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D.A.S.
D.A.S.
  • 148k
  • 3
  • 56
  • 190

specs:

  • Drop-in replacement of 3-terminal LM7805 or equivalent linear voltage regulator.
  • Guaranteed 1A output current
  • Wide input voltage range up to 4.5V to 17V
  • High efficiency, greater than 70% for loads greater than 1mA, peak efficiency achieved of 90% at 300mA load current.
  • Thermal shutdown and current limit protection

Thus if using only 300mA at 5V out or 1.5W , it must only dissipate 10% loss or 150mW. But if using 1A or 5W and spec is somewhere between 70% and 90% then losses might possibly be 20%+/-? or 1W +/-? then you must consider how to heatsink it with a thermal tape to avoid shorting the vias to a small heatsink or chassis.

But consider the 7805 linear regulator with a 7V input voltage drop from 12V to 5V , at 1A the load is 5W but the regulator loss is 7W !! , so this is much more efficient.

So depending on your application, you MAY need a heatsink with a nonconductive clip to ensure contact pressure to squeeze say a 3M thermal conductive tape soto some spec.on the datasheet.

but then again they are cheap so if 1W runs at 90'C= Tjcn for this area of copper, (assumption based on experience) it may burn your finger and not last as long, but did I say they were cheap?

specs:

  • Drop-in replacement of 3-terminal LM7805 or equivalent linear voltage regulator.
  • Guaranteed 1A output current
  • Wide input voltage range up to 4.5V to 17V
  • High efficiency, greater than 70% for loads greater than 1mA, peak efficiency achieved of 90% at 300mA load current.
  • Thermal shutdown and current limit protection

Thus if using only 300mA at 5V out or 1.5W , it must only dissipate 10% loss or 150mW. But if using 1A or 5W and spec is somewhere between 70% and 90% then losses might possibly be 20%+/-? or 1W +/-? then you must consider how to heatsink it with a thermal tape to avoid shorting the vias to a small heatsink or chassis.

But consider the 7805 linear regulator with a 7V input voltage drop from 12V to 5V , at 1A the load is 5W but the regulator loss is 7W !! , so this is much more efficient.

So depending on your application, you MAY need a heatsink with a nonconductive clip to ensure contact pressure to squeeze say a 3M thermal conductive tape so some spec.on the datasheet.

but then again they are cheap so if 1W runs at 90'C= Tjcn for this area of copper, (assumption based on experience) it may burn your finger and not last as long, but did I say they were cheap?

specs:

  • Drop-in replacement of 3-terminal LM7805 or equivalent linear voltage regulator.
  • Guaranteed 1A output current
  • Wide input voltage range up to 4.5V to 17V
  • High efficiency, greater than 70% for loads greater than 1mA, peak efficiency achieved of 90% at 300mA load current.
  • Thermal shutdown and current limit protection

Thus if using only 300mA at 5V out or 1.5W , it must only dissipate 10% loss or 150mW. But if using 1A or 5W and spec is somewhere between 70% and 90% then losses might possibly be 20%+/-? or 1W +/-? then you must consider how to heatsink it with a thermal tape to avoid shorting the vias to a small heatsink or chassis.

But consider the 7805 linear regulator with a 7V input voltage drop from 12V to 5V , at 1A the load is 5W but the regulator loss is 7W !! , so this is much more efficient.

So depending on your application, you MAY need a heatsink with a nonconductive clip to ensure contact pressure to squeeze say a 3M thermal conductive tape to some spec.on the datasheet.

but then again they are cheap so if 1W runs at 90'C= Tjcn for this area of copper, (assumption based on experience) it may burn your finger and not last as long, but did I say they were cheap?

Source Link
D.A.S.
D.A.S.
  • 148k
  • 3
  • 56
  • 190

specs:

  • Drop-in replacement of 3-terminal LM7805 or equivalent linear voltage regulator.
  • Guaranteed 1A output current
  • Wide input voltage range up to 4.5V to 17V
  • High efficiency, greater than 70% for loads greater than 1mA, peak efficiency achieved of 90% at 300mA load current.
  • Thermal shutdown and current limit protection

Thus if using only 300mA at 5V out or 1.5W , it must only dissipate 10% loss or 150mW. But if using 1A or 5W and spec is somewhere between 70% and 90% then losses might possibly be 20%+/-? or 1W +/-? then you must consider how to heatsink it with a thermal tape to avoid shorting the vias to a small heatsink or chassis.

But consider the 7805 linear regulator with a 7V input voltage drop from 12V to 5V , at 1A the load is 5W but the regulator loss is 7W !! , so this is much more efficient.

So depending on your application, you MAY need a heatsink with a nonconductive clip to ensure contact pressure to squeeze say a 3M thermal conductive tape so some spec.on the datasheet.

but then again they are cheap so if 1W runs at 90'C= Tjcn for this area of copper, (assumption based on experience) it may burn your finger and not last as long, but did I say they were cheap?