AC/DC Power Adapters: Input & Output Specifications

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Discussion Overview

The discussion revolves around the operation of AC/DC power adapters, specifically the apparent discrepancy between input and output current specifications in devices like laptop power bricks. Participants explore the principles of current transformation and power conservation in the context of electrical engineering.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions how the output current can exceed the input current in power adapters, citing specific input and output specifications.
  • Another participant explains that the input current is at a higher voltage, and when the voltage is reduced, the current can increase, referencing the relationship between voltage, current, and power (Watts).
  • A third participant mentions the presence of a transformer inside the adapter that converts high voltage AC to a lower voltage AC, which is then rectified to DC, and introduces the conservation of power principle.
  • Participants discuss the formula for power (P=I*V) and its implications for understanding input and output relationships.

Areas of Agreement / Disagreement

Participants express different levels of understanding regarding the operation of power adapters, with some agreeing on the principles of voltage and current transformation while others seek further clarification. The discussion remains unresolved regarding the specifics of how current transformation occurs in practical terms.

Contextual Notes

Some assumptions about the efficiency of power conversion and the role of transformers are not fully explored, and there is a lack of detailed mathematical derivation in the discussion.

habitat
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in regards to power adapters, mainly the brick-type ones that come with laptops and external hard drives, how is the output current greater than the input current?

for example, i see a laptop power brick with the following:
input 100-240V~1.7A 50-60Hz
output 18.5V == 3.5A 65W


how exactly does this work with a branch current method of analysis if the branch going into the power brick is 1.7A? i see that you can pull current from voltages, but if only 1.7A is going into the network, where are the other carriers coming from to bump up the output to 3.5A?

i've been trying to think this out but i can't arrive at anything. what i left off on is that 1 ampere = 1 coulomb per second. if the input is 1.7A = 1.7 coulomb per second operating at 50-60Hz, perhaps it actually collects enough current to drive 3.5A?

i'm a bit lost.


thanks!
 
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yes but its 1.7 A at the mains voltage 100 - 240V. But you then drop the voltage and you can increase the current. I don't have the formulae for showing the workings. Some one else is likely to be able to do that.

Same thing happens around the other way. I have a PSU that can supply for example... 12V at 5A. If I double the voltage, I'm going to 1/2 the current capability
That is. The PSU has a particular max Wattage rating V x I (current) = Watts if you change V then I is going to change to keep that max Wattage

Thats a basic way of looking at the problem :) I'm sure others could explain it in more detail :)

cheers
Dave
 
Welcome to PhysicsForums!

There's a transformer inside that converts the high voltage AC to lower voltage AC (which is fairly close to the output DC voltage). This lower AC voltage is then converted to the final DC voltage.

The formula for power, given voltage and current, is:
P=I*V

Since power is conserved (more or less, some of the energy is lost as heat during the transforming), you can find the input and output voltages:
P=I_{in}*V_{in}=I_{out}*V_{out}

More on this topic at:
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/transf.html
 
Great! thanks a lot!
 

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