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LED Power Problem

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69428scj

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Hi guys. I have a problem I was hoping you can help me with

I have built some tail lights for my car using 960 RED 5mm 6000mcd LED's per side (so a total of 1920) which all need to run at the same time for use as a taillight (with a lower output) and then full power for brake lights.

The problem I have is that I have installed the LED's on Strip board (long boring job) and now having problems finding a way of powering them. The use of a resistor is not great as (1) the amount of LED's running on one resistor drops the light output too low and (2) the resistor gets damn hot very quickly.

I would like to find a safe way of dropping the 12v-13.8v of the car battery down to a suitable high powered but 3.8-5v power for the LED's to run. I can use relays to switch them (no problems there) but need advice on a PSU for the lights.

Any ideas would be greatly appreciated.

Mick . . .
 
In a moment of clarity I just thought that I don't need a PSU, I just need a high power resistor of the right value.

What "ohms" resistor will I need and where do I get such a thing in the UK?

Mick . . .
 
im not the best when it comes to maths, but i would assume that you find the value of one resistor, say 383ohms resistance per led, and times that by how many leds you have.
 
hi,

Rseries = (Vsupply -Vled)/ Iled = ohms

Vsupply is the battery voltage
Vled is forward voltage drop across the LEDs, when it working
If you have two LED's in series then Vled is *2, if three leds it *3... etc

Iled is the current that you want thru the LED.

You get the last to variables from the LED datasheet.

Does this explain it ?
 
Last edited:
I must be lossing it!

Something seems a little off with my maths somewhere:

This is what I have:

Supply voltage = 12v
Forward Voltage = 1.8 - 2.2 (so we'll use 2)
Number of LED's = 1940 in parallel - none in serial so value = 1
Sum of above two lines = 2

Voltage - value = 10

ILED = 30

10/30 = 0.33333333333

I have the resistors needed to light ONE or a few LED's fine - but with the drain of 1940 LED's they are all a little dim and the resistor get damn hot. I have tried running several resistors in parallel but although the LED's do get a little brighter the resistors get even hotter over just a few seconds.

So if you have a load which is a single load comprising of 1940 5mm Red LEDS with typical values as below, what resistor and Watt rating would you use?

Emitted Colour : RED
Size (mm) : 5mm
Lens Colour : Water Clear
Peak Wave Length (nm) : 620~630
Forward Voltage (V) : 1.8 ~ 2.2
Reverse Current (uA) : <=30
Luminous Intensity Typ Iv (mcd) : Average in 5000
Life Rating : 100,000 Hours
Viewing Angle : 20 ~ 25 Degree
Absolute Maximum Ratings (Ta=25°C)
Max Power Dissipation : 80mw
Max Continuous Forward Current : 30mA
Max Peak Forward Current : 75mA
Reverse Voltage : 5~6V
 
You can use a switch mode regulator to regulate the current, with little dissipation and choose between 2 different currents. One lower other higher. You should choose a stepup regulator and connect all leds in series, this way they will have equal brightness.
 
Helder Ferreira said:
You can use a switch mode regulator to regulate the current, with little dissipation and choose between 2 different currents. One lower other higher. You should choose a stepup regulator and connect all leds in series, this way they will have equal brightness.

The problem is that I have 6 boards all wired in parallel - so basically I have 6 boards (3 for each side of the car) with 8 rows of 20 LEDS on strip board. I can rewire them so I have each row of 20 LED's fed individually with no major problems, but removing them all and wiring them in serial is a little more than I wish to do.
 
Its not advisable to connect leds directly in parallel, because of slight differences in them. One will glow different from another. To connect them in series with an individual resistor even of a very small value to "equalize" them. You can test this by connecting 2 different color leds in parallel and tri to light them with one resistor. About the 8 boards, if you could connect the 20 leds with a 10ohm smd resistor each, you have 8 boards with 20 led/resistors in parallel, now you connect the 8 boards in seris and use the step-up regulator network.
 
You can use this circuit, Changing R2 will cange the current Leds.JPG
 
69428scj said:
Something seems a little off with my maths somewhere:

This is what I have:

Supply voltage = 12v
Forward Voltage = 1.8 - 2.2 (so we'll use 2)
Number of LED's = 1940 in parallel - none in serial so value = 1
Sum of above two lines = 2

Voltage - value = 10

ILED = 30

10/30 = 0.33333333333

I have the resistors needed to light ONE or a few LED's fine - but with the drain of 1940 LED's they are all a little dim and the resistor get damn hot. I have tried running several resistors in parallel but although the LED's do get a little brighter the resistors get even hotter over just a few seconds.

So if you have a load which is a single load comprising of 1940 5mm Red LEDS with typical values as below, what resistor and Watt rating would you use?

Emitted Colour : RED
Size (mm) : 5mm
Lens Colour : Water Clear
Peak Wave Length (nm) : 620~630
Forward Voltage (V) : 1.8 ~ 2.2
Reverse Current (uA) : <=30
Luminous Intensity Typ Iv (mcd) : Average in 5000
Life Rating : 100,000 Hours
Viewing Angle : 20 ~ 25 Degree
Absolute Maximum Ratings (Ta=25°C)
Max Power Dissipation : 80mw
Max Continuous Forward Current : 30mA
Max Peak Forward Current : 75mA
Reverse Voltage : 5~6V

You are using your math with numbers without units (I know, your calculator has only numbers :D :D )

The voltage is in Volts but the led's current is in miliAmps:eek:

The resistor for one led should be

10 V / 30 mA = 10 V / (30 / 1000) A = 333 :eek:hm:

But 30 mA is the Max Continuous Forward Current, in a running car the voltage will be something like 13.5 to 14.5 V and you should add some safety margin - say calculate for a 15 V supply ==> 13 V across R ==> 433:eek:hm: (This 433 :eek:hm: resistor will dissipate 390 mW)


The problem is that I have 6 boards all wired in parallel - so basically I have 6 boards (3 for each side of the car) with 8 rows of 20 LEDS on strip board. I can rewire them so I have each row of 20 LED's fed individually with no major problems, but removing them all and wiring them in serial is a little more than I wish to do.

Try wiring the 4 (20 led) strips in series (which gives you a voltage drop of 8 V in the leds). The resistor for each series should be:

:delta: V = 15 - 8 = 7 V
I = 20 * 30 mA = 600 mA
R = 7 V / 600 mA = 11.6:eek:hm:
Power = (600mA)^2 * 11.6:eek:hm: = 4.2W in the resistor. Use at least a 10 W resistor.

Of course, you will not find a 11.6:eek:hm: resistor. You may use a 10:eek:hm: 10W resistor in series with a 1.5:eek:hm: 2W one.
 
ecerfoglio said:
You are using your math with numbers without units (I know, your calculator has only numbers :D :D )

The voltage is in Volts but the led's current is in miliAmps:eek:

The resistor for one led should be

10 V / 30 mA = 10 V / (30 / 1000) A = 333 :eek:hm:

But 30 mA is the Max Continuous Forward Current, in a running car the voltage will be something like 13.5 to 14.5 V and you should add some safety margin - say calculate for a 15 V supply ==> 13 V across R ==> 433:eek:hm: (This 433 :eek:hm: resistor will dissipate 390 mW)




Try wiring the 4 (20 led) strips in series (which gives you a voltage drop of 8 V in the leds). The resistor for each series should be:

:delta: V = 15 - 8 = 7 V
I = 20 * 30 mA = 600 mA
R = 7 V / 600 mA = 11.6:eek:hm:
Power = (600mA)^2 * 11.6:eek:hm: = 4.2W in the resistor. Use at least a 10 W resistor.

Of course, you will not find a 11.6:eek:hm: resistor. You may use a 10:eek:hm: 10W resistor in series with a 1.5:eek:hm: 2W one.

OK, I will rewire the boards to run a group of 20 LED's in series with each other - i.e. lots of groups of 20 LED's daisy chained together.

So this will mean that I have 48 groups of 20 LED's - so what resistor should be recommended for this application/format? I think it may be best to run each board (8 strips of 20) individually with a resistor per board. Does this make sensible logic?

Oh and by the way - thanks for the help so far guys. It is very much appreciated.

Mick . . .
 
hi

OK, I will rewire the boards to run a group of 20 LED's in series with each other - i.e. lots of groups of 20 LED's daisy chained together.

If the LED forward voltage is say 2V and you have a 12V supply, how can you connect 20 in series * 2V = 40V across a 12V supply and expect to work?

The maximum number in series across 12v for 2V LED's is 6, take one away to allow a series resistor 5 * 2 =10 at say 20mA

Rs= (12-10)/.02 = 100R
At 30mA
Rs= (12-10)/0.03 = 67R
 
Last edited:
69428scj said:
OK, I will rewire the boards to run a group of 20 LED's in series with each other - i.e. lots of groups of 20 LED's daisy chained together.

So this will mean that I have 48 groups of 20 LED's - so what resistor should be recommended for this application/format? I think it may be best to run each board (8 strips of 20) individually with a resistor per board. Does this make sensible logic?

Oh and by the way - thanks for the help so far guys. It is very much appreciated.

Mick . . .

With 48 groups in series you will need about 96 V :eek: which is more that the battery's 12 V :D

Each board of 8 strips, in series, will need about 16 V - still more than the 12 V hou have.

I recomend using series of 4 strips, making 2 series for each board (that is 6 series for each side of the car).

Each series should have its own 11.6 :eek:hm: 10 W resistor.

Be shure to allow for plenty of air flow arround the leds and resistors. each "lamp" will be dissipating up to 15 V x 600 mA x 6 = 54 W and can get very hot if enclosed.

As about half of this power will be in the leds ant the other half in the resistors. I would put the resistors outside the lamps, and if possible outside of the car´s trunk - Perhaps under the car, using plenty of heat shrink tubing to isolate every connection.
 
ericgibbs said:
hi

OK, I will rewire the boards to run a group of 20 LED's in series with each other - i.e. lots of groups of 20 LED's daisy chained together.

If the LED forward voltage is say 2V and you have a 12V supply, how can you connect 20 in series * 2V = 40V across a 12V supply and expect to work?

The maximum number in series across 12v for 2V LED's is 6, take one away to allow a series resistor 5 * 2 =10 at say 20mA

Rs= (12-10)/.02 = 100R
At 30mA
Rs= (12-10)/0.03 = 67R

Point taken - so I can run as many as needed in parellel but only up to 5/6 in serial?
 
I've tried to read through the posts and decipher the situation as best I can. As I understand it, and please correct me if I'm wrong, you have connected 960 LEDs in parallel to form a light cluster unit. You then hoped to use a single resistor to limit the current to all of the LEDs and you wanted to calculate the value of this resistor.

If we assume the LEDs take 20mA each when fully lit, then 960 of them in parallel will pass 19.2 Amps. The forward voltage drop across them is assumed to be 2V, and if we also assume the voltage supplied by the car is 13.2V (when engine running) then the resistor value would be;

11.2V / 19.2A = 0.58:eek:hm:

The resistor would consume 11.2V*19.2A = 215.04W of power.

So you're looking for a 0.58:eek:hm: resistor with a power dissipation capability of at least 215.04W. The picture I'm trying to paint for you here, is that this is not the best way to go about solving this problem. There are other issues with driving the LEDs this way as well, but I won't go into those because I think the resistor choice spells it out as a bad idea anyway!

If I were faced with this kind of engineering problem, I think I would want to power the LEDs with a pulse-width solution. In that way you can alter the on-time to give the required brightness. It would mean designing a switch-mode power supply to be honest.

Other than that, you could use 1920 current limiting resistors to limit the current to each single LED. That would be two resistors for each, which you would need to be able to switch between (to give you two levels of brightness).
If you try to use a single resistor to limit the current to a large number of LEDs in parallel, you will find that the forward voltage drop across each of them will be slightly different. Therefore, some will consume more current than others. If one of them fails, then the remaining LEDs will take on the extra current and the same situation will apply. Eventually another LED will give out, and then another and another until finally the problem has cascaded through ALL your LEDs and you'll have to replace the lot. Don't do it this way, it's not worth it.

Brian
 
I must admit 920 leds per lamp 6000mcd each has he says is a lot of brightness. (poor guy will be the one who's driving behind him :p ). My Toyota Prius has from stock, 7 leds per side with a little reflector and they switch on so fast and bright that it's impossible to ignore. Anyway if he uses the step-up regulator (a kind of PWM) solution he will cut down the dissipation in the resistors. The problem (I think) is that the led boards are ready and it's not easy to rebuild.


ThermalRunaway said:
I've tried to read through the posts and decipher the situation as best I can. As I understand it, and please correct me if I'm wrong, you have connected 960 LEDs in parallel to form a light cluster unit. You then hoped to use a single resistor to limit the current to all of the LEDs and you wanted to calculate the value of this resistor.

If we assume the LEDs take 20mA each when fully lit, then 960 of them in parallel will pass 19.2 Amps. The forward voltage drop across them is assumed to be 2V, and if we also assume the voltage supplied by the car is 13.2V (when engine running) then the resistor value would be;

11.2V / 19.2A = 0.58:eek:hm:

The resistor would consume 11.2V*19.2A = 215.04W of power.

So you're looking for a 0.58:eek:hm: resistor with a power dissipation capability of at least 215.04W. The picture I'm trying to paint for you here, is that this is not the best way to go about solving this problem. There are other issues with driving the LEDs this way as well, but I won't go into those because I think the resistor choice spells it out as a bad idea anyway!

If I were faced with this kind of engineering problem, I think I would want to power the LEDs with a pulse-width solution. In that way you can alter the on-time to give the required brightness. It would mean designing a switch-mode power supply to be honest.

Other than that, you could use 1920 current limiting resistors to limit the current to each single LED. That would be two resistors for each, which you would need to be able to switch between (to give you two levels of brightness).
If you try to use a single resistor to limit the current to a large number of LEDs in parallel, you will find that the forward voltage drop across each of them will be slightly different. Therefore, some will consume more current than others. If one of them fails, then the remaining LEDs will take on the extra current and the same situation will apply. Eventually another LED will give out, and then another and another until finally the problem has cascaded through ALL your LEDs and you'll have to replace the lot. Don't do it this way, it's not worth it.

Brian
 
Helder Ferreira said:
I must admit 920 leds per lamp 6000mcd each has he says is a lot of brightness. (poor guy will be the one who's driving behind him :p ). My Toyota Prius has from stock, 7 leds per side with a little reflector and they switch on so fast and bright that it's impossible to ignore. Anyway if he uses the step-up regulator (a kind of PWM) solution he will cut down the dissipation in the resistors. The problem (I think) is that the led boards are ready and it's not easy to rebuild.

Very true about the bright - but it has no reflectors and has a film infront of the LEDS for legal reasons in the UK so the light outpu so far is actually a little low IMO. I'll see what the final appearance is when I eventually get the boards sorted.

I will draw a little plan of what I have and see what wiring ideas you all have.

Mick . . .
 
I have drawn a very simple layout of what one of the six panels looks like, and what I guess I can do very simply to the links connections to alter the panel to a "20 group" in serial type.

The one on the left is what I have - i.e. 8x20 LED's all in parallel.

On the right is an example of what I could do to change it to a collection of 8 groups of 20 in serial.

I will obviously have to limit this pattern to just 5/6 runs of 20 per 12-15v feed though.

Mick . . .
 

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