Another question regarding mA

Mar 02, 2007 5 Replies

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AB 19 years ago

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.

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Peter Bennett 19 years ago

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 new newsgroup users info : http://vancouver-webpages.com/nnq GPS and NMEA info: http://vancouver-webpages.com/peter Vancouver Power Squadron: http://vancouver.powersquadron.ca

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Homer J Simpson 19 years ago

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

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ehsjr 19 years ago

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.

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Marra 19 years ago

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

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John Fields 19 years ago

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.

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