Exceptionally Good LED Power Tutorial
A seriously talented person wrote this article and it's for beginners! http://led.linear1.org/why-do-i-need-a-resistor-with-an-led/
Forum Discussion
The reason for not using current limiting resistors to drive LEDs.
Assume one 3.3V LED running off 12.5V at 20mA using a current limiting resistor.
Voltage drop across the resistor, 12.5V – 3.3V = 9.2V.
Resistor value, R=V/I, R= 9.2V/.02A, R = 460 Ohms.
Power dissipated in LED, P=VI, P=3.3V*.02A, P=.066Watts.
Power dissipated in resistor, P=VI, P=9.2V*.02A, P=.184Watts.
Total power from power supply, P=VI, P=12.5V*.02A, P=.25 Watts.
Efficiency = Pled/Pin, .066W/.25W, Efficiency = 26.4%! Not good.
If voltage goes up to 14V then:
Voltage drop across the resistor, 14V – 3.3V = 10.7V.
Resistor value from last calculation, R = 460 Ohms.
New current in LED and resistor, I=V/R, I=10.7V/460 Ohms, I=.023A.
Power dissipated in LED, P=VI, P=3.3V*.023A, P=.076Watts.
Power dissipated in resistor, P=VI, P=10.7V*.023A, P=.25Watts.
Total power from power supply, P=VI, P=14V*.023A, P=.322 Watts.
Efficiency = Pled/Pin, .076W/.322W, Efficiency = 23.6%! Not good.
From this it can be seen that the LED current will go from .02A with a 12.5V source to .023A with a 14V source. A current increase of .003A or an increase of 15%. Not too bad.
A current regulated power supply will hold the current to .02A and still have an efficiency of around 90%. Note it has to be a switching supply, a linear supply will be no more efficient.
So what happens if you put several LEDs in series, lets say 3 LEDs.
Total voltage across the 3 LED string, 3 * 3.3V = 9.9V.
Voltage drop across the resistor, 12.5V – 9.9V = 2.6V.
Resistor value, R=V/I, R= 2.6V/.02A, R = 130 Ohms.
Power dissipated in LEDs, P=VI, P=9.9V*.02A, P=.198Watts.
Power dissipated in resistor, P=VI, P=2.6V*.02A, P=.052Watts.
Total power from power supply, P=VI, P=12.5V*.02A, P=.25 Watts.
Efficiency = Pleds/Pin, .198W/.25W, Efficiency = 79.2%! Much better.
If voltage goes up to 14V then:
Voltage drop across the resistor, 14V – 9.9V = 4.1V.
Resistor value from last calculation, R = 130 Ohms.
New current in LEDs and resistor, I=V/R, I=4.1V/130 Ohms, I=.0315A.
Power dissipated in LEDs, P=VI, P=9.9V*.0315A, P=.312Watts.
Power dissipated in resistor, P=VI, P=4.1V*.0315A, P=.129Watts.
Total power from power supply, P=VI, P=14V*.0315A, P=.441 Watts.
Efficiency = Pled/Pin, .312W/.441W, Efficiency = 70.7%! Not as good as with 12.5V.
From this it can be seen that the LED current will go from .02A with a 12.5V source to .0315A with a 14V source. A current increase of .0115A or an increase of 57.5%. Not very well regulated with using a resistor and may well destroy the LEDs
A current regulated power supply will hold the current to .02A and still have an efficiency of around 90%. Note it has to be a switching supply, a linear supply will be no more efficient.
If you look at a voltage from 10.5V, a dead battery to 14.8V when charging the current regulation with a resistor is really bad and most likely will cause a disaster.
Hope I did not make a math error!
Assume one 3.3V LED running off 12.5V at 20mA using a current limiting resistor.
Voltage drop across the resistor, 12.5V – 3.3V = 9.2V.
Resistor value, R=V/I, R= 9.2V/.02A, R = 460 Ohms.
Power dissipated in LED, P=VI, P=3.3V*.02A, P=.066Watts.
Power dissipated in resistor, P=VI, P=9.2V*.02A, P=.184Watts.
Total power from power supply, P=VI, P=12.5V*.02A, P=.25 Watts.
Efficiency = Pled/Pin, .066W/.25W, Efficiency = 26.4%! Not good.
If voltage goes up to 14V then:
Voltage drop across the resistor, 14V – 3.3V = 10.7V.
Resistor value from last calculation, R = 460 Ohms.
New current in LED and resistor, I=V/R, I=10.7V/460 Ohms, I=.023A.
Power dissipated in LED, P=VI, P=3.3V*.023A, P=.076Watts.
Power dissipated in resistor, P=VI, P=10.7V*.023A, P=.25Watts.
Total power from power supply, P=VI, P=14V*.023A, P=.322 Watts.
Efficiency = Pled/Pin, .076W/.322W, Efficiency = 23.6%! Not good.
From this it can be seen that the LED current will go from .02A with a 12.5V source to .023A with a 14V source. A current increase of .003A or an increase of 15%. Not too bad.
A current regulated power supply will hold the current to .02A and still have an efficiency of around 90%. Note it has to be a switching supply, a linear supply will be no more efficient.
So what happens if you put several LEDs in series, lets say 3 LEDs.
Total voltage across the 3 LED string, 3 * 3.3V = 9.9V.
Voltage drop across the resistor, 12.5V – 9.9V = 2.6V.
Resistor value, R=V/I, R= 2.6V/.02A, R = 130 Ohms.
Power dissipated in LEDs, P=VI, P=9.9V*.02A, P=.198Watts.
Power dissipated in resistor, P=VI, P=2.6V*.02A, P=.052Watts.
Total power from power supply, P=VI, P=12.5V*.02A, P=.25 Watts.
Efficiency = Pleds/Pin, .198W/.25W, Efficiency = 79.2%! Much better.
If voltage goes up to 14V then:
Voltage drop across the resistor, 14V – 9.9V = 4.1V.
Resistor value from last calculation, R = 130 Ohms.
New current in LEDs and resistor, I=V/R, I=4.1V/130 Ohms, I=.0315A.
Power dissipated in LEDs, P=VI, P=9.9V*.0315A, P=.312Watts.
Power dissipated in resistor, P=VI, P=4.1V*.0315A, P=.129Watts.
Total power from power supply, P=VI, P=14V*.0315A, P=.441 Watts.
Efficiency = Pled/Pin, .312W/.441W, Efficiency = 70.7%! Not as good as with 12.5V.
From this it can be seen that the LED current will go from .02A with a 12.5V source to .0315A with a 14V source. A current increase of .0115A or an increase of 57.5%. Not very well regulated with using a resistor and may well destroy the LEDs
A current regulated power supply will hold the current to .02A and still have an efficiency of around 90%. Note it has to be a switching supply, a linear supply will be no more efficient.
If you look at a voltage from 10.5V, a dead battery to 14.8V when charging the current regulation with a resistor is really bad and most likely will cause a disaster.
Hope I did not make a math error!
