The high-powered LED’s have become very popular, especially in the flashlight applications. These are different variety of LED’s, but the basic concept is the same as the traditional LED’s.
These LED’s require higher current than the traditional ones. They can be from 300ma to several amps..
Here, we’ll see how to power them and see some of the options available for powering them.
It’s assumed that you are familiar with the basic electronic symbols and have knowledge in biasing traditional LED’s.
10W LED 3W LED
These are typical high-powered LED’s. They’re available as LED’s only, or ready-mounted onto a heat sink. The heat sinks are necessary for these LED’s due to their higher heat dissipation.
The white LED’s typically have about 3V to almost 4V drop across them at their rated biasing current.
They are available in various colors. It seems that colors other than white have lower power (at least at the time of this writing).
The brightness is rated in Lumens (lm), whereas the lower powerd LED’s are rated in Candelas (cd).
For example, the 10W LED is advertised as 975lm, where as the 3W LED is 180lm.
Something to Consider Before Diving In
Driving these LED require some thought to avoid damages/fire/injuries. Unlike low-powered LED’s, they can’t be easily driven with a power source and a resistor. The required current can be high enough that the power loss in the resistor becomes too much of a waste.
Example: the 3w LED is driven with 1A current. If you’re using a 5V supply, using Ohm’s Law, we can see what happens if resistor is use to drive this LED.
Calculate voltage across resistor:
Vsupp – Vled = 5V – 3V =2V (required across resistor)
Calculate resistance that would develop 2V at 1A:
Vres/I = 2V / 1A = 2Ω (Note that R and LED are in series, so current is the same as LED)
Calculate power dissipated in the resistor:
I x Vres = 1A x 2V = 2W
That’s 2W of wasted power just heating up the resistor.
How to Drive High Powered LED’s
An efficient method of driving these high-powered LED’s is to use a regulated current source. Similar to regulated voltage sources, there are many current sources available for driving LED’s. There are ready-made units available, where you simply connect the LED’s. You’ll have to match the regulated current of the unit and the LED current rating.
Here is a current regulator meant to go into MR16 sockets. It uses the PT4115 buck regulator. These are regulated at 700mA unmodified. The current can be adjusted by changing the feedback (FB) resistor, which is the one shown in the picture for the 100mA driver. The FB voltage across the sense resistor is 100mV, typical. In order to get 100mA, calculate the resistance using Ohm’s Law:
Resistance = 100mV / 100mA = 1 Ω
Similarly for the 700mA version,
Resistance = 100mV / 700mA = 0.143Ω (nearest standard value = 0.15Ω)
If you look closely, the FB resistor is a 0.15Ω (R150) surface mount resistor on the 700mA version.
The rectifier diodes (black rectangular devices) were removed to make soldering the supply wires easier. The diodes are not necessary for DC voltage applications.
This unit is good to 30V input. This means you can have up to about 7 or 8 power LED’s in series. However, the maximum current it can provide is about 1.2A. The FB resistor is small already at 700mA. Any smaller resistances may possibly result in inconsistent behaviors.
If you’re the ambitious type and still want to use the resistor method, here is something that works great: AMC7135. This is a surface mount, so it is best to design a PCB for it (not so great). Good news is that some ready-made circuits are available with various current ratings for different power LED’s.
It is rated for constant current of 300mA to 380mA (depending on options). To get 1A, you’ll need about three in parallel. It has a VDD pin that acts as an enable pin, so you can control that pin to turn ON/OFF the LED, or, it can be driven with a pulse to do strobing. Again, with this method, you must be careful of the power dissipated in the drivers.
No Heatsinks Needed?
There is a case where the heatsink may be omitted. If the LED is pulsed to only stay on for a fraction of a second, then heatsinking is not necessary. On top of that, you can use the resistor method with much lower power rating for the resistor.
For example, if the LED is ON for 20ms every 1 second, the DutyCycle is 2%. So, for a 3W LED with 5V supply, we’ve shown that we need 2W power resistor in series with the LED. At 2%, we calculate the average power:
Pave = [0.02(1A)]^2 * 2Ω (P = I^2 * R)
Pave = 0.0008W = 0.8mW
A typical 1206 surface mount resistors are rated at about 100mW.
Check out this exact application here as a strobe light, if you haven’t already. Everything remains cool to the touch!