If I have a device that draws 5 amps at 12 volts, I can use any 12 volt DC adapter that can provide at least 5 amps.

Why don't all DC adapters have the capacity to provide loads of amps!? If all DC adapters provided e.g. 1000 amps, we would only need to care about the voltage value.

Do too many amps make DC adapters bulky, inefficient, or expensive?

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    \$\begingroup\$ There's no such thing as a free lunch. \$\endgroup\$ – whatsisname Aug 27 '14 at 13:30
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    \$\begingroup\$ Why does a 500 HP engine cost more and weigh more than a 50 HP engine? \$\endgroup\$ – Olin Lathrop Aug 27 '14 at 13:33
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    \$\begingroup\$ Thank you all for your answers! I would upvote you all if I had the reputation... I also wondered if it was a safety thing? to restrict the current if there were a short-circuit or something. But I don't think that is correct...? \$\endgroup\$ – Rich Aug 27 '14 at 13:51
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    \$\begingroup\$ @Rich, safety definitely comes into play. If you ever drew an arc at 12V1000A (it's easy to draw an arc at 12V), that arc could potentially be much larger than a welding arc - not so much a direct life hazard, but certainly a fire hazard. Even if nothing else, it'd melt your cables into a puddle of molten metal. \$\endgroup\$ – TDHofstetter Aug 27 '14 at 14:12
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    \$\begingroup\$ But I enjoy lugging my 21 kg power supply around to charge my phone! \$\endgroup\$ – JYelton Aug 27 '14 at 16:36

The components that make up DC adaptors (inductors, transistors, capacitors, diodes, ect) are all rated for a certain current and/or power dissipation. Components that can handle 1000A vs. components that can handle 5A are orders of magnitude apart in cost, size, and availability.

For an example let's look at an inductor that could be used in a 1000A supply vs. a 5A supply.

Price: An inductor that can do 5A is $0.17 on digikey, an inductor that can do 200A is $400.

Size: The 5A inductor is 5mmx5mm and the 200A inductor is 190mmx190mm.

Availability: Digikey stocks well over 5,000 different inductors that can handle 5A. It didn't even have anything rated for more than 200A. It stocks only 7 that can do more than 100A.

Now repeat this experiment for all the components found in a common wall adaptor and you'll quickly get to the answer of your question.

To summarize: If you had two devices that needed 5A and 6A respectively would you rather buy something that costs in the thousands of dollar range and is larger than your bathtub so you could use it on both, or would you rather buy two palm sized adaptors for $30?

  • \$\begingroup\$ A 5kVA transformer is in the 500-1000$ range \$\endgroup\$ – Vladimir Cravero Aug 27 '14 at 14:13
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    \$\begingroup\$ The common wall adaptor you linked to seems to be a linear power supply. Switched-mode power supplies (basically all modern wall adapters) are even more complicated. \$\endgroup\$ – ntoskrnl Aug 27 '14 at 20:03
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    \$\begingroup\$ @ntoskrnl, you're right! I edited the link to show one that does. I think it was worth it because I used an inductor in my example and that one lacked the inductor! \$\endgroup\$ – ACD Aug 27 '14 at 20:08
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    \$\begingroup\$ @ACD I'm not so sure if you should use 2 palm sized PSUs to deliver 5A. Pushing 5A in a circuit with a PSU rated for less than half of that is certainly going to damage it (personal experience :P) \$\endgroup\$ – shortstheory Aug 28 '14 at 8:03
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    \$\begingroup\$ @shortstheory That's not what I meant but it was confusing so I clarified the summary. \$\endgroup\$ – ACD Aug 28 '14 at 12:31

There are several reasons, actually, including everything you mentioned:

There is only so much current

In the US, the average outlet is a 120V, 15A circuit. That means it can provide at most 1800W (P = V * I) (that is, power equals voltage times current). For a 12v circuit, that means there is only 150A available (1800W / 5v = 150A). To get a 12V, 1000A circuit, you would need a minimum of 100A supplied at the outlet - far more than it could provide. Obviously, a 5A or 10A circuit would fit well within the power capability of a standard outlet.

Energy transmission is inefficient

Even if the power were available, every single component, including wire, has some resistance in it. The more resistance, the lower the efficiency of the circuit. That means if you want to use a certain amount of power (say, to charge a cell phone), you have to pull more power than what you actually needed. If a circuit is 80% efficient - which is quite good, actually - then to provide 1000A, it would need to pull 1250A (1000/0.80=1250). Even at 95% efficiency, it would need to pull an extra 53A. Even worse, the rated efficiency only applies when the device is pulling power near the maximum. If your adapter can provide 1000A, but you're only using 5A, the efficiency at that power may be less than 1%, meaning that your device is using 5A, but the adapter itself is using 10A internally just to keep working.

Waste energy is heat

Wasted energy in this circuit would be almost entirely lost as heat. That means that for our 80% efficient charger, if it is charging at the full current, the lost current (250A) will be heating the air (and components) around it. That's roughly the same as a burner on an electric stove on full power - a lot of heat. Today's plastic adapters wouldn't last a minute!

Size matters

This link (scroll down to the table) shows that a 12 gauge wire (the usual wiring in homes) can transmit about 41A (using the "Maximum amps for chassis wiring" column). 12 AWG wire is about 2mm diameter. 6 AGW can transmit over 100A, but it's over 4mm thick. The thickest wire on the chart, OOOO, is almost half an inch thick (11.7mm), but can still only handle 380A. For 1000A, you would need a wire much thicker - as you could imagine, that would not attach to a phone very well!

Less is more

Often, devices and their adapters are matched on purpose. The adapter has been "tuned" to work with a specific current range, and using it at a much lower current than what it was designed for can make it much less efficient, or even damage the adapter over time.

Current is dangerous

While a high current source wouldn't necessarily mean that every amp would flow through the line, there are cases where even the possibility of providing high currents could be very dangerous. Most voltage adapters, high current or not, use some form of inductor - it helps reduce the "bumps" when converting from AC to DC. One way to think of inductors is that they add "inertia" to current, making quick changes very difficult. The adapter may work perfectly safely at high current while it is being used correctly, but if the plug is suddenly yanked out of the device, that 1000A current will continue to be 'pushed' through the connector by the inductor, causing dangerous (though short-lived) high-current, high-voltage sparking.

Even without inductance, if the adapter were to be shorted by water, metal, or another low-resistance substance, the resulting current would be powerful enough to instantly weld, boil, or burn whatever it touched. Licking the end of that wire could very well kill you. Making a high-current circuit safe is much more difficult than a low-current circuit, and thus much more expensive.

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    \$\begingroup\$ Touching a 12V power source capable of 1000A with dry skin is safe (have you tried a car battery?). Just don't touch it with your tongue like you would a 9V battery. Voltage sources north of 50…100V or so are dangerous if they are capable of anything more than a few milliaps. \$\endgroup\$ – ntoskrnl Aug 27 '14 at 19:57
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    \$\begingroup\$ I was going to upvote until I got to your Current kills section. That's just plain wrong. 1 A is way more than it takes to kill or seriously burn you (depending on where in your body the current flows). If a 12 V 1 A supply is safe to touch, than 12 V at anything more than 1 A is too. Please fix. \$\endgroup\$ – Olin Lathrop Aug 27 '14 at 20:16
  • \$\begingroup\$ Points 2 and 3 pertain to why building a 1000A supply to deliver 1000A is hard and doesn't really answer the question. Using a 1000A supply for a 5A load won't waste a ton of energy as heat or be inefficient to transmit. \$\endgroup\$ – ACD Aug 28 '14 at 17:40
  • \$\begingroup\$ @ACD: Sure it would waste a ton of energy. Inefficiencies usually have a component that is proportional to the actual draw, plus a component proportional to maximum draw. Even though the second is a smaller fraction, if you're overspec by three orders of magnitude, that could easily become dominant. \$\endgroup\$ – Ben Voigt Aug 28 '14 at 22:23
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    \$\begingroup\$ @ACD: Yes, for the same load, the massively overrated component will be less efficient in most cases. Not saying there are no exceptions, but high power components are physically larger almost without exception, which makes parasitics larger, leading to increased waste. For example, power FETs have much larger gates, meaning the input capacitance is higher. Other power components have longer leads than low-power SMD electronics, increasing inductance, even if those leads are thicker to keep resistance low. \$\endgroup\$ – Ben Voigt Aug 29 '14 at 15:43

A 12V adapter that can source 1000A would need to be connected to at least a 120V 100A supply or a 240V 50A supply, in either case much larger than your wall outlet can deliver.

  • \$\begingroup\$ If you think about it, the only sources that reliably deliver 12V @ 1000A are competition audio cars. Those take an entire V8 and a dozen alternators to deliver that, a puny little wall plug doesn't stand a chance (like you said). +1 \$\endgroup\$ – Bryan Boettcher Aug 27 '14 at 22:33
  • \$\begingroup\$ I have heard ideas where DC would be distributed to multiple devices, such as in this article. The idea is to strike a balance between DC from the generator and wall warts: distribute it along-side AC to a single house or small number of related buildings such as on a college campus. \$\endgroup\$ – user52164 Aug 28 '14 at 3:14
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    \$\begingroup\$ @Snowman, that is an idea, but it's a pretty bad one. The War of Currents should have taught us that... but maybe we've forgotten our Edison/Tesla history. Even Volkswagen and its famouse 6V electrical system should have been a good lesson, and a more recent one as well. \$\endgroup\$ – TDHofstetter Aug 28 '14 at 3:36

All of that. The most simple example is that the cable has to handle 5000 Amp. That's going to be a massive cable. I don't mean as thick as your arm or leg, it's worse than that.

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    \$\begingroup\$ Even with superfast fibre optic cable?! I joke, I joke. Thank you! \$\endgroup\$ – Rich Aug 27 '14 at 13:53

Just like anything else more power is larger, more costly and the components more expensive to build. Another element of 12 V is the amount of ripple (ac component) of the dc supply. So like anything else there are a number of elements that make a power supply decision and choices more complex.


the answer is very simple just the internal resistor of the adaptor i give you an exemple 1-open circuit (no current) only the voltahe of your adaptorenter image description here

2-the 5A case: theoriticly you get 5A if you have a load of 2.4ohm I=V/R I=12/2.4=5A with my simulation i got 4.998 which is close to 5A but the load is the internal resistor which is 2.4ohm enter image description here

3-the 1000A theoriticly you get 5A if you have a load of 0.012ohm I=V/R I=12/0.012=1000A which is the internal resistor for such internal resistor you should have a reactor not adaptor :) enter image description here


no matter what is the polarity of the adaptor, the relation will be the same l I l=l v l/R for example if we have a transformer ((most of the adaptor have a transformer inside)) than sure we have an ac current, i.e. the polarity will be change but the internal resistance will be the same (no change) and it will be effect on the current if you want to make a small internal resistance you should have a big transformer with big diameter on its coil to make a small resistance as possible, so the more biggest transformer (adaptor) the more low resistance and effective adaptor

I talk about the internal resistance if it is a dc adaptor (battery of electrochemical element), but if it is an ac (transformer og generator) it will be the impedance

  • \$\begingroup\$ This doesn't answer the question. If making a 12-V 1000-A supply is just a matter of reducing the internal resistance, then why don't we just do that? \$\endgroup\$ – The Photon Sep 25 '14 at 16:29
  • \$\begingroup\$ @ThePhoton - because reducing the internal resistance isn't easy or free. But conceptually, reducing the internal impedance (at a given load) is essentially what is done to make a higher current design. \$\endgroup\$ – Chris Stratton Oct 1 '14 at 16:52
  • \$\begingroup\$ @ChrisStratton, I know that. I'm asking the poster to improve their answer by including this information. \$\endgroup\$ – The Photon Oct 1 '14 at 17:20
  • \$\begingroup\$ That would depend on the internal details of the unspecified adapter - only speculation is possible. \$\endgroup\$ – Chris Stratton Oct 1 '14 at 17:23
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    \$\begingroup\$ @ChrisStratton, I think the other answers show that it is possible to give a useful answer. This answer doesn't usefully answer the question that was asked. If you think it does, you can give it a +1. In the meantime, I'm not sure why you're pissing on me for suggesting ways for m salim to improve his answer. \$\endgroup\$ – The Photon Oct 1 '14 at 17:40

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