Suppose you have a 12V supply and need to power a 5V device. If that
device is a simple resistive load like say an incandescent light bulb,
you can simply use the right size resistor to drop the voltage. If the
device is more complex and does not draw a constant current, the
resistor obviously won't work. Something like a 7805 would be the
easiest solution. In both cases you are wasting a lot of power. If the
first case over half the power is just generating heat in the resistor.
I assume the same, or close to the same, amount of power is lost in
the 7805 regulator.

If you needed to light two 5V light bulbs then you could hook them in
series and waste a lot less power by using a smaller resistor.

Now for my question, can you do something similar by using 2 7805
regulators hooked up in series to power 2 separate loads? Alternatively
and more what I want, can you also have the outputs hooked up in
parallel to power one 5V device?




--
Chris W
KE5GIX

"Protect your digital freedom and privacy, eliminate DRM,
learn more at http://www.defectivebydesign.org/what_is_drm"

Ham Radio Repeater Database.
http://hrrdb.com

On Thu, 10 Jan 2008 14:10:17 -0600, Chris W <1qazse4@cox.net> wrote:

>Suppose you have a 12V supply and need to power a 5V device. If that
>device is a simple resistive load like say an incandescent light bulb,
>you can simply use the right size resistor to drop the voltage. If the
>device is more complex and does not draw a constant current, the
>resistor obviously won't work. Something like a 7805 would be the
>easiest solution. In both cases you are wasting a lot of power. If the
>first case over half the power is just generating heat in the resistor.
> I assume the same, or close to the same, amount of power is lost in
>the 7805 regulator.
>
>If you needed to light two 5V light bulbs then you could hook them in
>series and waste a lot less power by using a smaller resistor.
>
>Now for my question, can you do something similar by using 2 7805
>regulators hooked up in series to power 2 separate loads? Alternatively
>and more what I want, can you also have the outputs hooked up in
>parallel to power one 5V device?

Sounds like you already have a grasp of the principles. In theory
with two regulators in series (and the ground returns isolated from
ground) each regulator would do its best to keep the voltage constant
to its load (barring oscillations or other anomalies).

But if one regulator/load is drawing less current than the other -
that regulator would have lots of input voltage to play with and the
other might not. So it would only work with very close unvarying
loads and enough voltage headroom to keep two regulators happy -
nothing that suggest higher efficiency.

Best bet is to use switching regulators to supply the loads from a
single source. National Semiconductor has a line of "simple
switchers" (now in their 3d or 4th generation) that are very easy to
apply and very efficient - and still available in packages you can
hand solder to without a magnifying glass.
--

Chris W wrote:
> Suppose you have a 12V supply and need to power a 5V device. If that
> device is a simple resistive load like say an incandescent light bulb,
> you can simply use the right size resistor to drop the voltage. If the
> device is more complex and does not draw a constant current, the
> resistor obviously won't work. Something like a 7805 would be the
> easiest solution. In both cases you are wasting a lot of power. If the
> first case over half the power is just generating heat in the resistor.
> I assume the same, or close to the same, amount of power is lost in the
> 7805 regulator.
>
> If you needed to light two 5V light bulbs then you could hook them in
> series and waste a lot less power by using a smaller resistor.
>
> Now for my question, can you do something similar by using 2 7805
> regulators hooked up in series to power 2 separate loads? Alternatively
> and more what I want, can you also have the outputs hooked up in
> parallel to power one 5V device?
>
The use of a "BUCK" regulator works well in these cases to save
power.
Look at the below example.

http://www.national.com/pf/LM/LM5005.html


http://webpages.charter.net/jamie_5"

On Jan 10, 3:10 pm, Chris W <1qaz...@cox.net> wrote:
> Suppose you have a 12V supply and need to power a 5V device. If that
> device is a simple resistive load like say an incandescent light bulb,
> you can simply use the right size resistor to drop the voltage. If the
> device is more complex and does not draw a constant current, the
> resistor obviously won't work. Something like a 7805 would be the
> easiest solution. In both cases you are wasting a lot of power. If the
> first case over half the power is just generating heat in the resistor.
> I assume the same, or close to the same, amount of power is lost in
> the 7805 regulator.
>
> If you needed to light two 5V light bulbs then you could hook them in
> series and waste a lot less power by using a smaller resistor.
>
> Now for my question, can you do something similar by using 2 7805
> regulators hooked up in series to power 2 separate loads? Alternatively
> and more what I want, can you also have the outputs hooked up in
> parallel to power one 5V device?
>
> --
> Chris W
> KE5GIX
>
> "Protect your digital freedom and privacy, eliminate DRM,
> learn more athttp://www.defectivebydesign.org/what_is_drm"
>
> Ham Radio Repeater Database.http://hrrdb.com

As others have said, use of a buck switching regulator will save on
the power waste.

Rather than connect two 5V regulators in series, it is possible (and
simple) to make a variable supply from any regulator. See the
datasheet for the LM317 regulator (essentially a 1.2V regulator) for
how to do it.

And rather than connect two regulators in parallel, a greater current
capacity can be realized by using a bypass transistor and a single
regulator. You might find a schematic in a regulator datasheet, or
perhaps through Google.

Regards,

Mark

On Jan 10, 12:10=A0pm, Chris W <1qaz...@cox.net> wrote:
> Suppose you have a 12V supply and need to power a 5V device. =A0If that
> device is a simple resistive load like say an incandescent light bulb,
> you can simply use the right size resistor to drop the voltage. =A0If the
> device is more complex and does not draw a constant current, the
> resistor obviously won't work. =A0Something like a 7805 would be the
> easiest solution. =A0In both cases you are wasting a lot of power. ...>
> If you needed to light two 5V light bulbs then you could hook them in
> series and waste a lot less power by using a smaller resistor.
>
> Now for my question, can you do something similar by using 2 7805
> regulators hooked up in series to power 2 separate loads?

There wouldn't be any point in using two regulators for loads like
light bulbs; a single regulator can power both, either a current
regulator
with two lamps in series, or a 10V regulator with two lamps in series,
or a 5V regulator with two lamps in parallel.

To get around the heat-generation problem, one can ALSO use a high
frequency switch (like, 1 kHz). The result, if one controls the
switch
duty cycle, is similar to voltage regulation. Because the heat in
the lamp filament is proportional to V-squared, a 5V lamp
can be driven by a 12V source with duty cycle 25/144.
Or a 10V pair of lamps can be driven by 12V chopped at duty cycle
100/144.

Chris W wrote:
> Suppose you have a 12V supply and need to power a 5V device. If that
> device is a simple resistive load like say an incandescent light bulb,
> you can simply use the right size resistor to drop the voltage. If the
> device is more complex and does not draw a constant current, the
> resistor obviously won't work. Something like a 7805 would be the
> easiest solution. In both cases you are wasting a lot of power. If the
> first case over half the power is just generating heat in the resistor.
> I assume the same, or close to the same, amount of power is lost in the
> 7805 regulator.
>
> If you needed to light two 5V light bulbs then you could hook them in
> series and waste a lot less power by using a smaller resistor.
>
> Now for my question, can you do something similar by using 2 7805
> regulators hooked up in series to power 2 separate loads? Alternatively
> and more what I want, can you also have the outputs hooked up in
> parallel to power one 5V device?
>
>
>
>
You got some good answers, but I'm not sure if
any explained why the series idea won't work.
You cannot hook up 2 7805 regulators in series:

Input---7805---7805---output

The first one will produce 5V output. That output
is too low to allow the second 7805 to work properly.
The 7805 needs a minimum of about 8 volts input.

Ed

On Jan 11, 2:37 am, ehsjr <eh...@bellatlantic.net> wrote:
> Chris W wrote:
> > Suppose you have a 12V supply and need to power a 5V device. If that
> > device is a simple resistive load like say an incandescent light bulb,
> > you can simply use the right size resistor to drop the voltage. If the
> > device is more complex and does not draw a constant current, the
> > resistor obviously won't work. Something like a 7805 would be the
> > easiest solution. In both cases you are wasting a lot of power. If the
> > first case over half the power is just generating heat in the resistor.
> > I assume the same, or close to the same, amount of power is lost in the
> > 7805 regulator.
>
> > If you needed to light two 5V light bulbs then you could hook them in
> > series and waste a lot less power by using a smaller resistor.
>
> > Now for my question, can you do something similar by using 2 7805
> > regulators hooked up in series to power 2 separate loads? Alternatively
> > and more what I want, can you also have the outputs hooked up in
> > parallel to power one 5V device?
>
> You got some good answers, but I'm not sure if
> any explained why the series idea won't work.
> You cannot hook up 2 7805 regulators in series:
>
> Input---7805---7805---output
>
> The first one will produce 5V output. That output
> is too low to allow the second 7805 to work properly.
> The 7805 needs a minimum of about 8 volts input.
>
> Ed

Perhaps he meant something more along these lines:

------
12V --o--------| 7805 |---- 10V out
| | |
| ------
| |
| |
| ------ |
`--| 7805 |--'
| |
------
|
|
-----
---
-


-- Mark

On Jan 11, 8:30 am, redbelly <redbell...@yahoo.com> wrote:
> On Jan 11, 2:37 am, ehsjr <eh...@bellatlantic.net> wrote:
>
>
>
> > Chris W wrote:
> > > Suppose you have a 12V supply and need to power a 5V device. If that
> > > device is a simple resistive load like say an incandescent light bulb,
> > > you can simply use the right size resistor to drop the voltage. If the
> > > device is more complex and does not draw a constant current, the
> > > resistor obviously won't work. Something like a 7805 would be the
> > > easiest solution. In both cases you are wasting a lot of power. If the
> > > first case over half the power is just generating heat in the resistor.
> > > I assume the same, or close to the same, amount of power is lost in the
> > > 7805 regulator.
>
> > > If you needed to light two 5V light bulbs then you could hook them in
> > > series and waste a lot less power by using a smaller resistor.
>
> > > Now for my question, can you do something similar by using 2 7805
> > > regulators hooked up in series to power 2 separate loads? Alternatively
> > > and more what I want, can you also have the outputs hooked up in
> > > parallel to power one 5V device?
>
> > You got some good answers, but I'm not sure if
> > any explained why the series idea won't work.
> > You cannot hook up 2 7805 regulators in series:
>
> > Input---7805---7805---output
>
> > The first one will produce 5V output. That output
> > is too low to allow the second 7805 to work properly.
> > The 7805 needs a minimum of about 8 volts input.
>
> > Ed
>
> Perhaps he meant something more along these lines:
>
> ------
> 12V --o--------| 7805 |---- 10V out
> | | |
> | ------
> | |
> | |
> | ------ |
> `--| 7805 |--'
> | |
> ------
> |
> |
> -----
> ---
> -
>
> -- Mark

I think this might work too, and they're in series:

------
12V --| 7805 |---- 10V out
| |
------
|
|
| ------
`--| 7805 |--o
| |
------
|
|
-----
---
-


BUT here is what is actually suggested in datasheets. Choose R1 and
R2 so that:

10V = 5V*( 1 + R2/R1) + Icom*R2

------
12V --| 7805 |--o-- 10V out
| | |
------ \
|com /
| \ R1
| /
| \
| |
`------o
|
\
/
\ R2
/
\
|
|
-----
---
-


-- Mark

"Chris W" <1qazse4@cox.net> wrote in message
news:OYuhj.8553$R55.6042@newsfe13.lga...
> Suppose you have a 12V supply and need to power a 5V device. If that
> device is a simple resistive load like say an incandescent light bulb, you
> can simply use the right size resistor to drop the voltage. If the device
> is more complex and does not draw a constant current, the resistor
> obviously won't work. Something like a 7805 would be the easiest
> solution. In both cases you are wasting a lot of power. If the first
> case over half the power is just generating heat in the resistor. I assume
> the same, or close to the same, amount of power is lost in the 7805
> regulator.
>
> If you needed to light two 5V light bulbs then you could hook them in
> series and waste a lot less power by using a smaller resistor.
>
> Now for my question, can you do something similar by using 2 7805
> regulators hooked up in series to power 2 separate loads? Alternatively
> and more what I want, can you also have the outputs hooked up in parallel
> to power one 5V device?
>

You can do the first thing, connecting two of them (although ehsjr's point
is a good one, and the 'drop out' of the 7805 regulators will cause a
problem with 12V).

However, the second one, connecting them in parallel, does not work. The
problem is that you need a ground reference, and the reference will be
different for the two regulators. There is no good way to hook them up in
parallel in the way you describe.

On the other hand, the industry has decided that the best way to do this
kind of thing is with a 'switched mode power supply', or SMPS. That circuit
uses devices that store energy (inductors) to offer power output at a lower
voltage using a larger voltage. The power efficiency can be in the 90% range
with good switchers. There are cheap chips that do most of the heavy lifting
for you, and some companies specialize in building those chips, like
http://www.linear.com. A linear regulator from 12V to 5V can only get around 42%
efficiency.

Regards,
Bob Monsen

On Jan 10, 10:14=A0pm, whit3rd <whit...@gmail.com> wrote:
> On Jan 10, 12:10=A0pm, Chris W <1qaz...@cox.net> wrote:
>
> > Suppose you have a 12V supply and need to power a 5V device. =A0If that
> > device is a simple resistive load like say an incandescent light bulb,
> > you can simply use the right size resistor to drop the voltage. =A0If th=
e
> > device is more complex and does not draw a constant current, the
> > resistor obviously won't work. =A0Something like a 7805 would be the
> > easiest solution. =A0In both cases you are wasting a lot of power. =A0..=
..>
> > If you needed to light two 5V light bulbs then you could hook them in
> > series and waste a lot less power by using a smaller resistor.
>
> > Now for my question, can you do something similar by using 2 7805
> > regulators hooked up in series to power 2 separate loads?
>
> There wouldn't be any point in using two regulators for loads like
> light bulbs; a single regulator can power both, either a current
> regulator
> with two lamps in series, or a 10V regulator with two lamps in series,
> or a 5V regulator with two lamps in parallel.
>
> To get around the heat-generation problem, one can ALSO use a high
> frequency switch (like, 1 kHz). =A0The result, if one controls the
> switch
> duty cycle, is similar to voltage regulation. =A0Because the heat in
> the lamp filament is proportional to V-squared, a 5V lamp
> can be driven by a 12V source with duty cycle 25/144.
> Or a 10V pair of lamps can be driven by 12V chopped at duty cycle
> 100/144.

Right. By switching an incandescent light or a heating element faster
than its thermal time constant, you get the efficiency of a switching
power supply. You don't need an inductor.