pls help - light on when armed

I regret that I must ask a question, but do not know where else to turn.

I need to build a circuit to blink a light when power is cut. This is for a motorcycle alarm that doesn't have a visual indicator of the alarmed status.

Using the multimeter, I have only been able to locate a +12v lead that is on when the alarm is off, and vice versa (I'll call this the alarm voltage lead). This is used by the alarm's perimeter sensor.

At first, I thought I could connect the positive lead of the LED to the battery's +12v and the negative lead to the alarm's voltage lead. When the +12v in the alarm lead drops to 0 (when armed), current flows and voila. But this doesn't work - it triggers the perimeter sensor and the alarm fires.

So I built an inverter circuit with two NPN transistors. It worked - but just learned that even when the alarm is off (no light blinking), it still drew current and drained the battery after only 2 weeks of inactivity. (The battery has a 10A capacity).

Here's the schematic that I used: **broken link removed**

Can someone offer up some help so I can find a solution to my problem?

Thanks,

lang

Use a inverter (NOT gate) from a 4069 IC to power the LED or to drive a one shot made with a 555 (Cmos version). The main consumer is the led in this case.