headlight alarm circuit

A useful thing. I'm glad my car already has it.

To limit the amount of current flowing through the circuit.

Yes.

Yes.

No.

Note that 12V is nominal. Probably closer to 14V when the engine is running.

Here's why everything is where it is. When the car is running the IGN AUX line is at 12V. When the lights are on the lights are at 12V too. Since both terminals of the buzzer are at 12V it doesn't buzz.

Now the resistor is required because the IGN AUX line is connected to the buzzer where ground is also connected. Without the resistor there would be a short between the IGN AUX line and GND blowing the fuse for the IGN AUX line rendering it useless.

Now you turn the car off. The IGN AUX line doesn't go to ground. It's simply disconnected from the battery and is floating. So in your arrangement the lights are at 12V and the IGN AUX line is connected to nothing. No buzzer.

In the circuit as specified when the IGN AUX line is disconnected then the other side of the buzzer is now connected to ground. If the lights are on then the buzzer has 12V on the lights side and GND on the other. So it starts buzzing.

BAJ

This is a common simple circuit to activate the buzzer when the lights are on. The diodes provide isolation. If the lights are on the buzzer will be activated from the lights plus through D1 and the resister. If the lights are on AND the ign is on the buzzer minus will be raised to plus and the buzzer will not have enough difference to buzz. If you tie the minus of the buzzer to the ign, the buzzer will not have a path when the lights are on. You don't want the ign plus tied to ground when the ign is on. Regards, Tom

The circuit assumes the the signal from the Ign-Aux is +12v without the 1k resistor there would be a short from +12V to Ground

The 2 diodes prevent the +12V from the Lights and from IgnAux from feeding each other thru the buzzer.

The circuit works as follows:

When the IgnAux is on there is 12V applied across the resistor

If the lights are turned on the buzzer doesn't sound because both sides of the buzzer are at the same volts

If you turn off the IgnAux current flows from the lights thru the buzzer and the 1k resistor causing the buzzer to sound

Dan

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Dan Hollands

1120 S Creek Dr Webster NY 14580 585-872-2606 snipped-for-privacy@USSailing.net

Found this circuit on the internet.. It is intended to be a headlight alarm (warning) to prevent one from leaving lights on by mistake.

My question is, what is the resistor for ? do you need it and do you need the diode from the ign aux line? Can't we wire the negative from the buzzer straight to ign circuit minus the resistor to ground and diode?

Help me undrstand the purpose of the added parts.

Thanks

CM

It may not work for some cars. Some car will ground the IGN AUX and other car it may be simply disconnected (floating). A quick way to check this is to turn the key to accessory or on and then use a meter to check for continuity from IGN AUX to ground. If you get zero, the above schematic might work. If you get infinity, you'll need to add another diode, 1k resistor, and ground connection as mentioned in the link in your first post.

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Ok thanks to the responses below. Reason I asked original question was that I have seen similar schematics wired as follows:

• headlights--->>>>| (diode)------+Buzzer-------IGN AUX

This was described as headlight power and ign aux 12 V (no buzzer)

Ign off, headlights on 12V will find ground path through ign aux circuit and buzzer will sound.

Does the above also make sense then? I realize there are many ways to accomlish the same thing.

CM