# Two questions on voltage regulators

Discussion in 'Electronic Basics' started by Jeff Dege, Jan 12, 2006.

1. ### Jeff DegeGuest

If have a circuit that needs 5V, and I have a battery that generates 9V, I
can wire a couple of resisters into a voltage divider, and have a 5V
supply. But the big "but" is that you're throwing away nearly half the
energy in the battery, dissipated as heat in the resisters.

Do voltage regulators do the same, internally? Do they limit the current
supplied externally by throwing away the excess? Or are they more
friendly to battery life?

Second - batteries come with two ratings - Volts and Amp-Hours. (Or
milliamp-hours in the usual sizes.)

Strictly by dimensional analysis, the energy stored in the battery is the
product of the two - Watts = Volts * Amps, so Volts * Amp-Hours =
Watt-Hours.

Suppose you have a 12V battery with a 60 mAH rating. If you plug it into
a DC circuit with 600 Ohms of resistance, you'll get a 20 mA draw. And
you can expect about 3 hours of battery life.

But suppose I have a circuit that has wants 3V and a 20mA draw, and I use
a voltage regulator to drop the voltage. It's still a 20mA draw, so
should I expect 3 hours of battery life? Or could I expect 12?

2. ### Dan HollandsGuest

Voltage regulators waste energy just like a resistor

AmpHrs is based on current coming from the battery. If 20mA comes out of the
battery it doesn't matter how much of reaches the load.

The only exception is if the regulator is dc to dc convertor - a small
switching power supply - This has the possibility of providing a lower
voltage with greater efficiency however the gain may be negligable at low
current levels.

--
Dan Hollands
1120 S Creek Dr
Webster NY 14580
585-872-2606

www.QuickScoreRace.com

3. ### Rich GriseGuest

There's still 20 mA coming out of the 12V battery, so, yes, you get
3 hours battery life - 3/4 of the energy is thrown away as heat.

That's why they invented the "switching regulator" - it turns the
supply on and off rapidly, storing the energy between pulses to feed
the load smoothly - that _does_ increase the efficiency, but you still
won't get 100%, but 80% to 95% are not uncommon.

Good Luck!
Rich

4. ### Jonathan KirwanGuest

You can, if the current through the two resistors is very much more
than the current needed by what is attached to the divider point.
Otherwise, not so good. Another often not so good way would be if
your 5V device current requirements are constant and then you can just
use a single dropping resistor.
You'd be throwing away a lot more than nearly half, using a divider.
Let's say your 5V device needs 10mA at 5V and that this 10mA is _very_
stable and doesn't fluctuate. If you simply use a dropping resistor,
you could get away with "nearly half" by using a (9v-5v)/10mA or 400
ohm resistor. But with a divider, and you might use any quiescent
current through it you want and calculate from there, you might pick a
divider current of 20mA, knowing that half of it will go through your
device. Somthing like this:

9V
|
|
\
20mA / R1
\ 4V/20mA
/ =200
|
| 5V
+----------,
| |
| |
\ \
10mA / R2 / 10mA
\ 5V/10mA \ Device
/ =500 /
| |
| |
gnd gnd

In this case, the power is 9V*20mA or 180mW. Your device needs
5V*10mA or 50mW. As you can see, 130mW is wasted. Way more than
half. It's only in the limit-case where R2 is infinity, that less
than 1/2 of the power is wasted.
They effectively have a dynamic R1 that changes its value according to
the time-dependent device current demands. Roughly speaking, R2 is
nearly infinite in linear regulator cases (almost.) So they do waste
power and they would, in the case you cite, waste almost half.
Not the excess current, as their is no excess current. But they throw
away the unneeded voltage. And to do that, they have to dissipate
power.
Nope. Not unless you look at switchers, which can put some of the
energy into magnetic or electric fields for repeated short times
instead of tossing it away as heat.
Energy is the watt-hours thing. Yes.
If that's close to the draw they used in coming up with the 60mAh
rating. Batteries do not have the same Ah-rating at all current draws
you might propose and manufacturers pick the better figure (the peak
much different, expect that rating to be optimistic.
With a linear regulator, yes. With a switcher, you'd probably get
better life than with the linear regulator. As always, mileage
varies.

Jon

5. ### Noway2Guest

You have some good advice here. When you search for switching
regulators, specifically look for DC-DC converters. A quick searrch on
google should take you to a number of companies that make devices that
will accomplish your goal for you. Based on your input, output, and
current requirements, I would be surprised if you can't find a device
in the < \$10 (US) range.

6. ### Guest

That's a class of devices I'd never heard of before, and they provide
some interesting options.

Consider the MAX619CPA - puts out a regulated 4.8-5.2V, from a 2.0-3.6V
input. Supplies 20mA, with an input of 2.0V, or 50mA with an input
And with a shutdown pin.

Quite a neat little device - lets you run TTL logic off of a pair of
primary cells, instead of those clunking 9-Volts. The next time I try
to build something in a palm-sized box, I'll very much keep them in
mind.

Thanks.

7. ### Roger DewhurstGuest

Try dropping the voltage across a string of diodes or LEDs.

I knew of someone who runs a 7 volt electric drill or similar tool off a 12v
car battery by dropping the voltage through a string of diodes.

R

8. ### Roger DewhurstGuest

What about pulse width modulation? You will waste no battery power that way
if the device will tolerate a square wave current supply.

R

9. ### Jasen BettsGuest

eg 7805, LM317: some of the same, they don't pass a significant waste current to
ground, but they do waste the energy in the extra 4 volts.
You'll get three hours with a series regulator like the LM317
a switching regulator will get you more than three (maybe 9 or 10) but they
cost a bit more.

Bye.
Jasen

10. ### Walter HarleyGuest

The advantage is that this drops a relatively more precise voltage (unlike
resistors, it drops about the same voltage regardless of current draw).

But it does still waste the excess as heat. The only way to lower a DC
voltage without wasting the excess is a switcher of some sort.

11. ### Jeff DegeGuest

Batteries don't provide precise voltages. And while electric motors are
fairly tolerant of imprecise voltages, digital circuitry isn't.

12. ### Byron A JeffGuest

You will have a 5V supply if and only if you draw current at some specific and
constant value. If the amount of current draw varies, then the voltage will
vary also. That's one of the major reasons why resistor voltage dividers
are not used as voltage regulators.
Linear voltage regulators work like that. They present a varaible resistance
and dissapate the excess energy as heat.
Not current.
Not linear regulators.
Be aware that a battery does have an internal resistance. This resistance
has an effect on the amount of power one can extract from the battery.
No necessarily. the AH rating is at a specific current draw. Many battery
manufacturers have a 20 hour draw. So you can expect to get 60 mAH if you
drew 3 ma for 20 hours.

However, when you change the current draw, the amount of energy you can
extract is vary based in the internal resistance of the battery. So if you
draw more power, more energy is dissapated as heat, and less is delivered to
the load. So in your example the battery would be exhausted well before the
theoretical 3 hours if you drew 20 mA from it.
But at a different voltage, so the amount of energy is different.
Nope. See above.
Definitely not!

If you are using a linear regulator then the other 9V is being dissapated as
heat. You can only draw 12V from the battery. So if your circuit is using 3V
@ 20 mA then 9V at 20 mA is being converted to heat.

OUCH!

And again because the 20 mA draw is nearly 7 times more than the nominal 3 mA
draw for the AH rating, you can expect your battery to peter out well before
3 hours.

So the real question is how to improve your performance. The best way is to
use a switching regulator, which delivers upwards of 95% of the energy to the
load instead of burning it up. The current draw will then change as 3V @ 20 mA
is the same as 12V @ 5 mA (Note that a switching regulator does not burn excess
energy as heat). So due to two factors (less heat dissapated and lower current
draw against the battery) your circuit will run longer. But it still won't be
3 hours.

BAJ