# How does power in "wall wart" power supplies work?

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?

• You're aware that a tiny 4W night light plugged into an outlet doesn't draw the same current as a 1500W toaster oven plugged into the same outlet, right? Thus, outlets do not force a given amount of current to flow. If a device is behind a 10:1 step down transformer, then if it draws 1A of current, only about 0.1A are drawn from the outlet.
– Kaz
Jul 8, 2013 at 23:41

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.

Yes, for a ideal power supply, power (current x voltage) in equals power out. Of couse nothing is ideal, so the electrical power out will be less than the electrical power in with the difference wasted as heat.

Perhaps a example will illuminate this. Let's say you have a 9 V 1 A wallwart. That is what it puts out. Under full load, the output power will be 9 W. Let's say the wallwart is 80% efficient, so that means it will take 11.3 W from the outlet and the remaining 2.3 W will make it warm. If the outlet is providing 110 V, then the wallwart will draw 11.3 W / 110 V = 102 mA from the outlet. Hopefully this addresses the power in / power out part of your question.

As for the safety aspect, you seem to have some misconceptions. First, the current rating of a constant-voltage power supply is only what it can put out, not what it will always force thru the load somehow. If you stripped the wires from the 9 V 1 A wallwart and held the two ends in your hand, nothing much will happen. That is because the resistance of your skin is high enough so that no dangerous currents will flow. The fact the the wall wart could have produced 1 A doesn't matter. Let's say the resistance of your hand between the two leads is 50 kΩ. That means 9 V / 50 kΩ = 180 µA will flow, which is not enough to harm your hand, probably not even enough to feel.

This seems dangerous, and also contradicts my impression that wall-power supplies like walwarts produce safe voltage/current.

This answer is meant to address the essential mistake that you're making which is that you're implicitly assuming that the load has scaled along with the step down but, it hasn't.

If your body represents a resistance $R$ and it becomes connected directly to the wall socket, the power delivered by the wall socket is proportional to $\dfrac{V^2_{AC}}{R}$

Now, if your body is connected to the output of the wall wart, where the voltage is 10 times less, the power delivered by the wall wart is proportional to $\dfrac{(V_{AC}/10)^2}{R}$.

That's 100 times less power! Where has your reasoning gone wrong?

The load on the wall wart side appears 100 times larger on the wall socket side.

So, it is still the case that the power is the same (ideally) it's just that, with the wall wart transformer between you and the wall socket, your body's resistance looks much larger to the wall socket.