LED lighting manufacturer, we introduce how to choose the resistor for LED lights

by:Sehon     2020-10-24

light emitting diode ( 领导) Is the indispensable key before deepening probe into electronic world. Whether you plan to use it for instructions, communications, lighting, or just to make the project more cool, all LED all have one thing in common: no current adjustment. If he does not limit the current, LED will eventually fails. Sometimes even lead to catastrophic results. In most cases, a series ( The appropriate size) Resistor, can deal with this problem. In high power application, you might also see current limiting power or current limiting IC. The following we will discuss how to use to deal with the current limiting resistor problem!

if you are not familiar with ohm's law, please let me clarify to you. Ohm's law defines the relationship of voltage, current and resistance, use the formula is expressed as V = I * R. In which any two known quantity, just don't know the amount convert this formula can be obtained. Case, the power supply voltage is 9 v, the LED forward voltage to 2. 4 v, rated current to 20 ma, and resistor resistance don't know.

in the series circuit, all components of the flowing current are the same. Therefore, the LED current to 20 ma, resistor with the same current. Another law is applicable to the series circuit, each component should be equal to the sum of the voltage drop of power supply voltage. This circuit, the power supply voltage is 9 v, the LED forward voltage to 2. 4V; With 9 v minus 2. 4 v resistor is obtained at both ends of the pressure drop is 6. 6V。 Then, using ohm's law to solve. Although the resistor value don't know, but known resistors at both ends of the voltage drop is 6. 6 V, current is zero. 02 ( 20 mA) 。

two values in ohm's law formula: 6. 6 V = 0。 02 A * R

conversion formula: R = 6. 6 V / 0。 02 A

get R: 330 & Omega; = 6. 6 V / 0。 02 A

now, resistance problems has been processing, is the last step. Not all 330 & Omega; Resistors are available, the rated power will need to be satisfied or more in this case the dissipation power. Please note: resistor power unit for W. Electrical energy formula for P ( W) =我( A) * u ( V) Using this resistor of 0. 02 and 6 A. 6 V power to accounting: P = 0. 02 A * 6。 Power dissipation of 6 V resistor is 0. 132 W ( 132 mW) 。

the commonly used power of the resistor value of & frac14; W, namely zero. 25 W, can work normally. Use higher resistor can be rated power, only capital is usually higher.

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