3rd POST: Is this the wrong forum to ask this in?

[Alan Kamrowski II]

Please understand that I have a very basic understanding of electricity, so
it may be quite likely that what I am thinking here is not doable.

I have a power brick that powers a garage door alarm. The power brick is
labeled 12VDC 100ma, but it actually measures 16.3 V to 17.0 V (it bounces
back and forth every second or so).

My original hope was to tap it into my alarm system panel because it is
battery backed up, but the 12.5V it was putting out was not enough and the
device was malfunctioning. I assumed this is because of the ~4 V difference
between it and the power brick.

So, my two questions are:

1. Is there an *easy* way to convert 12V up to 16-17V somehow so I could
just tap off my alarm system?

2. Is there a way I can put something small in line with the power brick
cable to keep the device from losing power for a very short amount of time
<=15s. Sort of like a mini Uninterruptable Power Supply. Could a capacitor
of some type do this? Ideas?

Thanks,

Alan

 

[Walter Harley]

Third post? I don't remember seeing the other two.

I have a power brick that powers a garage door alarm. The power brick is
labeled 12VDC 100ma, but it actually measures 16.3 V to 17.0 V (it bounces
back and forth every second or so).

Measures that when it's actually powering the alarm, you mean? Or measures
that when it's disconnected? Many (non-regulated) supplies will generate a
different voltage under load than not.

My original hope was to tap it into my alarm system panel because it is
battery backed up, but the 12.5V it was putting out was not enough and the
device was malfunctioning. I assumed this is because of the ~4 V difference
between it and the power brick.

Perhaps. Or it draws enough current that it loaded down your alarm system
panel and the voltage dropped too low.

1. Is there an *easy* way to convert 12V up to 16-17V somehow so I could
just tap off my alarm system?

Not really, depending on your definition of "easy". Certainly not easier
than buying another cheap UPS and plugging the power brick into it.

2. Is there a way I can put something small in line with the power brick
cable to keep the device from losing power for a very short amount of time
<=15s. Sort of like a mini Uninterruptable Power Supply. Could a capacitor
of some type do this? Ideas?

Yes. There were some threads on this group recently about using a capacitor
as a short-term UPS for DC supplies. You need to make sure there's a diode
between the supply and the capacitor, and a resistor to limit charge
current, and some other stuff; dig around and you'll find several postings
that showed schematics for how to do it, as well as some suggested values.
Make sure the caps you use are rated high enough voltage (e.g., use 25V
caps).

Or, just get a cheap UPS and plug the power brick into it. Depends on how
you value your time vs. your money.

 

[Ian Stirling]

Please understand that I have a very basic understanding of electricity, so
it may be quite likely that what I am thinking here is not doable.

I have a power brick that powers a garage door alarm. The power brick is
labeled 12VDC 100ma, but it actually measures 16.3 V to 17.0 V (it bounces
back and forth every second or so).

Typical of unloaded unregulated power supplies.
It will usually read aroung 15V under 10ma load, and 12V under 100ma or so.

My original hope was to tap it into my alarm system panel because it is
battery backed up, but the 12.5V it was putting out was not enough and the
device was malfunctioning. I assumed this is because of the ~4 V difference
between it and the power brick.

Maybe, maybe not.

2. Is there a way I can put something small in line with the power brick
cable to keep the device from losing power for a very short amount of time
<=15s. Sort of like a mini Uninterruptable Power Supply. Could a capacitor
of some type do this? Ideas?

You don't want to use just a capacitor.
And you'd need quite a big one.

Some more details would be handy to get the best solution.
What's the current drawn from the brick when it's idle?
When you connected it to the 12.5V supply, was the 12.5V supply still at
the same voltage, or did it droop as you were asking too much current?

 

[Alan Kamrowski II]

Walter,

Thanks for posting! Responses below:

Measures that when it's actually powering the alarm, you mean? Or
measures that when it's disconnected? Many (non-regulated) supplies will
generate a different voltage under load than not.
Perhaps. Or it draws enough current that it loaded down your alarm system
panel and the voltage dropped too low.

When it is powering the alarm that is what it measures. The draw of the
device I added bounces between 18ma-25ma. The voltage didn't drop after I
added it, but the device just didn't work well.

Not really, depending on your definition of "easy". Certainly not easier
than buying another cheap UPS and plugging the power brick into it.

I got you.

Or, just get a cheap UPS and plug the power brick into it. Depends on how
you value your time vs. your money.

Yes, time I don't have, if it was something easy, I might have done it!

Thanks,

Alan

 

[Alan Kamrowski II]

Hi Ian,

Thanks for posting too! Responses below:

Typical of unloaded unregulated power supplies.
It will usually read aroung 15V under 10ma load, and 12V under 100ma or

so.

I think it drops a bit, but it seems to stay in the 16.3v and above even
with a load on it (~25ma).

Some more details would be handy to get the best solution.
What's the current drawn from the brick when it's idle?

This is measuring the 110V input of the brick. 10.5ma when not plugged into
the device, 11.7ma - 12.7ma when the device is plugged in.

When you connected it to the 12.5V supply, was the 12.5V supply still at
the same voltage, or did it droop as you were asking too much current?

Yes, it stayed the same. The alarm panel says it can power up to 650ma, and
calculating the keypads I have, it doesn't have more than 200ma on it. I
was only adding 25ma and the voltage didn't drop.

Thanks for the help!

Alan

 

[Ian Stirling]

Hi Ian,

Thanks for posting too! Responses below:

Yes, it stayed the same. The alarm panel says it can power up to 650ma, and
calculating the keypads I have, it doesn't have more than 200ma on it. I
was only adding 25ma and the voltage didn't drop.

I would consider it bad design to not work at the nameplate voltage of the
power supply.
But anyway.

Several ways occur.
A "simple switcher" or similar DC-DC converter to boost the 12.5V.
You can buy isolated one watt DC-DC converters in an IC shaped box for
around $10US, in your local large electronics catalog.

Simply take a 3V output one, and wire the output in series with the 12V
input, to get a 15.5V output.

Going the other way, and assuming the thing can work with a volt drop from
the wall-wart, and it draws 25ma, and you need 10 seconds backup, then
that's .25 farads at 20Vdc working or so.
You'd want to arrange this so that it charges through a 500 ohm resistor and
diode, and discharges into the load through a diode.

A quick browse through my catalog shows the cheapest way is probably 4
5.5V 1F memory caps in series, with balancing resistors (these are needed
due to the horrible tollerance).
However, I'd gho witht the DC-DC conerter.

 

[Alan Kamrowski II]

Hi Ian,

Several ways occur.
A "simple switcher" or similar DC-DC converter to boost the 12.5V.
You can buy isolated one watt DC-DC converters in an IC shaped box for
around $10US, in your local large electronics catalog.

Simply take a 3V output one, and wire the output in series with the 12V
input, to get a 15.5V output.

I really appreciate your help.

This sounds like the way to go.

Can you tell me how you would output it in series ? Would this switcher
have a input for both - and +, and an output for - and + ? What would you
connect? Would the 3V+12V be 15V or 15.5V ?

Thanks!!

Alan

 

[Ian Stirling]

Hi Ian,


I really appreciate your help.

This sounds like the way to go.

Can you tell me how you would output it in series ? Would this switcher
have a input for both - and +, and an output for - and + ? What would you
connect? Would the 3V+12V be 15V or 15.5V ?

Yes, they've got positive and negative terminals.
Connect the input side to a 12V source, and the other terminals act like a
3V battery.

 

[Alan Kamrowski II]

Ian,

Yes, they've got positive and negative terminals.
Connect the input side to a 12V source, and the other terminals act like a
3V battery.

So, connect the input side to the 12V + and -. Then to connect the "3V
battery" in series, connect the original +12V to the - side of the 3V output
and the + side of the 3V output will become my 15V output along with the
original 12V ground?

Do you have any links to one of these online?

Thanks!

Alan

 

[Ian Stirling]

Ian,


So, connect the input side to the 12V + and -. Then to connect the "3V
battery" in series, connect the original +12V to the - side of the 3V output
and the + side of the 3V output will become my 15V output along with the
original 12V ground?

Do you have any links to one of these online?

Not to hand.
Though I believe http://www.digikey.com/ and http://www.farnell.com/
both seem to be likely candidates.
Look under "power supplies", for small devices in IC like packaging.

 

[Alan Kamrowski II]

Hi Ian,

Yes, they've got positive and negative terminals.
Connect the input side to a 12V source, and the other terminals act like a
3V battery.

I was thinking about this more and it seems to me that it is putting the
original source 12V in series with itself converted down to 3V to make 15V.
How can something be put in series with itself? For example, could this
same idea be applied to a single 1.5V battery somehow to make it 3V ?

Thanks,

Alan

 

[Uns Lider]

I was thinking about this more and it seems to me that it is putting the
original source 12V in series with itself converted down to 3V to make 15V.
How can something be put in series with itself? For example, could this
same idea be applied to a single 1.5V battery somehow to make it 3V ?

If you try to put something in series with itself, what you get is a short ;-)

If you put the power source through an isolated DC-DC converter, you can
then put the DC-DC converter's output in series with the power source.

There are a few approaches for going from a DC power source to a higher
voltage. They all involve the use of an oscillator, which operates switches
to direct the current flow across different paths during the several phases
the circuit iterates over thousands of times per second. The "switches" are
actually diodes or transistors, depending on the topology being used.

-- uns