5V regulator for batteries cheap efficient?

[[email protected]]

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...

 

[Mike Harrison]

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...

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

 

[richard mullens]

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

Replaced by HT7150-1
3 microamp typical quiescent. 6 max.

Ricoh has/had some LDO regulators with quiescent current around 1 microamp eg Rx5RL - see
http://www.ricoh.com/LSI/product_power/info/selectionguide.pdf

 

[[email protected]]

I'd estimate about 65 mA total.

 

[Ken Smith]

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.

 

[Ken Smith]

I'd estimate about 65 mA total.

65 mA is enough to make a switcher worth the work. If you don't need too
tight of regulation, you can use one chip to make both voltages. This may
save you a bit of power. Something like this:



--!!---------------+--->!---+----- +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.

 

[Rich Grise]

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

 

[Brian]

I'd estimate about 65 mA total.

65 mA is enough to make a switcher worth the work. If you don't need too
tight of regulation, you can use one chip to make both voltages. This may
save you a bit of power. Something like this:



--!!---------------+--->!---+----- +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
6 volts (4AA) a switcher would offer little, no?

 

[[email protected]]

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]

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]

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.

 

[[email protected]]

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.

 

[Brian]

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

In a battery circuit ground usually is false, no?

 

[Ken Smith]

[.. +/- 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

 

[Brian]

[.. +/- 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

I assume yoy mean boosting the voltage.

 

[Ken Smith]

[.. +/- 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.

[...]
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.

 

[Brian]

[.. +/- 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.

[...]
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?

 

[Ken Smith]

[.. +/- 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.