Cannara wrote:
Frank, please review your electrical theory. This is wrong:
“The paralell and series coupling was just to illustrate that the
internal resistance of a source/battery doesn’t influence the current
flow computations of a circuit. And the loop idea of the current flow
going round and round conveys a perpetuum mobile - but the battery is
being discharged by the work done in the circuit…”
What’s wrong about a battery being discharged when it powers a circuit,
Alex - ‘the battery’ refers to latter section of the ‘source/battery’…
The battery isn’t discharged by the O2 sensor, because its voltage,
generated as you say, drives a small current (about 0.8V divided by
[4000 + ECU input resistance Ohms]) into the ECU. That current varies
with how many working O2 sensors are in parallel, since each is about
4000 Ohms. So, two in parallel look like 0.8V battery with a 2000 Ohm
internal resistance; 3 in parallel look like a 0.8V battery with
4000/3 Ohms resistance, etc. Since the ECU has a finite input
resistance, it will add to whatever the combined source resistance of
the N O2 sensors is (4000/N) and so as N increases, the current driven
into the ECU’s circuit increases, necessarily. That may not be
something the ECU knows how to deal with correctly.
It’s somewhat uncommon to state that batteries in paralell increases
current flow in a circuit…
I think I have made it abundantly clear that batteries are not involved
in the O2 sensors current production…Any source/battery responds to
excessive current drain by a drop in voltage. In paralell batteries can
tolerate a higher drain before this happens, and in any case the O2
sensor maintains 0.8V untill oxygen shows up in the exhaust. Indicating
that the senor is not overtaxed and delivers all the current the ECU
requires. And two in series would make no difference…
Also, there is no perpetual motion – the 0.8V drives a current
determined by the full loop resistance from sensor wire through ECU
and back through chassis to the sensor’s case. That’s also why we
make sure the case is grounded to the body. The energy to do this
comes from the exhaust heat affecting the sensor element.
The ‘perpetuum mobile’ was to illustrate that the current returning to a
battery doesn’t recharge the circuit…
This is also untrue: “The O2 circuit is entirelyy different from,
say, the CTS where current is passed from the ECU through the sensor
and back to the ECU”.
The only difference is in the nature of the sensor – the CTS is
passive and is simply a temperature-sensitive resistor, so a source of
current is supplied by the ECU so the loop current or sensor voltage
can be used by the ECU’s decision circuitry for fuelling. The O2
sensor simply supplies the voltage to drive its loop through the ECU.
If a circuit with passive resistance sensor doesn’t differ from a
circuit with an active one I wonder what does. If you connect two CTSs
in paralell you most certainly halve the resistance - influencing curent
or voltage drop, whichever the ECU uses. While I doubt if the 02
sensor’s internal resistance is 4000 ohms, or whatever, when hot, I
cannot see it has has any bearing on the current flow in a paralell
set-up…
To the true dual exhaust with two sensors showing different voltages; I
think the ECU would go on leaning out as long as one sensor shows 0,8V,
but admittedly it could easily be that it it would fatten the mixture
until both show 0,8. Or the ‘high’ sinsor might go to ground through
the ‘low’ one - which would likely confuse the ECU…
I’ll keep going with this as long as necessary!
:]
Or someone take pity and ask us to go suck a lemon, Alex…:-))
Frank
xj6 85 Sov Europe (UK/NZ)===================================================
The archives and FAQ will answer many queries on the XJ series…
FAQs: http://www.jag-lovers.org/xjlovers/xjfaq/index.html
Archives: Jag-lovers Forums - Jag-lovers
To remove yourself from this list, go to Jag-lovers Forums - Jag-lovers.
// please trim quoted text to context only
