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The below circuit is 12v Input & 5V max@3Ampere output.

The work of the 7805 is only to open the base of the transistor.(and naturally regulate that voltage at the base to 5v)

What about efficency, how efficent is this circuit?

enter image description here

The transistor he uses is a 2N3055(15A)

As the 7805 has only to open the Base i guess this circuit is a little more efficent than the 7805 itself. If that is correct, using it for a battery powered project, could i use let's say something like a 2n2222 or 2n3904 to get a slightly more efficent powersupply (but decreasing the Amperage)?

Normally the 7805 is only 20%-21%(12v) efficent so if myproject needs 100mA the heat it needs to dissipate is around 2Watt

(i read that somwhere yesterday)

I ask because apparently the high efficent switching converter suffer from ripple & noise, but also cost more. And this could increase slightly the battery life & decrease the cost.

notes:

i could hook up multiple transistors and so get multiple 200mA/500mA regulated 5v sources, depending on the transistor. (2n2222=500mA,2n3904=200mA)

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  • \$\begingroup\$ This is a terrible idea. Adding the transistor as you show it reduces the efficiency of the circuit, because it increases the voltage drop from input to output. Furthermore, the output voltage will be poorly regulated. There's a much better circuit for using an external PNP pass transistor in every 78xx data sheet. \$\endgroup\$ – Dave Tweed Nov 2 '15 at 12:47
  • \$\begingroup\$ is the circuit with the pnp efficent? \$\endgroup\$ – cocco Nov 2 '15 at 12:50
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    \$\begingroup\$ No, there's nothing you can do to improve the efficiency of a linear regulator. But the PNP circuit at least preserves the ability of the 7805 to regulate the output voltage accurately. \$\endgroup\$ – Dave Tweed Nov 2 '15 at 12:52
  • \$\begingroup\$ thank you, i guess i need to use switching converters then. \$\endgroup\$ – cocco Nov 2 '15 at 12:53
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This circuit is no more efficient than just using the 7805 by itself. The 7805 uses BJTs just like you have here to "burn off" excess voltage. The efficiency of linear regulator is really dependent on the voltage that they have to reduce. Standard linear regulators use a Darlington pair of NPN transistors that requires at least 1.5v above the regulated voltage to operate. LDO or low drop out regulators use a single PNP and require about 0.7v above regulated voltage. At just above the operating voltage is where maximum efficiency is achieved because the minimum amount of voltage needs "burned off."

A basic power calculation is voltage drop time current.

In your case 12v down to 5v at 3A would give (12 - 5) * 3 = 21W. Your transistor would have to dissipate 21 watts.

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  • \$\begingroup\$ so basically that circuit just passes the load to the transistor.but the 2N3055 can dissipate 20Watt? \$\endgroup\$ – cocco Nov 2 '15 at 12:50
  • \$\begingroup\$ so best solution is switching converters? \$\endgroup\$ – cocco Nov 2 '15 at 12:53
  • \$\begingroup\$ @cocco yes, if you want efficiency then switching is the only way to go. \$\endgroup\$ – vini_i Nov 2 '15 at 13:04
  • \$\begingroup\$ A 2N3055 can provide both the current and the power you need, but be aware that you'll need a good heat sink on the 3055. And by good, I don't mean a clip-on. \$\endgroup\$ – WhatRoughBeast Nov 2 '15 at 15:45

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