Simon,
Many thanks for the detailed and informative answer - there's a load more to think about now

- I never thought it would be easy though!
Please bear in mind I have no experience of electronics and all of what follows is from what I have read on the internet this morning

Oh and I have no
experience of lambdas either, is it really possible to manually tune the mixture so that it is balanced between it's two extreme outputs?
QUOTE (simonr @ Aug 2 2008, 11:12 AM)

Your circuit would undoubtedly work with appropriate resistor values - although a single indication of rich/lean isn't going to be all that helpful when tuning. You need something to tell you when the output is in the middle. Perhaps you had intended to build three?
No... Only one: The lambda sensor that I have is a 'narrow band' Zirconia device which are (apparently) extremely difficult to 'balance'. Apparently (again) ECUs normally just repeatedly apply the logic: If it's rich - lean it off, if it's lean - richen it. (I guess it is implied that "if it's balanced, look again and it won't be"

) So I thought I can have the output from the circuit controlling a two way relay so that either one of two LEDs would light up and I could manually follow the same logic as above until it's flicking close enough.
QUOTE (simonr @ Aug 2 2008, 11:12 AM)

I would be more inclined to buy comparators than op-amps as they are more or less pre-packaged as the circuit you want.
The LM339 is a reasonable choice. They give a logic high if one output is at a higher voltage than the other and low if not. Use a preset potentiometer connected between your power rails (or better still a voltage regulator) with the wiper connected to one comparator input and the lambda sensor to the other.
The LM339 has 4 comparators on board which you could set up to give you LED's for rich, lean and OK. The output will drive LED's directly through a 330R resistor.
Right - I am with that. (BTW, aren't these things cheap! )
So I would preset the pot (note the use of jargon already

) to give the switchover voltage required (.5v as you said) and the output of that comparator would go on/off as the lambda voltage transited the reference voltage. And also if I then set up another comparator in the LM339 wired to the lambda and the pot-wiper, but the other way round to the first, then the output of that one would be the inverse of the first and I have what I wanted - with no need of a relay or assembling all of those other bits -
GREAT
Oh - one bit I don't get is "drive LED's directly through a 330R resistor." do you mean this:
CODE
Comp-Out---/\/\/\---(LED)---gnd
R1
330ohm
EDIT: Sorted this now - I found a really useful little
LED Resistor calculator and explanation.
QUOTE (simonr @ Aug 2 2008, 11:12 AM)

Alternatively just use a digital volt meter and adjust the engine for 0.5v reading.
That was my first thought - but then why keep it simple when I can complicate things and learn something along the way !)
QUOTE (simonr @ Aug 2 2008, 11:12 AM)

Is your sensor a 3 or 4 wire type with a built in heater? If not, you will need to heat it up for it to work reasonably. Just poking it up the exhaust pipe is unlikely to get it hot enough!
It's a four wire and if I build the circuit then I will mount it in the correct area - near the header - and have the LEDs where they can be seen as a bit of bling. Initially though, I was intending just wiring the lambda to my Fluke and sticking it up me pipe... I was wondering if (even with the heater) it would work there, but it wouldn't cost anything to try.
QUOTE (simonr @ Aug 2 2008, 11:12 AM)

To partially answer your actual question:
R1 & VR1 form a voltage divider and deliver a variable voltage to the - input of the amp. This is a voltage reference which the amplifier uses as it's zero point. It amplifies any difference from that voltage.
R2,3 & 4 set the gain of the amp (the 'Volume').
The output of an amp is actually a current which is proportional to the input. R3 & R4 turn that current into a measurable voltage. R4 is to insulate the amplifier from whatever you connect to it. Otherwise, the resistance of what you connect to the output will affect the gain.
Si
Brilliant! Thanks for that. Simple when you know eh?
I think (since rain has stopped any other sort of play) I will shoot over to Maplins now and get the bits. Ill let you know how I get on.
Rog