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'[EE]: 12V and 5V Power Supply'
2004\03\22@062517 by Lucian

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Hello,

I'm looking for a schematic for a dual 12V @ 3A, 5V @ 3A power supply
with battery backup and charging circuit (eventually). I have searched
the web and have found nothing similar (only for sale, but I need to
build it myself).

Does anyone have [seen] such a schematic and can share it with me ? :)

Hoping for any help from you,

Lucian

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2004\03\22@065249 by Jinx

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> Does anyone have [seen] such a schematic

Do you have any objection to a normal step-down transformer,
size for example, or would you want to try a switch-mode ?

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2004\03\22@070741 by Russell McMahon

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> I'm looking for a schematic for a dual 12V @ 3A, 5V @ 3A power supply
> with battery backup and charging circuit (eventually). I have searched
> the web and have found nothing similar (only for sale, but I need to
> build it myself).

Google for data sheets and application notes for an LM350 regulator.
This can be adjusted to 5v and 12v at 3A (one for each circuit) .

Basic information and datasheets here

       http://www.national.com/pf/LM/LM350.html

       There are a number of application circuits in this datasheet that
directly
       address your requirement.


The LM317 is the 1 amp equivalent - there are more application notes around
for the 317 than the 350.

Datasheet         http://aes.sdsu.edu/documents/LM317-D.pdf



Then come back and ask more questions :-)


       Russell McMahon

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2004\03\22@073931 by Jinx

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> LM350 regulator.
> This can be adjusted to 5v and 12v at 3A (one for each circuit) .

And a jacked-up 7812 for the battery charging ? Presumably
13.8V (plus steering diode V) for a SLA before the 12V regulator.
Perhaps even a 6V SLA as well for the 5V rail

Need more info please Lucian

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2004\03\22@074139 by Lucian

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I thought of using 2 LM1084 regulators (up to 5A for a maximum of 3A
load). I would prefer a linear power supply, because I think it is
cheaper, but don't know how to connect the back-up battery, in order to
be able to charge it when powered from the wall outlet.
If it is connected after the 12V regulator, with a diode, how could I
obtain 5V when the system is running on battery ? The 5V regulator
should be located in the schematic after the 12V one and the battery
connection point. But in this case, the 12V regulator should be up to 6A
(3A + 3A), isn't it ?
I don't know if I made myself clear, because my english is not so good,
but I hope someone of you will understand what I tried to explain. I
think this is am usual problem to deal when using dual voltage back-up
power supplies.

Lucian

{Original Message removed}

2004\03\22@075806 by David Bearrow

picon face
You can convert an old PC power supply. It has + and - 12V, + and - 5V, and
+3.5V. Do a google search for ATX power supply pinout. I use an old Dell
power supply as my bench supply. You have to put a 5W 8ohm resister on one
of the 5V leads and tywrap it to the case on the inside with heat sink
grease so it will stay turned on, they cut off automatically with no load.
You can even use the POST lead to power an LED letting you know its turned on.

At 05:24 AM 3/22/04, you wrote:
{Quote hidden}

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2004\03\22@080220 by Lucian

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I forgot to mention that a pc supply is not a solution, because I need
something smaller and built by myself. Something with regulators, but
with battery back-up...

Lucian

{Original Message removed}

2004\03\22@095559 by Jinx
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part 1 1855 bytes content-type:text/plain; (decoded 7bit)

> Something with regulators with battery back-up...

Just some drowsy thoughts.............(it's 3am. Flip, not again !)

Lucian, this would work as a basic linear setup to get 12V and
5V. However, there is a limitation on the 12V side. Because of
the diode, you'll need to use a low dropout regulator. That's 1.1V
for the LM1084 at 3A. Even with a Schottky, the battery cannot
fall below about 13.4V, if it's floated at 13.8V (assuming a lead
battery, varies slightly with temperature). Which doesn't give you
much capacity without losing regulation

I have similar applications, which need only a few 10s of mA back-
up power, and use a SPDT (ie a toggle) relay. The relay energises
to close the normally-open contacts when mains power is available,
otherwise battery voltage flows through the normally-closed contacts.
At least that makes an extra 0.6V because there's no diode. I'm
wondering if the input to the charger can be isolated another way,
perhaps with a semiconductor or relay on the input

Because of the droput problems and currents that you need, I think
you may have to go to a higher V batteries, perhaps a 24V for 12V
and a 12V for 5V ?

Maybe someone can suggest a step-up switcher instead

A junkbox way to get 13.8V is with a jacked-up 7812, that is one
that has the reference pin lifted 1.8V above ground. Better would
be with a specialised charger/monitor so it's not over-charged or
discharged it too far.

The other thing to watch out for is the wattage dissipated by the 5V
regulator. It will be dropping around 9V. A more efficient way would
be to have an additional lower voltage tap on the transformer and
a 6V battery as well




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part 2 1638 bytes content-type:image/gif; (decode)

2004\03\22@100806 by Byron A Jeff

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On Mon, Mar 22, 2004 at 02:40:47PM +0200, Lucian wrote:
> I thought of using 2 LM1084 regulators (up to 5A for a maximum of 3A
> load). I would prefer a linear power supply, because I think it is
> cheaper,

Cheaper may not be the issue, efficiency may be the controlling factor.
Let's talk linear. Probably OK for the 12V supply, but problematic for the
5V one because presuming that you have a 12V battery, the 5V regulator will
have to burn (waste) upwards of 21W of power at the 3A limit you're requesting.
There are two possible paths that make sense:

1) Use a 12V lead acid battery which you keep topped off (more on that later).
Use a LM1084 because it's low dropout for the 12V supply and use a switcher
like the LM2576 or LM2596 for the 5V supply. By switching the supply with the
greater voltage differential, you can conserve a lot of power.

2) Use a 6V lead acid battery which you also keep topped off. Use the LM1084
for the 5V supply (which now only will waste 3W or so of power) and use a
switching step up (LM2577 or LM2597) to get the 12V supply.

BTW you never stated the application. If it's a PC based system, then you may
want to consider ditching whatever needs 12V, like using a laptop HD instead
of a regular one. If you can get it to a single supply, then you can
pick a linear solution that efficient.

> but don't know how to connect the back-up battery, in order to
> be able to charge it when powered from the wall outlet.

Separate problem from the power supply. There's a bunch of info on battery
charging around, and a bunch in the archives. But the basic rundown for
sealed lead acid (SLA) is a follows:

1) Bulk: charge at constant current of C/4 where C is the Amp-Hour capacity
of the battery until the battery reaches a terminal voltage of 2.4V/cell
(7.2V for 6V, 14.4V for 12V SLA)
2) Absorption: charge at constant voltage of 2.4V/cell until the battery
draws current of less than C/50.
3) Float: charge indefinitly at 2.3V/cell.

Parts like the UC3906 (was Unitrode, now with TI) can handle this. Digikey
carries the part for $5.50 US. It'll solve all of your charging issues.

As a challenge, I'm taking on building my own chargers using a 16F88 and
a LM324 quad op-amp to drive a LM317/pass transistor to do all three charge
phases. Here's the nickel explanation:

The main charge line consists of the LM317/pass transistor connected to the
line positive voltage, the battery, and a 1 ohm sense resistor which is
connected to ground.

The sense circuitry consists of two opamps measure the voltage across the
resistor and across the battery terminals (dividing by 4 to reduce the up to
20 V terminal voltage down to 5V). Both of these opamps are connected to
the A/D inputs of the 16F88.

Control consists of buffering the low pass filtered PWM output of the 16F88
with an opamp (multiplied by 4) and then using that output to drive the ADJ
pin of the LM317.

So here's how it works: the PIC gets the battery voltage and C from the user.
Then charging starts:

1) Bulk: The LM317 is driven upwards until a voltage of C/4 is drawn across
the sense resistor. That state is maintained until the terminal voltages
reaches 2.4V/cell.
2) Absorption: Simply drive the LM317 so that 2.4V/cell is maintained. Watch
the voltage across the sense resistor until it drops to C/50.
3) Float: Set the LM317 to 2.3V/cell and maintain indefinitely.


> If it is connected after the 12V regulator,

Not after, before. The battery comes first in the circuit. So the line voltage
charges the battery, then the battery feeds the regulators.

>  with a diode, how could I
> obtain 5V when the system is running on battery ?

The battery feeds both regulators. Which is why one of them will need to be a
switcher.

> The 5V regulator
> should be located in the schematic after the 12V one

No. The 5V regulator's input should also be the battery. They are parallel
circuits, not series ones. ASCII Art:

Line Voltage -> Battery -> 12V regulator -> 12V regulated.
                  |
                  |-----> 5V regulator ->   5V regulated.

> and the battery connection point. But in this case, the 12V
> regulator should be up to 6A > (3A + 3A), isn't it ?

Asked and answered. Both are 3A parallel circuits connected directly to the
battery.

> I don't know if I made myself
>clear, because my english is not so good,

It's fine. I took 5 years of French and I couldn't hope to write it as well as
you write English.

> but I hope someone of you will understand what I tried to explain.

Quite clear.

> I think
> this is am usual problem to deal when using dual voltage back-up power
> supplies.

The usual problem is that while it's possible to match input and output
voltages pretty well for a single linear supply, it's problematic for
multiple linear supplies. And since you want battery backup, conserving power
is an issue.

BAJ
> > Lucian

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2004\03\22@164822 by Jinx

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Lucian, is the 12V negotiable ? Could it be, say, 10V ? 11V ?

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2004\03\22@165649 by Lucian

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First of all, thank those who answered to my question.
Jeff Byron, thank you for your explanations, I will take a serious look
tomorrow (now it is 23:54 in my time).
Jinx, I don't know what the difference would be it I had 10V instead of
12V, but unfortunately it is not negociable, I have to have 12V, or else
some comparators won't work...

Lucian
{Original Message removed}

2004\03\22@170452 by Jinx

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> Jinx, I don't know what the difference would be it I had 10V

Just thinking about getting more out of the battery or reducing
the differential if you went with a step-up switcher. But if you're
stuck with needing 12V, that's the way it's gotta be

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2004\03\22@173646 by Thomas C. Sefranek

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A 16 volt center tapped transformer using a full wave diode bridge is just
the thing!
You get better than 8 volts DC from the center tap, and better than 16 VDC
from the
positive output of the bridge.  (Negative goes to common ground.)

 *
 |  __O    Thomas C. Sefranek   spam_OUTWA1RHPTakeThisOuTspamARRL.NET
 |_-\<,_   Amateur Radio Operator: WA1RHP
 (*)/ (*)  Bicycle mobile on 145.41, PL 74.4

hamradio.cmcorp.com/inventory/Inventory.html
http://www.harvardrepeater.org

> {Original Message removed}

2004\03\22@183717 by Byron A Jeff

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On Mon, Mar 22, 2004 at 05:34:32PM -0500, Thomas C. Sefranek wrote:
> A 16 volt center tapped transformer using a full wave diode bridge is just
> the thing!
> You get better than 8 volts DC from the center tap, and better than 16 VDC
> from the
> positive output of the bridge.  (Negative goes to common ground.)

Battery backup? Two batteries?

BAJ

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2004\03\22@211208 by William Chops Westfield

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On Monday, Mar 22, 2004, at 14:34 US/Pacific, Thomas C. Sefranek wrote:

>>
>>> Jinx, I don't know what the difference would be it I had 10V

The problem is that getting a regulated 12V out of a 12V battery is a
relatively "hard" thing, requiring a switching topology of some sort
(and IIRC, switchers are easier when the output voltage ISN'T that
close to the input voltage.)  If you could get by with 10V, you could
use two simple linear regulators (depending on other circumstances, of
course.)

Are you allowed to use, say, 14V worth of battery?  Or is the 12V
requirement loose enough that you can get by with an unregulated 12V
direct from the battery, when the main power is off?

BillW

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2004\03\22@212623 by Russell McMahon

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Rather than approaching the specification by successive piecewise
approximation, why not spell out what is REALLY required?

Explaining what you really as clearly and fully as possible is a great help.
It doesn't matter if your English isn't perfect. Just do the best you can to
describe your requirement.

What is the use that the outputs are being put to?
What are the actually required voltages (eg MUST it be 12 or could it be
anywhere from eg 10 to 14)
What are the load levels on each supply? (average, peak, minimum, other
characteristics?)

Is this in a vehicle?
What is the power source (main?, car alternator, ??? ...)
What battery is it that you are talking about charging for backup?
What must be backed up? (12v?, 5v?, both?, how long?
Is the backup battery voltage fixed?
How fast do you want it to recharge in?
Have you got a battery in mind to use or can it potentially be any battery?
Is this a one off, small volume, large volume?

Is there anything else that is going to have to be explained before a full
solution can be arrived at?



       Russell McMahon


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2004\03\23@055734 by Lucian

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Ok, the source I need is for a security system I'm designing. if it
works well, it could be produced in high volumes.
The system is powered from the wall outlet to a 13.8V step-down
transformer, which feeds the regulators. This would be simple if I
didn't needed a battery backup. From the sensors point of view, the 12V
could be between 9V and 16V, but I have also some comparators which work
at 12V and if the voltage drops bellow, the result isn't accurate
anymore.
When the power is off, the battery must backup both the 12V and the 5V.
I thought of using 2 regulators, and when the power fails, a relay to
connect the battery to the 12V directly and to the 5V regulator.
Is this suitable ? I would also need a charging circuit for a lead-acid
battery, can you indicate me a good circuit ?

Lucian
{Original Message removed}

2004\03\23@065640 by Russell McMahon

face
flavicon
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> Ok, the source I need is for a security system I'm designing. if it
> works well, it could be produced in high volumes.
> The system is powered from the wall outlet to a 13.8V step-down
> transformer, which feeds the regulators. This would be simple if I
> didn't needed a battery backup. From the sensors point of view, the 12V
> could be between 9V and 16V, but I have also some comparators which work
> at 12V and if the voltage drops bellow, the result isn't accurate
> anymore.
> When the power is off, the battery must backup both the 12V and the 5V.
> I thought of using 2 regulators, and when the power fails, a relay to
> connect the battery to the 12V directly and to the 5V regulator.
> Is this suitable ? I would also need a charging circuit for a lead-acid
> battery, can you indicate me a good circuit ?

That's getting clearer. If the comparators are all that's making you need
12v then I would seriously consider looking closely at their circuitry and
seeing how you can reduce the voltage. Unless you are doing something VERY
high speed indeed you are unlikely to really require 12v. if this is a slow
speed (a few Hz to 10's of Hz switching rate) then it should be easy (famous
last words).

The reason I'm so keen to reduce the 12v rail is

- 12v gel cells are the norm in security systems. Cheapish, ruggedish,
lowish cost, highly available, well understood by security system users.

- Once you get below 12v you can charge the battery from the 13.8v supply
and also use the supply or battery to drive the regulators.

- Other batteries have better energy density but are generally dearer.

You could consider floating the battery at slightly above 12v and using that
as the 12v output voltage.
A 13.8v  AC supply will easily charge a 12v battery.
(13.8 x 1414 =~ 19v.)
OR was that 13.8V DC output?
If so, why?
(Lead acid batteries are usually floated at about 13.6v but gel and auto
cells are slightly different).

But first - tell us about your comparator circuit(s):

What are the input signals and their sources?
What are the outputs?
How fast do the signals change?
What is the purpose of the comparators?
What IC are you using?
What goes wrong with less than 12v supply?
Why?

It's quite possible that an LM324 or LM358 running on 5 volts will do what
you want.

Also, what do you use the 12v rail for apart from the comparators?
You mention sensors which need 9 to 16v. (PIR?)
Wouldn't it be nice to be able to run everything on 5 volts :-) ?

Also: How long does the battery backup need to last?
What is the expected current drain in backup mode?




       Russell McMahon

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2004\03\23@070054 by Jinx

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> From the sensors point of view, the 12V could be between 9V
> and 16V, but I have also some comparators which work at 12V
> and if the voltage drops bellow, the result isn't accurate anymore

Make a local, stable 12V just for the comparators. To guarantee
12V on battery power, use a simple voltage doubler (should give
you around 20V when battery is flat) followed by a regulator, for
instance, as the comparators presumably won't need many mA

> I would also need a charging circuit for a lead-acid battery, can
> you indicate me a good circuit ?

For best care of the battery, there are many charger/monitor ICs
around. For example, as Byron suggested, Unitrode's UC2906
or UC3906

http://focus.ti.com/docs/prod/folders/print/uc2906.html

Application notes include background theory and practical circuits

You'd find others to compare in catalogues or at retailers with on-line
product guides like Radiospares or Digikey

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2004\03\23@142754 by Byron A Jeff

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On Tue, Mar 23, 2004 at 12:57:02PM +0200, Lucian wrote:
> Ok, the source I need is for a security system I'm designing. if it
> works well, it could be produced in high volumes.
> The system is powered from the wall outlet to a 13.8V step-down
> transformer, which feeds the regulators. This would be simple if I
> didn't needed a battery backup. From the sensors point of view, the 12V
> could be between 9V and 16V, but I have also some comparators which work
> at 12V and if the voltage drops bellow, the result isn't accurate
> anymore.

OK. This helps.

Here's a key question: Why do the comparators work at 12V? As you know
comparators compare voltages. Using resistor dividers, possibly built into
an opamp divider, it should be possible to take the original 12V requirement
and cut it to something much less, such as 4V. Once you've done that, then
you can ditch the 12V requirement, still keeping the unregulated  higher
voltage supply to power the circuit. You then have:

1) 13.8V input supply feeding...
2) A 12V gel cell which serves as the input to...
3) A 5V switcher which then using dividers drives....
4) the comparators.

12V is then rendered unnecessary.

> When the power is off, the battery must backup both the 12V and the 5V.

Again using the comparators at a lesser voltage obviates the need for a
regulated 12V supply.

> I thought of using 2 regulators, and when the power fails, a relay to
> connect the battery to the 12V directly and to the 5V regulator.

Bad idea. It'll glitch when the power fails. The only way for this to work
effectively is for the battery to be the primary power source while actively
being charged from the line. I have a sunrise/sunset light controller that
is battery backed, and when developing I tried both ways. I never got the
relay method to work without the board resetting. But with the battery feeding
the regulator, it "switches" smoothly, because it doesn't switch.

> Is this suitable ? I would also need a charging circuit for a lead-acid
> battery, can you indicate me a good circuit ?

The UC3906 I alluded to earlier. However if you have a PIC onboard, there's
not necessarily a need for it. Also if your input voltage is only 13.8V
you can only float charge anyway as you need a minimum of 14.4V to
bulk/absorption charge.

BAJ

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2004\03\23@164832 by Russell McMahon

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flavicon
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> 12V is then rendered unnecessary.

He also has what appear to be nominally 12v sensors

> > ... From the sensors point of view, the 12V
> > could be between 9V and 16V,


       RM

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2004\03\24@054123 by Lucian

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First, thank you all who replied to me.

The comparators are feeded at the voltage provided as 12V, so if it is
10V, they are feeded at 10V, so I think it could be a voltage drop
bellow 12V.
You suggested that I put the battery first, after the transformer and
the rectifier bridge, if I understood well ? Shouldn't it be some
charging supervisory circuit there ?
It's not very clear to me how the schematic looks if I admit that I
don't need exacty 12V, but it can drop to 10V.
13.8V unrectified may power some sensors ? Isn't this dangerous for them
?

Lucian
{Original Message removed}

2004\03\24@055614 by Jinx

face picon face
> The comparators are feeded at the voltage provided as 12V

> 13.8V unrectified may power some sensors ? Isn't this dangerous
> for them ?

Can you tell us what the comparators and sensors are please, if
that's possible. I'm sure there's a circuit, probably using other parts
than those you're designing with now, that can be suggested for your
needs, and it's a good bet that the voltages required won't need to
be as rigid as you might think

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2004\03\24@062001 by Lucian

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The idea is that the system has to work with a variety of sensors
(powered at around 12V) and I don't use a particular sensor. It it about
PIRs, but can also be smoke detectors and all sort. I realised that the
voltage for the comparators is not an issue anymore, as I'm using a
voltage divider, so the 12V can be around 10V-14V without any problem.
Am I mistaking ?

Lucian
{Original Message removed}

2004\03\24@062830 by Byron A Jeff

face picon face
On Wed, Mar 24, 2004 at 12:40:03PM +0200, Lucian wrote:
> First, thank you all who replied to me.
>
> The comparators are feeded at the voltage provided as 12V, so if it is
> 10V, they are feeded at 10V, so I think it could be a voltage drop
> bellow 12V.

As Jinx has asked in another post, take some time and explain the purpose
of the comparators. Why are they in your circuit? What are their inputs?
What are their outputs?

The reason is that there's almost always another way to solve the problem.
But right now we don't have enough information to solve the problem.

> You suggested that I put the battery first, after the transformer and
> the rectifier bridge, if I understood well?

Yes.

> Shouldn't it be some charging supervisory circuit there ?

Maybe. If the output voltage of the transformer/bridge circuit is really 13.8V
you can simply float charge the battery without any charging supervisor.

However if the voltage is significantly higher than 14V under load, then at
the very minimum you need a regulator to 13.8V followed by a current limiting
resistor.

But it is probably better to match the transformer/bridge so that the battery
can be directly connected. Anything near 13.8V should be fine.

I do apologize for not being clear. I was addressing the point that in your
circuit that the battery comes before the regulators, not after

> It's not very clear to me how the schematic looks if I admit that I
> don't need exacty 12V, but it can drop to 10V.

I still don't know why it is that you need the 12V or the 10V. If you can
describe the purpose of the comparators, maybe it is possible to get this
down to a single regulated voltage source.

> 13.8V unrectified may power some sensors ? Isn't this dangerous for them
> ?

I don't know. What are the specifications on the sensors?

Also the battery will serve as a form of regulator, as batteries tend to pull
the input voltage towards its terminal voltage.

I am still looking for more information.

BAJ

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2004\03\24@070604 by Byron A Jeff

face picon face
On Wed, Mar 24, 2004 at 01:20:07PM +0200, Lucian wrote:
> The idea is that the system has to work with a variety of sensors
> (powered at around 12V) and I don't use a particular sensor. It it about
> PIRs, but can also be smoke detectors and all sort.

It would be helpful to know if they require a regulated 12V. But I'm almost
certain that alarm sensors can be run from a 13.8V battery voltage as they
all have regulators internal to the sensor itself.

> I realised that the
> voltage for the comparators is not an issue anymore, as I'm using a
> voltage divider, so the 12V can be around 10V-14V without any problem.

I'm still trying to resolve why it is that the comparators need anything
other than 5V to do their jobs.

> Am I mistaking ?

Still not enough information because you haven't yet told us why you are
using the comparators.

Let me give you an example. I have a wire loop for the doors. Because of the
length of the loop I power it with 12V. Instead of a comparator at panel,
I connect the loop across the LED in a optoisolator (using an appropriate
current limiting resistor), then drive the output transistor of the opto
from 5V.

Now the loop is powered from the unregulated line/battery voltage and the
optoisolator converts that voltage to 5V. Poof! only one regulator is needed.

You could do something like that if you can give more information on the
purpose of the comparators. If they are just there for level conversion, then
there are other ways to solve the problem.

The key point here is that you want 2 voltages: 13.8V unregulated, 5V
regulated. In our continuing discussion I still get the feeling that you want
to regulate the line/battery voltage, when it's probably unnecessary.

BAJ

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2004\03\24@070605 by Jinx

face picon face
> PIRs, but can also be smoke detectors and all sort

PIRs will work at 5V or more (lens selection is just as important
as operating voltage by the way), and most, if not all, domestic
Americium smoke alarms run on a 9V battery, so there's no
problem there. I'm sure you'd find a sensor for any application
that will run at 5V, or certainly under 12V

IMHO I would aim for all and any sensors working properly at or
below the battery's terminal voltage, whatever you determine that
to be. With a light load this could be around 11V. But you need to
have 12V/3A (and also 5V/3A if you intend to use just one battery)
available, presumably for alarm devices. Trying to draw this much
current from a nearly-flattened battery is going to cause a serious
dip in voltage. It is not good for lead batteries to go below 10.5 for
extended periods. You'll need to do some testing to find out how
low battery V can go before the required amperage is no longer there.
And then as best as you can, stay away from that limit

Operating temperature must be considered too. A cold battery has
a higher internal resistance (= lower Ah capacity) than a warm one.
At 0degC there'll be maybe 65-70% of the nominal 25degC Ah capacity,
and if you expect the unit to be outside in winter, you'll have to allow
for this with either more efficient high-load devices or with a bigger
battery

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2004\03\24@072929 by Lucian

flavicon
face
The purpose of the comparators is to do some line supervisory. If the
wire is cut or if it is shortened, the alarm goes on. Is there any
simpler solution ?
Most of the sensors accept voltages between 9V and 16V. So I think they
could be powered from 13.8V unregulated.
The battery could be charged with 13.8V with minimum circuitry (when it
has reached the maximum capacity, stop charging, and when it has
discharged under some value, disconnect it from the load). From the 13.8
unregulated, feed a 5V regulator. Did I understand correctly ?
The values for current (3A for each) are the maximum needed values, the
circuit won't necessary draw so much current.

Lucian
{Original Message removed}

2004\03\24@144253 by Byron A Jeff

face picon face
On Wed, Mar 24, 2004 at 02:29:24PM +0200, Lucian wrote:
> The purpose of the comparators is to do some line supervisory. If the
> wire is cut or if it is shortened, the alarm goes on. Is there any
> simpler solution ?

Sure there is. You're using a PIC which presumably has ADC. You can get an
exact value reading the wire voltage using the ADC. This completely eliminates
the need for a comparator, though you may still need an opamp circuit to
drop the voltage to a reasonable range.

The exact circuit you need is listed in Figure 3-17 of this Motorola
alarm system design document on page 44:

e-http://www.motorola.com/files/microcontrollers/doc/ref_manual/DRM008.pdf

They simply power the line with Vcc, which is presumably 5V. But even if
you powered it from the unregulated battery voltage, you only need very
minimal regulation. Here's what I'd do:

1) Create a minimal 9V supply using a low valued resistor (200 ohms) and a 9V
  zener.  Then tie 5K pullup resistors to that supply, represented by the
  R25 in the diagram.

2) Use 2.5k resistor for R1, 1K for RS and tie a 5V zener across the each
  alarm line between RS and the 5K pullup.

So here's that happens in the 4 states:

OK (armed): 1.5V line voltage 9V * 1K/(1K+5K)
    ALARM: 3.7V line voltage 9V * (1K + 2.5K)/(1K + 2.5K + 5K)
    SHORT: 0V
      CUT: 5V clipped by the 5V zener.

With that much separation it's easy to figure out the alarm state. And by
using the higher voltage (instead of the 5V regulated line), you can get
a full range of values from 0-5V. Finally if you have multiple doors/windows
in an alarm circuit, you can put different values of resistors across the
switches so you can detect exactly which door/window switch is open.

The PIC has the tools to pull this off, why not use them? No need for
comparators at all.

> Most of the sensors accept voltages between 9V and 16V. So I think they
> could be powered from 13.8V unregulated.

Right.

> The battery could be charged with 13.8V with minimum circuitry (when it
> has reached the maximum capacity, stop charging, and when it has
> discharged under some value, disconnect it from the load).

No need if the line voltage is 13.8V. You can charge indefinitely at that
voltage. So other than maybe a simple power shunt to keep the voltage below
13.9V or so (safety feature), you need no control at all between the line
voltage and the battery.

And you never want to to disconnect the battery from the load. That's the only
way not to have it glitch. Line feed battery, battery feeds regulator,
nothing ever switches.

> From the 13.8
> unregulated, feed a 5V regulator. Did I understand correctly ?

That last part yes.

> The values for current (3A for each) are the maximum needed values, the
> circuit won't necessary draw so much current.

Shouldn't matter as long as the regulator can handle it. Batteries are
designed to dump hundreds of amps of current if necessary.

Thanks for the discussion. It gave me a few minutes to formulate exactly
how to handle the alarm line issues for my alarm.

BAJ

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2004\03\24@150201 by Lucian

flavicon
face
Thank you for your information. I will consider it for redesigning the
circuit.

Lucian

-----Original Message-----
From: pic microcontroller discussion list
[.....PICLISTKILLspamspam@spam@MITVMA.MIT.EDU] On Behalf Of Byron A Jeff
Sent: 24 martie 2004 21:43
To: PICLISTspamKILLspamMITVMA.MIT.EDU
Subject: Re: [EE]: 12V and 5V Power Supply

On Wed, Mar 24, 2004 at 02:29:24PM +0200, Lucian wrote:
> The purpose of the comparators is to do some line supervisory. If the
> wire is cut or if it is shortened, the alarm goes on. Is there any
> simpler solution ?

Sure there is. You're using a PIC which presumably has ADC. You can get
an
exact value reading the wire voltage using the ADC. This completely
eliminates
the need for a comparator, though you may still need an opamp circuit to
drop the voltage to a reasonable range.

The exact circuit you need is listed in Figure 3-17 of this Motorola
alarm system design document on page 44:

e-www.motorola.com/files/microcontrollers/doc/ref_manual/DRM008.p
df

They simply power the line with Vcc, which is presumably 5V. But even if
you powered it from the unregulated battery voltage, you only need very
minimal regulation. Here's what I'd do:

1) Create a minimal 9V supply using a low valued resistor (200 ohms) and
a 9V
  zener.  Then tie 5K pullup resistors to that supply, represented by
the
  R25 in the diagram.

2) Use 2.5k resistor for R1, 1K for RS and tie a 5V zener across the
each
  alarm line between RS and the 5K pullup.

So here's that happens in the 4 states:

OK (armed): 1.5V line voltage 9V * 1K/(1K+5K)
    ALARM: 3.7V line voltage 9V * (1K + 2.5K)/(1K + 2.5K + 5K)
    SHORT: 0V
      CUT: 5V clipped by the 5V zener.

With that much separation it's easy to figure out the alarm state. And
by
using the higher voltage (instead of the 5V regulated line), you can get
a full range of values from 0-5V. Finally if you have multiple
doors/windows
in an alarm circuit, you can put different values of resistors across
the
switches so you can detect exactly which door/window switch is open.

The PIC has the tools to pull this off, why not use them? No need for
comparators at all.

> Most of the sensors accept voltages between 9V and 16V. So I think
they
> could be powered from 13.8V unregulated.

Right.

> The battery could be charged with 13.8V with minimum circuitry (when
it
> has reached the maximum capacity, stop charging, and when it has
> discharged under some value, disconnect it from the load).

No need if the line voltage is 13.8V. You can charge indefinitely at
that
voltage. So other than maybe a simple power shunt to keep the voltage
below
13.9V or so (safety feature), you need no control at all between the
line
voltage and the battery.

And you never want to to disconnect the battery from the load. That's
the only
way not to have it glitch. Line feed battery, battery feeds regulator,
nothing ever switches.

> From the 13.8
> unregulated, feed a 5V regulator. Did I understand correctly ?

That last part yes.

> The values for current (3A for each) are the maximum needed values,
the
> circuit won't necessary draw so much current.

Shouldn't matter as long as the regulator can handle it. Batteries are
designed to dump hundreds of amps of current if necessary.

Thanks for the discussion. It gave me a few minutes to formulate exactly
how to handle the alarm line issues for my alarm.

BAJ

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