Hopefully simple question about DC power supplies

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:
    formatting link

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.

Reply to
davidburgess99
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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

Reply to
Tom Biasi

(a) Get a USB cable, snip off the end that goes to the remote device, and find the two wires that have ground and +5V. You should be able to find the pin definitions from the web, and use an ohmmeter to figure out which pins go to which wires. There you are.

(b) Look around on the web for voltage regulator circuits. Pick one, build it, and connect it up to +12V. There you are.

I'd do (a), because the +5V is already taken care of, you just need to get your grubby paws on it. It's about an hour of time if your tools are in order and you may well be able to find the necessary donor cable at a grocery store, if not lying on the floor in the vicinity of your computer.

Doing (b) would take a good part of a day assuming you have the parts and the knowledge, its more work to make it look good, and the parts (voltage regulators and what not) require a drive to the nearest UK equivalent of a Radio Shack, if not mail order.

HTH.

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Tim Wescott
Wescott Design Services
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Reply to
Tim Wescott

2

Hi Tom,

Thanks for your quick response.

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

Reply to
davidburgess99

Hi Tom,

Thanks for your quick response.

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

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Reply to
Lord Garth

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

Reply to
ehsjr

.

Thanks Tim,

I am inclined to follow your advice and go for option (a). Of course, there are still the questions (from other posters) about power consumption, etc, which I'll try and provide some info on.

David

Reply to
davidburgess99

I am inclined to follow your advice and go for option (a). Of course, 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

Reply to
Phil Allison

Thanks for your quick response.

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

.
Reply to
Tom Biasi

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!

Reply to
davidburgess99

Good point. Sometimes they're kind enough to give you a visible smoke signal to let you know you screwed up.

The cable takes so little time to make up it may still be worthwhile to try (barring the smoke signal, of course).

--
Tim Wescott
Wescott Design Services
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Reply to
Tim Wescott

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.
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(I just typed 12V-5V into the e-bay search box and hit "go")
Reply to
Jasen Betts

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 drawing about 929 mA to provide 650 mA to the load. Add to that 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 detailed battery specs and more info about which device 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

Reply to
ehsjr

To produce 650 mA at 5 volts from a 12 volt supply using a 70% efficient regulator does not require 929 mA. 650 mA times

5 volts is 3250 mW. A 70% efficient regulator would need 4643 mW of input power to produce the 3250 mW of output power. Only 387 mA is needed at 12 volts to produce 4643 mW.
Reply to
Dan Coby

The common 3-terminal regulators are much more efficient (for one definition of "efficient") than you indicate. The LM7805 will only take about 8 mA for its own use. If it is supplying 650 mA to the load, it will only draw 658 mA from the 12 volt source, according to National's datasheet. It will, however, waste 7/12ths of the total power as heat, and only deliver 5/12ths to the load.

--
Peter Bennett, VE7CEI  
peterbb4 (at) interchange.ubc.ca  
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Reply to
Peter Bennett

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

Reply to
ehsjr

The discussion was actually about producing 650 mA at 5 volts from a

12 volt battery. Yes, that is a 'DC-DC converter'.

I used the term 'regulator' to keep things consistent with your previous discussion. A better choice of terms might be a switched-mode power supply. (See:

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)

Since you disagree with my choice of terminology, would you please provide a description or an example circuit of that is what you meant when you said: "For example, say the 5V regulator is ~70% efficient, and it is powering the 650 mA device. That means it is actually drawing about 929 mA to provide

650 mA to the load."

Where I come from, a circuit that is drawing 929 mA from a 12 volt supply (i.e. 11148 mW) and produces 650 mA at at 5 volts (i.e. 3250 mW) is only

29% efficient.

A final note: Even a simple linear voltage regulator like a 7805 (see:

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) would only need a little more than 650 mA from the 12 volt input to provide the desired 650 mA output. In doing so, it would be about 40% efficient which is better than your circuit. (AND YES, I USED THE TERM REGULATOR AGAIN.)

Reply to
Dan Coby

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 your idea of a DC-DC converter, providing the load 5 v 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.

Ok.

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

Reply to
ehsjr

I do not want to quibble about the definitions of the terms 'regulator' and 'DC to DC conversion'. The issue is not really relevant to the question from the original poster. I responded in the way that I did because the use of these terms seemed to be an attempt to disguise the problem in the efficiency calculation. Since your last response indicates that is not the case, I see no more reason to thrash this subject.

I suggest that you get back to the original issue upon which you were working: the amount of current that would needed to be drawn from the battery to provide the desired outputs and the resulting effect of that current upon the run time of the 12 volt, 4700 mAh battery. Except for the minor error in the efficiency calculation, you were providing a nice discussion of the issue.

Dan

Reply to
Dan Coby

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