1
\$\begingroup\$

I 'm planing to use a zinc–carbon 9 V battery to power up:

  1. the circuit that battery belongs to
  2. BeagleBoard-xM

I 'm thinking of using a breadboard, in order to power up both circuits. Beagleboard-xM's electrical specification is 5 V, >1.5 A. For this reason, I plan to use a buck converter to get this voltage. (e.g. LM2596S-ADJ). As I can see, there is a potentiometer in order to adjust the desirable output voltage.

My questions are:

  1. Except 9 to 5 V convertion, how can I adjust about 3 A output current?
  2. How voltage drop of the buttery that will occur over time, affects output voltage and current?
\$\endgroup\$
  • 1
    \$\begingroup\$ I doubt you'll be able to get 5V @3A from a carbon-zinc PP3 battery, regardless of the circuit, for any time at all. An alkaline PP3 battery will probably work (briefly). \$\endgroup\$ – Spehro Pefhany Mar 17 '14 at 15:45
  • \$\begingroup\$ 15 W is a lot to ask of that type of battery. \$\endgroup\$ – Olin Lathrop Mar 17 '14 at 15:49
  • \$\begingroup\$ What is energy? Watt is Power? What are milliamps? What are mAh and What is Ohm's Law because this shows the Best lack of awareness award for 2017 for battery matching to load requirements. en.wikipedia.org/wiki/Comparison_of_commercial_battery_types.. Try a LiPo pack. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 11 '17 at 23:33
7
\$\begingroup\$

You don't "adjust about 3 A output current". The job of the regulator between whatever power voltage you have and this Beagle thing is to provide a steady 5 V. How much current the load (the Beagle thing in this case) draws is up to it.

According to your specs, it can draw as much as 3 A. In that case, your 5 V supply has to be able to supply 3 A, but that doesn't necessarily mean it will always be supplying 3 A. That is probably a peak spec, and part of the time the load won't draw that much. That's fine though. Your supply will keep its output at 5 V as long as the load draws 3 A or less.

What you need, therefore, is a buck switcher that can take around 9 V in, make 5 V out, and can supply up to 3 A at that 5 V.

Note that you are asking for 15 W. That has to ultimately come from the unregulated supply (the 9 V battery in your case). A switching power supply doesn't make more power, only coverts it to a different voltage x current tradeoff. Let's say the switcher is 90% efficient. That means the 9 V source needs to be able to supply 15 W / 90% = 16.7 W. At 9 V that is 1.9 A. A ordinary "9 V" battery can't do that.

A better alternative for this kind of power might be a 12 V lead-acid battery. There are many switcher chips that can handle that voltage and make 5 V out. The 3 A output requirement will eliminate a bunch of them, but these things are still available, especially if you are willing to use a external switching element. Look around at offerings from ST, TI, and even Microchip at that voltage. Linear probably has some offerings too, but they tend to be more pricy.

\$\endgroup\$
  • \$\begingroup\$ Unfortunately, lead-acid are too heavy for my implementation. Could 2 (or more) lithium 3.7 V, 2000 mAh could make this work? Is there any alternative battery solution possible? \$\endgroup\$ – dempap Mar 18 '14 at 13:44
  • 1
    \$\begingroup\$ @demp: Check the datasheet and do the math instead of asking me to guess. With 2 cells, each needs to produce 8.3 W, which is 2.3 A at 3.7 V. Can these cells do that? Even if they can, can they do it long enough for them to be useful? I don't know since I don't have the datasheet. \$\endgroup\$ – Olin Lathrop Mar 18 '14 at 13:56
3
\$\begingroup\$

You don't "adjust" output current. Loads draw whatever amount of current they need, provided the power supply can deliver it. If your total load exceeds the buck converter's rating of 3A, then you will be overloading it. If your total load is less than 3A, then you need not adjust anything.

I'm not familiar with your battery, but keep in mind that the input power will be equal, minus efficiency losses, to the output power. For example, if the load requires 1A at 5V, that is 5W, thus the input at 9V would require more than 556 mA.

As the battery is depleted, the voltage drop will cause the buck converter to draw more current, which will hasten depletion. At some point the battery will be unable to provide the voltage needed by the buck converter.

Your battery manufacturer should have a datasheet that shows the battery performance characteristics, and from that you should be able to ascertain whether it can support the device and for how long.

Per the BeagleBoard-xM System Reference Manual, the current drawn by the BeagleBoard-xM is 350mA to 750mA at 5V (depending on whether it is powered over USB or not). However (page 136 of 164) it is recommended that a 1.5A or greater power supply be used:

It is recommended that a supply higher than 1.5A be used if higher current peripherals are expected to be used or if expansion boards are added. The onboard USB hub and Ethernet do consume additional power and if you plan to load the USB Host ports, more power will be required.

I would recommend measuring your current draw while using a bench power supply, and then determine if the battery can support it. The xM is designed to provide power to peripherals, so you have to take into account whether you or some end-user is going to plug in a USB device or something else that would exceed your power source's capabilities.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.