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I'm building a DIY security system for my home, but I'm using the existing sensors and siren that are already installed from my old security company.

So, I don't know the current draw of the siren or have a part number for it to look up the datasheet, but I'm attempting to make sure I have a sufficient transistor/relay to switch it on. All I know is that the siren is 12V (from measuring the alarm output of the previous system with a multimeter).

Is there a way I can measure the current draw of the siren without actually sending current into it? It blares at around 105dB, so I'd love to try to figure out a way to determine this without the deafening tone or attracting the attention of my neighbors.

I considered measuring the resistance of the siren, since it's probably just a piezo, and using Ohm's law, but I'm not sure that would give an accurate estimate without knowing what kind of other driving circuit, etc. is inside the siren.

Any suggestions?

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    \$\begingroup\$ Grab the part number and find a datasheet? \$\endgroup\$ – HikeOnPast Dec 18 '12 at 18:28
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    \$\begingroup\$ Thanks for the reply. However, my question is specifically how to find the current draw when a part number is not available. \$\endgroup\$ – Alan Christopher Thomas Dec 18 '12 at 18:36
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    \$\begingroup\$ Ah...missed that. Measure it or overprovision it. \$\endgroup\$ – HikeOnPast Dec 18 '12 at 18:42
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No, any static test could easily be quite a bit off. Without actually powering it, you don't know what the current draw will be. However, if this siren uses a motor, then then worst case current draw will be at startup. That's the value you want to design to anyway.

Non-motor types could possibly not draw much of anything until they got to a particular threshold. You really have to run it. Put some muffling around it and run it for a second or so.

Also note that what you need to design the relay or transistor switch to is both the worst case current draw and the average current. The parts needs to be rated for the worst case current, and not drop too much voltage during that time. The thermal design has to deal with the heat from the average current draw accross your switch.

All that said, its hard to imagine it will be more than a few amps at 12V. Get a 10A relay and be done with it.

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  • \$\begingroup\$ Ah, thanks. That's what I was afraid of. I'll wear some nice earmuffs and and something to muffle the device as much as possible. Or, see if I have any higher current relays lying around (the ones I have on hand are 500mA, which it's possible this siren falls below). \$\endgroup\$ – Alan Christopher Thomas Dec 18 '12 at 18:47

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