Re: Two questions on voltage regulators
- From: Jonathan Kirwan <jkirwan@xxxxxxxxxxxxxx>
- Date: Thu, 12 Jan 2006 19:25:23 GMT
On Thu, 12 Jan 2006 10:14:56 -0600, Jeff Dege <jdege@xxxxxxxxxxxxxx>
wrote:
>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.
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.
>But the big "but" is that you're throwing away nearly half the
>energy in the battery, dissipated as heat in the resisters.
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.
>Do voltage regulators do the same, internally?
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.
>Do they limit the current
>supplied externally by throwing away the excess?
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.
>Or are they more friendly to battery life?
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.
>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.
Energy is the watt-hours thing. Yes.
>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.
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
or near-peak) of the curve over current loading. So if your draw is
much different, expect that rating to be optimistic.
>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?
With a linear regulator, yes. With a switcher, you'd probably get
better life than with the linear regulator. As always, mileage
varies.
Jon
.
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- Two questions on voltage regulators
- From: Jeff Dege
- Two questions on voltage regulators
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