7805 Efficiency

Hello Guys,

I need a power supply of 5V 1-1.2A for a project which I am working on.

I am planning to use a 12V 1A Transformer with a 7805 as the Power Supply.

  • Will it work out…?
  • What will be the efficiency of the 7805…?
  • Will there be a variation in the Amperage…??

If you need the supply to give 1-1.2A then you need a transformer >1A. At the very least 1.5A, preferably more.
Don’t forget that a transformer will O/P AC, so you will need a bridge rectifier and smoothing caps.
The 7805 is rated at 1A so is not suitable if you need >1A. Also not advisable to run anything at, or close to its maximum rating.
If you were to draw 1A from the 7805, with a 7v drop across it (12v - 5v), you will be dissapating 7W, so you will need a fair sized heatsink, as well.
A better option would be to get a PSU with an O/P of at least 9v @ 2A, and then drop the voltage with a buck regulator module. There are plenty available on eBay.
Even better would be to get a plugpack (wallwart for those who live across the pond), with a 5v, >=2A O/P.


Better yet is to use an old cell phone charger, normally rated for 2A at 5V. You get plenty of options then, hacking a car charger into the circuit(12V to 5V), or a wall charger (110V or 220 depending on your location, to 5v), both extensively tested, and if removed from the bulky cases, relatively small.

Efficiency of any regulator is given by Pout/Pin
P = I*V, and for a linear regulator is Iin == Iout, so Efficiency = Vin/Vout
So for the nominal 12V in and 5V you’d have about 42%.
(Note that a 12V transformer will normally provide significantly more than 12V, so the actual efficiency is probably worse.)

Thank you for those replies,

The Datasheet of 7805 mentions the max current rating as 1.5A. But IamFof mentioned 1A. Which is correct..?

If I replace the 12V 1A transformer with a 9V 2A transformer, will I get the desired result..?

How are you heatsinking the 7805?

I have a couple of these cheap, dumb USB loads. It’s just a couple of wire-wound resistors with a switch so you can decide if you want to take 1A or 2A of current out. Each resistor is absorbing 5W of heat from the USB port, and when the surface temperature stabilizes in ambient air it’s hot enough to boil water, nearly 150oC. Those resistors are significantly larger than a TO-220 IC package, so if you try to put 7 watts into a 7805 you could probably make toast with it. You will need a heatsink and probably a fan on that sucker, much like this other USB load that is much better (and easier to handle) than the other one.

Or just use a switching regulator which won’t get nearly as hot.

How are you heatsinking the 7805?

I am using this heat sink as attached below.

Or just use a switching regulator which won’t get nearly as hot.

Can switching regulators provide high currents…?? Such as 1A or 2.5A…??

The heatsink in the link has a typical "degrees per watt" rating of 25C.

If you drop 12volt to 5volt, and draw 1.5Amp, then the heatsink will be (12-5)*1.5*25 + room temp = almost 300 C. That will fry the regulator and everything around it.

A switching buck converter like this one can do that without a heatsink. $1 clones on ebay. Leo..

Can switching regulators provide high currents...??

Yes. Switching regulators can generally provide HIGHER currents than a linear regulator, precisely because they have less of a heating problem. Digikey lists something like 1400 different modules providing 1.5 to 4A at 5V. Not many are competitive with Pololu's prices, though...