Date:         Tue, 3 Dec 2013 18:54:03 -0800
Reply-To:     Brett Ne <brettn777@GMAIL.COM>
Sender:       Vanagon Mailing List <vanagon@gerry.vanagon.com>
From:         Brett Ne <brettn777@GMAIL.COM>
Subject:      Re: Arduino and Vanagons
In-Reply-To:  <BLU177-W402E201EC4FF0F52AFD9C5E0D40@phx.gbl>
Content-Type: text/plain; charset=ISO-8859-1

On Tue, Dec 3, 2013 at 4:41 PM, James <jk_eaton@hotmail.com> wrote:

> I'm not surprised that your homemade voltage divider grounded out the Hall
> Effect sensor signal - typical sensor signal is under 10 milliamps at 5
> volts (though maybe VW used 12), and feeds an op-amp (think, very high
> input input impedance, maybe 1 Mohm).  Putting 13.3 kohm in parallel with 1
> Mohm diverted nearly all the signal away from the ECU.  The existing ISV
> control box probably uses an op-amp as an input stage, and our replacement
> will need to as well.  (The actual Hall Effect signal is around 30
> microvolts, typically, and has already been amplified inside the Hall
> Effect sensor's circuitry up to a 'useful' level.)
>

I figured that the Hall sensor assembly had some preconditioning of the
output signal; nice square pulses don't naturally result from rotating
electronics.  If the voltage on the signal turns out to be 5v or less, we
may not need an op-amp at all since the input pin on an Arduino is very
high impedance.


>
> The 10 kohm resistor should have been more than enough to protect the
> Arduino's inputs from harm.  (Remember, resistors are current limiting
> devices, and it's too high a current that damages a microcontroller, not
> just the voltage.)
>

Sometimes it's the voltage that causes harm and sometimes it's the current
induced by the voltage(heat dissipation problems) that causes the harm.  I
was concerned about the voltage rather than current because in input mode,
the microcontroller pin only draws about 1 microAmpere of current.


>
> It's fascinating to read your progress reports.  I predict that your
> oscilloscope check on the output of the Hall sensor will produce a square
> wave of about 5V, with an on period of about 50 microseconds, in a 2
> millisecond period.  (Duty cycle of about 3% on a 400-450 Hz square wave
> signal.)
>
>
Uhmmm...wait a minute...a 2 millisecond period means 500 pulses per
second...with two plugs firing with each crank rotation, thats 250
rotations per second...which translates into 60 x 250 = 15,000 rpm...which
means...that you've installed a Formula 1 engine into your Vanagon...Cool!!


--
Brett in Portland, OR
"Albert" '82 VanaFox I4 Riviera