Thread: Timer schematic help

The 12V is not coming from the PSU. It can come from the battery positive terminal or an empty appropriately-rated fuse slot that provides power all the time. The "Vacc" is coming from an ignition-switched power source, such as an empty fuse slot that is powered on/off by the ignition key state. The variable resistor is showing you can use varying resistors to adjust the brightness of the LED.

One thing that doesn't look right is how he's got the main 12V lead going to both the relay and the coil. Also, the relay's not grounded, which I believe relays usually are.

Originally posted by Dudah:
<STRONG>http://www.mp3car.com/usersites/hornet/poweroff.gif For this schematic I have a couple questions. Is the 12 V from the power supply or car battery? On the LED, what is the point of the pot? Do you need the DC-DC converter for this to work? If so, will this automatically shut off the computer or just shut it down? Thanks.

-Kev
by the way, I just got my computer for my car up and working! </STRONG>

This circuit looks like it will turn on the computer as soon as it gets that ignition signal. After it is on it will stay on as long as the ignition is on. When you turn the car off, it will take about 200 seconds for the relay to turn the car back off because the 40 uf cap keeping it on takes that long to discharge through the 5 meg ohm resistor. If you turn your ignition off, it looks like you can wait less than 200 seconds if you hit the power off button.

It's ok that the relay is not connected to ground. That just depends on the the circuit that provides the current to drive the relay. In this one, the transistor is providing the path to ground. If you used a P channel mosfet, you could have grounded one terminal of the coil of the relay, but the rest of the circuit would have to be different. The 2N7000 also has diode protection to protect against coil flyback voltages when the relay disengages.

Jeff_

yep it does work i get about 8 minutes of power on, after the key is removed, as i think i used a larger resistor, but it still works.

(Pops head out) Of course it works, I was the one who designed it Err, Jeff and hornet, you answered all the questions for me anyway, so I've got nothing more to say. (pops head back in)

Hey, Oms
Do you have detailed description of all electronic components in this circuit? (Name and type of diode; resistors-watt rating, variable resistor-type? relay type and amps rating for contacts; capacitor type...).
I looked all over the place and couldn't find it.

Thanks man

Sure, there's a detailed description on my original page:
http://www.astron.nl/~smirnov/poweroff.html
hornet - how about you provide a link from your pages, so people can get the details firsthand? Also, there's a typo on that diagram , so it really belongs together with my description.

sorry dude, shoulda thought of that.
Fixed now. What happens when u work on webpages at 2:30am, the only time i get spare time.

forgive me father for i have sinned...

Oms, what I don't understand is why you have the transistor going to the coil and then the +12V before the relay. Wouldn't you want what I call the "trip contact" in the relay hooked to the transistor so when the transistor gets energized it "trips" the relay on and thereby closing the circuit of the main +12V to provide power to the psu?

No, in fact the transistor controls the ground circuit. There's always +12V on one side of the coil, but normally the other side is not grounded, so no current goes through, so the relay stays closed. When you open ("energize") the transistor, it closes the ground circuit on the other side of the coil, allowing current to go through the coil, so the relay opens.

Trust me, it works. Just remember, I got "S" and "D" mixed up on the diagram. Check my link above for details.