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Calculate Parameters of Resistor and Its Power for One LEDAll of us attended school so I think everybody can remember the "Ohm's law" that is I=U/R. Then we will need an arithmetic formula to calculate power (P=IU). Using these formulas we can select necessary resistor for connecting light-emitting diodes toany power supply source.As an example we can take diode BL-L813UWC. Its parameters are: current rating is 30mA, voltage rating is 2.7 V.Besides the "Ohm's law" and power formulas there is so called reliability coefficient of a LED, it equals to 0.75. So you should multiply the useful current of the LED by this coefficient.According to Ohm's law, current (I) is directly proportional to the voltage (U) and inversely proportional to resistance (R). It implies that the resistance (R) will be equal to the proportion of voltage (U) to current (I). This is how we get calculation formula for resistance (R). Now you should add reliability coefficient (0.75) to this formula and multiply it by current (I).There is also a concept of LED voltage drop. Better to say this is a LED voltage which equals to 2.7 V in our case. We need to know this to calculate properly the voltage which needs to be reduced. That is Uf (the voltage that need to be set in the formula) makes the difference between Up (currently presented and where the LED has to be connected, let's take 12 V) and Ud (voltage drop on the LED, in this case it is 2.7 V). As current is measured in Amperes (A) and LED's current is measured in miliamperes (mA), we need to convert miliamperes into amperes (A). As 1 A equals to 1000 mA, then 30mА will be equal to 0.03А. This value should be used in the formula.So now we know the values we need. And we can calculate the necessary rating of the resistor according to the following formula:We insert the appropriate values into the ready formula:The rating of the resistor is known. But if it is impossible to find the resistor with this exact value, then we should take a resistor with any standard rating which is close to ours but exceeds it. In this case we can use a resistor of 430 Ohm. If the rating is lower than the calculated one, for example 390 Ohm then the LED's service life will be much shorter.The resistor's resistance we know. Let's calculate the power. The thing is that a resistor has such parameter as power. There are standard values of the resistor's power like 0.125W, 0.25W, 0.5W, 1W, 2W, 3W and others. The more the resistor's power is the thicker it is. If you install a resistor with power that is lower than the resistor gets then it will heat up very much and finally it can burn out. To avoid this you need to define the resistor's power in advance. It can be easily calculated with the formula.We need to know the following values: LED's current (0.03 A in our case), voltage in the circuit (12 V in our case) and voltage drop (2.7 V).Let's insert the values into the formula:The value we get is close to the standard one (0,5 W). The higher power parameters are the less it will heat up. So with the help of simple calculations we have found out that to connect LED BL-L813UWC (30mА; 2.7V) to the 12 V power supply source, we need a resistor with the rating of 430Ohm and the power of 0,5W.