Mosfet heating in current limiter circuit

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP





.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0;







up vote
2
down vote

favorite












I am designing a lead acid battery charger using Battery University for guidance. This is the circuit I made. Battery Charger



The power supply is from an Adjustable 12V power supply. I have set it to 14.4V. The mosfet is FQP55N06. The transistor is a BC547.
The current limit based on testing with multimeter is 1.24A.



If i connect a halogen bulb (12V 35W) of a motorcyle as load, the Mosfet limits the current but heats up rapidly. Why is that? (I connected a smaller 12V bulb which draws only 0.7A and the mosfet did not heat up)



If I use this circuit to charge a battery, and if it draws current upto the circuit limit, will the Mosfet heat up? If it will, how can I prevent it, while maintaining this current limit?







share|improve this question


























    up vote
    2
    down vote

    favorite












    I am designing a lead acid battery charger using Battery University for guidance. This is the circuit I made. Battery Charger



    The power supply is from an Adjustable 12V power supply. I have set it to 14.4V. The mosfet is FQP55N06. The transistor is a BC547.
    The current limit based on testing with multimeter is 1.24A.



    If i connect a halogen bulb (12V 35W) of a motorcyle as load, the Mosfet limits the current but heats up rapidly. Why is that? (I connected a smaller 12V bulb which draws only 0.7A and the mosfet did not heat up)



    If I use this circuit to charge a battery, and if it draws current upto the circuit limit, will the Mosfet heat up? If it will, how can I prevent it, while maintaining this current limit?







    share|improve this question






















      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      I am designing a lead acid battery charger using Battery University for guidance. This is the circuit I made. Battery Charger



      The power supply is from an Adjustable 12V power supply. I have set it to 14.4V. The mosfet is FQP55N06. The transistor is a BC547.
      The current limit based on testing with multimeter is 1.24A.



      If i connect a halogen bulb (12V 35W) of a motorcyle as load, the Mosfet limits the current but heats up rapidly. Why is that? (I connected a smaller 12V bulb which draws only 0.7A and the mosfet did not heat up)



      If I use this circuit to charge a battery, and if it draws current upto the circuit limit, will the Mosfet heat up? If it will, how can I prevent it, while maintaining this current limit?







      share|improve this question












      I am designing a lead acid battery charger using Battery University for guidance. This is the circuit I made. Battery Charger



      The power supply is from an Adjustable 12V power supply. I have set it to 14.4V. The mosfet is FQP55N06. The transistor is a BC547.
      The current limit based on testing with multimeter is 1.24A.



      If i connect a halogen bulb (12V 35W) of a motorcyle as load, the Mosfet limits the current but heats up rapidly. Why is that? (I connected a smaller 12V bulb which draws only 0.7A and the mosfet did not heat up)



      If I use this circuit to charge a battery, and if it draws current upto the circuit limit, will the Mosfet heat up? If it will, how can I prevent it, while maintaining this current limit?









      share|improve this question











      share|improve this question




      share|improve this question










      asked Aug 21 at 16:09









      Subin Roy

      134




      134




















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          4
          down vote



          accepted










          The mosfet heats up because your halogen bulb has a low resistance. 12V at 35W means it has a heated up resistance of 4 Ohms, and when its cold, the resistance may be 10x lower, at 0.4 ohms.



          If the load is in-between these two, at 2 ohms, and the current through it is 1.24A, the voltage across it is 2.48V. If the current through the other resistor is 1.24A, the voltage across it is 0.5828V. If the supply is set to 14.4V, then the voltage across the mosfet must be 14.4-(2.48+0.5828)=11.3V. At 1.24A, that means the mosfet is dissipating 14W of heat, which is huge. Even if the bulb is fully heated up with a resistance of 4 Ohms, the mosfet will still be dissipating over 10W of heat.



          If your 12V bulb is drawing 0.7A, its resistance is probably around 17 Ohms (assuming the voltage across it is 12V). This means that the voltage across the mosfet will be much lower, and so it wont be dissipating as much heat. Analyzing this actually gets a bit complicated, maybe impossible, since we dont actually know what the resistance of the bulb is at that moment. If you gave a power rating, we could calculate it in the same was as for the other bulb.



          If you charge a battery with this, it will probably be ok. Broadly speaking, the higher the voltage across the battery, the lower the voltage across the mosfet. If the battery voltage is above 11V (which it will be unless the battery is dangerously discharged), the maximum voltage across the mosfet will be around 2V, which at 1.24A is around 3W. The datasheet gives the junction to ambient thermal resistance as 62.5C/W, so it will heat up to >180C (and break quickly), so you'll need a significant heatsink.



          This heatsink has a thermal resistance of 9.6C/W. Combined with the 1.63C/W internal thermal resistance of the mosfet, you should get a temperature rise of around 35C, which isn't too bad, although is still pretty warm (assuming 20C ambient, this is 55C total, which is above the threshold for thermal burns for humans). If you add a fan, it will lower the temp even more.






          share|improve this answer





























            up vote
            2
            down vote













            While BeB00's answer is generally correct, there is a simpler explanation. Note that your halogen is rated for 35 watts at 12 volts. This implies a normal load current of 3 amps (close enough). When you force it to run at 1.24 amps, this clearly implies that there will not be 12 volts across it (since, if there were 12 volts, there would be 3 amps through it, right?). With less than 12 volts across the bulb, there must be more voltage across the FET, and since the FET/bulb combination are pulling 1.24 amps, there must be significant power being dissipated in the FET. You can check this simply enough: put a meter across the bulb, and then the FET, when operating.






            share|improve this answer






















              Your Answer




              StackExchange.ifUsing("editor", function ()
              return StackExchange.using("mathjaxEditing", function ()
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
              );
              );
              , "mathjax-editing");

              StackExchange.ifUsing("editor", function ()
              return StackExchange.using("schematics", function ()
              StackExchange.schematics.init();
              );
              , "cicuitlab");

              StackExchange.ready(function()
              var channelOptions =
              tags: "".split(" "),
              id: "135"
              ;
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function()
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled)
              StackExchange.using("snippets", function()
              createEditor();
              );

              else
              createEditor();

              );

              function createEditor()
              StackExchange.prepareEditor(
              heartbeatType: 'answer',
              convertImagesToLinks: false,
              noModals: false,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: null,
              bindNavPrevention: true,
              postfix: "",
              onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              );



              );













               

              draft saved


              draft discarded


















              StackExchange.ready(
              function ()
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f392083%2fmosfet-heating-in-current-limiter-circuit%23new-answer', 'question_page');

              );

              Post as a guest






























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              4
              down vote



              accepted










              The mosfet heats up because your halogen bulb has a low resistance. 12V at 35W means it has a heated up resistance of 4 Ohms, and when its cold, the resistance may be 10x lower, at 0.4 ohms.



              If the load is in-between these two, at 2 ohms, and the current through it is 1.24A, the voltage across it is 2.48V. If the current through the other resistor is 1.24A, the voltage across it is 0.5828V. If the supply is set to 14.4V, then the voltage across the mosfet must be 14.4-(2.48+0.5828)=11.3V. At 1.24A, that means the mosfet is dissipating 14W of heat, which is huge. Even if the bulb is fully heated up with a resistance of 4 Ohms, the mosfet will still be dissipating over 10W of heat.



              If your 12V bulb is drawing 0.7A, its resistance is probably around 17 Ohms (assuming the voltage across it is 12V). This means that the voltage across the mosfet will be much lower, and so it wont be dissipating as much heat. Analyzing this actually gets a bit complicated, maybe impossible, since we dont actually know what the resistance of the bulb is at that moment. If you gave a power rating, we could calculate it in the same was as for the other bulb.



              If you charge a battery with this, it will probably be ok. Broadly speaking, the higher the voltage across the battery, the lower the voltage across the mosfet. If the battery voltage is above 11V (which it will be unless the battery is dangerously discharged), the maximum voltage across the mosfet will be around 2V, which at 1.24A is around 3W. The datasheet gives the junction to ambient thermal resistance as 62.5C/W, so it will heat up to >180C (and break quickly), so you'll need a significant heatsink.



              This heatsink has a thermal resistance of 9.6C/W. Combined with the 1.63C/W internal thermal resistance of the mosfet, you should get a temperature rise of around 35C, which isn't too bad, although is still pretty warm (assuming 20C ambient, this is 55C total, which is above the threshold for thermal burns for humans). If you add a fan, it will lower the temp even more.






              share|improve this answer


























                up vote
                4
                down vote



                accepted










                The mosfet heats up because your halogen bulb has a low resistance. 12V at 35W means it has a heated up resistance of 4 Ohms, and when its cold, the resistance may be 10x lower, at 0.4 ohms.



                If the load is in-between these two, at 2 ohms, and the current through it is 1.24A, the voltage across it is 2.48V. If the current through the other resistor is 1.24A, the voltage across it is 0.5828V. If the supply is set to 14.4V, then the voltage across the mosfet must be 14.4-(2.48+0.5828)=11.3V. At 1.24A, that means the mosfet is dissipating 14W of heat, which is huge. Even if the bulb is fully heated up with a resistance of 4 Ohms, the mosfet will still be dissipating over 10W of heat.



                If your 12V bulb is drawing 0.7A, its resistance is probably around 17 Ohms (assuming the voltage across it is 12V). This means that the voltage across the mosfet will be much lower, and so it wont be dissipating as much heat. Analyzing this actually gets a bit complicated, maybe impossible, since we dont actually know what the resistance of the bulb is at that moment. If you gave a power rating, we could calculate it in the same was as for the other bulb.



                If you charge a battery with this, it will probably be ok. Broadly speaking, the higher the voltage across the battery, the lower the voltage across the mosfet. If the battery voltage is above 11V (which it will be unless the battery is dangerously discharged), the maximum voltage across the mosfet will be around 2V, which at 1.24A is around 3W. The datasheet gives the junction to ambient thermal resistance as 62.5C/W, so it will heat up to >180C (and break quickly), so you'll need a significant heatsink.



                This heatsink has a thermal resistance of 9.6C/W. Combined with the 1.63C/W internal thermal resistance of the mosfet, you should get a temperature rise of around 35C, which isn't too bad, although is still pretty warm (assuming 20C ambient, this is 55C total, which is above the threshold for thermal burns for humans). If you add a fan, it will lower the temp even more.






                share|improve this answer
























                  up vote
                  4
                  down vote



                  accepted







                  up vote
                  4
                  down vote



                  accepted






                  The mosfet heats up because your halogen bulb has a low resistance. 12V at 35W means it has a heated up resistance of 4 Ohms, and when its cold, the resistance may be 10x lower, at 0.4 ohms.



                  If the load is in-between these two, at 2 ohms, and the current through it is 1.24A, the voltage across it is 2.48V. If the current through the other resistor is 1.24A, the voltage across it is 0.5828V. If the supply is set to 14.4V, then the voltage across the mosfet must be 14.4-(2.48+0.5828)=11.3V. At 1.24A, that means the mosfet is dissipating 14W of heat, which is huge. Even if the bulb is fully heated up with a resistance of 4 Ohms, the mosfet will still be dissipating over 10W of heat.



                  If your 12V bulb is drawing 0.7A, its resistance is probably around 17 Ohms (assuming the voltage across it is 12V). This means that the voltage across the mosfet will be much lower, and so it wont be dissipating as much heat. Analyzing this actually gets a bit complicated, maybe impossible, since we dont actually know what the resistance of the bulb is at that moment. If you gave a power rating, we could calculate it in the same was as for the other bulb.



                  If you charge a battery with this, it will probably be ok. Broadly speaking, the higher the voltage across the battery, the lower the voltage across the mosfet. If the battery voltage is above 11V (which it will be unless the battery is dangerously discharged), the maximum voltage across the mosfet will be around 2V, which at 1.24A is around 3W. The datasheet gives the junction to ambient thermal resistance as 62.5C/W, so it will heat up to >180C (and break quickly), so you'll need a significant heatsink.



                  This heatsink has a thermal resistance of 9.6C/W. Combined with the 1.63C/W internal thermal resistance of the mosfet, you should get a temperature rise of around 35C, which isn't too bad, although is still pretty warm (assuming 20C ambient, this is 55C total, which is above the threshold for thermal burns for humans). If you add a fan, it will lower the temp even more.






                  share|improve this answer














                  The mosfet heats up because your halogen bulb has a low resistance. 12V at 35W means it has a heated up resistance of 4 Ohms, and when its cold, the resistance may be 10x lower, at 0.4 ohms.



                  If the load is in-between these two, at 2 ohms, and the current through it is 1.24A, the voltage across it is 2.48V. If the current through the other resistor is 1.24A, the voltage across it is 0.5828V. If the supply is set to 14.4V, then the voltage across the mosfet must be 14.4-(2.48+0.5828)=11.3V. At 1.24A, that means the mosfet is dissipating 14W of heat, which is huge. Even if the bulb is fully heated up with a resistance of 4 Ohms, the mosfet will still be dissipating over 10W of heat.



                  If your 12V bulb is drawing 0.7A, its resistance is probably around 17 Ohms (assuming the voltage across it is 12V). This means that the voltage across the mosfet will be much lower, and so it wont be dissipating as much heat. Analyzing this actually gets a bit complicated, maybe impossible, since we dont actually know what the resistance of the bulb is at that moment. If you gave a power rating, we could calculate it in the same was as for the other bulb.



                  If you charge a battery with this, it will probably be ok. Broadly speaking, the higher the voltage across the battery, the lower the voltage across the mosfet. If the battery voltage is above 11V (which it will be unless the battery is dangerously discharged), the maximum voltage across the mosfet will be around 2V, which at 1.24A is around 3W. The datasheet gives the junction to ambient thermal resistance as 62.5C/W, so it will heat up to >180C (and break quickly), so you'll need a significant heatsink.



                  This heatsink has a thermal resistance of 9.6C/W. Combined with the 1.63C/W internal thermal resistance of the mosfet, you should get a temperature rise of around 35C, which isn't too bad, although is still pretty warm (assuming 20C ambient, this is 55C total, which is above the threshold for thermal burns for humans). If you add a fan, it will lower the temp even more.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Aug 21 at 17:05

























                  answered Aug 21 at 16:45









                  BeB00

                  2,838619




                  2,838619






















                      up vote
                      2
                      down vote













                      While BeB00's answer is generally correct, there is a simpler explanation. Note that your halogen is rated for 35 watts at 12 volts. This implies a normal load current of 3 amps (close enough). When you force it to run at 1.24 amps, this clearly implies that there will not be 12 volts across it (since, if there were 12 volts, there would be 3 amps through it, right?). With less than 12 volts across the bulb, there must be more voltage across the FET, and since the FET/bulb combination are pulling 1.24 amps, there must be significant power being dissipated in the FET. You can check this simply enough: put a meter across the bulb, and then the FET, when operating.






                      share|improve this answer


























                        up vote
                        2
                        down vote













                        While BeB00's answer is generally correct, there is a simpler explanation. Note that your halogen is rated for 35 watts at 12 volts. This implies a normal load current of 3 amps (close enough). When you force it to run at 1.24 amps, this clearly implies that there will not be 12 volts across it (since, if there were 12 volts, there would be 3 amps through it, right?). With less than 12 volts across the bulb, there must be more voltage across the FET, and since the FET/bulb combination are pulling 1.24 amps, there must be significant power being dissipated in the FET. You can check this simply enough: put a meter across the bulb, and then the FET, when operating.






                        share|improve this answer
























                          up vote
                          2
                          down vote










                          up vote
                          2
                          down vote









                          While BeB00's answer is generally correct, there is a simpler explanation. Note that your halogen is rated for 35 watts at 12 volts. This implies a normal load current of 3 amps (close enough). When you force it to run at 1.24 amps, this clearly implies that there will not be 12 volts across it (since, if there were 12 volts, there would be 3 amps through it, right?). With less than 12 volts across the bulb, there must be more voltage across the FET, and since the FET/bulb combination are pulling 1.24 amps, there must be significant power being dissipated in the FET. You can check this simply enough: put a meter across the bulb, and then the FET, when operating.






                          share|improve this answer














                          While BeB00's answer is generally correct, there is a simpler explanation. Note that your halogen is rated for 35 watts at 12 volts. This implies a normal load current of 3 amps (close enough). When you force it to run at 1.24 amps, this clearly implies that there will not be 12 volts across it (since, if there were 12 volts, there would be 3 amps through it, right?). With less than 12 volts across the bulb, there must be more voltage across the FET, and since the FET/bulb combination are pulling 1.24 amps, there must be significant power being dissipated in the FET. You can check this simply enough: put a meter across the bulb, and then the FET, when operating.







                          share|improve this answer














                          share|improve this answer



                          share|improve this answer








                          edited Aug 22 at 0:38

























                          answered Aug 21 at 17:13









                          WhatRoughBeast

                          47k22770




                          47k22770



























                               

                              draft saved


                              draft discarded















































                               


                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function ()
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f392083%2fmosfet-heating-in-current-limiter-circuit%23new-answer', 'question_page');

                              );

                              Post as a guest













































































                              Comments

                              Popular posts from this blog

                              What does second last employer means? [closed]

                              List of Gilmore Girls characters

                              One-line joke