3
\$\begingroup\$

I have an AC adapter at home that reads (roughly) the following:

  1. uses 220-240v AC 50Hz at 0.2A
  2. produces 12V DC at 5A

I'm a little surprised that it uses 0.2 amps and produces upto 5. Here's what happens in my mind. Please stop and correct me whenever I'm wrong.

  • Amperes is the measure of current. It is measured in Coulomb/s: 1A = 1C/s
  • 1C is a fixed number of electrons (can't find the number again, what was it?)
  • Therefore, 1A is a fixed number of electrons that flow through a wire per second.

So, if 0.2C/s go in my adapter, how can 5C/s go out? Does this have something to do with AC versus DC?

As most writers below answered, these numbers are wrong. In reality, the adapter outputs 5v at 2A... my bad. However, the point is the same, it provides more Amps than it eats.

\$\endgroup\$
3
  • 1
    \$\begingroup\$ Related / Possible Duplicate: electronics.stackexchange.com/questions/12515/… \$\endgroup\$
    – zebonaut
    Apr 29 '13 at 15:39
  • 7
    \$\begingroup\$ More importantly, 220 V * 0.2 A is 44 W; 12 V * 5 A is 60 W. Where does the extra power come from? \$\endgroup\$
    – The Photon
    Apr 29 '13 at 15:51
  • 2
    \$\begingroup\$ Maybe you are holding the adapter upside down and the electrons are falling out? \$\endgroup\$ Apr 29 '13 at 17:48
9
\$\begingroup\$

What you are missing is that current flowing in one place can cause a different amount of current to flow elsewhere. What is really being conserved in this case is power, not current. You could just as well ask that since 220 V is going in and 12 V going out, where do the remaining 208 V go? That question is basically identical to the one you asked, just with voltage and current flipped around, and just as invalid.

The total power out has to equal the total power in, at least in the long term. Power is voltage times current. Either voltage or current by itself won't give you a useful picture. In this case, we have up to 200 mA in at 240 V, which is 48 W. What comes out is 12 V at 5 A, which is 60 W. Since this can't be, the specs above are inconsitant with themselves. Someone is either mistaken, being rather loose with the figures, or outright lying.

Lets work backwards to see what the input current really needs to be. 12 V at 5 A is very clearly 60 W out, so at least 60 W have to go in somehow. In reality, these things aren't 100% efficient, so a bit more than 60 W has to go in. The remaining power that doesn't come out electrically goes to heating the unit.

Let's say this power supply is 90% efficient, which isn't a bad number at all. In the 80-90% range is common. Something like 94% would be quite impressive. 60W / 90% = 67W, which is the power that must go in. 67W / 220V = 303mA, which is the input current required at 220 V due to conservation of energy alone.

Something is clearly wrong with your figures.

Added:

Once it became clear your figures were impossible, we got sidetracked by that and didn't get into much detail on your original question.

As I said briefly above, current in one place can cause more current in another place, even though the two current flows are not connected (totally different piles of Coulombs being moved around).

Another point you are missing is that the input current flows both into and back out of the device, and so does the output current. This power supply has basically 2 pairs of terminals, one for the input and one for the output. The 300 mA flowing into one of the input terminals comes back out the other input terminal. The same is true for the output current. The output current goes out the + terminal, thru your device that it is powering, and back in thru the - terminal. In the full power case, you have both 300 mA going in and out one pair of connections, and 5 A going in and out the other pair of connections.

\$\endgroup\$
1
  • \$\begingroup\$ Turns out my memory was failing on this one. I re-read it, and it's 220V 0.2A to 5v 2A, so I guess this is more honest. You brought a nice answer to that mistake however, thank you very much. \$\endgroup\$ Apr 30 '13 at 7:55
4
\$\begingroup\$

Think of a transformer as a gearbox. The rotational speed of the load will be some fraction or multiple of the rotational speed on the driven side, and the torque seen by the drive will be some fraction or multiple of the torque the load side. A "modern" wall power supply will in fact use something more like cross between a continously-variable transmission and an automatic transmission's torque converter, but the fact that fixed-ratio transformer can increase current while reducing voltage, or vice versa, shows that it's possible for a device to increase either current or voltage (but not both).

An analogy I like to use for a switching power supply is an arrangement of pipes where the "return" sides of the supply and load are connected, and where there is a heavy turbine, one side of which connects via selector valve to either the "high-pressure" side of the supply or to the "return", and the other of which connects via selector valve to either the "high-pressure" side of the load or the return. If there is a pressure imbalance between the two sides of the turbine, the rate of water flow will speed up or slow down proportional to the pressure difference.

If the supply-side valve rapidly switches between two sides of the supply, spending equal time on each, while the load-side valve stays on the high-pressure side of the load, then half of the water which goes through the load will have gone through the supply, while the other half will have bypassed the supply. If the pressure on the load is half the supply voltage, then the amount by which the turbine speeds up when connected through the supply will roughly balance the amount by which it slows down when the supply is bypassed. Thus, this arrangement will (roughly) double flow while halving pressure.

If the supply-side valve stays connected through the supply, but the demand-side valve switches between going through the load and bypassing it, again spending half the time in each position, then only half the water that goes through the supply will go through the load, but in order for the acceleration and deceleration on the turbine to balance, the load pressure will have to grow to twice that of the supply pressure.

\$\endgroup\$
3
\$\begingroup\$

The transformer does not violate Kirchhoff's current law in any way. It has four terminals. On the primary side 0.2A flows in one terminal and flows out of another. On the secondary side, a larger current flows in one terminal and out the other.

These currents of different magnitudes flow in separate circuits which are electrically isolated from each other, but magnetically coupled. The larger AC in the secondary circuit is magnetically induced by the smaller AC in the primary circuit.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.