Author Topic: CarPC project input circuit help.  (Read 788 times)

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Offline TbirdManTopic starter

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CarPC project input circuit help.
« on: August 18, 2018, 11:59:31 pm »
Hello, I am currently working on a project to replace the NAV headunit in my 2006 Legacy GT with an SBC and 7" touch screen. I would like to utilize some existing signals coming into the headunit plug (like park, ignition, reverse etc) and helpfully I found a service manual that has a pretty complete circuit diagram of the OEM headunit. I will need to shift the level of the signals from 12v to 3.3 and was looking into how best to do that and protect the SBC from transients and whatnot and decided to look into how the Pioneer engineers did it. what I found was variations on the circuit attached. I I was hoping I could get some help deciphering it and if it is worthwhile reproducing it on my project. the components are as follows:
Resistors/caps: obvious
D800: MA3220-H - Zener Diode; Silicon planar type; 22v
D807: DAN202U - Ultra High Speed Switching Diode
Q800: DTA124EUA - digital transistor

Without sounding too much like a novice, I assume the caps are for noise, the resistor is for current limiting and the zener is for voltage spikes? I am not really clear on the last one and am still working out the switching diode. Thanks for any help.
 

Offline JS

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Re: CarPC project input circuit help.
« Reply #1 on: August 19, 2018, 12:32:42 am »
To shift 12V input to 3.3V output just use a 3.3V zener (or 3.1) and a resistor. With that the protection is also there already, I'd suggest a 3.3V zener with a resistor that just makes it sit at 3.2V or 3.1V so you have some margin for the spikes. Then a resistor from there to the input pin would limit the current in case the voltage goes over by a little bit.

The bullet proof protection for high reliability could be optocouplers, but they are not required and for fast signals they can be a problem.

If you want you can use a transistor to shift the level as you suggested, but I'm not sure your circuit is optimal, it will invert the signal, is that what you want?

JS

If I don't know how it works, I prefer not to turn it on.
 
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Offline TbirdManTopic starter

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Re: CarPC project input circuit help.
« Reply #2 on: August 19, 2018, 12:58:43 am »
Thanks for the reply, I am curious where in the circuit the signal is inverted? I have traced it out on the schematic and the main microprocessor is definitely looking for an active high signal.
 

Offline floobydust

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Re: CarPC project input circuit help.
« Reply #3 on: August 19, 2018, 12:59:55 am »
Your Subaru NAV circuit uses many parts for a few good reasons.

It will never draw current from a 12V signal, which is an issue with battery-sourced (not ignition) signals.
Like brake lights and interiour lights. Some vehicles have +12V battery on the (closed) door switch line, so monitoring that would be a drain on the car battery all the time. The interiour light is a pullup resistor to +12V battery, on (-) doorswitch cars.

There is an automotive reverse-battery requirement, so a reverse-installed battery does not damage the circuit. This can lead to some oddball circuits with extra diodes.
Two capacitors are used instead of one to meet the ESD high voltage requirement.

Pic is a Kenwood NAV unit inputs.
IGN/ACC has an extra 6.2V zener D150 so it needs over 7V to activate.
PARK is your more typical digital input, C140 is big and filters PWM dashboard dimming pulses.
Both will drain a battery-sourced signal.
 

Offline JS

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Re: CarPC project input circuit help.
« Reply #4 on: August 19, 2018, 01:07:38 am »
Thanks for the reply, I am curious where in the circuit the signal is inverted? I have traced it out on the schematic and the main microprocessor is definitely looking for an active high signal.
Q800 is inverting the signal so check if you can use it like that or you need to add something else

JS

If I don't know how it works, I prefer not to turn it on.
 

Offline TbirdManTopic starter

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Re: CarPC project input circuit help.
« Reply #5 on: August 19, 2018, 01:28:51 am »
Thanks for the reply, I am curious where in the circuit the signal is inverted? I have traced it out on the schematic and the main microprocessor is definitely looking for an active high signal.
Q800 is inverting the signal so check if you can use it like that or you need to add something else

JS
Hmm, I wonder if that particular signal is active low. I will need to do some probing on the signals when I get a chance.
 


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