How do I connect 1watt or 3 watt LED

Thread Starter

andyqd

Joined Jul 4, 2008
1
How do I connect 1 watt or 3 watt LED or higher on a 14v source (in my car).....I know that you can connect a low power LED(5 mm) using a resistor....but a high power LED like 1watt onward....you need a driver for it...is that necessary? why cant i just connect a Resistor by itself? ....how can i design a driver that can handle....LED @ 200mA, Vf: 3.6 - 4.2....power source: 12-14 Volt
 

beenthere

Joined Apr 20, 2004
15,819
If you want to use a fixed resistor, figure it like you would for any LED - the voltage driving current is what's left after you subtract the LED Vf. That will be on the order of 50 ohms (standard values are 47, 51, and 56 ohms). The fun part is getting the resistor rated for at least 5 watts. If anybody still sells sand resistors, they would be pretty inexpensive (the body is a hollow rectangle of some ceramic. The wire element is inside and sealed in with a gritty composition that looks like sand).
 

SgtWookie

Joined Jul 17, 2007
22,220
If you use a fixed resistor for limiting current, then you will be wasting more power in the resistor than you will be using in the LED! This is due to Ohm's Law:
P = E * I, where
P = Power in Watts,
E = Voltage,
I = Current in Amperes
There's another problem, and that is the voltage in your vehicles' electrical system may vary widely between the battery being fully discharged (11.4v) to the alternator running at full blast, perhaps as high as 14.5v or even higher.

Let's say that your LED's current is 320mA @ 4v for the moment. That's 1.28 Watts.
Let's figure the vehicle's electrical system will put out a nominal 13.8v.
So, the limiting resistor will be:
Rlimit = (VoltageSupply - Vf(LED)) / I(LED)
Rlimit = (13.8 - 4) / 320mA
Rlimit = 9.8 / 0.32
Rlimit = 30.625
Now we need to calculate the wattage required:
P = EI
P = 9.8 * .32
P = 3.136 Watts
You'll consume 2.45 times as much power in the resistor as the LED. But, that's not all - we don't have REAL current regulation, because what happens when the alternator is trying to recharge a dead battery and is outputting 14.5v?
We're still going with the Vf of 4v, and 320mA is the target current.
14.5V - 4V = 11.5V across Rlimit, let's figure the current through the resistor.
I = E/R
I = 11.5/30.625
I = 375.5mA - oops, you just fried the LED! :eek:

So, do the whole series of calculations again for 14.5v. What happens when the ignition is off, and the battery voltage drops to 12.4v or so? The LED gets a lot less current, and is dim. You don't really want that, either.

You could use an LM317 as a constant current regulator. However, it's still a linear regulator, and will dissipate a lot of power (roughly 2.5 times the power used in the LED).

Wasting that power in the resistor or regulator means wasting the fuel that the engine has to burn to generate the electricity.
 

John Luciani

Joined Apr 3, 2007
477
Unless you use a switching converter you will be less than 30% efficient.
On-Semi, Maxim, Linear Technology all make switching converter ICs for LED
applications.

(* jcl *)
 

Audioguru

Joined Dec 20, 2007
11,249
Use two LEDs in series and in series with an LM317 current regulator.
Then the LEDs work perfectly and the LM317 does not waste much power.
 

conntaxman

Joined Oct 30, 2009
12
Audio; as you told the other person to hook the leds in series,Do you know of a cir. that would run 8 [3watt//FV=3.5 Current =700ma.Supply voltage =12vdc] I dont think that I could run all those in series. And I have been looking all over the net and asking around for a small cir. to run these.
Tks
John
 

cjdelphi

Joined Mar 26, 2009
272
If you use a fixed resistor for limiting current, You'll consume 2.45 times as much power in the resistor as the LED. But, that's not all - we don't have REAL current regulation, because what happens when the alternator is trying to recharge a dead battery and is outputting 14.5v?
We're still going with the Vf of 4v, and 320mA is the target current.
14.5V - 4V = 11.5V across Rlimit, let's figure the current through the resistor.
I = E/R
I = 11.5/30.625
I = 375.5mA - oops, you just fried the LED! :eek:

for almost all high powered LED's 375 vs 320 is just an extra little tickle as most LED's.. XP-RE and the XPG range from cree all handle 1amp of current, it's the heat you have to worry about....

So a 5watt 30 ohm resistor unregulated from the car would be easy, and if one day your car's alternator produced more voltages and did lead to current > 1amp then... well

the LED will be the last thing to be concerned about, i'd be worried about the battery / car electronics exploding :D
 

Audioguru

Joined Dec 20, 2007
11,249
Audio; as you told the other person to hook the leds in series,Do you know of a cir. that would run 8 [3watt//FV=3.5 Current =700ma.Supply voltage =12vdc] I dont think that I could run all those in series. And I have been looking all over the net and asking around for a small cir. to run these.
Tks
John
Eight 3W LEDs use 24W. If they are in series then you need 28V plus a few volts for the current regulator.
I have not seen a simple circuit that can stepup 12V to 31V at such a high power but maybe somebody designed one.
 

conntaxman

Joined Oct 30, 2009
12
How do I connect 1 watt or 3 watt LED or higher on a 14v source (in my car).....I know that you can connect a low power LED(5 mm) using a resistor....but a high power LED like 1watt onward....you need a driver for it...is that necessary? why cant i just connect a Resistor by itself? ....how can i design a driver that can handle....LED @ 200mA, Vf: 3.6 - 4.2....power source: 12-14 Volt
,
,
,Here is a link for you to connect a hp.led. to 14vdc,and their are other on DIY.
http://www.instructables.com/id/Super-simple-high-power-LED-driver/
 
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