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I have a choice of choosing 1x 9V Duracell battery or 4x AA Duracell batteries. I want to regulate the voltage to 5V with a 7805. From what I understand, the voltage regulators just converts the unwanted energy into heat to reach the regulated voltage.

My best guess for efficiency is using the four AA batteries, less wasted energy and higher mAh. Should I be looking into the drop off voltage for these different power sources?

My application is to power a MCU while data logging without having to change the batteries often. It would be best to have the highest mAh option, but with a maintained voltage.

Is there somewhere I can look to obtain more information on this? Or could I be enlightened?

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  • \$\begingroup\$ AA Alkaline = 1.6V new, 1V (or a bit less) exhausted. For 5V min out you need at least 5 cells. As many note, 805 needs almost 2V Vin to Vout drop so no good. Some LDOs allow < 0.1V drop. 5 x AAA Alkaline would do better than 1 x 9V PP3 battery. \$\endgroup\$ – Russell McMahon Mar 28 '13 at 2:03
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Neither is a good choice.

As you say, the 9 V battery will be quite inefficient. 9 V batteries are pretty poor in terms of energy density, and then you're going to waste 44% of what you do get in the regulator.

4 AA batteries will make about 6 V at best, which is well below what a 7805 regulator requires as input to make a reliable 5 V out. If you do go that route, you need a LDO (low drop out) regulator. There are plenty available that can regulate 6 V down to 5 V.

The efficiency is decent enough with the 4 AA batteries (83%), but the problem is that you won't get the full life from these batteries. Even a LDO needs some headroom to operate, figure 300 mV roughly, although you need to consult datasheets for the real figures. That means your 4 batteries must supply at least 5.3 V, which is 1.33 V per cell. There is still significant energy left in a normal AA cell at 1.33 V. The energy you do get will be used efficiently enough, but you'll leave a lot in the battery.

So what to do? There are several options:

  1. Don't use 5 V. Why do you want 5 V anyway? Modern microcontrollers run from 3.3 V, and so does most other circuitry you connect to it. These also tend to be more efficient in terms of computes/Joule, so you win on that account too. 4 AA into a 3.3 V LDO is less efficient at making the final target voltage, but you'll probably need less current and you'll be able to drain the batteries to their end.

  2. Use a small switcher. There are various small switcher chips intended for exactly this application. Microchip makes some nice ones I've used a few times, but I don't remember the part number off the top of my head. Check their selector guide. With a switcher, you don't have to match battery voltage and circuit voltage anymore. Now you can use 2 AA, which still contain more energy than one 9 V battery, you can use their energy efficiently, and you can suck them totally dry. This together with a 3.3 V circuit would be the best approach from a power standpoint. I have done this myself a few times.

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    \$\begingroup\$ OP - check out some of the small switchers here: sparkfun.com/categories/54 ... e.g. sparkfun.com/products/10968 \$\endgroup\$ – geometrikal Mar 28 '13 at 0:11
  • \$\begingroup\$ Good to hear Olin thinks Microchip makes nice switchers. I'm trying out a MCP16301T-E/CH, which is awaiting pickup at a FedEx depot! \$\endgroup\$ – Kaz Mar 28 '13 at 1:34
  • \$\begingroup\$ Thank Olin and others! I should've checked my datasheets before I wrote my post instead of relying off memory, thanks for the corrections! \$\endgroup\$ – user2066639 Mar 28 '13 at 6:43
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As you stated, the 7805, a linear regulator, turns a higher voltage into a lower one, with the difference being dissipated as heat. It is not very efficient, especially with higher voltages. A switching regulator is much more efficient.

As for using 4 AAs, assuming Alkaline 1.5v cells, and not NiCD/NiMH Rechargeable 1.2v, you would have 6 volts. 6 Volts is not enough for the 7805. A better option would be a comparable LDO (Low Dropout Regulator). You would want/need one that has a drop-out (Minimum required voltage difference between input and output) of less than 1 volt, because you also need to consider the batteries draining down to 5.6v (0.1v per battery) or less after some usage.

But the better question is, do you need 5v? 4 Rechargeable 1.2v Batteries give 4.8V, which is within most 5v ic operating range. Or use 3 regular batteries for 4.5 (or 2 for 3v) if your IC/Microcontroller has a wider input range. You might not even need a regulator, depending on your setup.

The other option would be powering your 5v microcontroller from 2 batteries (3v) by using a step-up regulator, many which have 75 to 90% efficiency.

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  • \$\begingroup\$ FWIW: Alkaline start at 1.5V ()actually just over 1.6V) BUT drop rapidly to about 1.4V and then progressively to 1V across life of cell. \$\endgroup\$ – Russell McMahon Mar 28 '13 at 2:01
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The 7805 needs 7 V or so input to regulate properly, which you won't get from four AA cells. An LDO regulator would be more sensible. I'd use the AA cells rather then a 9 V battery, they will last much longer.

Many modern MCUs don't need a regulator (it wastes power) and can run from two AA cells.

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  • \$\begingroup\$ Just as a note, some people do consider the 7805 an LDO. What a terribly subjective term :D. \$\endgroup\$ – Passerby Nov 3 '13 at 2:08
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When choosing batteries, one should strive to have an input voltage which, for the life of the battery, will either always be above or always be below the target voltage. If you use a good switching-power-supply chip (and associated components) you will get vastly better efficiency than with a 7805. If you need five volts, using a 9-volt battery with a decent switcher will allow you to run until the battery is basically depleted (0.9 volts per cell). If you're thinking about a 7805 for reasons of simplicity, you should consider instead using a three-pin switching-power-supply module which is designed to drop into places that would normally use a 7805. The 7805 is pretty well obsolete for just about any purpose; even if you want something simple and cheap, using some other linear regulator would probably double the life one could get from a 9-volt battery before a +5 supply started to sag; a switcher could probably extend things even further.

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