# Voltage Regulator Question

Discussion in 'General Electronics Discussion' started by Einstein Jr., Nov 4, 2014.

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Oct 21, 2014
2. ### Gryd3

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Jun 25, 2014
Batteries are typically charged directly with a constant current source.
If you plan to supply 5V to the phone and let it charge it's own battery then your regulator will work... but.
To determine a time to charge, you need to take the phone itself into account, as it will use power as it is charging the battery. You also need to consider the current you are supplying to the phone.
These factors will determine your charging speed.
(Also please note that a USB device... ie the charge port on a phone will typically only draw 500mA max without some sort of circuit to tell the device that more is supported. This will lengthen the charge time, even if your regulator can push out over an Amp. This regulator is a linear regulator, so be mindful of heat... the higher a voltage you put into it, the hotter it will get)

3. ### BobK

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Jan 5, 2010
What is the power source? You will need at least 7V to get 5V out.

Bob

4. ### Einstein Jr.

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Oct 21, 2014
Power source is 9v. If i use that regulator with 9v power source will it fry the phone because it is 1.5amps or will the phone avoid that?

5. ### Gryd3

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Jun 25, 2014
The device will only 'draw' what it is told.

6. ### BobK

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Jan 5, 2010
If you are planning on using the standard rectangular 9V battery, it will have enough energy for less then 1 charge, and you will waste nearly half of that in the regulator.

Bob

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7. ### Gryd3

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Jun 25, 2014
http://en.wikipedia.org/wiki/Nine-volt_battery#Technical_specifications
Will show you the capacity of different 9V battery chemistries.

It's common to have at least 2000mAh in a phone battery. (mine is a 2600mAh)
Your regulator will bring the voltage down to 5V... but the same current that flows into the phone will flow from the battery. They will not give you much of a charge.

Another way to look at it is...
9V x 1.2Ah (Lithium) = 10.8W stored in the battery.
(remember that the same current flows through the entire regulator... so)
5V x 1.2Ah (Result to phone) = 6W stored in the phone.
4V x 1.2Ah (regulator losses) = 4.8W lost as heat.

A little under half of the 'power' in the battery is lost in the regulator as BobK mentioned.
It will work, but is inefficient, and you will need to splurge on high end 9V batteries to give you any decent charge...

8. ### BobK

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Jan 5, 2010
And the capacity is way less at higher currents. The Alkaline that is listed at 565maH in Gryd3's table is less than 300mA hour when discharged at 500mA.

Bob

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