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Topic: current division (Read 177 times) previous topic - next topic

Liz0905

There is a matter of principle that I can not understand how the 5V's pin works. If I connect the Arduino to a 9V battery, suppose 600MAH. How is the current divided between all the pins and how is the 5V pin affected by the Arduino power source?

thank you!

CtrlAltElite

The Arduino has a linear voltage regulator which takes the 9V and burns off four volts at whatever current the AVR takes as heat.

The rest of your question, I didn't understand.

600mAh (note the spelling) is a measure of the capacity of the battery. In theory, the battery could supply 100mA for six hours, but this is unlikely. It can probably manage 10mA for sixty hours.
Has there been an outbreak of the stupid virus, and I didn't get the memo?

Liz0905

The Arduino has a linear voltage regulator which takes the 9V and burns off four volts at whatever current the AVR takes as heat.

The rest of your question, I didn't understand.

600mAh (note the spelling) is a measure of the capacity of the battery. In theory, the battery could supply 100mA for six hours, but this is unlikely. It can probably manage 10mA for sixty hours.
thank you,
i didn't understand the first part of your answer.

and about the capacity, why its unlikely?

Smajdalf

and about the capacity, why its unlikely?
It depends on the battery chemistry and (I guess) geometry. Battery supply less total current if the current draw is too high. Also if the application needs large current the battery is unable provide such current when it is nearly empty. "Too high" is a few mA for small coin battery but a few A for Li-Pol batteries.

CtrlAltElite

Suppose the AVR consumes 50mA running at 5V.
If you have a linear regulator, then the regulator device simply wastes, as heat,  ((9 - 5) x 50mA ) Watts
Has there been an outbreak of the stupid virus, and I didn't get the memo?

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