I'm new to electronics and am having trouble understanding how power supplies convert 120AC into a safe, manageable DC voltage. So far, I understand that there's a transformer to step the voltage down, then a rectifier, followed by an IC voltage regulator. However, any way I look at it, it seems that even if you step the voltage down 10x (via transformer), the current going into the rectifier is 10x larger than the wall outlet. This seems dangerous, and also contradicts my impression that wall-power supplies like wall warts produce safe voltage/current.
Am I correct in thinking that the power output from a wall wart is equal to the power output from an electrical jack (IV = IV)? If so, is getting shocked from an outlet equally dangerous to getting shocked from a wall wart? If not, what's missing from my analysis?
Best Answer
Power out of wall-warts is solely determined by the load you connect to it and not by the potential power that can be fed from the AC wiring.
Say it's a battery charger and it charges 4 x 1.5V batteries at 100mA - power out of the charger = power into the batteries = 6V * 0.1A = 600mW.
AC power into the charger is 600mW plus some small residual power related to how efficient the wall-wart is. Typically 700mW to 1W.
Power output from 120Vac might be 20A * 120V to power a kettle = 2.4kW. It might be a lot more but it is the load or appliance that dictates power taken.
Things like this are made safe because wires are insulated and circuits are isolated. Output voltages are kept low so that the effect of electric shock is barely felt.
Same with current - current drawn by a circuit is as per the load circuit demands and there are protection devices that inhibit the current beyond the intended use of the wall-wart - not to protect the user (because the low voltage does that) but to stop the electronics inside frying when someone shorts the output.
Wall-warts usually have AC line fuses inside them to prevent faults devloping into fires.