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I have a small problem with one of my projects. I am using a cheap adroid tablet and arduino for a home automation project. Now there is a problem with how to power this thing. Arduino will be powered with an external wall power supply (and a voltage regulator of course) because I need a higher current than a standard usb can provide. But that isn't really the problem. The problem is in powering the Android tablet. When acting as an USB host, the tablet won't charge from the USB port which really complicates things. So I looked inside and the battery is a standard 1 cell lithium battery connected with 2 wires. So for the test, I desoldered the battery and connected a regulated 5V power source (with three diodes wired in series so the voltage was about 3.9V) - and the thing worked perfectly. Android just showed the battery as full and that was it. Now for the problem - I would like to use the battery as an UPS. I know that I can't just wire the battery and the wall adapter together and hope that it won't blow up. But is there a simple way to isolate the battery from the power supply?
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Hmmm I just got a new idea. I have a few spare 2 cell lipos for one of my RC planes. I could lower the voltage from that by using a few diodes. Now if I connected a battery and a power supply at the same time which one would power the thing? Would it just draw power from both, or would it only use one of the power sources?
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If you ran both thru diodes, the source with the higher voltage would reverse bias the lower voltage source's diode and keep that source turned off.
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Perhaps relevant for your initial problem: Can you charge the tablet form a power jack? If so you can try to use a combined notebook power supply with additional USB power out. E.g. http://www.conrad.biz/ce/de/product/515004/VOLTCRAFT-NPS-90-USB-Notebook-Netzteil-Laptop-Netzteil-BASIC-LINE-12V14V15V16V18V185V19V195V20V21V22V-/ So you can drive the tablet by power jack and the Arduino via USB - if 5 V/1 A via USB fits your need.
« Last Edit: September 07, 2012, 10:55:48 pm by Clemens » Logged

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Perhaps relevant for your initial problem: Can you charge the tablet form a power jack? If so you can try to use a combined notebook power supply with additional USB power out. E.g. http://www.conrad.biz/ce/de/product/515004/VOLTCRAFT-NPS-90-USB-Notebook-Netzteil-Laptop-Netzteil-BASIC-LINE-12V14V15V16V18V185V19V195V20V21V22V-/ So you can drive the tablet by power jack and the Arduino via USB - if 5 V/1 A via USB fits your need.
Sadly I can't! It only charges through the USB connector.

If you ran both thru diodes, the source with the higher voltage would reverse bias the lower voltage source's diode and keep that source turned off.
So if I understand this correctly - if I have a 7.4V battery and a 9V power adapter, it would use the 9V power supply by default? But what would happen if I used a combination of a voltage regulator (L7805) and a few diodes to bring the voltage of the power source down. Because I need 5V for the arduino, and I also need 4V for the tablet.
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I think I would use 2 switching regulators like this
http://www.pololu.com/catalog/product/2103
and adjust to what ever levels you wanted. 
Bring the Arduino 5V in thru the USB  connector, other threads have suggested the voltage regulator used in recent arduino's does not handle being bypassed well and it can be damaged.
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Well I am not using the arduino board itself so the onboard regulator isn't really my concern. But either way, I can't use USB power alone because I need higher current to run a few servos.

I have a few LM317 regulators around - would that work? A quick schematic of what I was thinking about:

J1 is a 3.7V output for the tablet
J2 is 5V output for the arduino
If I wired things like this, would the power be only drawn from the DC power source? Or would it draw power from the li-po too?
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Power would only be drawn from the source with the higher voltage.
S0 9V wallwart when plugged in, and 7.4V battery when not.

Find a schottky diode, one with a lower drop, vs 1N4004 with 1.1V drop.
Example:
http://www.mouser.com/ProductDetail/ON-Semiconductor/MBRA210LT3G/?qs=sGAEpiMZZMuIUjt4yeP9c5NLLsTwJPGJdgqlfPLViFg%3d

I think you also want R1 between the Output and Adjust, not Input.
Don't forget caps as well.
http://www.ti.com/lit/ds/slvs044u/slvs044u.pdf
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That schematic was just a quick sketch - so there could be some errors there. And missing capacitors too.

But does it really matter which diode I use? After all, I will be using V-regs to lower the voltage even further. So it shouldn't matter if there is a drop at the diode. At least that is what I think...


I'll try this as soon as possible and we will see if it works
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Your call.
Datasheet says Vo - Vi needs to be 3V or more.
7.4V - (diode drop) - 3V = 4.4V -(diode drop).

So with 1V diode - can't get 4V out:
7.4-1-3 = 3.4V

With schottky diode:
7.4 - 0.4 - 3 = 4V.

You may  see 4V for a while when the LiPo is fully charged and sitting at 8.4V. (4.2V/cell)
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Arduino for Teens available at Amazon.com.

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