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.
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
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.
ecerfoglio said:You are using your math with numbers without units (I know, your calculator has only numbers)
The voltage is in Volts but the led's current is in miliAmps
The resistor for one led should be
10 V / 30 mA = 10 V / (30 / 1000) A = 333hm:
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 ==> 433hm: (This 433 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 = 11.6hm:
Power = (600mA)^2 * 11.6hm: = 4.2W in the resistor. Use at least a 10 W resistor.
Of course, you will not find a 11.6hm: resistor. You may use a 10 hm: 10W resistor in series with a 1.5 hm: 2W one.
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 . . .
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
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 = 0.58hm:
The resistor would consume 11.2V*19.2A = 215.04W of power.
So you're looking for a 0.58hm: 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). 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.
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