Usually, when we need to drive low power LEDs, we don’t care much about power losses. What we do we add a current limiting resistor, and that’s enough. For instance for 20mA LED we choose between 300Ω-1kΩ resistor when powering from 5V. But a different situation is with power LEDs. The currents are much more prominent here like 1A and more. Adding a resistor to limit the current isn’t an option, because power losses become significant. Here you need a constant current driver to drive it safely without wasting energy. It happens that I have Cree XR-E Q5 XLAPM-7090 LED lying around. It requires 3.7V driving voltage and can take up to 1A current.
There are several light intensities given at specific currents:
- 350mA: 107~114lm
- 700mA: 171.2~182.4lm
- 1000mA: 214 ~ 228lm
LED is placed on PCB with metallic bottom for heat dissipation. These things get hot and may be damaged without heatsinking. There are lots of specialized LED driver IC’s you can get. They all compete with price and efficiency. The main aim of all Led drivers is to ensure a stable current source. It has to be temperature independent so it would stay the same in different conditions. It is better to rely on special chips especially if your designed product must be reliable. But what if you want only to drive power LED without spending a dime on parts. In my situation, I want to drive LED at 0.3A to get decent light and still avoid using a heatsink. So I need to build a current source capable of providing 0.3A. In my drawer, I found BD911 power NPN transistor which I’m going to use. You don’t need that much as this transistor can handle up to 15A of current. Probably simplest constant current circuit is based on NPN transistor, and couple diodes are as follows:
In this circuit, two diodes at the transistor base provide a constant 1.4V (2×0.7V) voltage drop. Base-Emitter voltage drop is approximately Vbe=0.7V, and the rest 0.7V goes to R2 resistor. This becomes convenient because we can calculate its value using a simple formula:
If we want to drive LED with 0.3A, then we need R=2.3Ω. This is a current limiting resistor for the LED. It is going to dissipate 0.7W of power. To be sure I choose at least twice as much power resistor. Now how about R1? From the datasheet, we can find that the current gain is about 250. So to get 0.3A collector current, we need to supply base with 0.3A/250= 1.2mA. Having a base current, we can calculate R1. Don’t forget that diodes also require some current to operate properly. 1mA should be enough to provide a forward voltage drop. Then R1 can be calculated as follows:
R1=(VCC-Vdiodes)/(Ib+1mA) = (5V-1.4V)/(1mA+1mA)=1.8kΩ.
I could only find 2.6Ω current limiting resistor. So with it I can drive LED with 260mA current.
Lets build a circuit and see if it works. Assembled and powered with 5V supply.
Measured Led current is 240mA. The measured current is a bit smaller because the voltage drop on diodes were less than 1.4V and thus less voltage left for the emitter. Such current is sufficient to drive Led without a heatsink. If more current is provided it gets boiling.
Also, I measured the current draw from the power supply which is around 245mA. Let us see how efficient circuit is. LED voltage drop is 3.7V so power usage:
Ps =5V*0.245 = 1.225W
So we lose:
Ploss = 0.337W;
Or we can say that the circuit is 72% efficient:
N = PLED/Ps * 100% = 0.888/1.225*100% = 72%
For higher efficiency, it would be better to use MOSFET with low Rds resistance. Practically this circuit works stable in reasonable power voltage range. Powering from 5 to 15V shouldn’t be a problem. But with more supply voltage you dissipate more power with the transistor, and so you get less efficiency.