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This question already has an answer here:

How much can a power brick safely vary from what the device it's used with expects?

As an example, let's say I have a device that needs an AC=>DC power brick. The device expects 19V/1.58A, but the brick is missing. Now let's say I find a brick where the connector fits the device and is also 19V, but outputs 3.42A. How safe is this likely to be for the device? Will the device only draw the power it needs, or am I likely to blow a capacitor somewhere?

If that latter, this example the brick provides more than twice the current requested. How much closer would have I have to get to be reasonable? How much can voltage vary vs amps? Given different devices likely have more tolerances than others, is there a good rule of thumb to look for to know how much you cheat on matching power brick to device?

While the example does match a real situation, I'm also interested in the general case.

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marked as duplicate by laptop2d, Daniel Grillo, uint128_t, DoxyLover, dim Dec 8 '16 at 21:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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These bricks are constant voltage supplies, i.e. they're both spec'ed to deliver 19 V.

The 1.58 A and 3.42 A are specifications for the maximum current they can supply at that voltage – using less than the supply can offer must work.

Physically, a power supply can either define the voltage it offers, or the current it pushes through a load – never both, because for any load, one is a function of the other.

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  • \$\begingroup\$ What about part 2 of the question, though? Say I had a brick that was 18.5 volts. Clearly I'd void a warranty, but if it works well enough, how much change there could I get away with? \$\endgroup\$ – Joel Coehoorn Dec 8 '16 at 17:02
  • \$\begingroup\$ I can't tell you – that completely depends on how the device you're feeding with this works internally, and even if you told us what device it was, we wouldn't know, since none of us was involved in the design. However, 0.5 V of 19 V is something like 2.5 % error – and probably is OK. \$\endgroup\$ – Marcus Müller Dec 8 '16 at 17:04
  • \$\begingroup\$ Okay... so there's no rule of thumb here. Just try to be as close as humanly possible, and match exactly is by far the best option. Got it. \$\endgroup\$ – Joel Coehoorn Dec 8 '16 at 17:05
  • \$\begingroup\$ Most power inputs are +/- 5% at a minimum, 10% is more normal. However keep in mind that most DC power bricks are also going to be +/- 5% or so, generally they'll be over the specified voltage at low currents and under it at the higher end of the rated current. However ultimately it's up to the person who designed the system and how much margin they built in. \$\endgroup\$ – Andrew Dec 8 '16 at 17:05
  • \$\begingroup\$ It depends entirely on the specific design parameters of the individual device - some will happily operate from half the rated input voltage, some few will fail to operate at 95% of the rated input voltage. Most don't handle significant over-voltage well, but some few do - it's a design choice. \$\endgroup\$ – Ecnerwal Dec 8 '16 at 17:42
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Voltage is important. You may be able to get away with 18.5V, but it may take longer for the battery to charge up full. Or maybe the battery will never charge back up to 100%. Supplying a HIGHER voltage than the rated load can be assumed to always be dangerous. NOT recommended.

A load will draw only as much current as it needs. Having a source capable of MORE current is not a problem. You can plug a 5W night-light into the same power source that will supply a 1000W light.

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The answers here are mostly correct, although there are two things to consider when connecting a supply with significantly higher current capability: Most power bricks nowadays are switchmode converters, and many of them fail gracefully if they are overloaded by either cycling or limiting current.

So if the device to be powered by the 19V, 1.58A supply fails with a short circuit, the original supply is unlikely to be able to deliver more than like 2.5A, but the supply might go up in smoke if it is badly designed. On the other hand, the 3.5A supply will likely deliver 4A if it does not shut down on overload. In a cheap device without an input fuse, there is the possibility for the broken device to catch fire due to the higher power supplied into it. As long as the device is not broken, the stronger supply does most likely no harm.

Furthermore, there are devices with an input fuse that is quite close to the expected operating current of the device, and might be blown by the inrush current. These designs rely (on purpose or by chance) on the limited current supplied by the bundled power brick. A stronger brick might provide enough inrush current to blow the fuse. This gets more likely, if the output voltage of the replacement supply is also (even slightly) higher than the original supply.

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