Vled1 + Vled2 + Vled3 + Vled4
(The D1, D3 pair, or the D2, D4 pair drop the nominal
voltage - 13.8 - by ~1.2 volts)
Rdim is found experimentally. You use the value of
resistance that dims the LEDS to the brightness you
want.
The purpose of the TVS, R1 and C1 is to reduce transients
that might otherwise damage the LEDs.
Ed