low power switching circuit dc - Electronic Circuits Projects Diagrams Free

laverdure · 29th October 2009, 09:43 PM

Hi Guys,
I have a challenge here,,, I have been away from hardware electronics for awhile, now I am in software, I have a old Kenwood car alarm that I wish to use,,,, the siren had 3 leads, one -- ground, 2 -- constant power of 12vdc, and the other 3 -- switching voltage of 0-off, 1.01vdc on…. I somewhat remember doing this in college but how do I make the switching circuit???? Man I feel old

…. Thanks….

MikeMl · 29th October 2009, 11:56 PM

How about a pull-up resistor to +12V with a normally closed switch to ground. Tie the junction of switch and resistor to the input. Switch closed, alarm off. Switch open, pull-up resistor sources current into the input, pulling it above 1V, alarm makes noise.

laverdure · 30th October 2009, 12:32 AM

Do you have a quick and dirty schematic of that idea???

Thanks

laverdure · 30th October 2009, 03:32 PM

Let me give more background to this challenge. When the alarm is activated there is no voltage at the switching wire, when the alarm is tripped, i.e. open door or another sensor, the voltage at the switching wire is now approximately 1.01vdc ,,, do they make a relay, reed or solid state……anything….. , that will activate at 1vdc ????

Thanks

MikeMl · 30th October 2009, 03:41 PM

This is backwards compared to what you said in your first post. All you need to do is to wire a normally-open switch from the "alarm input" to ground. Closing the switch will make the alarm go off. This could be a switch similar to the one that lights the dome-light in your car.

Do you need to "reverse" the logic so that the application of a positive voltage turns on the alarm? If so, use an NPN transistor like a 2n3904 or 2n2222. Connect emitter to ground. Connect collector to the "alarm input", connect a 4.7K resistor to the base. Applying 3 to 12V to the other end of the base resistor will turn on the alarm.

laverdure · 30th October 2009, 09:00 PM

Would the 1vdc from the signal wire turn on the "switch"???
Sorry if I didn't clarify myself on the first post

Thanks

MikeMl · 30th October 2009, 09:10 PM

Quote:

Originally Posted by laverdure

Let me give more background to this challenge. When the alarm is activated there is no voltage at the switching wire, when the alarm is tripped, i.e. open door or another sensor, the voltage at the switching wire is now approximately 1.01vdc ,,, do they make a relay, reed or solid state……anything….. , that will activate at 1vdc ????

Thanks

Quote:

Originally Posted by laverdure

Would the 1vdc from the signal wire turn on the "switch"???
Sorry if I didn't clarify myself on the first post

Thanks

No, the switch shorts the "alarm input" to ground if closed. The switch is either totally open (∞Ω), or a short (zeroΩ). A relay contact can do that; so can an NPN transistor or NFet.

laverdure · 31st October 2009, 12:29 AM

Like This????

Boncuk · 31st October 2009, 02:29 AM

Try that one first.

It's much simpler.

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29th October 2009, 09:43 PM	#1
laverdure	low power switching circuit dc Hi Guys, I have a challenge here,,, I have been away from hardware electronics for awhile, now I am in software, I have a old Kenwood car alarm that I wish to use,,,, the siren had 3 leads, one -- ground, 2 -- constant power of 12vdc, and the other 3 -- switching voltage of 0-off, 1.01vdc on…. I somewhat remember doing this in college but how do I make the switching circuit???? Man I feel old …. Thanks….

29th October 2009, 11:56 PM	#2
MikeMl	How about a pull-up resistor to +12V with a normally closed switch to ground. Tie the junction of switch and resistor to the input. Switch closed, alarm off. Switch open, pull-up resistor sources current into the input, pulling it above 1V, alarm makes noise. __________________ Mike ML.

30th October 2009, 12:32 AM	#3
laverdure	schematic Do you have a quick and dirty schematic of that idea??? Thanks Last edited by laverdure; 30th October 2009 at 01:06 AM. Reason: poor spelling

30th October 2009, 03:32 PM	#4
laverdure	What about this? Let me give more background to this challenge. When the alarm is activated there is no voltage at the switching wire, when the alarm is tripped, i.e. open door or another sensor, the voltage at the switching wire is now approximately 1.01vdc ,,, do they make a relay, reed or solid state……anything….. , that will activate at 1vdc ???? Thanks

30th October 2009, 03:41 PM	#5
MikeMl	This is backwards compared to what you said in your first post. All you need to do is to wire a normally-open switch from the "alarm input" to ground. Closing the switch will make the alarm go off. This could be a switch similar to the one that lights the dome-light in your car. Do you need to "reverse" the logic so that the application of a positive voltage turns on the alarm? If so, use an NPN transistor like a 2n3904 or 2n2222. Connect emitter to ground. Connect collector to the "alarm input", connect a 4.7K resistor to the base. Applying 3 to 12V to the other end of the base resistor will turn on the alarm. __________________ Mike ML.

30th October 2009, 09:00 PM	#6
laverdure	Would the 1vdc from the signal wire turn on the "switch"??? Sorry if I didn't clarify myself on the first post Thanks

31st October 2009, 12:29 AM	#8
laverdure	Like This???? Attached Thumbnails

31st October 2009, 02:29 AM	#9
Boncuk	Try that one first. It's much simpler. Attached Thumbnails __________________ Proper Planning Prevents Piss Poor Performance