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simple circuit help pls

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darkshag

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Pls help if you can

I have a secondary output on my car alarm system when activated with the remote i give a (200mA) 12v negative output. (ie volt meter +ve probe conected to 12v +ve and the -ve probe conected to the output on the alarm when activated the volt meter shows 12v)

How ever i made a mistake and have damaged the output by over loading it :)

It now shows 0.44v on the volt meter.

I need help with a simple circuit which can take the 0.44v negative output from the alarm and operate a relay. so that i can continue to use the alarms output

pls help

stewart
 
it is normal to get a negative response when you connect your device to +ve and the output of the last circuit. For a positive response, change the +ve to -ve. in other words, connect + on your voltmeter to the output of the alarm and - on your voltmeter to common ground.

The only way I can see it being a constant value is if it is a DC output. It looks like your circuit does not come with an output coupling capacitor.

Check your voltmeter against a good power source. If you see a weird result, your voltmeter is screwed.

and if your circuit IS an alarm, you might want to connect a capacitor and a computer speaker in series and use that as your voltmeter. If you hear noise, you know your alarm works.

Once you get a true output value, look for a relay that can operate with the voltage.
 
Thanks for your fast reply

The alarm system is an 'off the shelf unit'. And it has two extra outputs.

maybe i got it wrong when i said a 'negtive' output perhaps i should have said 0v output.

ok when i connect the +ve probe of the volt meter to the alarm output and the -ve to common ground the volt meter will show 0.00v on both outputs.

when i connect the -ve probe of the volt meter to the alarm output and the +ve to the a constant 12v sorce (direct from the car battery). i get the following results

output 1 is ok the volt meters shows 12v (well to be exact 13.46v the battery has just been charged) and it will operate a relay.

output 2 is not ok because i conected to the boot lock and it draws more than 200mA (i know i should have used a relay but i measured the current draw wrong when i was testing it) any way the volt meter now shows 0.44v. before i overloaded the output it would show 12v (well what ever the voltage of the battery was at the time).

(if the output of the alarm was a +ve 0.44v i could just use a simple transistor and then a relay to utilise the output)

But because it is 0v output i need help for the circuit.

I hope this helps

stewart
 
darkshag said:
It now shows 0.44v on the volt meter.

Do the measurement again, but with a resistor of 1K to 10K across the voltmeter acting as loading.

If you still get 0.2V or more when the alarm operate, it might be possible to provide a circuit to operate a relay.

If there is no volt across the resistor when the alarm operates, then the output inside the alarm unit has open circuited and it is difficult to use this output signal to operate a relay.
 
darkshag said:
putting a 1k resistor across the volt meter gave me a reading of 0.33v

I suggest the following circuit which use an opamp to sense the small change in original output 2 voltage to operate a MOSFET. You can connect other circuit like the Boot Lock to the new output2.

Connect a meter at 0V and Pin1 output of LM358. Adjust the variable resistor slowly unitl the LM358 output rises to near 12V. Operate alarm and check that the voltage drops to near zero. Keep adjusting the VR until the voltage rises again. The set point for the VR is at the middle of this adjustment range.

By choosing suitable MOSFET, you can drive a load up to several amperes from new output 2.
 

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