Ohms law is a must in circuit design to calculate, Current, Resistance and Voltage in the different sections of a circuit. Ohms law simply represents the relationship between voltage, current and resistance. Let us see how Ohms law is applied to select the current limiting resistor for LED.
Ohms law can be expressed like this
V (Voltage) = I X R
I (Current) = V / R – in Amps
R (Resistance) = V / I – in Ohms
Suppose the input voltage is 8 volts, current is 2 Amps and resistor is 4 Ohms. Then
V = I x R = 2 x 4 = 8 Volts
I = V / R = 8 / 4 = 2 Amps
R = V / I = 8 / 2 = 4 Ohms
This is the basic method to calculate voltage, current and resistance in circuit design.
LED Series Resistor
A current limiting resistor is must for LED ,otherwise it will burn instantly. The value of the resistor must be proper to get sufficient brightness and to increase the life of LED. Different LEDs have different forward voltage drop across them. These are the voltage drops of general type LEDs
Red – 1.7 V
Green – 1.8 V
Yellow – 1.7 V
Blue – 3.1 – 3.6 V
White 3- 3.6 V
But this may slightly vary depending on the type and make of LEDs
Current through the LED must be between 10 to 25 Milli Ampere. Optimum current for good brightness is 20 mA. If the value of the resistor increases, current reduces followed by a reduction in the brightness. But this will increases the life of LED. If the value of resistor decreases, current increases followed by an increase in brightness.
Suppose the input voltage is 12 volts and the LED is standard diffuse type 5mm Red LED. It has a voltage drop of 1.7 volts. Fix the current through the LED as 20 mA (0.02 Amps)
Formula for selecting the resistor is
R = Vs – Vf / If
R is resistance in Ohms, Vs is input voltage , Vf is the Forward voltage drop of LED and If is the current through LED.
Out of the four values, we know three values
Vs = 12 V
Vf = 1.7 V
If = 0.02 A
So the value of resistor is
R = Vs – Vf / If = 12 V – 1.7 V / 0.02 A = 10.3 V / 0.02 A = 515 Ohms
So when we select 515 Ohms resistor, exactly 0.02 Amps or 20 Milli Ampere current flows through the LED. But 515 Ohms resistor is hard to get, so we can select either 470 Ohms or 560 Ohms. Then the current will be
I = V / R = 10.3 / 470 = 0.021 Amps or 21 mA.
I = V / R = 10.3 / 560 = 0.018 Amps or 18 mA .
So if 470 ohms resistor is used, LED gets 21 mA current and brightness slightly increases. If it is 560 ohms, LED gets only 18 mA so brightness reduces. So the exact value is 470 ohms. You can apply this formula for all LEDs.
White , Blue, and Pink LEDs
White, Blue and Pink LEDs can handle up to 40 mA current and the brightness will be sufficient only if 25 to 30 mA current is provided. More over the forward voltage drop of these LEDs is around 3 volts. So they need lower value resistor.
Vs- Vf / If = 12 V – 3 V = 9 V = 9 V / 0.03 A = 300 Ohms. Use 270 Ohms resistor
1 Watt White LED
The new type 0.5 and 1 watt White LED requires 100-350 mA current. Forward voltage drop is 3.6 V. So the value of the resistor must be very low. Since high current flows through the resistor, the Wattage of resistor must be ½ or 1 Watt otherwise, the resistor will heat up and burn. Suppose the current required is 250 mA, then the resistor is
R = Vs- Vf / If = 12 – 3.6 / 0.25 A = 8.4 / 0.25 A = 33.6 Ohms. Use 33 Ohms 1 watt resistor
If only 100 mA is required, then 8.4 V/ 0.1 A = 84 Ohms. Use 82 Ohms ½ W resistor
If 350 mA is required, then 8.4 / 0.35 = 24 Ohms. Use 27 Ohms 1 W resistor.
Note: All the calculations given above are based on 12 volt supply. If the supply voltage is 9, 6, 4.5 etc, use the formula Vs – Vf / If to get correct value of the resistor.