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Given a 12V 4A DC power supply that's connected from a wall outlet to a V59 Universal LCD Controller, can the voltage be split from this power supply to power another device with 5V 1200mA power requirements? If so, how exactly should the wiring be connected and what other components such as resistors might be needed?

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    \$\begingroup\$ Yeah, just use a voltage regulator, like the 7805. You can find more information by Googl'ing it. \$\endgroup\$ Commented Apr 22, 2017 at 15:41
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    \$\begingroup\$ @user95482301, did you notice that the OP may need up to 1.2A at 5V? That is beyond the abilities of a 7805. Also, it is probably not a good application for any linear regulator, because the power dissipation in the regulator will be 1.2A * (12-5V) = 8.4W. A switching regulator would be a better choice. \$\endgroup\$
    – user57037
    Commented Apr 22, 2017 at 15:53
  • \$\begingroup\$ @mkeith, "Output Current up to 1.5 A" datasheet I agree that a switching regulator would be better. \$\endgroup\$ Commented Apr 22, 2017 at 16:08
  • \$\begingroup\$ In that case, the best is of course just to buy a "dc dc buck converter" from aliexpress, for $3 incl. shipping. \$\endgroup\$ Commented Apr 22, 2017 at 16:10
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    \$\begingroup\$ @sean2078, Yes. \$\endgroup\$ Commented Apr 23, 2017 at 17:43

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You need a 12V to 5V DC-DC step-down converter (also look for: switching regulators). You can find plenty of them in the usual electronics parts vendors.

You could also consider a linear voltage regulator, but that would mean a maximum power dissipation in it of about \$ P_d = (12-5) \cdot 1.2 = 8.4W \$, which will (almost surely) either fry it or drive it into thermal shutdown. Also, it's very inefficient power-wise. Not recommended at all.

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Use a high efficiency D.C.-D.C. Converter, these devices are used in hobby remote controls where they are termed a battery eliminator circuit (BEC). Forget about 7805, too much waste heat. No other components needed., 12v in, 5v out, very reliaable.

Eg http://www.dx.com/s/Bec?PriceSort=up

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