I have a laptop adapter that takes in 120V AC and converts it to 20V DC. The adapter is rated for 230W (11.5A 20V.) Knowing conservation of energy and such, this means that the adapter takes in 120V at 1.92A from the wall.
That confuses me. How can it provide more amperes than it takes in? Clearly, it would be insane inefficiency for it to draw in 120V 11.5A and then provide 20V 11.5A (230W vs 1380W).
What is the mechanism? Is there a current booster?
You're misunderstanding what the AC adapter is actually doing. In your mind, you think it is taking in some amount of current, and then has to "push" that same current back out. That's not really what's happening.
The AC adapter fundamentally is taking in energy. It takes in that energy at a certain rate, for example 1 J of energy per second, which we call power with units of J/s.
So the correct way to think about it is that it can't "push out" any more energy than it is taking in. So if it's taking in 120 V AC rms at 2 A, then it is taking in up to 240 W of power (P=IV) if it's running at it's max (this simplifies it a bit because there is apparent vs real power, but ignore that for the moment). So we know that it can't output more than 240 W of power. However, the form of the output power can be any combination of current I and voltage V as long as the product of them (P=IV) is not more that 240 W of power. So it is totally acceptable to output 12 A at 20 V if the output power is exactly the same as the input power (again simplifying here because there are inefficiencies).
The AC adapter is really converting energy into a different format. There is no rule that input current must match output current for there to be energy conservation.
The specific mechanism through which the input power is converted to an output power with higher current (but lower voltage) is a transformer.
But what is the mechanism? Is there a current booster?
Switching regulators (a.k.a. switchmode power supplies or SMPS in short) can provide isolation as well as good efficiency at various power levels. For a 20V/11.5A (230W) off-line regulator, an efficiency of higher than 92% even with an internal active power factor correction wouldn't surprise me today. Note that PFC (power factor correction) is still part of conversion here. Without that, the real power you measure would still be 240W but the current consumption from the mains would be, like, 4 Amps which means ~500 VA (volt-amps).
I will not dive into the details of operation principle of SMPSs here as Internet provides millions of resources. There are numerous different methods (topologies) for power conversion, they have their own advantages and disadvantages for different applications and conditions.
$$ P = VI $$ $$ P_{IN} = 120 \times 2 = 240 \ \text W $$ $$ P_{OUT} = 20 \times 11.5 = 230 \ \text W $$
No real problem there. The numbers will be somewhat approximate. I wouldn't expect the efficiency to be that high.
I wouldn't expect the efficiency to be that high. with CrM PFC along with LLC converter as DC-DC these numbers are not surprising. Today you can get 98% from each, making a total of 96%.
\$\endgroup\$
Commented
Mar 11 at 9:46
But what is the mechanism? Is there a current booster?
Think of it in simple terms; a power transformer converts the power input to a power output but, the output voltage can be lowered at the same time that the output current is increased. It does this with a high energy efficiency hence, there is not much wasted heat produced.
There are a few more subtleties involved but, a power transformer is the basic principle.
How can it provide more amps than it takes in? and " I assume there's a transformer in there."
Yes, your assumption is correct. Typical way how all laptop adapters are designed is this:
AC110-240V -> DC rectifier 150-300V -> high-frequency DC-AC converter (aka "switcher") -> transformer (high-frequency power transformers do have small size) -> rectifier -> DC out.
That's why you can see the output current vastly exceeds the input current.
