I want to add auto unlock to my ford fiesta(UK) which has positive Door Locks.
I have looked at the Door locks section of the 12Volt install bay.
Do i connect it up according to the 'Constant to Momentary Output' relay diagram?
i.e. Relay Pin: 86 - Switched 12V+
Pin: 87 - Fused 12V+
Pin: 85 - Ground via a 1K uf capacitor & a 10K fixed resistor
Pin 30 - momentary 12V(+) to door Unlock wire
The Diagram in the Install bay also shows a connection between pins 86 & 85 with a diode pointing towards pin 86
Why Do I need to make this connection?
Also should I not add a diode on the switched 12V(+) wire with the diode pionting towards pin 86, hence preventing 12volts from flowing back?
Any help appreciated
regards.
Sungura
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(;-)
The diode across the coil is normally put there to supress the big voltage spike that occurs when the relay switches off. Without it, if you have some electronics like a transistor controlling the relay, it could possibly cause damage by exceeding the maximum voltage.
In this particular application, it prevents the relay from briefly switching on again if your 12V control signal happens to switch back to ground rather than an open circuit. Without the diode to bypass the coil, it might energize the coil in the reverse direction for an instant until the cap discharges. If your control signal just opens up when it switches off, you don't need the diode at all.
You might want to be careful using this method depending on what you're planning on using for the switched line, because of the current that gets quickly discharged back through the diode from the rather large cap. Though this would work, I'd use a transistor for the timing control so you can use a much smaller cap for timing; 1000uF isn't as cheap or easy to get as a 1uF, and with a transistor driver, the timing won't be dependent on the particular relay you use.