LM317 - Calculate dissipation due to voltage dropWhy don't people tend to use voltage dividers or zeners in...

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LM317 - Calculate dissipation due to voltage drop


Why don't people tend to use voltage dividers or zeners in front of linear regulatorsWhich kind of regulator/battery setup should I use for 5 volt microcontroller and transmitter circuitry for optimum performance?Internal thermal limit of LF00 voltage regulatorsHow to calculate heatsink requirements? Explanation of the K/W unitDoubt on the Voltage Regulator Application with LiPo BatteryHow to limit current for voltage regulator/decrease power dissipation?Is it okay to use a device whose junction temperature is more than operating temperature?sizing voltage regulatorHigh voltage linear regulator (with pre-regulator)Voltage regulator with heatsink gets overheated






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}







6












$begingroup$


If I use a linear voltage regulator as LM317:




  • Input voltage = 24 V

  • Output voltage = 5 V (so a voltage drop of 19 V)

  • Mean load = 480 mA (peak load = 700 mA)


From the datasheet, I read that maximum operating temperature = 125°C, junction = 150°C



How I can calculate if this component is thermally suitable, instead of a switching regulator?










share|improve this question











$endgroup$








  • 1




    $begingroup$
    Even if it's "thermally suitable" you're still wasting almost four times as much power as you're actually using. I'd use a switcher unless you have an overwhelming reason not to.
    $endgroup$
    – Finbarr
    20 hours ago












  • $begingroup$
    @Finbarr "for times as much power" is not really relevant, it's the power you lose that matters. If the OP had a load consuming 4.8mA, then a linear regulator could have been absolutely the way to go.
    $endgroup$
    – Dmitry Grigoryev
    15 hours ago


















6












$begingroup$


If I use a linear voltage regulator as LM317:




  • Input voltage = 24 V

  • Output voltage = 5 V (so a voltage drop of 19 V)

  • Mean load = 480 mA (peak load = 700 mA)


From the datasheet, I read that maximum operating temperature = 125°C, junction = 150°C



How I can calculate if this component is thermally suitable, instead of a switching regulator?










share|improve this question











$endgroup$








  • 1




    $begingroup$
    Even if it's "thermally suitable" you're still wasting almost four times as much power as you're actually using. I'd use a switcher unless you have an overwhelming reason not to.
    $endgroup$
    – Finbarr
    20 hours ago












  • $begingroup$
    @Finbarr "for times as much power" is not really relevant, it's the power you lose that matters. If the OP had a load consuming 4.8mA, then a linear regulator could have been absolutely the way to go.
    $endgroup$
    – Dmitry Grigoryev
    15 hours ago














6












6








6


2



$begingroup$


If I use a linear voltage regulator as LM317:




  • Input voltage = 24 V

  • Output voltage = 5 V (so a voltage drop of 19 V)

  • Mean load = 480 mA (peak load = 700 mA)


From the datasheet, I read that maximum operating temperature = 125°C, junction = 150°C



How I can calculate if this component is thermally suitable, instead of a switching regulator?










share|improve this question











$endgroup$




If I use a linear voltage regulator as LM317:




  • Input voltage = 24 V

  • Output voltage = 5 V (so a voltage drop of 19 V)

  • Mean load = 480 mA (peak load = 700 mA)


From the datasheet, I read that maximum operating temperature = 125°C, junction = 150°C



How I can calculate if this component is thermally suitable, instead of a switching regulator?







temperature heat thermal linear-regulator power-dissipation






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 17 hours ago









SamGibson

11.7k41739




11.7k41739










asked 20 hours ago









SingedSinged

684




684








  • 1




    $begingroup$
    Even if it's "thermally suitable" you're still wasting almost four times as much power as you're actually using. I'd use a switcher unless you have an overwhelming reason not to.
    $endgroup$
    – Finbarr
    20 hours ago












  • $begingroup$
    @Finbarr "for times as much power" is not really relevant, it's the power you lose that matters. If the OP had a load consuming 4.8mA, then a linear regulator could have been absolutely the way to go.
    $endgroup$
    – Dmitry Grigoryev
    15 hours ago














  • 1




    $begingroup$
    Even if it's "thermally suitable" you're still wasting almost four times as much power as you're actually using. I'd use a switcher unless you have an overwhelming reason not to.
    $endgroup$
    – Finbarr
    20 hours ago












  • $begingroup$
    @Finbarr "for times as much power" is not really relevant, it's the power you lose that matters. If the OP had a load consuming 4.8mA, then a linear regulator could have been absolutely the way to go.
    $endgroup$
    – Dmitry Grigoryev
    15 hours ago








1




1




$begingroup$
Even if it's "thermally suitable" you're still wasting almost four times as much power as you're actually using. I'd use a switcher unless you have an overwhelming reason not to.
$endgroup$
– Finbarr
20 hours ago






$begingroup$
Even if it's "thermally suitable" you're still wasting almost four times as much power as you're actually using. I'd use a switcher unless you have an overwhelming reason not to.
$endgroup$
– Finbarr
20 hours ago














$begingroup$
@Finbarr "for times as much power" is not really relevant, it's the power you lose that matters. If the OP had a load consuming 4.8mA, then a linear regulator could have been absolutely the way to go.
$endgroup$
– Dmitry Grigoryev
15 hours ago




$begingroup$
@Finbarr "for times as much power" is not really relevant, it's the power you lose that matters. If the OP had a load consuming 4.8mA, then a linear regulator could have been absolutely the way to go.
$endgroup$
– Dmitry Grigoryev
15 hours ago










2 Answers
2






active

oldest

votes


















8












$begingroup$

What you need to know is the Junction-to-Ambient thermal resistance. There is a table for that on page 4 of the DATASHEET.



enter image description here



It has the values for each of the packages.



So, you know your voltage drop (19V), you know the required current (480-700mA). With this information, you can now find out your power dissipated (P=IV) and use this value to see how much your IC will heat up.



For an example, lets assume you have a TO-263 package. You calculate your power to be 3.68W. You see that the thermal resistance of this package is 38°C/W, thus the temperature will rise by 139.8°C. Now, while you may think that is fine, because it is under 150°C, you also need to add on the ambient temperature of the environment. Assuming this is 25°C, this will give you a total of 164.8°C. This now exceeds the maximum.



There are other factors involved, such as the current drawn by the device itself, not just your load, some environmental factors etc, but this is the easiest way to calculate what your temperature could be. You can use this method for any IC, not just the LM317, and you should find all the information here for you to calculate this yourself.



Further reading






share|improve this answer









$endgroup$













  • $begingroup$
    I calculated the mean power in the regulator as 19V * 480mA as 9.12W with a temperature rise (TO220 package) of 214C.
    $endgroup$
    – Peter Smith
    20 hours ago






  • 4




    $begingroup$
    @PeterSmith I deliberately missed out the exact calculations for OP so they can work it out for themselves
    $endgroup$
    – MCG
    20 hours ago






  • 1




    $begingroup$
    You are assuming no heat sink. With a heat sink and a series resistor to dump some of the power, it has a chance. Unfortunately, convection calculations for heat sinks are much more complicated than a conducted heat analysis. When I was working, I would let an ME do the thermal analysis. At home, with lot of experience by trial and error, I know about how big a heat sink needs to be.
    $endgroup$
    – Mattman944
    20 hours ago






  • 3




    $begingroup$
    @Mattman944 Correct. I am assuming no heat sink. Why would I include a heatsink when the question was not asking about that? The whole point of this answer was to make it simple and allow OP to calculate it themselves. They will realise they need a heatsink (actually easier to just use a switching IC) then they can ask another question about heatsinks
    $endgroup$
    – MCG
    18 hours ago












  • $begingroup$
    1. Wot? No TO3 ? :-) 2. A heatsink is an integral part of such a design - whether it is the one on the device used (the metal and/or plastic overall packaging) alone or that plus any added one.3. A series resistor is also an extremely good idea idf they insist on doing this.
    $endgroup$
    – Russell McMahon
    3 hours ago



















4












$begingroup$

You need to know the package. Then calculate the temperature rises. Thermal calculations are analogous to electrical calculations, power dissipated is analogous to current. Thermal resistance is analogous to electrical resistance, temperature is analogous to voltage.



Rough calculations show that you will need a really good heat sink to keep the case temperature reasonable. Some of these parts have a large metal pad on the bottom designed to be soldered to a large copper pad on the PWB.



Another trick to use is to put a resistor in series with the input power to dump some of the power.



enter image description here






share|improve this answer








New contributor




Mattman944 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    8












    $begingroup$

    What you need to know is the Junction-to-Ambient thermal resistance. There is a table for that on page 4 of the DATASHEET.



    enter image description here



    It has the values for each of the packages.



    So, you know your voltage drop (19V), you know the required current (480-700mA). With this information, you can now find out your power dissipated (P=IV) and use this value to see how much your IC will heat up.



    For an example, lets assume you have a TO-263 package. You calculate your power to be 3.68W. You see that the thermal resistance of this package is 38°C/W, thus the temperature will rise by 139.8°C. Now, while you may think that is fine, because it is under 150°C, you also need to add on the ambient temperature of the environment. Assuming this is 25°C, this will give you a total of 164.8°C. This now exceeds the maximum.



    There are other factors involved, such as the current drawn by the device itself, not just your load, some environmental factors etc, but this is the easiest way to calculate what your temperature could be. You can use this method for any IC, not just the LM317, and you should find all the information here for you to calculate this yourself.



    Further reading






    share|improve this answer









    $endgroup$













    • $begingroup$
      I calculated the mean power in the regulator as 19V * 480mA as 9.12W with a temperature rise (TO220 package) of 214C.
      $endgroup$
      – Peter Smith
      20 hours ago






    • 4




      $begingroup$
      @PeterSmith I deliberately missed out the exact calculations for OP so they can work it out for themselves
      $endgroup$
      – MCG
      20 hours ago






    • 1




      $begingroup$
      You are assuming no heat sink. With a heat sink and a series resistor to dump some of the power, it has a chance. Unfortunately, convection calculations for heat sinks are much more complicated than a conducted heat analysis. When I was working, I would let an ME do the thermal analysis. At home, with lot of experience by trial and error, I know about how big a heat sink needs to be.
      $endgroup$
      – Mattman944
      20 hours ago






    • 3




      $begingroup$
      @Mattman944 Correct. I am assuming no heat sink. Why would I include a heatsink when the question was not asking about that? The whole point of this answer was to make it simple and allow OP to calculate it themselves. They will realise they need a heatsink (actually easier to just use a switching IC) then they can ask another question about heatsinks
      $endgroup$
      – MCG
      18 hours ago












    • $begingroup$
      1. Wot? No TO3 ? :-) 2. A heatsink is an integral part of such a design - whether it is the one on the device used (the metal and/or plastic overall packaging) alone or that plus any added one.3. A series resistor is also an extremely good idea idf they insist on doing this.
      $endgroup$
      – Russell McMahon
      3 hours ago
















    8












    $begingroup$

    What you need to know is the Junction-to-Ambient thermal resistance. There is a table for that on page 4 of the DATASHEET.



    enter image description here



    It has the values for each of the packages.



    So, you know your voltage drop (19V), you know the required current (480-700mA). With this information, you can now find out your power dissipated (P=IV) and use this value to see how much your IC will heat up.



    For an example, lets assume you have a TO-263 package. You calculate your power to be 3.68W. You see that the thermal resistance of this package is 38°C/W, thus the temperature will rise by 139.8°C. Now, while you may think that is fine, because it is under 150°C, you also need to add on the ambient temperature of the environment. Assuming this is 25°C, this will give you a total of 164.8°C. This now exceeds the maximum.



    There are other factors involved, such as the current drawn by the device itself, not just your load, some environmental factors etc, but this is the easiest way to calculate what your temperature could be. You can use this method for any IC, not just the LM317, and you should find all the information here for you to calculate this yourself.



    Further reading






    share|improve this answer









    $endgroup$













    • $begingroup$
      I calculated the mean power in the regulator as 19V * 480mA as 9.12W with a temperature rise (TO220 package) of 214C.
      $endgroup$
      – Peter Smith
      20 hours ago






    • 4




      $begingroup$
      @PeterSmith I deliberately missed out the exact calculations for OP so they can work it out for themselves
      $endgroup$
      – MCG
      20 hours ago






    • 1




      $begingroup$
      You are assuming no heat sink. With a heat sink and a series resistor to dump some of the power, it has a chance. Unfortunately, convection calculations for heat sinks are much more complicated than a conducted heat analysis. When I was working, I would let an ME do the thermal analysis. At home, with lot of experience by trial and error, I know about how big a heat sink needs to be.
      $endgroup$
      – Mattman944
      20 hours ago






    • 3




      $begingroup$
      @Mattman944 Correct. I am assuming no heat sink. Why would I include a heatsink when the question was not asking about that? The whole point of this answer was to make it simple and allow OP to calculate it themselves. They will realise they need a heatsink (actually easier to just use a switching IC) then they can ask another question about heatsinks
      $endgroup$
      – MCG
      18 hours ago












    • $begingroup$
      1. Wot? No TO3 ? :-) 2. A heatsink is an integral part of such a design - whether it is the one on the device used (the metal and/or plastic overall packaging) alone or that plus any added one.3. A series resistor is also an extremely good idea idf they insist on doing this.
      $endgroup$
      – Russell McMahon
      3 hours ago














    8












    8








    8





    $begingroup$

    What you need to know is the Junction-to-Ambient thermal resistance. There is a table for that on page 4 of the DATASHEET.



    enter image description here



    It has the values for each of the packages.



    So, you know your voltage drop (19V), you know the required current (480-700mA). With this information, you can now find out your power dissipated (P=IV) and use this value to see how much your IC will heat up.



    For an example, lets assume you have a TO-263 package. You calculate your power to be 3.68W. You see that the thermal resistance of this package is 38°C/W, thus the temperature will rise by 139.8°C. Now, while you may think that is fine, because it is under 150°C, you also need to add on the ambient temperature of the environment. Assuming this is 25°C, this will give you a total of 164.8°C. This now exceeds the maximum.



    There are other factors involved, such as the current drawn by the device itself, not just your load, some environmental factors etc, but this is the easiest way to calculate what your temperature could be. You can use this method for any IC, not just the LM317, and you should find all the information here for you to calculate this yourself.



    Further reading






    share|improve this answer









    $endgroup$



    What you need to know is the Junction-to-Ambient thermal resistance. There is a table for that on page 4 of the DATASHEET.



    enter image description here



    It has the values for each of the packages.



    So, you know your voltage drop (19V), you know the required current (480-700mA). With this information, you can now find out your power dissipated (P=IV) and use this value to see how much your IC will heat up.



    For an example, lets assume you have a TO-263 package. You calculate your power to be 3.68W. You see that the thermal resistance of this package is 38°C/W, thus the temperature will rise by 139.8°C. Now, while you may think that is fine, because it is under 150°C, you also need to add on the ambient temperature of the environment. Assuming this is 25°C, this will give you a total of 164.8°C. This now exceeds the maximum.



    There are other factors involved, such as the current drawn by the device itself, not just your load, some environmental factors etc, but this is the easiest way to calculate what your temperature could be. You can use this method for any IC, not just the LM317, and you should find all the information here for you to calculate this yourself.



    Further reading







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 20 hours ago









    MCGMCG

    6,66431850




    6,66431850












    • $begingroup$
      I calculated the mean power in the regulator as 19V * 480mA as 9.12W with a temperature rise (TO220 package) of 214C.
      $endgroup$
      – Peter Smith
      20 hours ago






    • 4




      $begingroup$
      @PeterSmith I deliberately missed out the exact calculations for OP so they can work it out for themselves
      $endgroup$
      – MCG
      20 hours ago






    • 1




      $begingroup$
      You are assuming no heat sink. With a heat sink and a series resistor to dump some of the power, it has a chance. Unfortunately, convection calculations for heat sinks are much more complicated than a conducted heat analysis. When I was working, I would let an ME do the thermal analysis. At home, with lot of experience by trial and error, I know about how big a heat sink needs to be.
      $endgroup$
      – Mattman944
      20 hours ago






    • 3




      $begingroup$
      @Mattman944 Correct. I am assuming no heat sink. Why would I include a heatsink when the question was not asking about that? The whole point of this answer was to make it simple and allow OP to calculate it themselves. They will realise they need a heatsink (actually easier to just use a switching IC) then they can ask another question about heatsinks
      $endgroup$
      – MCG
      18 hours ago












    • $begingroup$
      1. Wot? No TO3 ? :-) 2. A heatsink is an integral part of such a design - whether it is the one on the device used (the metal and/or plastic overall packaging) alone or that plus any added one.3. A series resistor is also an extremely good idea idf they insist on doing this.
      $endgroup$
      – Russell McMahon
      3 hours ago


















    • $begingroup$
      I calculated the mean power in the regulator as 19V * 480mA as 9.12W with a temperature rise (TO220 package) of 214C.
      $endgroup$
      – Peter Smith
      20 hours ago






    • 4




      $begingroup$
      @PeterSmith I deliberately missed out the exact calculations for OP so they can work it out for themselves
      $endgroup$
      – MCG
      20 hours ago






    • 1




      $begingroup$
      You are assuming no heat sink. With a heat sink and a series resistor to dump some of the power, it has a chance. Unfortunately, convection calculations for heat sinks are much more complicated than a conducted heat analysis. When I was working, I would let an ME do the thermal analysis. At home, with lot of experience by trial and error, I know about how big a heat sink needs to be.
      $endgroup$
      – Mattman944
      20 hours ago






    • 3




      $begingroup$
      @Mattman944 Correct. I am assuming no heat sink. Why would I include a heatsink when the question was not asking about that? The whole point of this answer was to make it simple and allow OP to calculate it themselves. They will realise they need a heatsink (actually easier to just use a switching IC) then they can ask another question about heatsinks
      $endgroup$
      – MCG
      18 hours ago












    • $begingroup$
      1. Wot? No TO3 ? :-) 2. A heatsink is an integral part of such a design - whether it is the one on the device used (the metal and/or plastic overall packaging) alone or that plus any added one.3. A series resistor is also an extremely good idea idf they insist on doing this.
      $endgroup$
      – Russell McMahon
      3 hours ago
















    $begingroup$
    I calculated the mean power in the regulator as 19V * 480mA as 9.12W with a temperature rise (TO220 package) of 214C.
    $endgroup$
    – Peter Smith
    20 hours ago




    $begingroup$
    I calculated the mean power in the regulator as 19V * 480mA as 9.12W with a temperature rise (TO220 package) of 214C.
    $endgroup$
    – Peter Smith
    20 hours ago




    4




    4




    $begingroup$
    @PeterSmith I deliberately missed out the exact calculations for OP so they can work it out for themselves
    $endgroup$
    – MCG
    20 hours ago




    $begingroup$
    @PeterSmith I deliberately missed out the exact calculations for OP so they can work it out for themselves
    $endgroup$
    – MCG
    20 hours ago




    1




    1




    $begingroup$
    You are assuming no heat sink. With a heat sink and a series resistor to dump some of the power, it has a chance. Unfortunately, convection calculations for heat sinks are much more complicated than a conducted heat analysis. When I was working, I would let an ME do the thermal analysis. At home, with lot of experience by trial and error, I know about how big a heat sink needs to be.
    $endgroup$
    – Mattman944
    20 hours ago




    $begingroup$
    You are assuming no heat sink. With a heat sink and a series resistor to dump some of the power, it has a chance. Unfortunately, convection calculations for heat sinks are much more complicated than a conducted heat analysis. When I was working, I would let an ME do the thermal analysis. At home, with lot of experience by trial and error, I know about how big a heat sink needs to be.
    $endgroup$
    – Mattman944
    20 hours ago




    3




    3




    $begingroup$
    @Mattman944 Correct. I am assuming no heat sink. Why would I include a heatsink when the question was not asking about that? The whole point of this answer was to make it simple and allow OP to calculate it themselves. They will realise they need a heatsink (actually easier to just use a switching IC) then they can ask another question about heatsinks
    $endgroup$
    – MCG
    18 hours ago






    $begingroup$
    @Mattman944 Correct. I am assuming no heat sink. Why would I include a heatsink when the question was not asking about that? The whole point of this answer was to make it simple and allow OP to calculate it themselves. They will realise they need a heatsink (actually easier to just use a switching IC) then they can ask another question about heatsinks
    $endgroup$
    – MCG
    18 hours ago














    $begingroup$
    1. Wot? No TO3 ? :-) 2. A heatsink is an integral part of such a design - whether it is the one on the device used (the metal and/or plastic overall packaging) alone or that plus any added one.3. A series resistor is also an extremely good idea idf they insist on doing this.
    $endgroup$
    – Russell McMahon
    3 hours ago




    $begingroup$
    1. Wot? No TO3 ? :-) 2. A heatsink is an integral part of such a design - whether it is the one on the device used (the metal and/or plastic overall packaging) alone or that plus any added one.3. A series resistor is also an extremely good idea idf they insist on doing this.
    $endgroup$
    – Russell McMahon
    3 hours ago













    4












    $begingroup$

    You need to know the package. Then calculate the temperature rises. Thermal calculations are analogous to electrical calculations, power dissipated is analogous to current. Thermal resistance is analogous to electrical resistance, temperature is analogous to voltage.



    Rough calculations show that you will need a really good heat sink to keep the case temperature reasonable. Some of these parts have a large metal pad on the bottom designed to be soldered to a large copper pad on the PWB.



    Another trick to use is to put a resistor in series with the input power to dump some of the power.



    enter image description here






    share|improve this answer








    New contributor




    Mattman944 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    $endgroup$


















      4












      $begingroup$

      You need to know the package. Then calculate the temperature rises. Thermal calculations are analogous to electrical calculations, power dissipated is analogous to current. Thermal resistance is analogous to electrical resistance, temperature is analogous to voltage.



      Rough calculations show that you will need a really good heat sink to keep the case temperature reasonable. Some of these parts have a large metal pad on the bottom designed to be soldered to a large copper pad on the PWB.



      Another trick to use is to put a resistor in series with the input power to dump some of the power.



      enter image description here






      share|improve this answer








      New contributor




      Mattman944 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      $endgroup$
















        4












        4








        4





        $begingroup$

        You need to know the package. Then calculate the temperature rises. Thermal calculations are analogous to electrical calculations, power dissipated is analogous to current. Thermal resistance is analogous to electrical resistance, temperature is analogous to voltage.



        Rough calculations show that you will need a really good heat sink to keep the case temperature reasonable. Some of these parts have a large metal pad on the bottom designed to be soldered to a large copper pad on the PWB.



        Another trick to use is to put a resistor in series with the input power to dump some of the power.



        enter image description here






        share|improve this answer








        New contributor




        Mattman944 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






        $endgroup$



        You need to know the package. Then calculate the temperature rises. Thermal calculations are analogous to electrical calculations, power dissipated is analogous to current. Thermal resistance is analogous to electrical resistance, temperature is analogous to voltage.



        Rough calculations show that you will need a really good heat sink to keep the case temperature reasonable. Some of these parts have a large metal pad on the bottom designed to be soldered to a large copper pad on the PWB.



        Another trick to use is to put a resistor in series with the input power to dump some of the power.



        enter image description here







        share|improve this answer








        New contributor




        Mattman944 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        share|improve this answer



        share|improve this answer






        New contributor




        Mattman944 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.









        answered 20 hours ago









        Mattman944Mattman944

        1314




        1314




        New contributor




        Mattman944 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.





        New contributor





        Mattman944 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






        Mattman944 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






























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