Pull up or pull down?

[Imagewerx]

This will be basic electronics to most people on here,but it's something that always confuses me.

I've got a timer that's triggered by the central locking lock pulse on my car.The central locking is just a 12 volt motor that spins one way to lock the car and the other way to unlock it.Both the wires rest at ground while one goes to +12 volts to lock and the other one goes +12 volts to unlock.My timer is triggered ok by the 12 volt lock pulse (about 1 second),but when I unlock there's a corresponding 1 second pulse on the lock wire of about 0.4-0.5 volts,which is enough to trigger the timer again.

Can I use simple pull up or pull down resistors to prevent it from triggering when I unlock,or is there more to it that that? Yes I know I could do it easily enough with a relay,but I really want to keep this fully solid state.

 

[Mike odom]

A pull down might work, depending on how much energy is in the phantom pulse. You could use a series diode to drop .7v, so the signal would need to get above .7v to trigger the timer.

 

[Imagewerx]

Thanks for the answer.I tried a series diode and it didn't make any difference.How would I go about working out what value pull down resistor I'd need to use?

 

[KeepItSimpleStupid]

What does your input circuit look like? It might be possible to clamp it. It's possible to change to a non-retrigerable timer too.

 

[Imagewerx]

It looks like this.....


I'm not using R2 though as it would only trigger with a -ve pulse with it still in circuit.

 

[Mike odom]

try a 10k TR1 base to emitter (ground).

 

[KeepItSimpleStupid]

You can also try a number of diodes in series with trigger in. Each diode will give you about 0.6 V as a voltage drop. 4 would give you about 2.4 V. You can replace with a Zener diode if it works.

You can also try another transistor arranged like TR2 and R7 that pulls the VE trigger low when the pulse is occuring.

The 555 timer has a general nasty habit of triggering when you don't want it too, like on power up.

 

[rumpfy]

Maybe your problem is solved now, but for the record, you say R2 is not used.
If so, the base of TR1 is floating.
Noise immunity is always an obscure problem and the first rule is that the base emitter circuit needs to be maintained at a low impedance level. Also, the triggering level of your timer needs to be set to require a certain voltage level to trigger it.
This is why logic circuits have guaranteed levels of 'maximum voltage for a 'low'; and minimum voltage for a 'high'.
Post 6 is a good way of setting the impedance to a lower value than you currently have. As well, you can use a shunt capacitance to lower the AC impedance of the base circuit.
The way TR1 is set up, the input voltage required to trigger is only about 0.5 volt. Also, I believe the odd half volt pulse is generated from capacitive coupling between the lock and the unlock wires in the cable harness.
To set the triggering level for the timer at say 6 volt, then use a base to emitter shunt resistor of 4K7 ohm. Use say 1 Kohm if you want to raise the triggering point even higher.
From what you present, R2 is redundant anyway. Maybe even wire up TR1 like you have the base of TR3 wired.

 

[Dick Cappels]

Keep in mind that the signals your circuit is looking at are probably very dirty and probably include "flyback" pulse from the motor turning off. Maybe low-passing the signal will help.

 

[Imagewerx]

Ooooopppssss sorry for the late reply,didn't get an email to say there'd been any more responses to this topic.

Re the post above.I thought that as well,so one of the first things I tried was a TVS,which didn't work.

Tried the first suggested 10K which didn't work,neither did the 4K7.The 1K how ever did work and now it only starts the timer when it locks.This is wired up to extend the lock pulse to give me total close when I lock the car.So now not only does it lock the doors,it also winds the windows up.

 

[rumpfy]

With your original, when one looks in to the base circuit, there is effectively an open circuit (high impedance) because there is no current flowing in to the base.
Using a TVS would not solve the problem because it too, looks like an open circuit at voltages below the conduction threshold.
In your post #1 you say you measured the voltage on the base at 0.4 to 0.5 volt. I believe what is happening, is that there is enough current coupled to the feed wire to raise the base voltage/current to the transistor conduction level. So the voltage you read was the interference energy coupled to the base. With the base disconnected, you would see a much higher voltage on the wire.
The nature of the coupling is capacitive, and a shunt capacitor to ground would be effective too, but it is always better to make the circuit as stable as possible and the resistor will do that. For some circumstances one wants to raise the impedance level as much as possible, but in this case, there is plenty of battery energy available; so the low value resistor is a good choice.
Hope this helps.

 

[Imagewerx]

Ok thanks for the explanation rumpfy,it kinda makes sense to me.