O2 voltz

G. Scott Ponton gscottp at ix.netcom.com
Mon May 17 06:50:05 GMT 1999


Ok guys I have sat back and waited for awhile for you all to get this
straight. The sensor "compares the amount of oxygen in the exhaust stream
with the amount in the outside air.  The reason it doesn't seem to work this
way is two or three fold.

1. The sensor needs to be at a "reasonably" stable temp before its output is
"reasonably" stable.

2.The sensor only works in a very narrow A/F ratio. As the amount of oxygen
climbs above a certain point it can't drop the voltage any lower and the
oppisite is just as true. The output voltage is also limited by the design
of the sensor. After a point removing more oxygen can't effect a voltage
change.

Much has been said and made of the "wide band" sensors too. After reading
much on the "wide band" sensor it works exactly the same way as a "normal"
sensor. The difference is through outside influences, either a standard
sensor or electronics atached to the output, are used to create a "current
pump" when the sensor cross stoich the "outside" control device switches the
current in the sensor so that it breaks oxygen atoms away from the oxidized
gases in the exhaust. By know at what rate this happens chemically we can
use the output of the sensor to "roughly" ( better than what we started
with) determine the amount of oxidized fuel (and other such nasties as NOx)
in the exhaust stream.

Notice the word normal used to desribe combustion as we currently use it on
a daily basis. This doesn't include off the wall or experimental engines
which use some sort of modifided combustion cycle. Why not you ask? This is
going to get very long if I have to get into everything that can effect
this. Simply put:

Within the range of "normal" combustion, a little either side of stoich, the
leaner the mixture the greater the amount of free oxygen in the exhaust
stream as there isn't enough fuel to combine with all the oxygen in the
cylinder. As the mixture is enrichened the amount of free oxygen drops for
exactly the opposite reason.

If you go to the point of misfire all bets are off. Although a lean cylinder
will tend to dump more free oxygen into the exhaust than a rich one as until
the mixture is extremely rich some part of the air charge ingested will be
used as the fuel attemps to oxidize ( burn) whereas in a extremely lean
cylinder next to nothing oxidizes as our current ignition system cannot
produce the energy needed to properly promote combustion of these mixtures.

   Testing a sensor with argon and other similar gases doesn't work very
well unless they are in an enclosure with only the tip of the sensor exposed
to the "inert" gas. I won't try to get into ther chemistry involved at this
moment as I will have to research the chemistry of inert gases before I can
explain it in those terms. At this point my chemistry is rusty to say the
least. At least that part of it which doesn't deal with the above as I use
this nearly every day.

OK let the flames begin!! LOL :-)

Scott




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