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My application requires me to drop voltage from 12v to 6v or any other lower voltage.

I have used resistors but they are not efficient for prolonged use as they tend to heat up and burn my fuse. I want to know whether there are any other IC or a simple circuit which 'll be small in size, inexpensive, efficient and does not heat up much to cause problem.

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  • \$\begingroup\$ AC or DC voltage? If AC, phase difference matters? If DC, will the voltage be fixed or varying? You have to specify more details. \$\endgroup\$ Sep 15 '14 at 7:14
  • \$\begingroup\$ How much current are you going to draw? \$\endgroup\$
    – Simon
    Sep 15 '14 at 11:20
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    \$\begingroup\$ a repeat of electronics.stackexchange.com/questions/127525/… \$\endgroup\$ Sep 15 '14 at 14:24
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I presume this would be for a power supply? If the voltage drop from 12v to 6v releases too much heat, then I would recommend using a DC to DC converter of some sort. A linear regulator is basically just a glorified resistor, so converting 12V to 6V would make a linear converter 50% efficient. A switching power supply can be 70% to 90+% efficient, and as a result will produce far less heat. A buck power supply would be the way to go, and you can either build one yourself with an off-the-shelf controller chip or just buy a complete module.

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you can use regulator for this with heat sink and you have 6volt constant in output.

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    \$\begingroup\$ Just like alex said in his answer, "a linear regulator is basically just a glorified resistor". You still have heat equal to the power loss = current * (voltage drop). \$\endgroup\$
    – Ben Voigt
    Sep 15 '14 at 13:40
  • \$\begingroup\$ so you can use LED or motor with good efficiency so Additional power convert to light or mechanical energy not heat. \$\endgroup\$
    – Panda
    Sep 15 '14 at 15:04
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    \$\begingroup\$ if you do that, then the resulting voltage will not be well-defined. And you will still be converting some of the difference to heat. Maybe this is OK, but generally you want a power supply that doesn't change voltage with changing conditions (e.g. mechanical load, or more commonly just variations in current draw). A DC to DC converter will convert power in to power out with a very stable output and (relatively) little loss. For a step down, you will get more current out than the converter will draw. For 12v to 6v, a 1A load on the output might draw 550 mA on the input (~90% efficient). \$\endgroup\$ Sep 15 '14 at 22:07

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