I have a 4.2V 10A battery and i want to use it to charge my mobile . But i cannot use 10amps so how can i get 1 amp from battery without losing much power.

 

hi,
If your mobile requires only 1Amp, it will only draw 1Amp from the 10Amp source.
E

 

But if the phone is expecting 5V it might turn its nose up at 4.2V and refuse to charge at all.

 

..... and even if it accepts 4.2V it may refuse to draw more than a trickle of current unless it can negotiate with the charger to get more.

 

Maybe someone else can comment on this but I think most
phones wants a narrow range regulated input to supply the
onboard phone charger circuit.....?

Regards, Dana.

 

If you have a 4.2V /10A power supply, putting a 4.2Ω resistor in series with the power supply will result in the current never exceeding 1A.

It also will result in your load getting some voltage lower than 4.2V.

 

Thanks for your replies
I was directly going to connect the battery to the usb , will it still draw only 1 amp.
And how do i know how much current a device is going to take like a resistor or a home made motor ,etc.
Thanks again for your replies.

 

Thanks for your replies
I was directly going to connect the battery to the usb , will it still draw only 1 amp.
And how do i know how much current a device is going to take like a resistor or a home made motor ,etc.
Thanks again for your replies.

You use Ohm's law if the load is resistive.

I = V / R

A motor is not a pure resistor. It will draw more current on startup and when loaded and less current when free-running.

 

Maybe someone else can comment on this but I think most
phones wants a narrow range regulated input to supply the
onboard phone charger circuit.....?

Regards, Dana.

My Samsung cell phone states ...
USB Max Voltage = 5.25v
USB Min Voltage = 4.40v
other cell phones may vary
It is possible that a 4.2 volt battery does absolutely nothing when plugged in to the USB port of the cell phone.
A 5 Volt Boost Converter is the solution.

 

If you have a 4.2V /10A power supply, putting a 4.2Ω resistor in series with the power supply will result in the current never exceeding 1A.

It also will result in your load getting some voltage lower than 4.2V.

Adding in 4.2 Ohm resistor is not a solution to this problem.
Adding 5 Volt Boost Converter is the solution.

 

Adding in 4.2 Ohm resistor is not a solution to this problem.
Adding 5 Volt Boost Converter is the solution.

I am not offering a solution.
I am trying to get the TS to understand Ohm's law and how power supplies behave.

 

example ... batrery powered 5 Volt USB Charger ...
https://www.sparkfun.com/products/14411

 

I am not offering a solution.
I am trying to get the TS to understand Ohm's law and how power supplies behave.

I am offering a solution.
I am making it clear that adding a 4.2 ohm resistor in series with the Battery & Cell Phone is not a solution.
In fact, the cell phone (load) will not draw more than 1 amp from the 4.2 volt battery,
therefore adding a 4.2 ohm resistor is useless in this situation and actually makes the problem worse.

 

I am offering a solution.
I am making it clear that adding a 4.2 ohm resistor in series with the Battery & Cell Phone is not a solution.
In fact, the cell phone (load) will not draw more than 1 amp from the 4.2 volt battery,
therefore adding a 4.2 ohm resistor is useless in this situation and actually makes the problem worse.

Your solution is a viable one.
What I suggested is NOT a solution and will not work!

 

@mvas, TS wants to know how much current his mobile will take.
Can you explain that to him?

 

The charging circuit in a phone needs an input of 5V so that it can work properly to limit the current to 1A and limit the fully charged voltage of the phone's battery to 4.20V.

A 4.2V battery is Lithium which is very dangerous unless it is charged properly. Look In Google about it.

 

Sorry, I was interrupted by a thunderstorm.
If you directly connect a Lithium battery to 5.0V USB then it has nothing to limit the charging current which might destroy the USB. If the USB limits the current without damage then the lithium cell's voltage will exceed its 4.20V absolute maximum then explode and catch on fire. Don't doo dat.

You wrongly want to use your 4.2V/10A battery to charge your phone, but the charging circuit in your phone needs 5.0V so the phone will not fully charge.
But your 4.2V/10A battery voltage runs down as it is used. It might actually be 3.5V which will not be enough for charging the phone.

You are trying to make a "powerbank" so it can charge your phone. Did you look at what is in it?
1) A charging circuit to charge its battery probably from 5V USB or from a 5V or 6V adapter.
2) A voltage stepup circuit so that its output is always 5.0V even if its battery voltage has dropped to about 3.0V.
3) A voltage regulator so that its output is never higher than 5.0V.
4) A circuit to disconnect its battery when its voltage has dropped lower than 3.0V.

 

Right you are. TS needs to buy a "powerbank".

I was trying to get TS to understand the typical newbie question = How can I get 1A from a 10A battery?

TS needs to understand Ohm's Law and power consumption.

10A @ 5V = 50W

If I only take 1A @ 5V = 5W, where do the other 45W go?
(Question is for only TS to answer.)

 

The starter motor in my car draws a current of 400A when it is cold. But the clock in the car uses the same battery but it draws only 0.01A, not the entire 400A. The battery does not "force" 400A into every load. Does the Thread Starter know why?

 

The starter motor in my car draws a current of 400A when it is cold. But the clock in the car uses the same battery but it draws only 0.01A, not the entire 400A. The battery does not "force" 400A into every load. Does the Thread Starter know why?

No actually !!