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O2 Simulator

2/2/2015

26 Comments

 
Since turbocharging my IS300 this summer, I've been plagued with P0420 & P0430 codes -- Catalyst System below Efficiency. I tried simulating the output with my Innovate Wideband MTX-L, but after many months and configurations, I've given up. So I built a simulator based on a 555 Timer IC. I will install the new simulator tonight and report back w/ the results. Hopefully CEL free!

You can find the original schematics and instructions here. 
http://mkiv.com/techarticles/oxygen_sensor_simulator/index.html
Picture

R1  100 K Ohm
R2  1 M Ohm
R3  100 K Ohm
R4  10 K Ohm
C1  4.7 uF Electrolytic
C2  22 uF Electrolytic
D1  1.7v@20mA LED
D2  1.7v@20mA LED

This is meant to fool the ECU into thinking the post-cat O2 sensors are working and the catalyst is within spec. 
IMPORTANT NOTE: If you build this, make sure you do not leave PIN 4 on the 555 timer open (floating).  This will cause operation to stop and/or reset with even the slightest EMF or input voltage variation.  Instead, connect PIN 4 to Input Power. I connected to PIN 8 and it works flawlessly. 


The only other improvement would be a smoothing capacitor on the output - just to make the signal more of a SINE wave as opposed to the square wave.
26 Comments
Jared
5/27/2015 10:34:39 am

Did the O2 simulator work on your IS300? Thanks

Reply
Jon @ Chippernut
5/28/2015 12:05:16 am

Yes - it worked fantastic. In this example I built two simulators on one board, but you only need one. You can drive both O2 signals by splitting the signal and inputting both to the ECU. Works great. Been CEL free since install.

Reply
Jared
6/1/2015 03:16:03 pm

Just built mine like you advised, it works great. One last question, did you weather proof it some how or did you just tuck it in the ecu box? Thanks!

Reply
Jon @ Chippernut
6/2/2015 12:11:35 am

I ended up putting mine in the same box as the shift light Arduino -- which is a small project box located under my dash. You could weatherproof it and put it under the hood though, or I like your idea of tucking it in the ECU box (if it'll fit). Just keep it away from excess heat and water/mud. Glad you got it to work!!

Reply
Kelly
10/27/2015 12:28:19 pm

I'm not very savvy when it comes to stuff like this, but I'm very interested. How would I go about figuring out what I'm looking at, or how to properly execute this? My CEL is currently on, and I am all for a cheap fix. Thanks.

Reply
Jon @ Chippernut
10/27/2015 12:52:22 pm

Hi Kelly,

I'll put together a better diagram and schematic with parts list -- I see the original website I had posted is no longer active. I'll remedy that. Check my.is - i'll post more about your post there too.

Reply
Joel1212
7/23/2016 07:11:42 pm

Did u ever post details on my.is? O2 sims are going for big bucks and it would be great to build one myself but like the previous poster I need more info n parts list

Farmer john
5/26/2016 11:38:34 am

What do you do with prong 5? And are the led lights needed?

Reply
Jon @ Chippernut
5/26/2016 12:57:29 pm

I believe you just leave pin #5 disconnected. And yes, the LED's function as diodes and provide a small load to help it function correctly. Hope this helps
- jon

Reply
Zack
12/21/2016 10:06:25 am

Is this a plug and play type component or do i need to open up the ECU and do some cutting?
-thanks-

Reply
Jon @ Chippernut
12/21/2016 01:41:36 pm

Hi Zack,
This is sorta plug-n-play. You will have to do some wiring depending on your plans. You will have to cut the O2 wire going from Ox sensor to the ECU (you can do this either at the plug or at the ECU harness). And splice in the signal from the O2 Simulator.

You likely have another circuit the ECU is looking for and that's the Ox sensor heater circuit. You can leave the O2 sensor plugged in to satisfy the ECU and either leave it in the exhaust stream, or tuck it away somewhere in the engine bay - although this isn't recommended as it gets VERY HOT. The best alternative is to splice a high-wattage resistor on the Ox heater wires coming from the ECU. Refer to your service manual for the correct ohms.

Something like this: https://www.amazon.com/Amico-Aluminum-Housed-Wirewound-Resistors/dp/B00899WR32/ref=sr_1_5?ie=UTF8&qid=1482349267&sr=8-5&keywords=high+wattage+resistor


Hope this helps
- jon

Reply
Dave
12/4/2017 08:57:28 pm

Where can I get the equations for this?

Reply
Gary Charlton
2/14/2018 11:23:39 am

Would the input positive be switched or constant?

Reply
Jon @ Chippernut
2/15/2018 08:16:51 pm

It can be either - I have it wired up to constant in my car, it draws very little current, but it will drain the battery slowly over time. Switched input would probably be better. if you connect it to a power circuit that's powered when the key is in the "ON" position. Then it's only running when the vehicle is running.

Reply
DJ
4/8/2018 03:41:49 pm

Does the control voltage (pinout #5 need to be connected to anything?

Reply
Jon @ Chippernut
4/19/2018 07:46:06 pm

No - you can leave this open. Hope this helps. Sorry for the delayed response.

Reply
Randell T gribben
4/26/2019 03:53:07 pm

Question, on the positive side, how do you reduce the voltage from 12 volts to 1.5 volts that I see here, 12 volts directly to the first led, would get it instantly
Perhaps I reading this wrong

Reply
James
8/17/2019 07:43:32 pm

So there is 12v across the supply rail(12v) and there is a 10k ohm resistor to groud plus the diode ((LED). There is 0.6v across a diode typically (this is the voltage required for forward conduction of a P-N) junction.

So there is no regulation of the 12v but it does mention an operating region up to 15v so without me doing up a mock simulation in multisim I’m assuming it works not too bad.

Reply
Randell T gribben
4/27/2019 06:07:11 pm

The input is from the evil to sendor/ 555. And output from 555 to ecu?

Reply
DJ
10/18/2019 12:49:52 pm

Hi Jon, after building the circuit, I have a 6.9v signal coming from the output, the only issue is it not flipping every 3.3sec like originally designed, any help would be appreciated!

Reply
John
6/23/2020 03:33:54 am

This looks great I want to try this as I'm getting the efficiency error. I tried putting a capacitor along with a resistor on the blue and white wires which is supposed to smooth out the output but for some reason for me it creates a higher voltage. Can you kindly tell me how this works? how is it taking the 15-12v input and converting it to 0.1-0.9V in the waveform that O2 sensor does?

Also, looking at the diagram: do I connect +12v supply from ignition to the top right where it says +12..+15V? and ground from the car to ground? I'm assuming the OUT on the right goes to where the blue wire from the O2 sensor usually goes to right? and I just share the same ground back to white wire?

I currently have an O2 sensor plugged in and the two black wires I believe go from the car to the sensor to heat it or something. And the white and blue go from the sensor back to the car with the waveform voltage created by the O2 sensor. Could I just completely disconnect this and just use the connectors blue and white wire and de-pin the black wires? Will the car detect a fault if heat isnt being provided to the O2 sensor?

Thank you

Reply
Jon @ Chippernut link
6/24/2020 09:34:49 pm

Hi John! This circuit uses a 'voltage divider' R3 and R4 to bring the voltage to a max of 1v. Adding a capacitor across these lines changes the voltage output because the capacitor has a lower resistance than the resistor. Check out this link and put the values for R3 and R4. http://www.ohmslawcalculator.com/voltage-divider-calculator

Instead of wiring the capacitor in parallel between the output and ground (across the blue and white wires) you might want to wire it in series between the "out" junction and R4.

Yes - Battery + or 12V goes to the input labeled 12v...15v. And ground is ground as you described. The "OUT" should only go to the signal wire (all cars have different colors, sounds like either white or blue in your case) the other signal wire which is ground can just be left open. I don't think you need to tie that to ground because it's tied to ground inside the ECU. Try just leaving it open and see if that works. If not you'll get a check engine light and that can be easily solved.

The heater wires are monitored by the ECU and will set a check engine light if not connected. You can leave an old O2 sensor plugged in, though I would recommend a high-wattage resistor. This way you can get the exact ohms needed for your ECU (check the repair manual) and mount them anywhere. I would at least 100W resistors though, and mount them carefully. they got HOT!


Reply
John
6/27/2020 11:27:13 am

Thank you. So you suggest first just connect +12v from ignition to the in on the top right and connect ground from the car to the ground at the bottom of the diagram. Then connect the output only to the signal wire that usually goes from the o2 sensor to the ECU. So basically don't let the signal wire from o2 sensor go to ECU instead the output from this goes there. Ane the ground wire from the o2 sensor you suggest to cut it and leave it disconnected and see as the ECU should already be grounded?

Also, what is the waveform supposed to actually look like for a good efficient catalytic converter? Is it supposed to be something like it stays at 0.5-0.6v for 6 or so seconds then goes down to 0.0 or something then goes back up to 0.5 etc?

John
6/27/2020 01:14:02 pm

Is the waveform supposed to look something like this: https://i.imgur.com/1ONNU4S.png

John
6/27/2020 02:28:34 pm

I've just quickly drawn up how i'm understanding to wire it up. Is it supposed to be something like this? https://i.imgur.com/Ey9oHbh.png my blue wire on the o2 sensor is the + signal wire I believe.

Bobby link
1/5/2021 07:04:19 pm

Interesting thoughtss

Reply



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