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Calculation of LED current limiting resistors    (#80228)

 

When connecting LED to higher DC voltage, it is necessary to add a series resistor in order to limit the LED current to a safe level. This is to avoid damaging the LED due to excessive current.  In the following short discussion, we shall demonstrate by way of examples how the value of the resistor can be calculated (in different situations) using simple math & Ohm's Law.

Example (1) :    Connecting a common red LED of to DC 6V

Red LED of normal brightness typically has forward voltage of around 1.5V. In this example, we set current to 10 mA.

The voltage (VR) across the resistor R   is :  VR = Vs - VLED = 6V - 1.5V = 4.5V
By Ohm's Law the resistor value R = VR / I = 4.5V / 0.01A = 450 Ω
For convenience we will select a 470 Ω  standard resistor which is easier to get the the market. With this resistor the actual current would be I = VR / R = 4.5V / 470 Ω  = 0.0096A (or 9.6mA)

What size of resistor to use ?   The wattage of resistor can be calculated using the equation  P = V2 / R
Here we have  P =  VR2  / 470 Ω = (6 - 1.5)2 / 470  = 0.043W.  So, both 1/4W & 1/6W resistors can be used
       ( 1/4W=0.25W & 1/6W=0.167W. Both are larger than 0.043W)

Example (2) :    Connecting  3 super bright white LED  to DC 12V 

The voltage of white super bright LED is around 3V.  Let us set the LED current to 30 mA which is quite standard for super bright LED

The voltage across the resistor R  is :   VR = Vs - (3 x VLED ) = 12V - (3 x 3V) =  3V
By Ohm's Law the resistor value   R = VR / I = 3 /  0.03A =  100 Ω

What size of resistor to use ?   The wattage of resistor can be calculated using the equation  P = VR2/ R
   P =  VR2  / 100 Ω = (12 - 3x3)2 / 100  = 0.09W.  You can use either a 1/4W  or 1/6W resistor
  

Example (3) :    Connecting 9 super bright blue  LED  to DC 12V 

Blue & white super bright LED has the same forward voltage which is around 3V, so we will be using the same resistor (100 Ω) from example(2) 

Here we  connect  3 strings of LED in parallel each consists of 3 LED and a resistor.
Since the current for each string is 30mA (ref to example (2) above), the total current I is therefore equal to  3 x 0.03A  = 0.09A or 90mA.

You can have more LED by adding additional strings of LED but you have to make sure that the DC supply (or battery) is able to provide the increased current

 

NOTE :  The brightness of the LED is proportional to the current that flows thru it. That is, the higher the current, the brighter the LED but it should be limited within a safe level to avoid damaging or shortening the life of the LED.    


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