Will linear voltage regulator step up current?












11












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I have a regulated 9 volt 300mA power supply I want to step it down to 5 volt using Linear Voltage Regulator LM7805 , I want to know how much current can I can draw at 5 volts, will it be 300mA or will it be close to 540mA, since power = voltage * current.






power-supply linear-regulator







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  • $begingroup$
    With a 9V*0.3A=2.7W supply you can only achieve >90% efficiency with an SMPS to store energy and transfer with rapid switching.
    $endgroup$
    – Sunnyskyguy EE75
    yesterday
















11












$begingroup$


I have a regulated 9 volt 300mA power supply I want to step it down to 5 volt using Linear Voltage Regulator LM7805 , I want to know how much current can I can draw at 5 volts, will it be 300mA or will it be close to 540mA, since power = voltage * current.






power-supply linear-regulator







share|improve this question







New contributor




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







$endgroup$












  • $begingroup$
    With a 9V*0.3A=2.7W supply you can only achieve >90% efficiency with an SMPS to store energy and transfer with rapid switching.
    $endgroup$
    – Sunnyskyguy EE75
    yesterday














11












11








11


2



$begingroup$


I have a regulated 9 volt 300mA power supply I want to step it down to 5 volt using Linear Voltage Regulator LM7805 , I want to know how much current can I can draw at 5 volts, will it be 300mA or will it be close to 540mA, since power = voltage * current.






power-supply linear-regulator







share|improve this question







New contributor




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







$endgroup$




I have a regulated 9 volt 300mA power supply I want to step it down to 5 volt using Linear Voltage Regulator LM7805 , I want to know how much current can I can draw at 5 volts, will it be 300mA or will it be close to 540mA, since power = voltage * current.






power-supply linear-regulator




power-supply linear-regulator






share|improve this question







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Jayant Pahuja 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 question







New contributor




Jayant Pahuja 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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share|improve this question






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asked yesterday









Jayant PahujaJayant Pahuja

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Jayant Pahuja is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • $begingroup$
    With a 9V*0.3A=2.7W supply you can only achieve >90% efficiency with an SMPS to store energy and transfer with rapid switching.
    $endgroup$
    – Sunnyskyguy EE75
    yesterday


















  • $begingroup$
    With a 9V*0.3A=2.7W supply you can only achieve >90% efficiency with an SMPS to store energy and transfer with rapid switching.
    $endgroup$
    – Sunnyskyguy EE75
    yesterday
















$begingroup$
With a 9V*0.3A=2.7W supply you can only achieve >90% efficiency with an SMPS to store energy and transfer with rapid switching.
$endgroup$
– Sunnyskyguy EE75
yesterday




$begingroup$
With a 9V*0.3A=2.7W supply you can only achieve >90% efficiency with an SMPS to store energy and transfer with rapid switching.
$endgroup$
– Sunnyskyguy EE75
yesterday










3 Answers
3






active

oldest

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28












$begingroup$

No. A linear regulator works by burning off excess voltage as heat, therefore current in equals current out. The linear regulator is essentially throwing away the excess energy in order to regulate, rather than converting it to the output. You need a switching regulator if you want to take advantage of power in equals power out in order to convert a high input voltage, low input current into a lower output voltage, higher output current.



$P_{in} = P_{out}$



but for a linear regulator it looks like this:



$V_{in} times I_{in} = (V_{out} times I_{out}) + [(V_{in} - V_{out}) times I_{out}]$



The last term in square brackets is the excess voltage being converted to heat. If we expand and simplify the right hand side, a bunch of things cancel out and we get:



$V_{in} times I_{in} = V_{in} times I_{out}$



Therefore:



$I_{in} = I_{out}$






share|improve this answer










New contributor




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






$endgroup$





















    10












    $begingroup$

    No, it won't step up current. You can think of a regulator as a resistor that adjusts it's resistance to keep the voltage stable.



    However, you can buy DC to DC converters that 'boost' the current. But DC to DC converters are usually called by what they do to the voltage, not the current.



    A boost converter 'boosts' or steps up the voltage from a lower voltage to a higher one (at the expense of current and a small loss in power)



    A buck converter or step down converter takes a higher voltage into a lower one (with potentially more current than is on the input of the converter, also with a small loss)



    They actually make 78XX series DC to DC converters that are drop in compatible with linear regulators that buck or boost voltage.






    share|improve this answer









    $endgroup$









    • 2




      $begingroup$
      "drop in compatible" - whee, thanks! that's an improvement a hobbyist like can easily overlook
      $endgroup$
      – quetzalcoatl
      15 hours ago



















    1












    $begingroup$


    since power = voltage * current.




    It is true when applied to both sides of devices that transform electricity (as AC transformers, or more sophisticated devices known as "DC-DC converters"). These devices do transform voltages/currents, so if the output voltage is lower, the output current might be higher. Keep in mind that these devices do the transformation with certain efficiency (80-90%), so the "output power" = "input power" x 0.8 practically.



    In the case of linear regulators it is not true, the regulators don't "transform", they just regulate output by dissipating the excess of voltage (drop-out voltage) in its regulating elements (transistors). Therefore whatever current comes in, the same current goes out, and even a bit less, since the regulation takes some toll. For example, the old LM7805 IC will consume within itself about 4-5 mA for its "services", so if your input is strictly 300 mA, you might get only 295 mA out.






    share|improve this answer











    $endgroup$













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      3 Answers
      3






      active

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      3 Answers
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      active

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      active

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      28












      $begingroup$

      No. A linear regulator works by burning off excess voltage as heat, therefore current in equals current out. The linear regulator is essentially throwing away the excess energy in order to regulate, rather than converting it to the output. You need a switching regulator if you want to take advantage of power in equals power out in order to convert a high input voltage, low input current into a lower output voltage, higher output current.



      $P_{in} = P_{out}$



      but for a linear regulator it looks like this:



      $V_{in} times I_{in} = (V_{out} times I_{out}) + [(V_{in} - V_{out}) times I_{out}]$



      The last term in square brackets is the excess voltage being converted to heat. If we expand and simplify the right hand side, a bunch of things cancel out and we get:



      $V_{in} times I_{in} = V_{in} times I_{out}$



      Therefore:



      $I_{in} = I_{out}$






      share|improve this answer










      New contributor




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






      $endgroup$


















        28












        $begingroup$

        No. A linear regulator works by burning off excess voltage as heat, therefore current in equals current out. The linear regulator is essentially throwing away the excess energy in order to regulate, rather than converting it to the output. You need a switching regulator if you want to take advantage of power in equals power out in order to convert a high input voltage, low input current into a lower output voltage, higher output current.



        $P_{in} = P_{out}$



        but for a linear regulator it looks like this:



        $V_{in} times I_{in} = (V_{out} times I_{out}) + [(V_{in} - V_{out}) times I_{out}]$



        The last term in square brackets is the excess voltage being converted to heat. If we expand and simplify the right hand side, a bunch of things cancel out and we get:



        $V_{in} times I_{in} = V_{in} times I_{out}$



        Therefore:



        $I_{in} = I_{out}$






        share|improve this answer










        New contributor




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






        $endgroup$
















          28












          28








          28





          $begingroup$

          No. A linear regulator works by burning off excess voltage as heat, therefore current in equals current out. The linear regulator is essentially throwing away the excess energy in order to regulate, rather than converting it to the output. You need a switching regulator if you want to take advantage of power in equals power out in order to convert a high input voltage, low input current into a lower output voltage, higher output current.



          $P_{in} = P_{out}$



          but for a linear regulator it looks like this:



          $V_{in} times I_{in} = (V_{out} times I_{out}) + [(V_{in} - V_{out}) times I_{out}]$



          The last term in square brackets is the excess voltage being converted to heat. If we expand and simplify the right hand side, a bunch of things cancel out and we get:



          $V_{in} times I_{in} = V_{in} times I_{out}$



          Therefore:



          $I_{in} = I_{out}$






          share|improve this answer










          New contributor




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






          $endgroup$



          No. A linear regulator works by burning off excess voltage as heat, therefore current in equals current out. The linear regulator is essentially throwing away the excess energy in order to regulate, rather than converting it to the output. You need a switching regulator if you want to take advantage of power in equals power out in order to convert a high input voltage, low input current into a lower output voltage, higher output current.



          $P_{in} = P_{out}$



          but for a linear regulator it looks like this:



          $V_{in} times I_{in} = (V_{out} times I_{out}) + [(V_{in} - V_{out}) times I_{out}]$



          The last term in square brackets is the excess voltage being converted to heat. If we expand and simplify the right hand side, a bunch of things cancel out and we get:



          $V_{in} times I_{in} = V_{in} times I_{out}$



          Therefore:



          $I_{in} = I_{out}$







          share|improve this answer










          New contributor




          Toor 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








          edited yesterday





















          New contributor




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









          answered yesterday









          ToorToor

          61619




          61619




          New contributor




          Toor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          New contributor





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






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

























              10












              $begingroup$

              No, it won't step up current. You can think of a regulator as a resistor that adjusts it's resistance to keep the voltage stable.



              However, you can buy DC to DC converters that 'boost' the current. But DC to DC converters are usually called by what they do to the voltage, not the current.



              A boost converter 'boosts' or steps up the voltage from a lower voltage to a higher one (at the expense of current and a small loss in power)



              A buck converter or step down converter takes a higher voltage into a lower one (with potentially more current than is on the input of the converter, also with a small loss)



              They actually make 78XX series DC to DC converters that are drop in compatible with linear regulators that buck or boost voltage.






              share|improve this answer









              $endgroup$









              • 2




                $begingroup$
                "drop in compatible" - whee, thanks! that's an improvement a hobbyist like can easily overlook
                $endgroup$
                – quetzalcoatl
                15 hours ago
















              10












              $begingroup$

              No, it won't step up current. You can think of a regulator as a resistor that adjusts it's resistance to keep the voltage stable.



              However, you can buy DC to DC converters that 'boost' the current. But DC to DC converters are usually called by what they do to the voltage, not the current.



              A boost converter 'boosts' or steps up the voltage from a lower voltage to a higher one (at the expense of current and a small loss in power)



              A buck converter or step down converter takes a higher voltage into a lower one (with potentially more current than is on the input of the converter, also with a small loss)



              They actually make 78XX series DC to DC converters that are drop in compatible with linear regulators that buck or boost voltage.






              share|improve this answer









              $endgroup$









              • 2




                $begingroup$
                "drop in compatible" - whee, thanks! that's an improvement a hobbyist like can easily overlook
                $endgroup$
                – quetzalcoatl
                15 hours ago














              10












              10








              10





              $begingroup$

              No, it won't step up current. You can think of a regulator as a resistor that adjusts it's resistance to keep the voltage stable.



              However, you can buy DC to DC converters that 'boost' the current. But DC to DC converters are usually called by what they do to the voltage, not the current.



              A boost converter 'boosts' or steps up the voltage from a lower voltage to a higher one (at the expense of current and a small loss in power)



              A buck converter or step down converter takes a higher voltage into a lower one (with potentially more current than is on the input of the converter, also with a small loss)



              They actually make 78XX series DC to DC converters that are drop in compatible with linear regulators that buck or boost voltage.






              share|improve this answer









              $endgroup$



              No, it won't step up current. You can think of a regulator as a resistor that adjusts it's resistance to keep the voltage stable.



              However, you can buy DC to DC converters that 'boost' the current. But DC to DC converters are usually called by what they do to the voltage, not the current.



              A boost converter 'boosts' or steps up the voltage from a lower voltage to a higher one (at the expense of current and a small loss in power)



              A buck converter or step down converter takes a higher voltage into a lower one (with potentially more current than is on the input of the converter, also with a small loss)



              They actually make 78XX series DC to DC converters that are drop in compatible with linear regulators that buck or boost voltage.







              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered yesterday









              laptop2dlaptop2d

              25.4k123279




              25.4k123279








              • 2




                $begingroup$
                "drop in compatible" - whee, thanks! that's an improvement a hobbyist like can easily overlook
                $endgroup$
                – quetzalcoatl
                15 hours ago














              • 2




                $begingroup$
                "drop in compatible" - whee, thanks! that's an improvement a hobbyist like can easily overlook
                $endgroup$
                – quetzalcoatl
                15 hours ago








              2




              2




              $begingroup$
              "drop in compatible" - whee, thanks! that's an improvement a hobbyist like can easily overlook
              $endgroup$
              – quetzalcoatl
              15 hours ago




              $begingroup$
              "drop in compatible" - whee, thanks! that's an improvement a hobbyist like can easily overlook
              $endgroup$
              – quetzalcoatl
              15 hours ago











              1












              $begingroup$


              since power = voltage * current.




              It is true when applied to both sides of devices that transform electricity (as AC transformers, or more sophisticated devices known as "DC-DC converters"). These devices do transform voltages/currents, so if the output voltage is lower, the output current might be higher. Keep in mind that these devices do the transformation with certain efficiency (80-90%), so the "output power" = "input power" x 0.8 practically.



              In the case of linear regulators it is not true, the regulators don't "transform", they just regulate output by dissipating the excess of voltage (drop-out voltage) in its regulating elements (transistors). Therefore whatever current comes in, the same current goes out, and even a bit less, since the regulation takes some toll. For example, the old LM7805 IC will consume within itself about 4-5 mA for its "services", so if your input is strictly 300 mA, you might get only 295 mA out.






              share|improve this answer











              $endgroup$


















                1












                $begingroup$


                since power = voltage * current.




                It is true when applied to both sides of devices that transform electricity (as AC transformers, or more sophisticated devices known as "DC-DC converters"). These devices do transform voltages/currents, so if the output voltage is lower, the output current might be higher. Keep in mind that these devices do the transformation with certain efficiency (80-90%), so the "output power" = "input power" x 0.8 practically.



                In the case of linear regulators it is not true, the regulators don't "transform", they just regulate output by dissipating the excess of voltage (drop-out voltage) in its regulating elements (transistors). Therefore whatever current comes in, the same current goes out, and even a bit less, since the regulation takes some toll. For example, the old LM7805 IC will consume within itself about 4-5 mA for its "services", so if your input is strictly 300 mA, you might get only 295 mA out.






                share|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$


                  since power = voltage * current.




                  It is true when applied to both sides of devices that transform electricity (as AC transformers, or more sophisticated devices known as "DC-DC converters"). These devices do transform voltages/currents, so if the output voltage is lower, the output current might be higher. Keep in mind that these devices do the transformation with certain efficiency (80-90%), so the "output power" = "input power" x 0.8 practically.



                  In the case of linear regulators it is not true, the regulators don't "transform", they just regulate output by dissipating the excess of voltage (drop-out voltage) in its regulating elements (transistors). Therefore whatever current comes in, the same current goes out, and even a bit less, since the regulation takes some toll. For example, the old LM7805 IC will consume within itself about 4-5 mA for its "services", so if your input is strictly 300 mA, you might get only 295 mA out.






                  share|improve this answer











                  $endgroup$




                  since power = voltage * current.




                  It is true when applied to both sides of devices that transform electricity (as AC transformers, or more sophisticated devices known as "DC-DC converters"). These devices do transform voltages/currents, so if the output voltage is lower, the output current might be higher. Keep in mind that these devices do the transformation with certain efficiency (80-90%), so the "output power" = "input power" x 0.8 practically.



                  In the case of linear regulators it is not true, the regulators don't "transform", they just regulate output by dissipating the excess of voltage (drop-out voltage) in its regulating elements (transistors). Therefore whatever current comes in, the same current goes out, and even a bit less, since the regulation takes some toll. For example, the old LM7805 IC will consume within itself about 4-5 mA for its "services", so if your input is strictly 300 mA, you might get only 295 mA out.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 3 mins ago









                  Volker Siegel

                  5641417




                  5641417










                  answered yesterday









                  Ale..chenskiAle..chenski

                  27.6k11865




                  27.6k11865






















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