RC Low-pass filter for O2 sensor

MisterBill2

Joined Jan 23, 2018
6,696
It's to avoid having to replace the catalytic converter
OK, that makes things a lot simpler. If adding the simple filter to the oxygen sensor output solved one problem mut seems to have made the voltage too high, I am wondering about the ground return connection. To be specific, is the capacitor ground tied to the sensor output ground? If the cap is at all leaky and the ground points at different voltages then that is your problem. If not, then more thinking is required.
 

Wolframore

Joined Jan 21, 2019
1,748
Try the mounting extender. by taking the sensor out of the path of exhaust it has the same effect without messing with the circuit.
 

Thread Starter

john2k

Joined Nov 14, 2019
87
So I changed the 4.7uF capacitor to a 1uF capacitor and when I monitor the live voltage on that sensor it's reading at 1.27V so only gone down by 0.01V. I've kept the 1Mohm resistor. I'm wondering if the resistor needs to be higher? The opposite bank sensor displays at about 0.07V and if I remember correctly before adding the resistor and capacitor this one was reading at about 0.11V so why is it going so high? What exactly is the ECU looking for? I'm assuming not a flat line voltage, is there a way to maybe just add a circuit to the two wires going to the ECU that just sends the voltage to the ECU?
 

Reloadron

Joined Jan 15, 2015
5,604
Changing caps or resistors will only change the RC time constant so what you see does not surprise me. Something you can try to reduce the voltage is a resistor to ground across the cap. Maybe we can divide the voltage down a little.

Ron
 

Thread Starter

john2k

Joined Nov 14, 2019
87
Changing caps or resistors will only change the RC time constant so what you see does not surprise me. Something you can try to reduce the voltage is a resistor to ground across the cap. Maybe we can divide the voltage down a little.

Ron
So right now from the O2 sensor towards the ECU I first have a 1Mohm resistor on the positive wire, then the positive leg of the 1uF capacitor then to ECU pin. And on the negative wire I have the negative pin of the capacitor. So you are saying put a resistor between the positive and negative wire? After the capacitor I take it? And what kind of resistor and will it generate a lot of heat while it's dividing the voltage ?
 

Reloadron

Joined Jan 15, 2015
5,604
The negative pin of the cap should be ground? What I am saying is we can try placing a resistor across the cap to ground. Maybe 1 Meg which should give us a 2:1 divider network.

Ron
 

MisterBill2

Joined Jan 23, 2018
6,696
The negative pin of the cap should be ground? What I am saying is we can try placing a resistor across the cap to ground. Maybe 1 Meg which should give us a 2:1 divider network.

Ron
The resistor will not generate a noticeable amount of heat. it is supposed to drop the output voltage a small amount. What it actually does will be interesting.
 

Reloadron

Joined Jan 15, 2015
5,604
This should be what you currently have:

RC Network for O2 Sensor.png

Try placing a second 1 Meg resistor across Vout and lets see what it gets us. This should half (2:1 Divide) the voltage. The added resistor can also be decreased in value which will increase the voltage to the ECM. You will have the smoothing.

Ron
 

Reloadron

Joined Jan 15, 2015
5,604
The resistor will not generate a noticeable amount of heat. it is supposed to drop the output voltage a small amount. What it actually does will be interesting.
Heat? I never mentioned heat. With 1 volt and 1 MEG I don't see any current to speak of. The proposed idea looks good on paper. :) I am not real sure how it will play out.

Ron
 

Thread Starter

john2k

Joined Nov 14, 2019
87
Heat? I never mentioned heat. With 1 volt and 1 MEG I don't see any current to speak of. The proposed idea looks good on paper. :) I am not real sure how it will play out.

Ron
So just to be clear I should put the extra 1Mohm resistor across like the pic below? it won't create a short across the two lines? The 1Mohm resistors I have are 1/4W Carbon Resistors, that won't matter will it? I may have to try this on Saturday as it gets dark outside by the time I get home after work on weekdays :)

 

Reloadron

Joined Jan 15, 2015
5,604
So just to be clear I should put the extra 1Mohm resistor across like the pic below? it won't create a short across the two lines? The 1Mohm resistors I have are 1/4W Carbon Resistors, that won't matter will it? I may have to try this on Saturday as it gets dark outside by the time I get home after work on weekdays :)

That's fine, just make it like the picture. :)

Ron
 

Thread Starter

john2k

Joined Nov 14, 2019
87
Since, i'm not going to be able to try it out until Saturday, I thought if I get a AA battery that's 1.5V and put a resistor across the + and - that should essentially be simulating the halfing of the voltage idea no? So I first measured the AA battery with a DMM and it read 1.60V, when i put a 1Mohm resistor across the + and -, it made no difference on the DMM reading. It was still 1.60V. However, if i moved the resistor inline on the + line then the voltage dropped by 0.07V. Every additional 1Mohm resistor I add in series makes the voltage drop a further 0.07V. So if it's not working in this test, will it really work by putting the resistor across the positive and ground on the O2 sensor?
 

Reloadron

Joined Jan 15, 2015
5,604
Here is what you should have:
O2 Smooth 1.png

Ignore the Pulse. What I did to simulate a 1.5 volt battery was create a pulse which goes from 0 to 1.5 volts after one second. The below is what we get.

O2 Smooth 2.png

We know from past post that a 1.5 Meg resistor along with a 1 uF capacitor make up a 1,500,000 Ohm * .000001 (R*C) time constant filter. We also know it will take about 5 RC time constants for the cap to charge to the applied voltage or in this case % * 1.5 seconds = 7.5 seconds. The cap will charge to about 63% of the applied voltage the first RC time constant then slowly continue to charge. Now as drawn above we added another resistor, another 1.5 MEG so now R1 and R2 form a 2:1 voltage divider so Vout will be 1/2 of Vin and as we can see Vout is about 750 mV or 0.750 volt. That is what you should see on the bench. From the time Vin is applied the meter should jump up pretty quick in the first 1.5 seconds and then slowly increase until we get about 750 mV. I pause the pulse in the above only so the trace is more pronounced for clarity. Think of V1 as the O2 sensor out.

Ron
 

Thread Starter

john2k

Joined Nov 14, 2019
87
Ooops yeap sorry you are right i was measuring the battery output on the wrong end :oops:

But yes i've tested this concept with a AA battery and a 1Mohm resistor then a 1uF cap and then another 1Mohm resistor across the cap legs and the output of the AA battery was measuring at 0.78V. So i'll try this on the weekend and report back. Thank you.
 
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