Question: I have a car cigarette extension cord with a USB out on it as well. The USB out though will not charge my phone, which is a bit annoying. I had a look at the circuit inside in the device and it looked something like this:

![Power supply circuit](https://i.imgur.com/faPZh30.png)

I'm guessing the reason the phone won't charge is the 78L05 chip is only rated for 100mA. Therefore I was thinking of changing it to a 7805 chip, which is rated for 1A. Would this idea stand up to my reasoning? Or am I likely to damage any equipment I connect if I go down this root?

Also, if I go about this, will I also need to do something with the capacitor and the resistor on the board?

Edit:

Thanks for all the answers.  

I didn't think of the heating problems I might have in my scenario above.  

Its looking like the best solution for the device is the one Nick Alexeev gave below if I'm going to stick with the device I have.

I have also changed the diagram above as the resistor is actually 360 ohm.

Also, I am including a image of the device itself below as requested to show exactly what I was talking about.

![Extension Cord Device](https://i.imgur.com/BkWeRoW.jpg)

Answer: 
A linear regulator such as 7805 would work, if you heatsink it adequately.  

Suppose, that the 7805 is outputting 1A\*\* at 5V, which is 5W. It draws 1A at 14.5\*\*\* from the supply, which is 14.5W. Where does the remaining 9.5W go? It's lost as heat. To keep 7805 cool, the heat needs to be dissipated. That's why you need to do the thermal analysis (think the heatsinking through).

\*\* 1A is the max current, which your 7805 is rated for. Worst-case current in this scenario.  

\*\*\* 12V supply in a car can be as high as 14.5V sometimes.

Commercial cigaret plug chargers, which I've seen had buck converters (switch-mode). They are more complex than linear, but they have 85% to 95% efficiency. Which means that in case of 1A at 5V, you would need to dissipate only 0.5W or so.

There are self-contained buck modules, which were designed for drop-in replacement of 7805. Such as [this one](http://www.mouser.com/ds/2/281/oki-78sr-56393.pdf). (There are many more modules with suitable specs, so don't get bent of that particular one.)
The basic concept is sound, although a bit profligate. You want to worry a little bit about heat-sinking the 7805, since for every 5 mW out, you're burning 7 mW in the regulator heat sink (because current out essentially = current in, and you're dropping 7V across the regulator). You'd do a lot better with a 12V-to-5V stepdown switcher.

The capacitor is there to smooth out ripple and noise on the input to the regulator. I would add a 1 uf TANTALUM (!) across the 7805 output, to suppress spurious oscillation of the regulator IC. I've seen that happen on a friend's homemade linear supply; adding the capacitor fixed it. (Supposedly, current-gen 78xx regulators don't need it, but it is CHEAP insurance.)

What color LED is it, and how much current does it require? That's what sizes the resistor.