LED Drivers are used to drive High Power LEDs that require more than 100 mA current. A typical 1 watt LED consumes around 100-300 mA current while a 10 watt LED consumes 600 mA to 1 Amps current. So a resistor to drop the current is not sufficient since it heats up when high current passes. Moreover, the high power LEDs require a Constant Current to give steady brightness. So LED Drivers are used for this purpose. A typical LED driver has two sections. An SMPS converts 220V AC to DC and a Constant Current Generator Circuit provides Constant current to LED. But LED Drivers are costly units and the cost range from Rs. 150- Rs.600. We can make a Low cost LED driver just for Rs.15. Try it.
We are going to use the Variable voltage regulator IC LM317 as a “Constant Current Generator”. The Constant current generator is a circuit that provides a Constant current at the output irrespective of the input voltage changes. LM317 is the versatile Voltage Regulator IC commonly used to make Variable power supply.
It has three pins. The first pin is the Adjust pin, second is the output pin and third is the input pin. In Variable power supply circuits, the Adj pin is connected to the ground through a fixed or variable resistor and to the output pin. If we connect the Adjust pin to the output pin only with a resistor, we can make LM317 as a Constant current generator. This trick is used here. See image below
Here the Adj pin is connected to the output through the resistor R. So the output current will be constant depending on the value of R. The adjust pin has a Reference voltage of 1.25 volts. So the current output is calculated using the formula
I = 1.25 V / R in Ohms
Suppose we are providing 1 Amps current as input and we are using 1 Ohm resistor as R. Then the current will be
I = 1.25 V / 1Ohm = 1.25 Amps. Since we are giving 1 Amps as input, output will be 1 Amps.
Suppose we are using 10 Ohms resistor as R, then
1.25V / 10 Ohms = 0.125 Amps or 125 Milli Amps.
So just use this formula to select the required current. For 1 watt LED, select 10 Ohms to 4.7 Ohms resistor which will provide 125 mA to 265 mA current. For 10 watt LED, use 2 or 3 Ohms resistor to get 625 or 416 mA current. Remember, the resistor should handle the high current. So use 1 or 2 watts resistor for 1 watt LED and 5 or 10 watt resistor for 10 watt LED.
We have to consider the input voltage also. The 1 watt LED is rated 3.6 volts and 10 watt LED 12 volts. If we use 5 volt in the circuit for 1 watt LED, it will not light fully. So to get the value of the Input voltage, use the formula
V in Minimum = VLED + 1.30 + VDO
Vin = Input voltage
VLED = Forward voltage of LED
VDO = Drop out voltage of LM 317. It is 2.5V generally.
1.30 = The Reference voltage of LM317 is generally 1.25V. But it can go up to 1.30V. So take the higher value.
Now let us see the case of 1 watt LED. Its Forward voltage (VLED) is 3.6V.
So 3.6V+1.30V +2.5V = 7.4V.
That is, we have to give more than 7.4 volt and 500 mA as input. A 9 Volt battery is ideal for the circuit.
For 10 watt LED, the condition changes. Its Forward voltage is around 10 volts. Then
10V + 1.30V + 2.5V = 13.8V. So we must give minimum 14 V and 1 Amps current as input.
Now let us see the circuit.
You need the following
Now the connections
1. Connect the 10 Ohms resistor between the Adj ( pin 1 ) and Output ( pin 2 ) of LM317
2. Connect the Anode ( + ) of LED to the Adjust pin ( pin 1 )
3. Connect the Cathode ( – ) of LED to the Negative of Battery Snap
4. Connect the Positive of Battery Snap to the Input ( pin 3 ) of LM317.
Connect the battery. LED will light using 125 mA current till the battery voltage remains above 7 volts. The brightness will be steady.
Remember, LM317 becomes too hot when high current passes through it. So use a good heat sink applied with Heat sink compound to make close contact between the heat sink and IC. Then only heat transfer takes place smoothly. Similarly , use a heat sink for the LED.
Note : Use only Battery or Transformer power supply or correct rated SMPS for high power LEDs. Capacitor power supply is not ideal to get full brightness since the current is less than 100 mA.