Simple Two-Transistor Motorcycle Alarm


Posted on Feb 5, 2014

When one of the switches is closed - the base of Q1 is connected to ground through D1 & R2. This switches Q1 on - and it in turn switches Q2 on. Q2 connects the positive side of the relay coil to the supply line. The relay energizes - and the siren sounds. = If R1 had a high value - moisture on the switches might be enough to lower the voltage - a


Simple Two-Transistor Motorcycle Alarm
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t the junction of R1 & R2 - to the point where Q1 would switch on. Using a 1k resistor for R1 eliminates this problem. = R2 limits the current through the emitter-base junctions of Q1 and Q2. It also controls the speed at which the capacitor charges. C1 is part of the latching circuit that keeps the relay energized after the trigger-switch has been re-opened. I wanted the relay to latch quickly - so that even a brief closing of the trigger-switch would activate the alarm. That`s why R2 is a 1k resistor. It charges C1 very quickly. = When the trigger-switch is re-opened - it`s the charge stored in C1 that keeps the transistors switched on. So the relay remains energized and the siren continues to sound. It will go on sounding until the charge in C1 falls to a level where it can no longer keep the transistors switched on. At this point - the relay will drop out and the siren will stop. = How long this takes to happen depends on the size of the capacitor and the value of R3. I selected both - by trial and error - to give a delay of about a minute. There is no point in trying to calculate values. The precise time it takes for the siren to stop depends on the characteristics of the actual components used. If you want to increase the time it takes for the siren to stop - use a larger value capacitor. If you want to decrease the time it takes for the siren to stop - reduce the value of R3. = Q1 & Q2 are wired together to form a...




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