My question is pretty simple. we all know a wideband senses oxygen and outputs an air/fuel ratio. So ambient air is 20.9%oxygen, now lets say we burn a little bit of fuel in that air and get 15% oxygen. With a simple formula, we can derive A/F ratio. My understanding of a wideband becomes blurry when we approach and pass stoich (14.7AFR). Theoretically, at 14.7 AFR we have a perfectly balanced chemical equation, right? All of the hydrocarbons and o2 molecules will turn into h2o and co2. How can the wideband read anything richer than 14.7, since there should be no oxygen in the exhaust? Is it making an assumtion about burn efficency, obviously there will be some o2 and hydrocarbon molecules that didn't combust, but this would be a major flaw in the accuracy of a wideband if it was aproximating this.

Thanks in advance for your replys.

 

Does this help?

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by EE_Chris &raquo;</TD></TR><TR><TD CLASS="quote">Does this help? </TD></TR></TABLE>

very good article. it really didn't answer my question as to how the sensor tells a difference between rich and richer conditions though.

 

"The atmosphere's oxygen content is approximately 20% while the oxygen content
of the exhaust gases are more like 1-2 %. The oxygen ions travel from the higher
concentration in the atmosphere to the lower concentration in the exhaust. This
flow of oxygen ions through the zirconia produces a voltage that is fed to the
ECM. The greater the difference in concentration between the atmosphere and the
exhaust the greater the oxygen ion flow and the greater the voltage. For instance
when the exhaust goes lean, the exhaust has a higher concentration of oxygen, the
flow of oxygen ions is less, and the voltage produced is less. When the exhaust
goes rich, the exhaust has a lower concentration of oxygen, the flow of oxygen
ions is more, and the voltage produced is higher."

This is for a regular narrowband O2 - but the principle still applies as described in section 18.4 and 18.4.1.

Maybe I'm misunderstanding your Q....but that article gives a pretty clear explanation of how the WBO2 works....it even gives you an explanation of exactly what happens in particular with a rich condition (18.4.1).

 

18.4.1 does describe how the o2 sensor reads rich conditions the reduced oxygen content.

But my thinking is that in a perfect combustion chamber, there will be no oxygen in the exhaust whether it is at 9:1 AFR, or 11:1 AFR, there still should be no oxygen in the exhaust. how can a wideband differentiate between 9:1 and 11:1 afr?