# Hopefully simple question about DC power supplies

Discussion in 'Electronic Basics' started by [email protected], Nov 11, 2008.

1. ### Guest

Hi,

I'm hoping that this is a simple question to answer. Although I'm not
a technical ignoramus, I know very little about power or cabling, so I
would like some advice with what I'm trying to do!

What I need is:

* PORTABLE (i.e. non-mains) 5V DC power into 'Device A'
and
* PORTABLE (i.e. non-mains) 12V DC power into 'Device B'

What I have at my disposal (so far) is:
* A mains adapter for 'Device A' that seems to output 5V at up-to 2A
via a small-ish DC power jack (3mm outside diameter, I think)
* A mains adapter (arriving in the post tomorrow) for 'Device B' that
should output 12V at up-to 500mA via a suitably-sized DC power jack
* A 'rechargeable laptop battery pack', that can output 12V DC via a
number of different jacks, and 5V DC via a USB cable. (details here:
http://www.maplin.co.uk/Module.aspx?ModuleNo=48490)

What I originally wanted to do was to power both devices from the
'rechargeable laptop battery pack'. But the only 5V output from that
is via a USB cable, which will not fit the device that needs a 5V
input.

I cannot use either of the mains adapters, so I need to get DC power
from the 'rechargeable laptop battery pack' (or some equivalent).

So what should I do? I have a few ideas, but I'm not sure if they are
any good:

(a) Could I somehow construct a USB-to-DC power jack for the 5V power?
I'm not sure such a thing exists to buy, but how difficult would it be
to construct?

(b) Could I split the 12V output to two destinations (1) 'Device B'
directly; and (2) 'Device A' via some way of converting it to 5V? How
would I do this?

I'm sure what I'm trying to do shouldn't be too difficult. But I'm
just not sure how to do it, and I don't want to damage my devices or
start a fire or electrocute myself. So any advice would be
appreciated!

Thanks,

David

P.S. In terms of getting hold of kit to do this, I'm UK-based, so
ordering from UK companies would work best for me.

2. ### Tom BiasiGuest

Lets cut out some of the words
You have a 12 volt battery that you need to power two devices, one with 12
volts the other with 5 volts.
Obviously both devices will stay with the battery and you need a means to
charge the battery.
It would help if you weren't so cryptic and named the devices and your
intent.

Tom

3. ### Guest

Hi Tom,

The problem here is that I can be criticized for both providing too
much information (hence your need to "cut out some of the words") and
for providing insufficient information (i.e. being "cryptic")! I had
tried to find the right balance with describing my problem, but
obviously I didn't get it quite right. Sorry.

I'll try to distill it down to what I think the core problem/question
is.

I have a need for a 5V DC feed via a small DC jack, and a 12V DC feed
via a slightly larger DC jack. But what I have (at the moment) is a 5V
DC feed via a USB cable, an a 12V DC feed via a (hopefully suitable)
DC jack.

The problem is how to get the 5V DC feed via a jack.

Do I:
(a) Try to convert the USB connector to a DC jack? If so, how?
(b) Try to split the 12V feed into two, and somehow convert half of it
to 5V? If so, how?
(c) Do something else?

As background info, I am trying to setup surveillance of my garage,
where stuff seems to be regularly getting pilfered and generally
messed around with. So, 'Device A' is a small handheld DVR (with 5V
'charging' input), 'Device B' is a small CCD camera, which takes 12V.

Thanks,

David

4. ### Lord GarthGuest

Hi Tom,

The problem here is that I can be criticized for both providing too
much information (hence your need to "cut out some of the words") and
for providing insufficient information (i.e. being "cryptic")! I had
tried to find the right balance with describing my problem, but
obviously I didn't get it quite right. Sorry.

I'll try to distill it down to what I think the core problem/question
is.

I have a need for a 5V DC feed via a small DC jack, and a 12V DC feed
via a slightly larger DC jack. But what I have (at the moment) is a 5V
DC feed via a USB cable, an a 12V DC feed via a (hopefully suitable)
DC jack.

The problem is how to get the 5V DC feed via a jack.

Do I:
(a) Try to convert the USB connector to a DC jack? If so, how?
(b) Try to split the 12V feed into two, and somehow convert half of it
to 5V? If so, how?
(c) Do something else?

As background info, I am trying to setup surveillance of my garage,
where stuff seems to be regularly getting pilfered and generally
messed around with. So, 'Device A' is a small handheld DVR (with 5V
'charging' input), 'Device B' is a small CCD camera, which takes 12V.

Thanks,

David

Hi David,

Many CCD cameras can operate from an unregulated supply. My experience
is with the small cameras used to observe store fronts. Do check your
camera
for its power requirements as an AC input to a DC jack would prove
disastrous.

The cameras we use do not specify a polarity to the connection point on the
rear
which means the regulator circuitry is inside the camera itself. A typical
installation
is with a cable we call 'siamese' because it consists of a coax for the
video and a
two conductor 16 gauge stranded cables connected to the insulation of the
coax.

Most of the power supplies are full wave bridge with a few filter
capacitors. Some
of the cameras have been connected in reverse with respect to the others
(say the
red wire was placed on the left rather than the right). The camera operated
perfectly
this way.

You do not mention the current requirements of your equipment but that is
important
as well as the voltage (and polarity if your camera requires DC). A
voltmeter will
show the voltage drooping if your current demand is beyond the capability of
the
supply.

It is possible to supply power over the same coax as the video signal and
there are
adapters made to do this...a simple capacitor will block the DC but pass the
video.

You can also opt for an IP camera that supports PoE or Power Over Ethernet.
This
will allow the camera to be powered over the same CAT5 as the video plus you
can
view the image on any browser and most of these cameras come with a software
DVR that supports motion detection. Further, they can email you when motion
is
detected or in some cases, text your mobile phone. A Trendnet TV-201 is
such a
camera. A good price is from www.provantage.com

5. ### ehsjrGuest

We don't know how much current either your 5 volt or your
12 volt device draws, nor the capacity of your laptop
battery, nor how much current it can provide via the
usb cable.

That said, the general hookup is this:

Battery
-----
| +|---+---------------------> +12V to 12V device
| | |
| 12v | | -------------
| | +---| 5V regulator|---> +5V to 5V device
| | -------------
| -|-------------------------> - to both devices
-----

The 5V regulator could be a "linear regulator" which
wastes power but is easy and cheap to get parts for
and build, or a "DC-DC converter" which is harder to
build/get parts for and more expensive. YMMV

Not being mains connected is going to defeat you, in any
event. The setup will work only as long as the battery
does not need to be recharged. You will need to charge the
thing on a regular basis, but without knowing how much
current will be drawn and what the battery capacity is,
we can't say how often that will be. If the adapter
for the 5V device (which you said puts out up to 2A) is
any indication, you probably won't get 1 day from the
setup before you need to recharge. That guess is as
exact as the spec you gave, so don't make any wagers
based on it!

Ed

6. ### Guest

Thanks Tim,

there are still the questions (from other posters) about power
consumption, etc, which I'll try and provide some info on.

David

7. ### Phil AllisonGuest

<>

there are still the questions (from other posters) about power
consumption, etc, which I'll try and provide some info on.

** If you use the same battery to power both devices - then you will wind
up with a common negative for both DC supplies. Each device will then work
but they may not tolerate being interconnected via the video leads without
malfunction or damage.

AC adaptors ALWAYS have floating outputs and many equipment makes RELY on
this being the case.

..... Phil

8. ### Tom BiasiGuest

Hi Tom,

The problem here is that I can be criticized for both providing too
much information (hence your need to "cut out some of the words") and
for providing insufficient information (i.e. being "cryptic")! I had
tried to find the right balance with describing my problem, but
obviously I didn't get it quite right. Sorry.

I'll try to distill it down to what I think the core problem/question
is.

Notice I had a after "cut out some of the words".
I was just trying to boil down you situation. Usually don't try to give a
scientific dissertation of what you want just say what you want to do.
Others have supplied info so I need not say more.
Best Regards,
Tom

..

9. ### Guest

I've tried to dig out some power figures...

* DVR operation current: 650mA. Although I've seen conflicting
information here.
* camera operation current: 90mA
* battery capacity: 4700mAh (80Wh energy, apparently)

Does this mean I'll get somewhere in the region of 6 hours out of it?
It's a long time since I studied all this stuff at school!

10. ### Jasen BettsGuest

very easy to build, take a USB cable and chop one end off and fit a
suitable plug. google for the pinout (the outermost contacts are the
power ones)

2A could be an issue. USB is only rated for 1/4 of that (500mA), check
what the documentation on the 'rechargeable laptop battery pack' says,
you may have to ask maplin to look it up for you.
you would use a regulator.

you could start with circuit on the LM7805 datasheet (which may need
a large heatsink and will waste more energy than it ptovides to the
5V device) or buy a prefabricated 12v to 5V converter. something like
this one perhaps.
(I just typed 12V-5V into the e-bay search box and hit "go")

11. ### ehsjrGuest

Well, no. 12 volts at 4700 mAh would be equal to 56.4Wh,
not 80 Wh. (To figure watts, multiply amps times volts)

Can't figure the current taken from the battery by the
650 mA or the 90 mA without knowing which runs on 12 volts,
and which runs on 5 volts, because the 5V regulator will
waste some power. There's nasty little things to consider,
like the discharge curve for the battery and the regulator
efficiency.

For example, say the 5V regulator is ~70% efficient, and it
is powering the 650 mA device. That means it is actually
the 90 mA used by the 12 volt load, and you're using 1019 mA
every hour. So in very general terms, it would last a little
over 4 hours. But it can get a lot worse. Battery capacity
ratings usually are for 20 hour discharge. If that is true of
your battery, then the 4700 mAh rating is based on a current
draw of about 235 mA - and you're drawing over 4 times that
in this example. Generally that means you'll get a lot less
than 4700 mAh.

Tha above is not to give you any real numbers - it is just
a general outline. To get better numbers, you need the
uses which voltage and current.

In any event, whatever the detailed specs, you'll need to
charge the battery at least 4 times daily based on the
specs you've posted so far. Since it is not to be mains
connected, you'll need at least 2 batteries - one to power
the equipment while the other is charging. And you'll
need 4 to 6 battery changes each day, minimum, for 24 hour
coverage. That is impractical. You need to get mains
connected. Even if you use equipment that draws _far_
less current, you still have a problem. Say you can get
equipment that draws a total of 100 mA. You'll need to
charge the battery every 2 days. Guaranteed you won't
keep that up for long. The only possible option I see,
other than mains connection, is equipment with very low
current draw, powered from a battery that is connected
to a solar charger. I have no idea if that is practical
for you. You might consider using a much bigger deep
cycle battery, but you would still need to charge it
periodically. It will be heavy and awkward to move and
take a long time to charge, so solar power seems like
a better option.

Ed

12. ### ehsjrGuest

The discussion was of a regulator, not a DC-DC converter.
Your description is of a DC-DC converter, which you
incorrectly call a regulator. They are not the same
thing.

Ed

13. ### ehsjrGuest

Would you call a 10.769 ohm (approximate) resistor in series
with the 12V battery and the load a DC-DC converter? Are
you calling a 7805 a DC-DC converter? Both seem to fit
at 650 mA.

Both a resistor and a linear regulator convert some of the
electrical energy to heat energy. That is not a DC to DC
conversion. The electrical energy that is not converted to
heat is passed to the load.
No, because it is incorrect. I totally ignored the
input - output voltage difference and looked only
at current. Stupid mistake.
Yes.

The 7805 _is_ a regulator, and that is the term that should be used.
Shouting is not needed.

Ed