Is it possible to charge the device "stronger" (in terms of [A], not [

Hello,
I guess that's a lame question, but I'm a lame
- does charging the device with a charger with a higher rated current than the original (dedicated) charger risk damaging the device?

To be more specific - I have an eye on a wall charger with a USB output, which has an output voltage of 5V, current .. 2.1A. Can I use it to charge a kindle, whose original charger has 5V and 850mA?

You can use it, but as you write yourself, it has a higher current. At a voltage of 5V, one gives a maximum of 2.1 A and the next 850 mA

The use of such a charger will not damage the device, but will shorten the life of the batteries.

agent8888 wrote:

At a voltage of 5V, one gives a maximum of 2.1 A and the next 850 mA

The use of such a charger will not damage the device, but will shorten the life of the batteries.

You will give me the source of this folk wisdom

marek216 - and what you don't like about my statement?
The markings on the device say unequivocally - 5V 850mA - i.e. you always have 5V until you connect a device that has a current demand greater than 850mA.

Above this current the voltage drops !!!!!!!!!!

Added after 2 [minutes]:

and if you think that having a 5V 2000mA power supply and a 5V 500mA power supply - you will charge the batteries in the same time without any consequences (for the cells) then .... good luck.

Move the head :!:

A more powerful charger will not damage the cells, because the system in the device is responsible for the current and thus the charging time :!:

A more powerful charger will allow you to charge the cells in the optimal time without overheating the charger itself.

not every device has a charging system - in most cases, the charger itself is responsible for it ...

agent8888 wrote:

not every device has a charging system - in most cases, the charger itself is responsible for it ...

Again, I ask you to provide the source of your wisdom because such nonsense has not been in this section for a long time.

Buddy agent8888.
In the case of having two voltage sources with a different value of the nominal current, then receiver this source is connected to it is responsible for the current flow through it. If you used a charger with a lower current before, and now it will be with a higher current, nothing bad will happen to the device. The current from the more powerful charger will be taken the same as from the weaker one, but as mentioned by the brand 216, the charger will heat up less.

Thank you for your comments, gentlemen.
Maybe I will add my 3 grolsche.
I mean this:
- current intensity depends on voltage (5V) and resistance - device (Ohm's law). In both cases they are the same and the amperage should be the same. In such a situation, the rated current of the charger would be something of a "high limit" that could possibly be reached if the connected device pulls so much.

But I asked a question because relying on my "peasant reasoning" seems risky to me. For example, there may be an internal resistance of the charger (impedance) which may prevent some "overcharge / overheating" when using this lower current charger. And I would not like to destroy the device.

I think that speculation based on "folk wisdom" is not enough for us here, I am counting on someone who simply knows

The current supplied on the charger should in theory correspond to the charger performance - that is, the maximum current that can be taken from the charger to the device without harm to the charger, the fact that it will have a higher efficiency does not bother you in any way.

It was already - but I will repeat - practically all mobile electronic devices have a built-in charge controller and it manages the process. It selects the currents so as not to overheat the cells. Of course, when the cells are not heated, it gives the cells more current and in some cases a "more efficient" charger may shorten the charging time.
This applies to "intelligent" devices with complex energy management chips - primitive devices usually have a current limitation - but this device still decides what current is drawn from the charger.

Why, then, are chargers with a low current, e.g. 500 mA, produced at all? For questionable material savings? Because yes, it would be a simple rule for the buyer - the more mA, the better, and the weak ones would be forced out of the market ...

Because this is exactly enough. A compromise between production costs, i.e. the price, and the average charging current and charging time.
You also need to remember about the "maximum charging current" for the cells depending on the design and capacity - it's not that if you have 2A at your disposal, this will be the charging current - this current will be limited by the temperature and the "maximum charging current" (often pulsed through a complicated algorithm) - so in practice, 500mA will flow from your two-amp charger anyway. However, when the device has a "reserve" of battery capacity (e.g. sufficiently high capacity) and at the same time the temperature is low, the controller may "want" to increase the charging current - only it must have this reserve of current on its power supply - on a classic charger (e.g. this 500mA) will not pick up and load with "what is".

It's very simple - those with higher efficiency cost more - where it is not justified they are used cheaper - accountants rule the world - and every penny with the millionth scale of loader production matters.

Hello,
I have a question on 2 sides.
Can I charge the router with a weaker charger, namely 5V / 0.7A,
and should it be 5V / 1A?

majakpl wrote:

Can I charge the router with a weaker charger, namely 5V / 0.7A,
and should it be 5V / 1A?

If turned off, it will take longer to charge, but will charge.
Not necessarily when it is switched on - it depends on the power demand of the device during operation.

An analogous situation is described in post # 16 and # 17.