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Does resistor placement change power dissipation in simple LED circuit?


How does the current limiting resistor for an LED affect current and voltage drops?Understanding current, voltage and resistancePower booster 12V to 36V for LEDWhy did my LED resistor burn while lighting four LEDs in series?What Wattage does my resistor need to have to be save to dim an LED strobe module?12V DC LED and motor circuit - overheatingLED forward voltage is a range, so how do you calculate resistor value?Driving an LED with resistor directly from 3.3v GPIO pin of a microcontrollerIs there a point in adding a 1 Ohm resistor to this LED circuit?Resistor Selection to retain same brightness in LED PWM circuit






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3












$begingroup$


Does the power dissipation in an LED and resistor change depending on the resistor placement?



Let's assume the following:




  • Supply Voltage: 3.3v (DC)


  • LED Forward Voltage: 2.1v


  • Resistor: 330 Ohm

I think my problem is that I'm thinking that the circuit "starts" at 3.3v, then goes through the resistor which restricts current and then goes through the LED which drops voltage. Alternatively, if I place the LED before the resistor, then the LED will drop the voltage (3.3v - 2.1v = 1.2v) before it reaches the resistor so the resistor is dissipating less heat?










share|improve this question









$endgroup$


















    3












    $begingroup$


    Does the power dissipation in an LED and resistor change depending on the resistor placement?



    Let's assume the following:




    • Supply Voltage: 3.3v (DC)


    • LED Forward Voltage: 2.1v


    • Resistor: 330 Ohm

    I think my problem is that I'm thinking that the circuit "starts" at 3.3v, then goes through the resistor which restricts current and then goes through the LED which drops voltage. Alternatively, if I place the LED before the resistor, then the LED will drop the voltage (3.3v - 2.1v = 1.2v) before it reaches the resistor so the resistor is dissipating less heat?










    share|improve this question









    $endgroup$














      3












      3








      3





      $begingroup$


      Does the power dissipation in an LED and resistor change depending on the resistor placement?



      Let's assume the following:




      • Supply Voltage: 3.3v (DC)


      • LED Forward Voltage: 2.1v


      • Resistor: 330 Ohm

      I think my problem is that I'm thinking that the circuit "starts" at 3.3v, then goes through the resistor which restricts current and then goes through the LED which drops voltage. Alternatively, if I place the LED before the resistor, then the LED will drop the voltage (3.3v - 2.1v = 1.2v) before it reaches the resistor so the resistor is dissipating less heat?










      share|improve this question









      $endgroup$




      Does the power dissipation in an LED and resistor change depending on the resistor placement?



      Let's assume the following:




      • Supply Voltage: 3.3v (DC)


      • LED Forward Voltage: 2.1v


      • Resistor: 330 Ohm

      I think my problem is that I'm thinking that the circuit "starts" at 3.3v, then goes through the resistor which restricts current and then goes through the LED which drops voltage. Alternatively, if I place the LED before the resistor, then the LED will drop the voltage (3.3v - 2.1v = 1.2v) before it reaches the resistor so the resistor is dissipating less heat?







      power resistors basic






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked May 31 at 13:18









      SofaKngSofaKng

      182128




      182128




















          2 Answers
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          6












          $begingroup$

          The order of the components does not make any difference. The voltage is diveded on the two components in the same way, in your case 2.1V on the LED and 1.2V on the resistor. This results in 1.2V/330Ohm = 3.6mA through your series circuit. So the power dissipation is also the same in both configurations.






          share|improve this answer











          $endgroup$




















            6












            $begingroup$

            What matters is the voltage across the terminals of the resistor (as you would measure with a meter connected directly across the resistor). It doesn't matter how the voltage gets there.



            So, no it doesn't matter which order they're connected when they're in series, because the magnitude of the voltage across the resistor will always be the supply voltage minus the LED forward voltage.



            Whether you measure the voltage across the resistor as positive or negative doesn't matter either, since the power dissipation of the resistor is the square of that voltage divided by the resistance.






            share|improve this answer











            $endgroup$













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              6












              $begingroup$

              The order of the components does not make any difference. The voltage is diveded on the two components in the same way, in your case 2.1V on the LED and 1.2V on the resistor. This results in 1.2V/330Ohm = 3.6mA through your series circuit. So the power dissipation is also the same in both configurations.






              share|improve this answer











              $endgroup$

















                6












                $begingroup$

                The order of the components does not make any difference. The voltage is diveded on the two components in the same way, in your case 2.1V on the LED and 1.2V on the resistor. This results in 1.2V/330Ohm = 3.6mA through your series circuit. So the power dissipation is also the same in both configurations.






                share|improve this answer











                $endgroup$















                  6












                  6








                  6





                  $begingroup$

                  The order of the components does not make any difference. The voltage is diveded on the two components in the same way, in your case 2.1V on the LED and 1.2V on the resistor. This results in 1.2V/330Ohm = 3.6mA through your series circuit. So the power dissipation is also the same in both configurations.






                  share|improve this answer











                  $endgroup$



                  The order of the components does not make any difference. The voltage is diveded on the two components in the same way, in your case 2.1V on the LED and 1.2V on the resistor. This results in 1.2V/330Ohm = 3.6mA through your series circuit. So the power dissipation is also the same in both configurations.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited May 31 at 13:51

























                  answered May 31 at 13:25









                  jusacajusaca

                  1,546625




                  1,546625























                      6












                      $begingroup$

                      What matters is the voltage across the terminals of the resistor (as you would measure with a meter connected directly across the resistor). It doesn't matter how the voltage gets there.



                      So, no it doesn't matter which order they're connected when they're in series, because the magnitude of the voltage across the resistor will always be the supply voltage minus the LED forward voltage.



                      Whether you measure the voltage across the resistor as positive or negative doesn't matter either, since the power dissipation of the resistor is the square of that voltage divided by the resistance.






                      share|improve this answer











                      $endgroup$

















                        6












                        $begingroup$

                        What matters is the voltage across the terminals of the resistor (as you would measure with a meter connected directly across the resistor). It doesn't matter how the voltage gets there.



                        So, no it doesn't matter which order they're connected when they're in series, because the magnitude of the voltage across the resistor will always be the supply voltage minus the LED forward voltage.



                        Whether you measure the voltage across the resistor as positive or negative doesn't matter either, since the power dissipation of the resistor is the square of that voltage divided by the resistance.






                        share|improve this answer











                        $endgroup$















                          6












                          6








                          6





                          $begingroup$

                          What matters is the voltage across the terminals of the resistor (as you would measure with a meter connected directly across the resistor). It doesn't matter how the voltage gets there.



                          So, no it doesn't matter which order they're connected when they're in series, because the magnitude of the voltage across the resistor will always be the supply voltage minus the LED forward voltage.



                          Whether you measure the voltage across the resistor as positive or negative doesn't matter either, since the power dissipation of the resistor is the square of that voltage divided by the resistance.






                          share|improve this answer











                          $endgroup$



                          What matters is the voltage across the terminals of the resistor (as you would measure with a meter connected directly across the resistor). It doesn't matter how the voltage gets there.



                          So, no it doesn't matter which order they're connected when they're in series, because the magnitude of the voltage across the resistor will always be the supply voltage minus the LED forward voltage.



                          Whether you measure the voltage across the resistor as positive or negative doesn't matter either, since the power dissipation of the resistor is the square of that voltage divided by the resistance.







                          share|improve this answer














                          share|improve this answer



                          share|improve this answer








                          edited May 31 at 13:54

























                          answered May 31 at 13:25









                          Spehro PefhanySpehro Pefhany

                          217k5166447




                          217k5166447



























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