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10-07-2013 #1
Seeing as you already have a smooth DC supply, then all you need to reduce the voltage is use 2 resistors to create a "potential divider". Values can be calculated using Ohms Law. Or you can cheat and use this online calculator: Potential Divider Calculator
Not sure where you're getting your 12V supply from, but if it's from the vehicle battery, then beware that it is greater than 12V, particularly if the engine is running as you will then see around 14V+ from the alternator/charging circuit.
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10-07-2013 #2
Cheers birchy, yes the car voltage will be 14.4v so i have used your link to calculate that i need a 620k and 20k Resistor if the voltage is 14.4v and the output was say 0.45v
As for getting my supply voltage there is normally 4 wires to an O2 sensor and two are the heater wires that are the voltage of the car i.e 14.4v when running, i will tap into this supply. looking at the schematic on that Potential Drive Calc will the resistors get hot or blow if i wire them like it suggests and take my feed from between the two?
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10-07-2013 #3
To be honest, I think you'd be better off using a small AA battery or similar for the 0.5V signal. It will certainly be a lot safer as you don't want to be shorting out a battery with 300A+ peak current!
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10-07-2013 #4
On further thinking...I don't quite understand what you are trying to do here? If you have a suspect O2 sensor (i.e. the output signal voltage is not within tolerance), then it's faulty. If it's not faulty, then you can use the (0.5V?) signal from it to test the ECU or whatever.
Might be worth making yourself a little variable power supply. Basically all you'd need is a 9V PP3 battery and a potentiometer.
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