5V regulator for batteries cheap efficient?

I know I can't have everything, but I'm going to ask for it anyway.

The circuit is 8 or so opamps, comparators, and logic circuits driven
by +/-5 V. It will run on batteries. Is there a comparatively cheap
(double the price? maybe triple) but more efficient regulator than the
7805/7905 for this application? i don't want the batteries throwing
all their energy into heatsinks and running out in 5 minutes. if
switching, it's got to have decent noisefreeness. noise above the
audio band is fine.

Any suggestions? the lm2940 looks promising, but you folks probably
know better...

 

I'd estimate about 65 mA total.

 

[Ken Smith]

In article <1108185958.959874.84180@l41g2000cwc.googlegroups.com>,
<u035m4i02@sneakemail.com> wrote:

I know I can't have everything, but I'm going to ask for it anyway.

The circuit is 8 or so opamps, comparators, and logic circuits driven
by +/-5 V. It will run on batteries. Is there a comparatively cheap
(double the price? maybe triple) but more efficient regulator than the
7805/7905 for this application? i don't want the batteries throwing
all their energy into heatsinks and running out in 5 minutes. if
switching, it's got to have decent noisefreeness. noise above the
audio band is fine.

Any suggestions? the lm2940 looks promising, but you folks probably
know better...

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


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--
kensmith@rahul.net forging knowledge

 

[Rich Grise]

On Fri, 11 Feb 2005 21:25:58 -0800, u035m4i02 wrote:

I know I can't have everything, but I'm going to ask for it anyway.

The circuit is 8 or so opamps, comparators, and logic circuits driven
by +/-5 V. It will run on batteries. Is there a comparatively cheap
(double the price? maybe triple) but more efficient regulator than the
7805/7905 for this application? i don't want the batteries throwing
all their energy into heatsinks and running out in 5 minutes. if
switching, it's got to have decent noisefreeness. noise above the
audio band is fine.

Any suggestions? the lm2940 looks promising, but you folks probably
know better...

Use stuff that's not so persnickety about its supply voltage, like CMOS.
I've never heard of an opamp that can run off +-5V that couldn't run off
+-6.

IOW, lose the regulators, and redesign the circuit to work off raw battery
power.

Good Luck!
Rich

 

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?

 

[richard mullens]

u035m4i02@sneakemail.com wrote:

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?

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.

 

[Ken Smith]

In article <B7KdnUEfS6h7xY3fRVn-rA@centurytel.net>,
Brian <brian@w3gate.com> wrote:
[... switcher circuit ..]

Wouldn't the added efficiency depend on what his input voltage is? If only
6 volts (4AA) a switcher would offer little, no?

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.

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kensmith@rahul.net forging knowledge

 

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.

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.

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.

He needs plus and minus 5V this means two batteries, a false ground,
or a

switcher of some type.

Input is likely two 9V or so.

 

[Ken Smith]

In article <n4-dnSxDCb3rW4zfRVn-qA@centurytel.net>,
Brian <brian@w3gate.com> wrote:

[.. +/- 9V batteries ..]

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

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

In a battery circuit ground usually is false, no?

Yes effectively


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[Ken Smith]

In article <TZydnZyl5qab6InfRVn-pg@centurytel.net>,
Brian <brian@w3gate.com> wrote:

"Ken Smith" <kensmith@green.rahul.net> wrote in message
news:cuvlvb$i5n$1@blue.rahul.net...
In article <8JmdnaefCoMeu47fRVn-sw@centurytel.net>,
Brian <brian@w3gate.com> wrote:

[.. +/- 9V batteries ..]

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

[...]
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


0.5/0.9 = 0.5555

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.

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