Power two devices at the same time using one only battery.

I know,I have the USB port that gives the 5V,but the power that I get from it's not enough to power the Beagle Board with the devices attached. Or it's not fixed and it is not able to do it. Instead,if I get 5V from the selectable voltage,it's enough. The problem is that If I do it I can't power any other device. Any workaround ?

Why you say that It is a low power device ?

Why if I get 5V from the USB port,I haven't the enough power and if I get it from the selectable voltage,I do ? Tekkeon's technicians say : The DC OUT port can output up to 4A and the USB port is rated at 5V and up to 2A.

That wasn't on the data sheet I saw and a standard USB port is limited to 0.5 amps.  It does say that it is rated for 58 Watt Hours.  12V at 5A is 60 Watts which means you won't get quite an hour out of it before you have to charge it again.

There are 2 ICs available in market for stabilizing voltage to 12 volt and 5 volt.

One is 7812 and another is 7805. There are 3 pins. Pin number 1 is for input, Pin number 2 is for Ground and Pin number 3 is for Output.

7812 will give an output of 12 V and 7805 will give an output of 5 V. Both input and output will be DC.

Initial input voltage can be set to 15 V as you have the option to change the voltage and you will get your desired output from 7812 and 7805.

I need to stabilize the voltage of 5V. Can you give me the link for the 7805 ?

ohhh I see that this is a circuit that should be soldered on the board ?

7805 has only 3-4 extra components needed. Also there is a possibilty that your VGA board has 7805 on it somewhere.

You may be able to use a Y USB cable. That provides double output.

Double output to make what ? Please let me understand better. I would like to know what's the reason why the Beagle Board + a lot of devices attached to the 5V regulated voltage of the battery works,but not on the +5V of the USB port on the same battery,even if the manufacturer of the battery says that the USB port is rated at 5V and up to 2A. This is the point that I need to understand. The power given by the USB port seems to be enough. You say that I need to use a Y USB cable to stabilize it ?

What happens if I get this adapter :

http://cgi.ebay.it/HB200-ALIMENTATORE-SWITCHING-5-12V-2A-3-PLUG-PER-6-/400227633038#ht_4475wt_901

and :

a) I configure the voltage of the battery let's say to 15V
b) and I make a little modification to the switching adapter,like you see on the picture.

Could it work ? What do you think ?

 

mod.png

Neither 7805 nor 7812 are switching regulators.   They are linear regulators which supply 1A.  

What, exactly, do you need?  You mention 12V/5A, but then ask if a 12V/2A will work.  

I'm already using the 5/12V switching adapter that I've mentioned. It's perfect because with that single adapter I can power the beagle board and the vga board at the same time. The problem is that it gets the power from the wall,instead I need to get it from the battery. The idea is to replace the wall with the battery. Excuse me for my ignorance,I'm trying to imagine a solution.

What I need is to power on the beagle board and the vga board with one only battery,without to make solderings.

Double output means double of the official USB output (500 mA  per port).

That mains to 12 V thing may work but the quality  is often mediocre.

There are several USB hubs that can output upto 2 Amp per port.  You need to buy a powered USB hub.

This is not a problem. If you say that my idea could work,I need to know which kind of cable I need to use to connect the switching adapter to the battery.

Crocodile clips or brass battery posts.

To run from a battery (and not the AC main) you need a DC-DC converter.   Take a look at Mouser (http://www.mouser.com) or Newark (http://www.newark.com/dc-dc-converters) or other suppliers.  

What I'm looking for is a 12V / 5A Switching Power Adapter to use with the battery,not from the AC-DC. The battery with which I want to do the operation is the Tekkeon MP3450i R2. Please kindly note that with this battery you can configure the voltages choosing from 5V to 19V.

Let me paraphrase your requirements:
You want to use the MP3450 by itself as the source of power for both boards.
You need:

a) VGA board that wants from 8.5 to 12V
b) the Beagle Board XM that wants 5V

So, to use the MPP3450 you will have to set it for 8.5 to 12v (I will call this "9v" hereafter) output in order to power the VGA board.  That is, unless you do something even more ambitious.  That takes care of the VGA board then.

For the Beagle Board you need 5v which could be had in a few ways starting from the 9v output:
For just about any of these, you waste less power with the least output voltage from the MP3450.
1) connect the 9v to the beagle board through a dropping resistor.  Depending on how variable the current is on the Beagle Board, this may work.  Just select a resistor value to yield 5v.  Not a great solution but perhaps "good enough".  
2) connect the 9V to the beagle board through a 4v zener diode with adequate power rating and heat sink.
3) connect the 9v to the beagle board through a series chain of 5 or 6 silicon diodes.
4) connect the 9v to the beagle board through a voltage regulator chip as has been suggested.
5) similarly connect the 9v to the beagle board through a switching DC-DC converter with 5v output.

In any case you're going to have some soldering to do in creating a "special" connection from the MP3540 output.

@marshall
sure you come from the analogic world, your recomendation for getting 5V for powering a digital device are scary...
you cannot ever regulate tension depending on charge like using series of direct diodes.. even worst with a zener diode that need a permanent draining current in order to work while continuosly draining the battery w/o sense...
cascading dc-dc converters can have serious side effects, remember those converters are really switching power supply, cascading them could easilly lead to oscillations and burning things...
you haven't said a word about ground loops....

@marietto2008
the USB connector is going to provide the ammount of current the factory says and not what the USB standard says. If the company is a reputable company saying their USB conector can give 2 A you can take that for granted...
if you cannot get the ammount of current you need from the USB port forget the thing (it cannot be done with your reqs), and please be careful with some advices....

I have two easy choices :

a) to use an HDMI enabled board with a matching screen
b) to use two batteries

I choose the option b.

from an electrical point of view a) is the best option...

why ? If I choose to use one only battery it means that I have to take current from both battery's ports (5V and the regulated voltage port). Does not this action consume the current faster than using two batteries ?

for option A it means that I will get 8.4V from the battery to power the VGA board and 5V from the USB port to power the beagle board with the devices attached....

errata corrige :

8.4V from the regulated voltage port to power the HDMI board and 5V from the USB port to power the Beagle board + devices attached...

Instead,now I'm using two batteries. To power the VGA board,this :

http://www.bixnet.com/5v7libapa.html


and to power the Beagle board,this :

http://www.tekkeon.com/products-mypowerall.html   (model : MP3450i R2)

from an electrical point of view is not good having more than one switching power supply arround...
the best you can do is just powering everything from an unique device.

switching power supplies are not toy batteries that you can arrange the way you want, there are several considerations as ground loops, feedback, oscillation, etc that make then better used as single devices...

can you clarify to me one point,please....from the point of view of the system's autonomy : is it greater if I use one battery or two ?

it would be better with just one converter...

Converter ? what it means ?

you cannot power the thing "directly" with a "real" battery then you need a device converting from
ac-to-dc  (powered from your electrical outlet at your home) or dc-to-dc powered from some battery; both are converters...

I don't understand your last sentence. I'm using succesfully two batteries to power everything and you say that I haven't real batteries ?

ok, are you using DC batteries ?  Eveready or Duracell things like that?

What ? I don't need that kind of batteries. As I said I'm using two portable batteries.

To power the VGA board,this :

http://www.bixnet.com/5v7libapa.html


and to power the Beagle board,this :

http://www.tekkeon.com/products-mypowerall.html   (model : MP3450i R2)

I see okok
those are rechargable batteries that won't last much if your screen is big but yes you can use that can of batteries but you know the 2nd one mus have some dc-dc conver on it in order to provide more than one tension...

if you can power everything with just the 2nd one it would be better if not use both batteries....

Exactly. What I would like to do is to use only the battery n.2 because it is strong,but there is no way.

@fmarshal on ID:36343874 explained very well that I have to do any kind of soldering on the board,but I don't want.

using both batteries is really an ugly alternative... anyway...

Please choose the best option here,from an electrical point of view :

a) HDMI board + 8.9" screen and the battery n. 2
b) VGA board + 8" screen and battery n.1 and n. 2

no way...  ID:36343874 does not explain clearly what to do at all.
I already commented on that

a) HDMI board + 8.9" screen and the battery n. 2

ok...

Use whatever works for you.  I would be looking for a larger battery/power supply because I don't want to carry extra things around.  I don't know how portable you want this to be but it sounds like you're going to have to recharge the battery (batteries) fairly often.

I don't know what "analog" has to do with it.....  just volts and amps are required.  How many bits resoltuion on those??

you cannot ever regulate tension depending on charge like using series of direct diodes.. .

Do you mean "voltage" = "tension"??  OK.
Actually both will work nicely enough.
In the case of series diodes there will be around 0.7v forward bias drop per diode, so 5 or 6 to drop 4v from 9 to 5.  The voltage drop will be pretty stable at any appreciable current.  The power dissipation in the diodes will be 4/9 in comparison to the device's 5/9.   So, if the 5v device draws 1A then 0.7W per diode.  That's pretty reasonable for any number of diodes.  The diodes can be arranged in the wire so no soldering on any boards.  Most devices will work with +/-5% to +/-10% variation in the supply voltage.  Once you decide on the number of diodes the voltage should not vary much at all.

even worst with a zener diode that need a permanent draining current in order to work while continuosly draining the battery w/o sense..

No.... you put the zener diode in series with the load to get a fixed voltage drop.  The load current biases the diode.   No battery drain without load.  One possible disadvantage: the one diode has to dissipate all the power.  As in the 1A case, it would be 4W.  Also can be in the wire.

@Davebaldwin : can you suggest a larger battery ?

@fmarshall :  you say that using the diodes no soldering is necessary ?

@fmarshall

I don't know what "analog" has to do with it.....  just volts and amps are required.
this is too simplistic when dealing digital equipment, unfortunatelly there are other considerations when using switching power sup-lies or converters that you completelly overlooked...

Actually both will work nicely enough.
again you oversimplify things, your method with direct diodes is never used on digital equipment, because:
1) "voltage" dependends on the load
2) it is not efficient. (important if battery powered)
3) if the digital load goes to "sleep" it drastically drops the current demand, then  the forward bias per diode is not 0.7 but much less than that then you can easilly roast digital gear for over voltage

No.... you put the zener diode in series
1) the zener curve is also depending on charge, -> no stable "voltage"
2) uneficient from an energy point of view...
3) if the digital load goes to "sleep" it drastically drops the current demand, then  the zener reverse bias is not the zener "voltage" but much less than that, then you again can easilly roast digital gear for over voltage


botom line: when you power digital gear forget about diodes,
and if you also need efficiency forget about linear regulators  7805/7812 etc....

I don't see anything exactly like you're talking about.  Duracell makes 'PowerPacks' at 300, 450, and 600 watts that are shown in their ads as powering laptops and LCD screens.  The 300 Watt unit is $129.

With 140$ I bought the portable Tekkeon MP3450i R2 with the selectable voltages (5 to 19V). The Duracell powerpacks are not portable at all. Don't forget that I'm trying to make a device that can be used also when you are out of home,something like a tablet.

The Duracell powerpacks are battery powered AC inverters that you can take anywhere.  They are used to Generate AC power.

ok,but :

1) how much big they are
2) how they can be attached to power the Beagle Board and the VGA/HDMI board ?

About 12x9x8 inches, 14 pounds.  The datasheet is here: http://www.duracellpower.com/documents/tech-specs/DS20070621_duracell-pp300.pdf  One 12VDC output plus two 120VAC outputs which could power a 5VDC power supply.  It says the 12VDC output can be used to jump start a car.  It does have a lead-acid battery in it.

12 x 9 x 8 inches = 30.5 x 23 x 20 cm x 6.4 kg. not portable at all,man.

If you say so.

and I think that I don't need all the power generated by the duracell 300 power pack. I can give up to some power to achieve portability.

ok boys. The best solution is to change the beagle board XM with the Tegra 2 soc :)