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The supply cable that came with my PSU has a 13A fuse fitted inside it's plug. However on the side of the PSU it says it's cable of producing 30A of current at 5V. How is that possible when it only has 13A to play with?

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What power supplies do is a conversion of power, not current.

This means you take primary power (\$P_{pri} = V_{pri} I_{pri}\$) from the mains network and you get secondary power (\$P_{sec} = V_{sec} I_{sec}\$) at the output. The power at the output can never be greater than the power at the input:

$$P_{pri} > P_{sec}$$

Substituting the above equations, we get:

$$V_{pri} I_{pri} > V_{sec} I_{sec}$$

Let's solve for \$I_{sec}\$:

$$I_{sec} < \frac{V_{pri}}{V_{sec}} I_{pri}$$

You see that the "trick" is hidden within the relationship of the input and output voltages. If the output voltage is lower than the input voltage, the output current may well be higher than the input current.

With the numbers you give (and assuming you live in the UK because this is where fuses are often found in mains plugs):

$$I_{sec} < \frac{240V}{5V} 13A$$

$$I_{sec} < 624A$$

This is just meant to prove the point. The number is very (!) much theoretical and way beyond the real numbers, because (i) the fuse has a rating much higher than the RMS current used by your power supply, usually around a factor of 5 to 10, because it must not blow during inrush events at turn-on, (ii) the power supply's output current may be limited by other factors and (iii) the power supply has some losses.

This relationship can also be explained using the waterfall model, where voltage equals height and current equals the size of the stream. On the input side, you can imagine having a very high waterfall with not that much water, on the output you have a big river with a lot of water but not much height that pushes the water towards the ocean.

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There are a couple answers posted that point out the limit between the input and output of a PSU is actually power, not current. This is true; you can't get more power out than you put in.

But maybe you're still puzzled by the current. An amp is a large number of electrons per second. If you have only 13 amps coming into the box and 30 amps going out, it seems like you'd run out of electrons in the box pretty quick.

Here's a diagram with 3 PSUs that might clear up what's actually happening.

  1. Imaginary PSU-- what you might think was happening in a PSU. How can 13 A turn into 30 A?

  2. A PSU with a transformer, vastly simplified. There are two separate loops of current with no conductors connecting them.

  3. A switchmode PSU, also vastly simplified. The current flows in two overlapping loops. How exactly the electrons are persuaded to flow in this path involves a very fast switch and an inductor, but I've left those out to make the current paths clear.

enter image description here

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    \$\begingroup\$ +1 for freehand circles (and more!) \$\endgroup\$ – JYelton Apr 4 '11 at 17:45
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    \$\begingroup\$ +1 for clear freehand drawings, rather than "proper" CAD drawings that are often harder to understand. \$\endgroup\$ – davidcary Apr 5 '11 at 14:18
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    \$\begingroup\$ Downvoted because the explanation is a little bit too simplified. It does nothing to explain the crucial point about the voltage being different and total power etc... If I knew "nothing" about electronics, I'd probably be puzzled by that diode in the switch mode supply. Do diodes magically create current? etc. I simply think your explanation does more to confuse than it does to enlighten. \$\endgroup\$ – nitro2k01 May 3 '11 at 17:46
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    \$\begingroup\$ +1 because the downvotes were unfair. True this is oversimplified, but he made that clear and it is correct at the level of detail it goes into. If you downvote for leaving out information, you'd have to downvote everyone who didn't answer at all. \$\endgroup\$ – Olin Lathrop Oct 25 '11 at 12:30
  • \$\begingroup\$ +1 because this is actually a very good example of leaving out the details that are important to answer a different question, but distract from the simple and direct question that was actually asked. Doing this well takes skill. \$\endgroup\$ – RBerteig Apr 29 '13 at 19:09
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Look at the power. It can take 230 * 13 = 3kW from your outlet, but is only giving you 5 * 30 = 150w out.

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It presumably has a transformer - 1 amp at 240 volts on the input equates to 240 amps at 1 volt on the output, ignoring losses.

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    \$\begingroup\$ many power supplies today are switching power supplies. They do not have transformers. However the relationship between current and voltage still holds ture \$\endgroup\$ – Jim C Apr 4 '11 at 12:27
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    \$\begingroup\$ Switchers do have transformers. The same basic math applies, you just use higher frequencies and other waveforms of voltages and currents in the transformers, which is the reason why the transformers in switchers are much, much smaller. But they're there. The question can be answered no matter if it is a 50 or 60 Hertz transformer or a switch-mode power supply. \$\endgroup\$ – zebonaut Apr 4 '11 at 12:34
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    \$\begingroup\$ Isolating switchers have transformers. Non-isolating switchers could use inductors, especially if a single output is all that is required. Or even capacitors eg IC7660. \$\endgroup\$ – Martin Apr 4 '11 at 12:46
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    \$\begingroup\$ If it is a switcher, it must be isolated from the mains with a transformer for safety reasons. \$\endgroup\$ – Leon Heller Apr 4 '11 at 14:10
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    \$\begingroup\$ There isn't an isolation requirement if the power supply is primary-only. You can use a buck converter between rectified AC and a primary-referenced supply that the user isn't directly interacting with, and regulatory agencies won't have a problem with it. \$\endgroup\$ – Adam Lawrence Apr 4 '11 at 14:55

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