![]() ![]() Substituting these values in the above formula, we get the power dissipated by the resistor. ![]() So, the current through the Series Resistor is The current through the Resistor is same as the current through the LED as they are series. So, the Voltage Drop across the Series Resistor is We know that supply voltage is 12V and Voltage drop across LED is 3.6V. I RES is the current through the Resistor. Here, V RES is the voltage drop across the resistor and The Power Rating of a Resistor can be calculated using the following formula. Power Rating of a Resistor specifies the value of power that a resistor can safely dissipate. ![]() Now that we have calculated the resistance of the series resistor, the next step is to calculate the power rating of this resistor. Since there won’t be a 280Ω Resistor, we will use the next big resistor i.e. Substituting these values in the above equation, we can calculate the value of Series Resistance as ![]() Therefore, V S = 12V, V LED = 3.6V and I LED = 30mA. In our simple LED Circuit consisting of a single LED, we have used a 5mm White LED and a power supply of 12V.Īs per the datasheet of the 5mm White LED, the Forward Voltage of the LED is 3.6V and the Forward Current of the LED is 30mA. I LED is the desired current through the LED. V LED is the voltage drop across the LED and Here, V S is the Source or Supply Voltage The value of the series resistor can be calculated using the following formula. So, selecting the right resistor with the right wattage is very important. Connecting a small LED to a 12V Supply would burn the LED and you can see the magic smoke instantly. The important component (other than the LED of course) is the Resistor. The following image shows the setup of single LED connected to a 12V Supply and a current limiting series resistor. The circuit diagram for this circuit is shown below. We will try to turn ON a single 5mm White LED using a 12V Supply. The first circuit in the simple LED Circuits is a single LED Circuit.
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