Would T-junction help low current situation

I have a doorbell that uses 18V. I'd like to put a relay between so that I can record when doorbell rang (the relay is wired to a computer).

Unfortunately the relay seem to introduce too much resistance as the doorbell won't ring:

(not working) +-------+ | | | +-------------->| Relay coil |------> gnd Bell | | | | | | | +--------- --------------------< Vcc | | / +-------+ / Switch /

Would this help? +-------+ | | | +------+------>| Relay coil |------> gnd Bell | | | | | +---------------------------> gnd | | | +--------- --------------------< Vcc | | / +-------+ / Switch /

Reply to
Test
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Your second approach won't work either (you may have just drawn it incorrectly).

You need to only connect the relay coil across (in parallel with) the bell. The relay will get power when the bell does.

BELL |----| | |----------------------GND | |----RELAY------| |----| | |--------SWITCH-----VCC

Bob

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Reply to
BobW

How does a relay coil with ground on each end of it ever get energized? If the relay is rated for something close to

18VAC, connect it across (in parallel with) the bell. If it is an AC relay rated for a lower voltage, you may need to add a series resistor to burn up the extra voltage. If it is a DC relay, you might use a bridge rectifier to convert AC to DC.
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Regards,

John Popelish
Reply to
John Popelish

The problem is that I can't get to the Vcc wire between the switch and bell. It is inside the wall.

+-------+ | | (wall ends) | +-------------->| Relay coil |------> gnd Bell | | | | | | (wall starts) | +--------- --------------------< Vcc | | / +-------+ / Switch /
Reply to
Test

Well, then you'll have to go back to your original idea and either use a relay with a very low coil resistance or an opto-coupler. This way, you'll only lose a couple of volts across the bell.

One other idea comes to mind -- get a Sawz'all.

Bob

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Reply to
BobW

Try sensing the bell current with a transistor. See above:

You might need a resistor from B-E to shunt some of the bell current, but if you use something like an MJE170 or other medium size power transistor it should be able to handle about 1 amp. Don't forget to add a diode across the relay coil and also perhaps one across B-E. If you have an AC bell then a similar method will work, but it would need some extra components. You might also be able to use a SSR or optoisolator, with a bridge rectifier if AC, and an appropriate current shunt resistor.

Paul

Reply to
Paul E. Schoen

I can

Or, you can wire the doorbell button straight to the relay coil, so that (when the button is pressed) it activates the relay and one pair of contacts rings the bell, while another pair of contacts informs your computer.

If you ground the 'common' terminal of an SPDT relay, the one branch can have a pullup resistor and your computer, the other can complete the bell circuit. With multipole relays, it's even easier.

This way, you're sensing the doorbell switch position, which is a simple task, rather than the high current of an operating doorbell.

Reply to
whit3rd

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First things first.

Is the bell AC or DC, how much current does it draw, and what wiring can
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Reply to
John Fields

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