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For my "prolonging life of old batteries" project I decided to try a switching DC-DC converter. I've got a couple of Recom's R-7805s (datasheet: http://www.recom-international.com/pdf/Innoline/R-78xx-0.5.pdf ), which are capable of 5V at 500mA output. It is choosen over linear regulator, because I want to lose less energy to heat.

However, when connecting 8 old AAA batteries (about 8.6 V) in series to Vin, the converter works fine only with approx. > 100 ohm loads (50 mA current output), and already 50 ohm causes oscillations (I have tried 100, 50 and 10 ohms), because input voltage drops below the minimum. (I am not using blocking diode on the output, if this matters.)

The question is, how can I deduce transfer characteristics of the converter from the datasheet, needed for further circuit analysis? Is my approach somehow flowed, and if not, am I better off with getting values empirically instead of trying to make calculations first?

My current understanding is that at this rate I will need to connect 40 batteries to get at least 250mA, which means I need to add a second converter, because this one is max 34V.

A subquestion I have is how to combine two converters in parallel. My guess is I need blocking diodes to prevent current going back into R-7805. Any hint on this is also welcome.

I have already asked about a different strategy, but I want to try step down approach first, because I need relatively high current.

Also, I do not have a good understanding of internal resistance of old batteries yet. I made some experiments, which showed about 20% of the nominal energy can be had from 1.3V used AA battery (about 500mAh), however, the current should not be too high, certainly not 250 mA. As I am concerned with efficient power transfer, textbooks are recommending matching impedances. I can make more experiments to have some idea of the battery internal resistance, but I'd liked to see if I can get the converter's input/transfer characteristics from specs, regarding input resistance given the 5 V, 250 mA (ideally 5 V 500 mA) DC load.

UPDATE: Some measurements:

  • 1M (scope's input): 7.8 V input, 5 V output
  • 1k load: 7.4 V input, 5 V output
  • 100 ohm load: 6 V input, 5 V output
  • 47 ohm load: oscillating, hard to measure.

After heavy load, input drops to 7.4-7.6 V and need some time to recover.

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  • \$\begingroup\$ Do you have an input capacitor in place of 4.7 - 10uF? I've used the same modules before and got 500mA no problems, although had caps on the output as well from memory about 1000uF because of the peak currents drawn by the particular device I was driving. \$\endgroup\$ – PeterJ Oct 12 '13 at 8:26
  • \$\begingroup\$ Yes, I have 4.7 uF capacitor. \$\endgroup\$ – Roman Susi Oct 12 '13 at 8:33
  • \$\begingroup\$ @PeterJ the module is not probably a problem. I guess, it's batteries. Thus, the question how to make calculations right and how to combine two converters. Figures (need for 40 batteries) are very rough approximations. \$\endgroup\$ – Roman Susi Oct 12 '13 at 8:52
  • \$\begingroup\$ sounds that way I'd missed the part about old batteries and was assuming fresh. \$\endgroup\$ – PeterJ Oct 12 '13 at 8:58
  • \$\begingroup\$ @PeterJ Are you sure about getting 500 mA from 8 fresh AAA alkaline batteries? Or how many batteries you were using? And no, I have not used supercapacitors at the output \$\endgroup\$ – Roman Susi Oct 12 '13 at 9:01
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  1. For a black-box type of converter like these linear regulator replacements, black-box empirical analysis of the transfer functions is about all you'll be able to do. Gain-phase analyzers are expensive but allow you to directly measure things like loop stability, plant response, input impedance and audio susceptibility. I wouldn't advocate $8k or more to characterize a 3-terminal regulator. You've already hit on one of the potential gotchas - lack of headroom caused by the input sagging under load.

  2. With buck converters, paralleling involves some external OR-ing circuitry (be it a diode or a MOSFET) to prevent backfeed. Also, if load sharing is required, some form of balancing (forced current sharing or droop sharing) is also needed. Adding diodes will work but there's always the possibility that there will be no load sharing until one of the converters starts dropping out. I doubt you'd be able to make these 3-terminal devices share well given the low current rating without wasting lots of power implementing external droop (and even then, variations in the OR-ing diode drop could swamp out everything).

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The recom device needs a minimum of 6.5v to produce a 5v output. According to your measurements, with 100 ohm load on the 5v out, the input voltage has dropped to 6v and this is the problem. Try new batteries.

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  • \$\begingroup\$ The whole idea is to apply used batteries. So, maybe I will try to add more of them. \$\endgroup\$ – Roman Susi Oct 12 '13 at 13:51
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    \$\begingroup\$ It'll certainly be better with a higher input supply voltage. The current draw will be smaller. Aim for a couple of volts below max for the recom module. The bigger the better. \$\endgroup\$ – Andy aka Oct 12 '13 at 14:15
  • \$\begingroup\$ Yes. With 24V ("cheated" by adding two 9v batteries) the output is much better. I think, the output is now about 200 mA, but I do not happen to have 2W rated resistors (those not rated high enough overheat in a matter of seconds), so tested with a small motor. \$\endgroup\$ – Roman Susi Oct 12 '13 at 15:10
  • \$\begingroup\$ It's a bit like a solar panel - you've got to take just enough from the battery so that it is meaningful power but not too much so that it grinds to a wasteful halt. How far are you going to take this? There are switcher chips (Linear tech) that can work up to well over 100V. They're not difficult to wire-up either. \$\endgroup\$ – Andy aka Oct 12 '13 at 15:58
  • \$\begingroup\$ I am exploring possibilities at the moment (I have not forgot step-up/Joule thief solution referenced in the question, but I have no clue how to combine sources). My original application idea was to make 500mA USB charger for mobile devices. \$\endgroup\$ – Roman Susi Oct 12 '13 at 17:07

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