# AC/DC Laptop adapter power losses

I have a laptop power adapter shown in the picture below.

It is written there

170Watt

INPUT: 100-240V ~ 2.5A

OUTPUT: 20V 8.5A

I believe for AC Input they mean Root mean square voltage/current.

for output it is clear ($$W=U \cdot I$$) or $$170W = 20V \cdot 8.5A$$.

But I do not understand it for input. If it is working for minimal input voltage 100V, then minimum power should be $$100V \cdot 2.5A = 250W$$ or for 240V $$240V \cdot 2.5A = 600W$$.

Even for minimum 100V power is 250W, then where the difference between input and output 250W-170W=80W goes? I mean from energy conservation law output power should be equal to input power minus losses or $$Output Power = Input Power \times Energy Conversion Efficiency$$

Does that mean my adapter has 68% efficiency on 100V and 28% on 240V? I believe Energy Conversion Efficiency for modern AC/DC adapters should be quite high around 80-90% or I am wrong? This adapter is quite new. I bought it in January 2020 and it was a new one and a new Laptop model.

My question is: Why there is so big difference in input and output power or what is wrong with my calculations?

Lenovo ADL170NLC3A has an Energy Star V (ES 2.0) rating. Energy Star states to get that rating for nameplate output power ($$P_{no}$$) > 45W, the Minimum Average Efficiency in Active Mode ≥ 87%.

At nominal conditions of 170W, 20V @ 8.5A, and a minimum efficiency of 87% implies 195W input. So worst case, the ac charger consumes 25W.

The AC adapter is a switching power supply converting 100V to 240V ac at 50Hz or 60Hz to 20V DC @ 8.5A. The 2.5A corresponds to the peak ac draw, not sustainded draw.

Typically, the actual current drawn by the laptop depends on the charge on the battery. This means efficiency changes. When the battery is uncharged, output from charger will be maximum and efficiency will be at a low of 87%. As battery charges, demand on charger will be less as battery voltage increases. Efficiency should increase slightly because the 25W lost to switching power supply will change slower than load from battery. Educated guess, efficiency > 90%.

This is a rating label, not a measure of device efficiency. That power brick is definitely not putting out 80 watts of heat because it would melt in short order.

You should expect that the device will operate within the expected input voltage, not draw more than the rated input amps, and will not put out less than the rated output power under those conditions.

The 2.5 A current is the peak in charging up the internal high-voltage DC capacitors for the switching power supply.

In normal running the input power should be $$P_{in} = \frac {P_{out}} {efficiency}$$ where efficiency should be somewhere between 80 - 95%.