Not bad to put a phone that consumes 750 mA charging from the device, giving A 2?

There is a charge from your phone Lenovo P780. Gives 5V 2A. Is any other device, for example, e-book, which its own network does not have a charge, and usually charge from computer USB, where if I'm not mistaken, 5V, 0.5 a Fraught that I'm going to charge the book from this charge? Or the device itself will take as much as you?
September 19th 19 at 13:35
4 answers
September 19th 19 at 13:37
Solution
Yes, you can. In terms of replacement power sources requirements are:
  • the voltage must match is made, you and there and there is 5 volts
  • the power supply must be greater than or equal to the power of the consumer.
Power is calculated as the product of the current (Amps) voltage (volts). As the voltage is matched, it is permissible to connect any power source, issuing 0.5 A or more.

There is also a third option to stabilize the voltage. So in this case we are dealing with USB, the stabilizer is present in both food sources, so this is no problem.

I brought the General conditions. For devices with the USB connection (not including OTG) rules are:
USB can provide at least 0.5 A (USB-OTG - 0.15 A)
In order to get more, the device needs to "negotiate" with the host. In the USB3.0, for example, it should indicate its energy efficiency class. In the Junior standards are applied various tricks with resistors and the like - so the charger could tell the consumer that he can give more than the standard 0.5 A. otherwise, the device should not take more than 0.5 A

That is why, if you confuse the chargers from smartphones, they can start to charge more slowly or not at all charged, even if the current is stated on the charger, meets.
In my opinion, it's not so simple. Let's say I have two devices, each with its saradhi: one is designed for 1A, and the second to 2A. But they both can safely be charged from the computer - 0.5 A. That is, the power of the consumer is not fixed. If he did not give 2A to which it is targeted, it would still not burn a USB port, and be content with what we have. Moreover, if we change the charging devices, too, nothing terrible happens. Most likely it protection are on both devices. - Josiah.Homenick52 commented on September 19th 19 at 13:40
: It's worth noting that this logic works up until the stabilizer may give the required voltage. If the connected device is too powerful and the voltage at the source drops below the required, the battery will not. - kaden_Reichert81 commented on September 19th 19 at 13:43
September 19th 19 at 13:39
Secure.

Gives 5V 2A.
Gives a maximum of 2A.
When charging the phone will give 0.5 A.
September 19th 19 at 13:41
no
Justify. - Josiah.Homenick52 commented on September 19th 19 at 13:44
no, not harmful - kaden_Reichert81 commented on September 19th 19 at 13:47
: here is my wording of the question played a role (the word "fraught" meant "bad"). For the replies thanks. - Missouri.Hoeg commented on September 19th 19 at 13:50
September 19th 19 at 13:43
2 options:
1)will always negotatiate
2)when something will happen trouble(lead free solders, reduction of Assembly b elements..) and it will begin to recharge
The accelerated demise of the battery and/or gadget.

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