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Can lamda sensor 4.9 be used without the special LSU-IC?

Engineering Asked on July 16, 2021

I was looking into using Lambda Sensor LSU 4.9 to know the lambda value. From the datasheet, its looks though this sensors cannot be used without a special IC such as Bosch lambdatronic LT4. Can we get any information out of the sensor by connecting only 12v to the heater+ and heater-. What would be the minimum circuit to get some information from this sensors?

One Answer

No you can't. The LSU4.9 is a wideband sensor, and so differs from the old school narrow-band zirconia sensors that just produced a voltage. The wideband incorporate an oxygen pump along with a zirconia element to make it a wideband. That pump must be controlled for proper working, as well as the temperature which must be kept within limits. You can't just put 12V on the heater, it will destroy the sensor. The heater has to be throttled. And you can't just only use the zirconia element within the sensor to use it as a narrowband.

Driving the 4.9 requires a quite sophisticated construction. The LT4 uses the CJ125 IC, and even without the LT4, you will need that IC. The CJ125 IC is specially designed and made by Bosch to allow users to use the LSU sensors. And it is quite adequate for that purpose.

Some manufacturers succeeded in making their own circuitry without the CJ125, some with more success than others. But unless you're capable of producing your own IC's, you'll have to use the CJ125. And even then correctly using it is certainly not easy. A lot of peripheral hardware is needed to use the CJ125. At that point, you could best just use the LT4 or a similar tool. Innovate and AEM are manufacturers that also offer tools to read out the 4.9. Both use the CJ125 IIRC.

14Point7 offers a cheap PCB(~$40) that closed-loop heats and reads the sensor, and gives you the lambdavalue and temperature in either an analog voltage or via I2C communication. I have used it and it works fine. I think it's the closest you'll come to successfully reading the 4.9 yourself.

Correct answer by Bart on July 16, 2021

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