sucksqueezebangblow
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| posted on 6/7/10 at 11:24 AM |
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How to Wire LEDs for 12 Volts?
Hi,
Any electronics whizzes on here that can tell me how to wire LED's together to work on 12 Volts?
I want to make up a rear light from discreet (individual) LEDs. For a single LED you would normally wire it in series with a resistor, but say if I
have LEDs with a forward voltage rating of 2.5 V can I just wire 5 of them in series and put 12 Volts across them?
I want to avoid losses across resistors (and heat) if I can and make the unit as efficient as possible.
Better to Burnout than to Fade Away JET METAL ~ AndySparrow © 
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trextr7monkey
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| posted on 6/7/10 at 11:28 AM |
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Why bother withresistors when Rapid now sell 12v high output LEDs?
atb
mike
http://www.flickr.com/photos/14016102@N00/ (cut and paste this dodgey link)
Our most recent pics are here:
http://s129.photobucket.com/albums/p211/trextr7monkey/
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blakep82
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| posted on 6/7/10 at 11:29 AM |
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if you wire them in series, and 1 blows you lose the lot. not the best way.
in parallel the same voltage is applied to each led, so you should need the same in line resistor you would have for 1 led
________________________
IVA manual link http://www.businesslink.gov.uk/bdotg/action/detail?type=RESOURCES&itemId=1081997083
don't write OT on a new thread title, you're creating the topic, everything you write is very much ON topic!
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hobbsy
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| posted on 6/7/10 at 11:35 AM |
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Also bear in mind you often see up to 14V when the engine is running. I was lazy and used one of the those LED rain lights without a resistor (rated
at 11 ish volts) and it popped a load of them. Had about 4 left working when I replaced it (and fitted a resistor this time!)
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sucksqueezebangblow
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| posted on 6/7/10 at 12:02 PM |
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quote:
Originally posted by hobbsy
Also bear in mind you often see up to 14V when the engine is running. I was lazy and used one of the those LED rain lights without a resistor (rated
at 11 ish volts) and it popped a load of them. Had about 4 left working when I replaced it (and fitted a resistor this time!)
What rating of resistor did you use (Ohms/Watts)
Better to Burnout than to Fade Away JET METAL ~ AndySparrow ©
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sucksqueezebangblow
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| posted on 6/7/10 at 12:03 PM |
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quote:
Originally posted by blakep82
if you wire them in series, and 1 blows you lose the lot. not the best way.
in parallel the same voltage is applied to each led, so you should need the same in line resistor you would have for 1 led
Doesn't the resistor drop a lot of voltage and create a lot of heat?
Better to Burnout than to Fade Away JET METAL ~ AndySparrow ©
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sucksqueezebangblow
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| posted on 6/7/10 at 12:14 PM |
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quote:
Originally posted by trextr7monkey
Why bother withresistors when Rapid now sell 12v high output LEDs?
atb
mike
Hi Mike,
I just checked, They have inbuilt resistors!
Better to Burnout than to Fade Away JET METAL ~ AndySparrow ©
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Madinventions
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| posted on 6/7/10 at 01:00 PM |
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Here's the formula required to work out the resistance value.
First, gather some values:
Vin: 14V (battery voltage max)
Vf: 2.5V (forward voltage of LED)
If: 20mA (0.02A) (how much current you want to put through the LED)
n: Number of LEDs.
Remember ohms law - V/I=R...
R = (Vin-(Vf*n))/If
So, say you want to put 3 LEDs in series, and run from 14V:
R = (14 - (2.5 x 3)) / 0.02A
= (14 - 7.5) / 0.02A
= 6.5 / 0.02A
= 325 ohms.
330ohms would be a suitable value here.
To calculate the power rating of the resistor, use the formula:
P=VxI
The resistor above will have 6.5V across it and will pass 0.02A, so P = 6.5 * 0.02 = 0.13W. A standard 0.4W resistor would be fine.
Hope this helps.
Ed.
Mojo build diary: http://www.madinventions.co.uk
Solo music project: Syrrenfor http://www.reverbnation.com/syrrenfor
View my band website:
http://www.shadowlight.org.uk
http://www.eastangliankitcars.co.uk/
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02GF74
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| posted on 6/7/10 at 01:23 PM |
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LEDs do not need any drop resistor if the current can be precisely controlled. Too small a current and the LED is dim, too much and it goes pop. The
resistor is there to stop the popping.
But power is wasted (as heat) in the resistor which is not acceptable for high power LEDs (emitters) so these use electroinc drivers that supply the
emitter with power; they also work on a wider range of input voltages.
Since we do not have a precise current source in a car but a constant-ish voltage source we need to make do with that.
The manufacturer supplies the volttage/current data so once we have decided on the current through the LED, preferably below the max it can take, we
know the voltage across it. Using that value of voltage we work out the value of the drop resistor to work on the 14 V car electrical system.
Using a single drop resistor per LED is more expensive and wasteful of power but as ^^^^ said, only one LED will blow if t here is a problem in that
circuit. I am not convinced this is a problem since all the LEDs will be wired to the same point so are all likely to blow.
You can connect 2, 3 or 4 LEDs in series and providing there is a couple of volts head room, use one drop resistor.
I have fitted LED lamps to instruments consisting of 4 LEDs in series that work quite happily.
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02GF74
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| posted on 6/7/10 at 01:31 PM |
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back to original question.
quote:
Originally posted by sucksqueezebangblow
Hi,
Any electronics whizzes on here that can tell me how to wire LED's together to work on 12 Volts?
I want to make up a rear light from discreet (individual) LEDs. For a single LED you would normally wire it in series with a resistor, but say if I
have LEDs with a forward voltage rating of 2.5 V can I just wire 5 of them in series and put 12 Volts across them?
I want to avoid losses across resistors (and heat) if I can and make the unit as efficient as possible.
as per my previous response, by fitting no resistor you are at the mercy of the car voltage to either blow the bulbs or for them to be too dim plus
may be subject to fluctuations in brightness depending on where the LEDs are on their operating voltage/current curve.
Wire up 4 LEDs to give head room of 4 volts (14 V - 4 x 2.5 V across LED) then divide by your current, typically 20 mA for 5 mm LED.
Resistor is 200 ohm.
You can go to lower resistor, e.g. 180 ohm if you want brighter LEDs as your alternator output will be closer to 13.5 V plus you can go up to 25
mA.
[Edited on 6/7/10 by 02GF74]
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sucksqueezebangblow
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| posted on 6/7/10 at 02:23 PM |
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Aaahhhh! Thanks guys, this is making more sense.
Just to get it clear in my head, what would happen if I used 5 LEDs on a 12.5V circuit (just for the purposes of illustration)
Would the following calculations be correct;
R=(12.5V-(2.5Vx5))/0.02A
=(12.5-12.5)/0.02
=0/0.02
= 0 Ohms
What is the downside of not using a resistor (if my calculation is correct), Current fluctuation?
Better to Burnout than to Fade Away JET METAL ~ AndySparrow ©
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matt_gsxr
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| posted on 6/7/10 at 03:02 PM |
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the downside of not using a resistor is that there is nothing in the circuit to keep the current down.
V=IR
so
I = V/R
where V = the voltage left after you have taken away the voltage drop across the LEDS.
So,
Bat - 5*2.5V
12.5V - 12.5V = 0V (maybe fine)
12.6 - 12.5 = 0.1V
I = V/R
I = 0.1 / 0 = infinity = blown LEDs
Matt
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02GF74
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| posted on 6/7/10 at 03:36 PM |
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quote:
Originally posted by sucksqueezebangblow
What is the downside of not using a resistor (if my calculation is correct), Current fluctuation?
kinda. the voltage-current for resistor is linear but not for a diode - exponential IIRC
let's say you made my circuit, 4 LEDs and 200 oh resistor.
If car battery voltage is 12 V, current in LEDs wil be 10 mA.
If engine is running, car battery goes up to 14 V so current in LEDs will be 20 mA. (actually it will be less due to the voltage across the LED
increasing so less volts across the resistor so lower current in LEDs).
Worst case it has doubled.
Below is voltage/current gaph for a diode; let's just look at the green line.
Lets say the voltage across a LED is 3 V when connected to 12 V so current is about 6 mA.
Now engine runs so voltage across LED goes to 3.5 V, the current is well off the graph - guessing around 50 mA so has gone up by 9 times due to the
non linear relationship.
Result is that LEDs pops and no lights 
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trextr7monkey
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| posted on 6/7/10 at 03:41 PM |
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quote:
Originally posted by sucksqueezebangblow
quote:
Originally posted by trextr7monkey
Why bother withresistors when Rapid now sell 12v high output LEDs?
atb
mike
Hi Mike,
I just checked, They have inbuilt resistors!
If you are wanting more resistors  give us a a shout- loads here.
atb
Mike
http://www.flickr.com/photos/14016102@N00/ (cut and paste this dodgey link)
Our most recent pics are here:
http://s129.photobucket.com/albums/p211/trextr7monkey/
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sucksqueezebangblow
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| posted on 6/7/10 at 03:53 PM |
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Fantastic, this is really good stuff.
Final question. How do I calculate what the minimum head room voltage should be, or what the minimum resistance I should use is. For example (for the
LEDs I have bought)
(13.3-(2.1x6))/0.03=30 Ohm with 0.9V headroom
or
(13.3-(2.1x5))/0.03=100 Ohm with 2.1 V headroom
Can I get away with 30 Ohm or are the LEDs still likely to pop?
Better to Burnout than to Fade Away JET METAL ~ AndySparrow ©
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02GF74
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| posted on 6/7/10 at 04:12 PM |
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really need to see the voltage/current graph for the LEDs you are using.
If you are using 5 mm white LEDs, it will not be disimilar to that shown on page
6here
let's say you are going for 20 mA, then Vf is 3.5 V.
You cannot connect 4 LEDs in series since for that current, the voltage drop across them will be 14 V leaving no head room at all (chances are you
alternator puts out 13.5 V anyway - I use 14 V to be on the safe side).
You select the numebr of LEDs to be less than your minimum voltage, 12 V so that is 3 in series = 10.5 V.
max voltage is 14 V so head room is 3.5.
divide this by 20 mA gives 175 ohm
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sucksqueezebangblow
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| posted on 6/7/10 at 04:56 PM |
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This is the bit I don't quite follow (I'm not very bright when it comes to wiggly amps). If the resistance, currents and voltages are
calculated at the max possible voltage (14V) and if the voltage drops, it appears to me that the currents will drop also, and each will drop by a
similar factor. So the voltage across the resistor will drop and so will the voltages across the LEDs. So why the need for headroom (other that enough
to warrant a resistor to control the current). Sorry if I'm being thick!
Better to Burnout than to Fade Away JET METAL ~ AndySparrow ©
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germanguy
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| posted on 6/7/10 at 07:40 PM |
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sorry to stick my 5 pence worth in but
don`t forget you are using a motorbike alternatow which runs through through a voltage reg
you will find that your charging voltage differs from your engine rpm and load ie with driving lights on you might only get 12.8 volts but without
the lights u`ll get 14.4 volts with a resistor in the cirquit you will be less likly to pop the led`s
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JF
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| posted on 6/7/10 at 09:20 PM |
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If you want to go really secure and same brightness whatever the input voltage is. Set up a regulated power supply, either still putting the leds in
series. Or get one with the right voltage out that the leds you want to use require.
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02GF74
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| posted on 7/7/10 at 07:08 AM |
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quote:
Originally posted by sucksqueezebangblow
If the resistance, currents and voltages are calculated at the max possible voltage (14V) and if the voltage drops, it appears to me that the currents
will drop also, and each will drop by a similar factor.
Not strictly true since the voltage/current relationship for a LED is not linear - see graph above..... but we'll continue
.
So the voltage across the resistor will drop and so will the voltages across the LEDs. So why the need for headroom (other that enough to warrant a
resistor to control the current). Sorry if I'm being thick!
The headroom isn't required if you can control the current precisely.
Again referring to the graph. (from memory, you have voltage on the horizontal axis and current on the vertical axis)
If say you decided run the LEDs at 20 mA, look at vertical axis, find 20 mA and move right until you hit the line, then go down to see the voltage
across the LED. If you have a power supply that can give that voltage and not other under all conditions, then you don't need any
resistor.
What you have is a variable voltage supply ranging from 12 to 14 ish volts. If you did the same for the voltage - read horizontal axis first then go
up, hit the line and go left to read the current, the range of amps for the 2 voltage above will mean either that the current is too big = dead LED or
too low = dim LED.
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sucksqueezebangblow
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| posted on 7/7/10 at 09:06 AM |
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I think I follow; the resistor acts like a crude sort of voltage regulator, reducung (but not eliminating) the fluctuation in voltage as the
battery/alternator varies between 12 and 14 volts?
I can get a 12V fixed voltage regulator for about 70p so I'm thinking (from all the good advice you guys have given me) that I should use 5 x
2.1V LEDS, a 12 V regulator and a resistor to drop the remaining 1.5V. Does that sound like the optimum set up? Or would it be better to go for a 9V
or 10V regulator?
Better to Burnout than to Fade Away JET METAL ~ AndySparrow ©
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Tipster69
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| posted on 7/7/10 at 10:39 AM |
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http://www.reuk.co.uk/Zener-Diode-Voltage-Regulator.htm
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02GF74
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| posted on 7/7/10 at 11:09 AM |
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quote:
Originally posted by sucksqueezebangblow
I think I follow; the resistor acts like a crude sort of voltage regulator, reducung (but not eliminating) the fluctuation in voltage as the
battery/alternator varies between 12 and 14 volts?
hmmm not really. It protect the LEDs from getting too much current by stealing the voltage.
Consider this smple circuit, one LED and one resistor connected to 5 V.
Let's say when current in LED is 10 mA, the volts across it is 2.5 V. You know what value resistor you need as you are already expert in
calculating them (it is 250 ohm).
So there is 10 mA flowing through both the LED and resistor and the voltage is the same scross them: 2.5 V.
Now let's say the voltage supplyincreases by 0.4 V and assume that it is shared equally by the LED and resistor.
You end up with 2.7 V across the LED. but since the current-voltage relation for LED is non linear - a small change in volts results in big change
in current, the current now is 20 mA.
This will cause a much bigger drop in voltage across the 250 ohm resistor (would be 5 V)
This situation cannot happen since the voltage across both components would exceed the supply voltage.
What happens is that increase in voltage across the resistor "steals" voltage from the LED so the voltage across he LED is less, which in
turn means less current flows through it.
[Edited on 7/7/10 by 02GF74]
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JF
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| posted on 7/7/10 at 01:56 PM |
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Well your voltage is lower then 12V when the engine is off. Certainly with lights on.
I think I'd opt for a 10,5V regulated supply using 4 led series with a resistor to take up the last 2,1V.
Although if your supply is accurate and stable you could just as well use series of 5 leds. But I'd run some tests with the supply to check it
is accurate and stable accross variable loads and voltage inputs.
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sucksqueezebangblow
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| posted on 7/7/10 at 04:18 PM |
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Excellent, thanks guys, I think I got it in the end! Time to get the breadboard out (if that is what they still call them these days) and do some
experimenting! I think I might invest in a variable voltage bench power supply as well to test the circuits thoroughly.
Better to Burnout than to Fade Away JET METAL ~ AndySparrow ©
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