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Another question regarding mA

As I mentioned in previous posts, I have this alarm panel that uses

12VDC to power a set of output contacts for an optional sounder device. The only requirement is that you don't exceed 120 mA current draw @12VDC for these contacts. Thanks to the help I've gotten here, I was able to measure the draw from all of my sirens, as well as some small strobe lights. Now here's my question. My lowest draw siren draws 50mA. That leaves me another 70mA free on that circuit. I have a low draw strobe assembly (strobe light w/ little circuit embedded under it) that I'd like to add to the siren, but it draws. 140mA @ 12VDC...too much still.

Is there anyway I can modify the strobe so that it will draw only 70mA rather than 140? Maybe with a resistor or something? Before you say it, I know that there is the possibility that the strobe won't fire or be pretty dim...but...I'd like to try it anyway. Thanks again for the great advice I've gotten here. Bob G.

Reply to
AB
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It would be easier, and probably better, to use the alarm output to operate a relay, and use the relay contacts to apply power to the sirens and strobe.

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

You could change the charging/firing resistor to slow the strobe down to less than half speed.

Reply to
Homer J Simpson

Assuming the power supply in the alarm is capable of supplying the current the strobe needs, use that for the following circuit:

+12 ----+----Strobe----+ | | o | S1 / /c o---[470R]---| BC337 \\e | Gnd -------------------+

It's about 20 cents worth of parts - 1 resistor and

1 transistor. When the alarm contact (S1) closes, the transistor conducts and the strobe operates.

If the alarm contact is normally closed instead of normally open, use this circuit:

+12 ----+----Strobe----+ | | [470R] | | /c o------------| BC337 S1 | \\e o | | | | | | | Gnd ----+--------------+

If you have NPN transistors on hand, give one of them a try - almost any NPN will work fine.

If your alarm power supply cannot provide the current your strobe needs, use a separate 12V DC wall wart supply.

Ed

Reply to
ehsjr

You need a constant current source to achieve this. Just google it.

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Reply to
Marra 

--
The strobe resistance:


           E      12V
     Rs = --- = -------- ~ 85.7 ohms
           I     0.14A


The total resistance required to drop 12V at 70mA:


           E      12V
     Rt = --- = ------ ~ 171.4 ohms
           I     0.07


The resistance you need to add, in series, between the strobe and
the supply in order to limit the current to 70 mA:


     Rx = Rt - Rs = 171.4R - 85.7R = 85.7 ohms

 
The closest standard 5% resistor is 82 ohms, so it\'ll drop the
current to:

             E             12V
     I = --------- = ------------- = 0.072A
          Rs + Rx     85.7R + 82R


And will dissipate:


     P = I²R = 0.072A² * (85.7R + 82R) = 0.869 watts,


so I\'d use at least a 2 watt resistor in order to keep it from
getting _real_ hot.
Reply to
John Fields

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