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About boosting battery volts? Answered

 If i use an XL6009 module to boost up my battery for example from 3.7 to 5 volts
how it will effect on the battery
my battery will drine much faster ?  will stay the same ? 

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Best Answer 2 years ago

To maintain maximum efficiency, look at the datasheet for the buck converter or data for the module you are using to find an efficiency over load curve. Typically maximum efficiency is achieved at 40% to 60% load, and is generally something between 85% to 98%, but it can easily fall to < 65% under full load (due to ohmic losses and high currents) or very little load (where the quiescent current of the driver chip becomes significant compared to the output power.)

If you are driving a resistive load with greater voltage (due to the booster) then that load will draw proportinally more current, which means much more power, and that multiplied to the losses of the converter (1 - efficiency) is how much power is pulled from the battery. Use the watt-hour rating to approximate how long it will last.

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NivA2

2 years ago

Thanks for the answers guys !

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Answer 2 years ago

And thanks for best answer! If you have any other questions or concerns feel free to ask! :)

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rickharris

2 years ago

To add to what the others said:

Your battery has a finite amount of power available measured in watt hours. It can deliver that power as Voltage and current. volts x current = watts so increasing voltage or current will reduce the time the power is available for.

So right voltage - less time for a given load current.

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iceng

2 years ago

When you boost 3.7 to 5 V your battery current increases by (5/3.7) / Efficiency....

In other words, you pay more current draw for raising the voltage which translates to a shorter battery capacity

ie time per charge referred to as watt-hours or milliwatt-hours...

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Downunder35m

2 years ago

Depends on the load and efficiency of your converter.

Modern ones work with 90% or more, so the real loss will be your load.