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Topic: USB Power Backup (Battery) (Read 1 time) previous topic - next topic

ZereoX

Today, I purchased a new fancy school bag that included a Universal Charging Battery (Solar Powered) and I wanted to know if it would be powerful enough to run my Arduino Uno?

Specs:
Li-ion 3.6V 1200mAh
IN: DC 5V 400mAh
OUT: DC 5V 450 mAh

RuggedCircuits

5V at 450mA is plenty powerful. I'm guessing the spec is really 450mA and not 450mAh? The former is a measure of how much current it can supply, the latter a measure of energy storage capacity (of batteries e.g.)

Do you have a link to this backpack? It sounds like a really neat product.

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ZereoX


I'm guessing the spec is really 450mA and not 450mAh?
Do you have a link to this backpack? It sounds like a really neat product.


The back does say mAh.
Here is a picture of the unit:

RuggedCircuits

Well, then I'm confused. The battery is 3.6V 1200mAh...I don't know where the other mAh numbers are coming from.

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ZereoX

Is it possible since the Battery is designed for recharging Mobile devices it's output is only 5V/450mAh instead of 1200?

rabits

It is too powerfull for simple runing arduino dependig what yo wish arduino to do and how long time

ZereoX


It is too powerfull for simple runing arduino dependig what yo wish arduino to do and how long time


Are you saying it will fry the Arduino if plugged in?

weirdo557

#7
Feb 07, 2011, 07:38 am Last Edit: Feb 07, 2011, 07:43 am by weirdo557 Reason: 1
woah there, if its a regulated 5 volt output then nothing will be in danger of being fried. and i think the conversion in mAh comes from the boost converter. to get 5 volts from 3.whatever you need to drop the current capacity, so more amp draw on the battery in exchange for more voltage. so long as the energy is the same (minus conversion losses) it works out.

[edit: the input of 400mAh makes no sense however]

ZereoX


woah there, if its a regulated 5 volt output then nothing will be in danger of being fried. and i think the conversion in mAh comes from the boost converter. to get 5 volts from 3.whatever you need to drop the current capacity, so more amp draw on the battery in exchange for more voltage. so long as the energy is the same (minus conversion losses) it works out.

[edit: the input of 400mAh makes no sense however]


Thanks. I think the Input might have to do with it being Charged by USB or by the Solar Panel.

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