5V output circuit

For efficiency, you would want to use a switching regulator for both routes. The 'down' switcher, removed the need for the inductor, but remember that you would only be able to discharge the batteries to just over 1.25v, before you lose the ability to deliver 5v, which will give less utilisation of the available power in the cells. Realistically, the boost regulator, done with an IC, designed to do this, should quite easily get over 90% efficiency, and use ost of the available power in the cells as well. However remember that board design round a switching regulator, needs to be carefully done... The battery in the last Ipod Ilooked at, was only a 150mAHr cell, and this was quoted to give 3 hours typical life (giving a consumption of perhaps only perhaps 50mA to 60mA - given the extra losses at the '3 hour' rate, versus the power normally quoted at the '10 hour' rate). The circuitry drew about 400mA, _when charging_ the battery.

Best Wishes

Reply to
Roger Hamlett
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It's hard to downregulate 6 V to 5 V because there's not enough drop. If you had five AA's, giving 7.5 V, a 7805 chip would do the whole job. (I suggest adding a 0.1 uF capacitor on the input and output sides.) That gives you a 3-component circuit.

Reply to
mc

I should add that if you use a 7805, be sure to disconnect it from the battery when it's not in use; the same goes for other voltage regulator chips; they consume a small amount of power of their own.

Reply to
mc

But it's a pennywise and pound foolish solution because you burn up a significant amount of power in heat.

A stepdown switcher will have a limited number of parts too. Less than

  1. It will have way better efficiency of conversion which means that the batteries will last longer.

Another suggestion is that unless size is of the utmost importance, that you'd be better off using D cell batteries. They have much more capacity than AAs.

BAJ

Reply to
Byron A Jeff

20 years ago it might have been hard. Now, LDOs ("low dropout regulators") are as common as dirt.

But the thing to hold on to here is that a nominal 1.5V battery has to discharge to about 0.8V before it is considered "discharged", that is, before its full energy capacity is used. That's almost a 2:1 ratio. A linear-regulator based solution that can still provide 5V at the end of the discharge cycle must therefore start out with around 10V, half of which it must drop as heat.

So a linear regulator is *never* going to be an efficient approach to this problem.

An aside to Roger H: respectfully, although there are simple switcher chips capable of 90% efficiency, that's with good magnetics, good PCB design, and the like, and it's with input voltage at optimum (that is, 90% is usually the peak of the efficiency curve, not the mean). The OP appears to be a beginner, and it seems unreasonable to expect him to get the magnetics and PCB design just right, at least the first time. Thus my earlier estimate of

70%.
Reply to
Walter Harley

Yes. I'd say that these days, if you follow the guidelines published, with many of the chips, something in the order of 80-85%, is not 'unreasonable' to expect for somebody with at least general electronic competence, but 90%, would require care, and 70%, is a much 'safer' realistic figure, especially for use as a guideline. With care, you can get a mean over 90% though. I have a switcher used in a somewhat similar application, and this manages just over 93% on test in the first production units, over the top

1/3rd of the discharge curve, and only drops below 90%, as the battery voltage goes below 1v.

Best Wishes

Reply to
Roger Hamlett

So, how do I make this 70-90% efficent circuit?

Reply to
jasonbot

You already got probably as much of an answer as you'll get on this group. The answer, from MC, was:

But you didn't like that, because it seemed too complex. Unfortunately, switching regulator design *is* complex, as power supply circuitry goes.

Sorry, I doubt someone is going to hand this one to you on a silver plate, for you to "marvel at your own design" as you put it. I think your choices are either to plunge in, or to buy it from someone else. The former will perhaps be more rewarding; the latter will be much less expensive, in the long run, and will be done sooner and be more reliable.

Reply to
Walter Harley

To elaborate on this a little: In order to actually post a diagram, I'd have to breadboard the circuit myself first, and to do that, I'd have to get one of these Maxim chips and a suitable inductor. Or preferably several, to try. Then I'd have to specify exactly what you should buy. Notoriously, inductors have lots of properties besides inductance (they also have current carrying capacity, Q, resistance, and a strange interaction of current with inductance) so you generally want to do more than just get the same inductance that someone else used; they're often specified by make and model.

I could spend $5 on parts, wait for them to arrive, and do it. Eventually I'll probably do so, since I need to get experience with switching power supplies.

But the reason I was able to give you a 3-component, inefficient but reliable, 7805-based circuit off the top of my head is that I've build hundreds of similar circuits. I don't have that level of experience with swtichers.

Reply to
mc

Well, thanx everyone for all the help. I'll get to it next week. I'll post pics when i'm done!

Reply to
jasonbot

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