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I'm basically loading up enough batteries to run a USB hub and a webcam. The USB hub is rated at 5 volts. Right now I've connected 8 batteries (these batteries, if relevant) in series and in parallel to output 5.3 volts. Will the extra .3 volts cause problems for the USB hub? The 7805 voltage regulator requires a minimum of 7 volts to output a steady 5 volts, so I was wondering if I should increase the battery voltage and do it that way? (I've been trying to avoid this because I figured there would be significant power loss, is that so?)

The application is a little 6 volt car/robot with a webcam/raspberry pi/wifi USB.

I realize that the hub will probably draw more amperage than my 8 batteries can currently support, so I plan on stacking more batteries if necessary.

Also, I'm thinking of powering the USB hub/raspberry pi by separate battery packs. Is there any reason to connect the grounds to each other since they are on separate circuits?

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    \$\begingroup\$ First, can the hub handle 5.3? That depends on the sensitivity of the electronics within. If 5.3 is your open circuit voltage with those batteries, it may be even closer when a load is applied. Find out how much current the hub will draw, simulate a load, and see what the voltage with load is. Second, if the hub draws more amperage than your batteries can provide, stacking more won't help you (unless you mean in parallel, in which case you have to double the number of batteries). Third, yes, they should have common ground. Why have them floating at different levels? \$\endgroup\$
    – krb686
    Apr 27, 2014 at 0:49

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There are many solutions for 5V output. USB can typically take at least 5.25V, which means that 5.3V is pushing it perhaps too far for some very sensitive devices.

However, you don't need a 7805 for getting to 5V stable. There are many LDO regulators with 0.5A - 1.5A current ability that have much lower drop-out. A LF50AB for example can go from 5.3V to 5.0V up to 500 mA (which is the max draw on a standard USB 2.0 port.)

Another option is, as has been suggested, a switching converter of some sort. These are more efficient than linear regulators in many cases, although when going from 5.3V to 5.0V with a linear regulator, you're already a 94% efficiency, which is great in that case. A particular regulator I like because it's cheap and reasonably high current and efficiency is the Murata OKI 78SR. It takes 7-35 Volts in, and outputs 5.0V up to 1.5A, with the same pinout as a 7805, but uniformly high efficiency.

Finally, batteries are typically higher voltage than indicated when fresh, and lower voltage than indicated when depleted. Thus, using linear regulators is often not the right choice. Putting all the batteries in series and using a switching buck regulator (as suggested above) is a good way to get around this. Putting all the batteries parallel and using a boost converter is also often acceptable -- and for Lithium chemistries, easier to charge correctly. If your batteries start above the target voltage and end below the target voltage, you will need a buck/boost, SEPIC, or similar converter that can work with the input being both above, and below, the input voltage. Pololu has some ready-made ones, although they are not terribly high-current rated, and these topologies are often less efficient than buck-only or boost-only.

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  • \$\begingroup\$ I wish later versions of the USB spec had included a proviso that any device may draw just about as much current as it wants, up to a fairly high voltage-dependent limit, when the bus voltage is between 5.1 and 5.5 volts, provided that it instantly stops drawing that much current if/when the voltage falls. That would have made high-current charging really easy--just say that a charging supply should output 5.3 volts when the current draw is within whatever its limit happens to be, and sag linearly beyond that. Such a proviso... \$\endgroup\$
    – supercat
    Apr 28, 2014 at 16:05
  • \$\begingroup\$ ...would have been compatible with earlier specifications (since devices have always been required to tolerate up to 5.5 volts), and would have avoided the compatibility issues that seem to plague USB chargers these days (it's hard to know which devices can be charged at useful speed via which chargers). \$\endgroup\$
    – supercat
    Apr 28, 2014 at 16:08
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The USB specification is +5V, ±10%, which means that anything from +4.50V to +5.50V will be fine, including your +5.33V battery pack.

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The proper solution for 5V power is a Lithium cell with a boost converter for most power density. Instead of rolling your own, I would look at a pack made to charge USB devices. This will give you a run time more space efficient than a group of eneloops.

If anything on the separate systems talk with each other, you generally need to use a common ground. Otherwise, they don't have a common reference.

If you have the hub and all devices, you would be a good idea to sacrifice a usb extension cable to break the 5V line and measure current from of your entire setup. This, combined with your run time, will give you a battery capacity. That is much cheaper than continually purchasing larger packs, as you find you don't have enough.

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    \$\begingroup\$ ^^ As Joe said, why don't you try a 3.7V lithium cell, and a 70 - 90% efficiency boost converter. pololu.com/product/798 \$\endgroup\$
    – krb686
    Apr 27, 2014 at 0:55

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