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I have a battery pack consisting of 4x1.2V AA Sanyo NiMH Rechargeable batteries used to charge a Beagleboard device. I want their discharge output to not exceed 5V. But when they're fully charged, the voltage output is over 5V (it triggers the overvoltage detector of the Beagleboard). What is a simple (and relatively small) intermediate component I can place between the battery pack and the Beagleboard to limit the voltage to under 5V?

donquixote
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You need a voltage regulator.

Unfortunately, you've spec'ed batteries which are awfully close to the operating voltage of your system. If you can use 5 batteries instead of 4 (giving you 6.something to 5V, instead of 5.something to 4V) , a low-dropout linear regulator will be a simple, easy solution. The standard 7805 has too high a dropout for this purpose, but there are other pin compatible regulators; you'll want a TO-220 to dissipate the power that the Beagleboard can draw at full charge.

If you must use 4 batteries, you need to dissipate the excess voltage as heat through a MOSFET when the voltage is greater than 5V, and turn the MOSFET on if the voltage is less than 5V. You are running slightly out of spec when you're below 5V, though I'm not sure what the absolute minimum voltage is for the board.

If you want the best solution possible, a buck-boost regulator would get you optimum efficiency for voltages slightly above and slightly below 5V. You could even run it off a single battery, or from voltages much higher than 5V. However, this is an expensive and complicated solution. I'd recommend just using 5 batteries and an LDO.

Kevin Vermeer
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  • If the dev board will function with less than 5 V (e.g. 3.3-5.0 V), then you could just as well power it with the 4 cells and an LDO. That said, however, the archaic 7805 may misbehave (draw an abnormal amount of quiescent current) if you are operating near the regulation point. Modern LDOs should not have this problem, but as always, check the datasheet. – Nick T Feb 24 '11 at 01:55
  • @Nick T - The 7805 is NOT a LDO (Low-Drop-Out regulator). As long as you use a real LDO, you should be fine. – Connor Wolf Jun 13 '11 at 21:15
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You need a "voltage regulator". The standard answer to "I just want 5 volts with no hassles" is to use a 7805 voltage regulator chip, but they "eat" 2 volts, so you need at least 7v of batteries.

You have two good options:

  1. Use a "low dropout" 5v regulator (eg LM117). The datasheet will contain an example circuit.

  2. Just go to a junk store and buy a car cellphone charger for an obsolete phone. These are dirt cheap and typically contain an efficient 5v "switch mode" regulator capable of delivering up to 1 amp. (Motorola 34063 is the usual regulator device inside these).

Christopher Biggs
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  • Hey unixbigot, thanks for your answer. Regarding (2), I actually tried one that delivered 1A but apparently that wasn't enough to power it. 2-3A would probably be ideal, but I can't find a ready-made charger that offers that combination of voltage and current. If you know any or find any, I'd be glad to accept your contribution in the [Battery-powering up a Beagleboard XM](http://electronics.stackexchange.com/questions/9772/battery-powering-up-a-beagleboard-xm) question. – donquixote Feb 24 '11 at 01:12
  • The beagleboard xm manual specifies the power consumption on page 26, it says it uses 750mA. Maybe you have more stuff plugged into the expansion port? – Christopher Biggs Feb 24 '11 at 03:17
  • You can use LM317 which can suply 3A output current instead of LM117. @unixbigot Beagleboard's itself uses 750mA but as you say this does not include peripheral devices. – Emre Yazici Dec 08 '11 at 20:44
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You use a voltage regulator to keep a power supply voltage at a fixed level. Linear regulators require an input voltage higher than the output voltage, often a few volts, less with LDOs (Low DropOut). So that's no good. Then there are switching regulators. They are a bit more complex, but more efficient. They exist as "buck" (input voltage higher than output voltage) or "boost" (input voltage lower than output voltage).
But there's also a "buck/boost" which can handle both situations. Since your input voltage is near the output voltage, with fresh batteries higher than 5V, with partly drained batteries lower than 5V, the buck/boost regulator is the way to go. The Linear LTC3785 requires quite a few external components, but gives you an efficiency to 95%.

stevenvh
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Maybe a 5V Zener diode + resistor will be good enough?

Dave
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    This will just burn up the excess charge in the batteries until it no longer conducts, and then you'll have a series resistor in your supply line that will cause the input voltage to be less than the battery voltage. This is not a good idea. – Kevin Vermeer Feb 23 '11 at 22:48
  • I could have sworn that H&H showed that as a solution when I was in school. :) Thank you for the clarification! – Dave Feb 24 '11 at 03:53
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Could you permanently place a 5.6V 3W zener (maybe make one from a TL431 and a power transistor) over the battery pack, so it woill be charged only up to that level? But note that I am not sure how your charger will respond to this, and you will need a charger for the battery pack, not for the individual cells.

Wouter van Ooijen
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