5V regulator for batteries cheap efficient?

I'd estimate about 65 mA total.

How much does your product draw..? If more than a few tens of mA then only a switcher will give you significantly better efficiency. If a few mA, then look at the Holtek HT7150 - quiescent a couple of microamps, and takes input up to 24V. About US$0.30 from memory. Also look at OnSemi 78LC/FC series - similar efficiency but lower input range

switcher will give you

- quiescent a couple

Replaced by HT7150-1

Ricoh has/had some LDO regulators with quiescent current around 1 microamp eg Rx5RL - see

If the input voltage is over 10V and the current is low enough, a switched capacitor converter may the way to go.

--!!---------------+--->!---+----- +5V ! ! ! ! [L] --- ! ! --- ! ! ! (+)-----[L]----+---+--!!-----+-->!--+--+--------+------ GND ! ! ! Floating / switcher ! --- Battery ! chip [L] --- ! ! ! (-)------------+-------------+------+------------------ -5V

You may even be able to find the inductors almost all as one part. IIRC. J.W. Miller has a multiwinding inductor.

The input side inductor could be part of that plus a small independant choke in series.

Wouldn't the added efficiency depend on what his input voltage is? If only

Actually, this is the best idea yet. There is nothing in the circuit that needs exactly 5 V, except a comparator that I want to replace anyway. So this will save parts and be the most efficient.

....

Or will it? Will running it off higher battery voltage actually be less efficient or will the batteries last longer this way?

your devices. But you will have saved the power used by the regulator.

But will the battery last longer? I guess since the chips are optimal at +/-10 V I should just like it the way it is.

loss. If it is plus and minus 30V a switcher is a good answer. Somewhere between the two, the efficiency of the two methods are equal.

or a switcher of some type.

Input is likely two 9V or so.

Run it off a higher voltage, and a greater current will be drawn by your devices. But you will have saved the power used by the regulator.

In article , Brian wrote: [... switcher circuit ..]

Yes: If his input is already plus and minus 5V a switcher must be a net loss. If it is plus and minus 30V a switcher is a good answer. Somewhere between the two, the efficiency of the two methods are equal.

an No:

He needs plus and minus 5V this means two batteries, a false ground, or a switcher of some type.

Then I would lose the regulator and run straight from the batteries.

In a battery circuit ground usually is false, no?

At low currents, straight off the battery makes sense. At higher current, a switcher would greatly increase the battery life.

Yes effectively

I assume yoy mean boosting the voltage.

No, with a 9V battery making 5V with a bucker will extend the battery life quite a bit when the current requirments are enough that the overhead to run the chip is not important in the total system power.

Unfortunately, there are no large P-MODFETs that can be certain to be biased to a low on resistance by 5V. Parts like Supertex's TP0604 would be a good choice for the pass element of a bucker at modest currents.

You can also "abuse" a booster swither chip to make a minus side bucker. Since the OP is making +/-5V from a pair of 9V batteries, you could do this twice and wire the resulting 5V circuits in series. The 9V batteries would not have any common connection.

Would the added efficiency come from the devices drawing less current at 5V than they do at 9V?

No, is because:

P= I * V

P9 = 0.1A * 9V = 0.9W

P5 = 0.1A * 5V = 0.5W

So if the switcher is more than 56% efficient, the switcher starts saving you power. If the switcher is 90% the product should run 62% longer with the switcher.