Is a 7805 Regulator With Small Heatsink Safe for 1A iPhone Charging From 12V Car Battery?

So I want to power an iPhone off a car battery. It takes 5volts in and the car battery gives off around 12v. I want to use a 7805 to step the voltage down to 5volts. If my understanding of how the 7805 works is correct, and my iPhone pulls 1amp, then the 7805 will dissipate 1amp * (12v - 5v) = 1amp * 7v = 7 watts to be dissipated.

I have a small heatsink for the 7805 but I'm worried it's not enough. This is pretty much the exact heatsink I have: http://www.radioshack.com/product/index.jsp?productId=2102856

The heatsink says it can dissipate 5 watts so am I in trouble since the 7805 is going to dissipate 7 watts in my use case?

Thanks!

I think you should look at a switching regulator (buck) configuration instead. It is more efficient. You will need more components but I think it's a better alternative. If you want go to Pololu.com and see what they have to get an idea of what you need.

The wattage ratings are based on specific conditions. Usually free air at 20 degrees C or 68 degrees F.
There are lots of variables here.

Is the heatsink in free air or enclosed in a case? In a case it will be much hotter.

Is the ambient air 20 degrees C?. In the engine compartment or under the dash in the sun, on a warm day or near the heater hoses it could be much hotter.

The car voltage will be 12.6V with a charged battery and no load. When the engine is running the alternator will put out 14 to 15v, so I'd design for 15v, not 12v.

You may be able to put two heatsinks on the 7805, one on each side, though this is still marginal. You could also use two regulators in series. There are circuits on the web showing how to generate 10v from a 7805. Then put the 10v into a regularly configured 7805. At 15v each regulator will drop 5v. If your ambient temperature is high or the airflow is poor I'd still use two heatsinks on each 7805..

Well you're right. I hooked it up and the heatsink got pretty hot pretty quick. I decided to try your advice of using a switching regulator and thought "where can I find one....?" and then it came to me! My car adapter for my iPhone that uses the cigarette lighter. I cracked it open and sure enough, there was an IC inside. I soldered my wires directly to the cigarette lighter connector and hooked those wires directly up to the car battery. I plugged the phone in and voila, it charged, and the iPhone adapter didn't even get a little warm after a few mins of use.

In case you're wondering why I'm doing this, I like on the coast in CT and I'm about to get hit by a massive hurricane and lose power, and I want to be able to charge my iPhone from a spare car battery I have lying around.

I also just realized I can use this to power my netbook with it's cigarette lighter adapter in the same fashion. So at least I'll have a phone and a netbook if we lose power!

If I want to power the netbook and the iPhone at the same time, they should be in parallel, NOT series, correct?

Yes, parallel.

Also (though it's mute, now, because you sound like you have solved your problem), but just in case, another trick is to put a power resistor in series with the linear regulator. The power resistor will drop the voltage (thus leaving less for the regulator to deal with) and dissipate some of the power. But, it has to be chosen with care, in your case, if the max current will be 1A, then determine how much voltage you want to drop across the resistor [Vr = Vmax - Vdo - Vreg where Vr is the voltage across the resistor, Vmax is the maximum input voltage (can be as high as 14 to 15V as Stephan indicated), Vdo is the dropout voltage of the Regulator IC (around 2 volts for the 7805), and Vreg is the regulated output voltage (5V for the 7805). R = Vr/1A

Thus: Vr - 15 - 2 - 5 = 8V and R = 8/1 = 8Ohm. P = E**2/R (that's the voltage squared over the resistane or: P = 8V**2/8Ohm = 8Watts so use a 15 Watt resistor.