One thing to note when using unregulated wall-bricks to drive linear regulators (a common pattern) is that the amount of heat produced by the regulator will be the product of the current drawn and the amount by which the input voltage of the regulator exceeds the output. If one is trying to get +5V from a wall-brick which outputs 12 volts at 1mA but only 7 volts at 500mA, then when only 1mA is being drawn from the regulator, the regulator will have to dissipate 7 volts (the difference between 12 and 5) times 1mA, i.e. 7mW. If the wall-brick voltage didn't drop as current increased, heat dissipation would increase proportional to current; at 500mA, the regulator would have to dissipate 7V times 500mA, which is 3.5 watts. If the wall-brick voltage drops as current increases, however, this will reduce the power the regulator has to dissipate (e.g. if a 500mA load drops the voltage to 7V, then it would only have to drop (7V-5V)*500mA, which is to say one watt, not 3.5.