# Another question regarding mA

Discussion in 'Electronic Basics' started by AB, Mar 2, 2007.

1. ### ABGuest

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
Bob G.

2. ### Peter BennettGuest

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

3. ### Homer J SimpsonGuest

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

4. ### ehsjrGuest

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

5. ### MarraGuest

You need a constant current source to achieve this.

6. ### John FieldsGuest

---

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.