Re: SI reset tool information - here's the goods.

[Aaron Lung <alung@xxxxxxxxx>]

bohnen@domain.elided (Aaron Bohnen) writes:
>
> 2. a directional jumper lead with both a diode and a resistor. I'd sug>> My personal version of this tool uses a 600+ Volt [3] Amp high speed
>> fast recovery Motorola part . . . [and] a resistor in series with this
>> diode. I suggest a 470 Ohm resistor which will limit the potential
>> current flow from the reset pin to any other pin of under 11 mA in
>> case of a misconnection of the tool.
>
>While a desire for more durability is laudable when designing other
>things - bridges, for example :) - in this instance, a 600V 3A diode
>is probably overkill.  Much of the car's electronics is already diode
>protected to at least 10 ma.  In 12v circuits, a 470 ohm resistor might
>allow too much current.
>
>If you were to omit the diode and use a 1.5K resistor, your
>circuit would be simpler and might protect just as well against
>misconnections.  You might consider a higher resistance value in light
>of some of the ISO/TC 22 standards, but I would not bother, since you can
>only use the tool when the car is quiescent, electrically speaking.
>But perhaps an EE will comment.  My PhD is in geophysics.


OK, I'm an EE here :-)

 
I've been using a $0.01 1/4W 100ohm resistor as my SI
jumper for 10yrs, and all works honky dory for my 'oil changes' 
and "inspection I's and II's".  If and when it 
breaks, I just get another one from my junk box.  470ohms 
should work fine also, but that's about as high as I'd go.

Without seeing actual motronics schematics (I don't think
any normal person like us has), I can reasonably surmise the pin 
that's being pulled to ground is an input to some TTL/CMOS gate 
input with a light pullup to 12V.  It will never be a direct 
connection to 12V.  The amount of current you'd typically need 
to sink will be in the low single-digit milliamps, and never
anything that would fry anything assuming Bosch 
does what normal design engineers would do for this 
application.  [I guess I could actually go and measure
the current to be more credible]

Therefore, all you need to do is short it to ground or
through a low ohms resistor such as 100ohms so you don't
arc weld anything or blow out an output driver if you shorted 
the wrong pins.

one could argue using a higher wattage resistor so you
don't fry the thing if accidentally tied across power
pins.  in that case, a $0.02 1/2W, 470ohm unit will be perfectly 
fine.  But let's face it, most of us aren't going to blindy poke
the thing in any hole that fits :-)

A diode/resistor combination might not always work reliably.
It certainly won't hurt anything.
The diode/resistor pair or high ohms resistor can have a 
forward drop high enough to not be seen as a low
by the motronics.   IMHO, I see no need for a diode.

With regards to Peake Tools, etc, those are most likely
nothing more than a fancy one-shot timer-driven open-collector
transistor across the 2 pins no matter what they say is inside.  
Not worth $50 in my opinion.  Most of the cost is probably 
associated with the nice plastic housing with the connector 
pins.

aaron
88 528e

      ||        ||                         Aaron Lung
      ||        ||                         Hardware Design Engineer 
     ||||      ||||                        RWBU Engineering 7500
 ..:||||||:..:||||||:..                    alung@domain.elided
cisco Systems                              408-526-6043 (W)
170 West Tasman Drive                      408-526-7064 (F)
San Jose, California 95134

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