Hello Thiaguez,
Consider these things:
The 7805 regulator has to drop the input voltage from 12 (or about 14 if you are using it in a vehicle, as you mentioned cigarette lighter) to 5 volts. The power that it has to dissipate is this voltage drop multiplied by the current going into it. Check it and see the watts.
These regulators have built-in protection. If the load draws excessive current the regulator heats up until it reaches a point where it will shutdown. Remove the load and it will come to life as soon as it cools off. You can place a dead short across the output and it will do the same thing without damage.
About the only thing that will damage one is the application of a reverse voltage across its output. Say, for example, you apply 14 volts across the output while there is 12 volts across the input. To safeguard against this it is wise to place a diode, like a 1N4001, across it with the cathode connected to the input pin and the anode to the output pin.
The 1000uF capacitor on the output is overkill. A 10uF is adequate. In fact if you use a very large value capacitor in this position the regulator will try to charge it with a high initial current which will cause it to shut down. When the load discharges the capacitor the regulator will start up and go through the cycle again ... and again.
Some times, depending on the length of the leads between the unregulated source and the regulator, it can go into self-oscillation and overheat and then shutdown. In this case place a .01uF from input and output to ground.
Good luck!
Trini