# 7805 efficent circuit?

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?

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)

• 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. – Dave Tweed Nov 2 '15 at 12:47
• is the circuit with the pnp efficent? – cocco Nov 2 '15 at 12:50
• 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. – Dave Tweed Nov 2 '15 at 12:52
• thank you, i guess i need to use switching converters then. – cocco Nov 2 '15 at 12:53