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I was wondering what the effects of using a lower rating power rated supply for device would be. For example using a 1.52A supply at 19V to power a device where the charger that comes with the device is 4.74A at 19V.
Is doing this likely to damage the device or will it just take longer to charge.

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    \$\begingroup\$ When I was a kid I connected a 60W car headlight to a 12V/300mA transformer. This setup has worked for weeks before it ended with the infamous magic smoke. \$\endgroup\$ – Federico Russo Jun 14 '11 at 13:35
  • \$\begingroup\$ I find it hard to believe it worked, meaning the headlight lit up like it was intended to do. It may have "worked" only in the sense that nothing blew up - until it did of course. So really it didn't work at all. \$\endgroup\$ – Olin Lathrop Jun 14 '11 at 14:11
  • \$\begingroup\$ I have been borrowing a laptop charger that is an amp or two less than the one that came with it it does not seem to be heating up any more than usual and and just takes slightly longer to charge \$\endgroup\$ – Hugoagogo Jun 14 '11 at 21:23
  • \$\begingroup\$ @Olin Lathrop: it worked, but I guess the lamp only lit rather dim. (The weeks were not continuous!) \$\endgroup\$ – Federico Russo Jun 15 '11 at 6:43
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Normally the device you are trying to charge will not "know" the charger is underrated. So it will not reduce it's current (ampere) usage.

Under high load components in the charger will overheat and fail. All chargers have protection against high loads. There are two kinds of protection:

  1. The charger will die permanently because some fuse or component melted safely.
  2. Protection will kick in and shut down the charger (temporarily).

Without protection the overheating of components could cause a fire or shortcut in the charger.

Today there are mobile devices available which communicate (with resistance for example) with the charger. When the data pins of the USB charger are shorted, the device knows it is connected to a special charger and it will draw 1A. Otherwise it will just draw 500mA. Several Android devices and the iPhone use this.

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  • \$\begingroup\$ Out of curiosity, given that IIRC the USB spec requires USB devices to tolerate 5.5 volts, I wonder what would have happened if Apple and other vendors had simply designed their charger to source 5.4V and their equipment to ramp up their current draw as long as the supply was over 5.2V, and reduce their current draw if the supply fell below that (including logic to avoid oscillation)? Figure that devices that aren't expecting to supply lots of current won't have a voltage that high, and that drawing excess current will cause a supply's voltage to dip (signalling devices to reduce current)? \$\endgroup\$ – supercat Sep 19 '11 at 17:52
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You seem to be confusing two things, operating the device and charging a internal battery.

However, in either case using a power supply with a order of magnitude less current capability is very unlikely to work. The device will try to draw more than the little current the power supply can produce, which will either collapse its output voltage, cause the supply to go into current limit mode if it has such a thing, or perhaps cause it to actively shut down. Either way, the output voltage won't be 9V or anything close, so the device won't work.

Lower voltage is unlikely to damage most devices even if they don't operate correctly. Obviously the device must be able to deal with brownout for at least some small time since no power supply can go instantly from 0 to full output voltage.

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  • \$\begingroup\$ The device in particular that this question is about says it is short circuit protected is this what you are talking about \$\endgroup\$ – Hugoagogo Jun 14 '11 at 21:55
  • \$\begingroup\$ That means the power supply won't get damaged. It will turn down its output voltage or completely shut down to protect itself. However, that doesn't mean the device it's trying to power will run. Since the power supply output voltage will go low when the device tries to draw more current than it can supply, the device is very unlikely to work correctly. \$\endgroup\$ – Olin Lathrop Jun 14 '11 at 22:06
  • \$\begingroup\$ Is it likely to damage the device in the long term (it seems to function) \$\endgroup\$ – Hugoagogo Jun 15 '11 at 0:06
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It won't damage the device. But it is likely to damage the transformer or SMPS, if it does not have over-current, over-temperature or over-power protection. Sometimes supplies have a thermal fuse which prevents a fire, but when it blows it's not easy to replace it (ever tried to take one of those supplies apart?)

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