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Want to remove/disable your rear O2 sensor without a CEL? Here's how..

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Old Jun 27, 2008 | 08:12 PM
  #316  
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From: Tri-Cities, WA // Portland, OR
Originally Posted by mrfred
If you've got a current meter, the ECU looks for a current between 0.17 A and 10.5 A across the heater when the heater circuit is engaged, and then it looks for a voltage of > 2.0 V across the heater when the circuit is not engaged. The current seems easy enough to measure. Not sure where the ECU is measuring the voltage (wouldn't seem to be across the O2 sensor), so not sure how to measure that.
Assuming the heater power is 12-14V, then the resistor needs to be at least 1.1 ohm and less than 80 ohms. dexmix's 50 ohm, 10 watt resistor should be perfect because the current will be in the right range, and the total power that has to be dissipated is only 4 watts.
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Old Jun 27, 2008 | 08:43 PM
  #317  
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Originally Posted by Jack_of_Trades
I suggest running two 50ohm, 10 watt resistors from RadioShack in parallel to create a 25ohm resistor. The current is split between the two resistors to minimize how hot they'll get. Each resistor will have around 4 watts of power being dissipated through it, less than half of its maximum rating.
You could try this. i have been using it for a few months now and its working great. its also good to have 2 resistors in parallel to help with the heat.
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Old Jun 27, 2008 | 11:26 PM
  #318  
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Originally Posted by andrewzaragoza
You could try this. i have been using it for a few months now and its working great. its also good to have 2 resistors in parallel to help with the heat.

2 50Ohms in parallel will heat up the same as a single 50 ohm.



the only difference between the 2 configurations is that you will draw twice the current (over half an amp)
I would prefer to keep the current as low as possible on the stock wiring harness, but still enough to trick the ecu.

Last edited by EvoBroMA; Jun 27, 2008 at 11:30 PM.
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Old Jun 28, 2008 | 10:27 AM
  #319  
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From: Opelika,AL
Originally Posted by dexmix
2 50Ohms in parallel will heat up the same as a single 50 ohm.



the only difference between the 2 configurations is that you will draw twice the current (over half an amp)
I would prefer to keep the current as low as possible on the stock wiring harness, but still enough to trick the ecu.
Two 50 ohm resistors will draw the EXACT same current as one 25 ohm resistor. Resistors heat up more on an exponential scale as you reach their wattage limit so having the 2 10 watt resistors in parallel will indeed be less hot than one 10 watt, 25 ohm resistor.

The reason we are achieving a 25 ohm load rather than a 50 ohm load is because it is far closer to the resistance of the stock O2 sensors heater circuit. The Average NB O2 sensors heater circuit is roughly 12 ohms which means the dual 50 ohm resistor setup will draw half the current through the stock wiring harness than the factory sensor would.

Current draw at 14 volts:
Stock sensor draws = 1.2 amps@12 ohms
two 50 ohm resistors in parallel draw = 0.56 amps
one 50 ohm resistor draws = .28 amps
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Old Jun 28, 2008 | 12:03 PM
  #320  
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I've been using one 50 ohm 10 watt for a few months now without problems. I had never considered running parallel resistors though, lol.

For my set-up I spliced a new plug set-up off of the original O2 heater leads (leaving them intact) and enclosed the resistors in a project box being careful to keep them away from the plastic & project board. That leaves the whole assembly something I can unplug & remove (to reinstall the stock O2) without needing to make any changes.

After reading about the dual resistor idea I did just added the second resistor to my set-up. They come as a 2-pack (Radio Shack pn: 271-133) so I had a spare.

Last edited by erioshi; Jun 29, 2008 at 10:33 PM.
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Old Jun 28, 2008 | 12:57 PM
  #321  
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Originally Posted by erioshi
I've been using one 50 ohm 10 watt for a few months now without problems. I had never considered running parallel resistors though, lol.

.
10 W is twice what's needed, a good engineering practice. You can always add more, just do not use less
https://www.evolutionm.net/forums/sh...1&postcount=80

Last edited by Zeitronix; Jun 28, 2008 at 04:33 PM.
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Old Jun 28, 2008 | 01:24 PM
  #322  
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Originally Posted by Zeitronix
ECU needs to see O2 sensor heater current to keep CEL happy. When O2 heater current falls below certain level ECU assumes that O2 sensor heater is not functional or disconnected. To simulate O2 heater current you need to use adequate load. 47 ohm resistive load keeps ECU happy in most cases. I=U/R so at 12V the load pulls 0.25A of current and at 14V 0.3A. P=U*I so at 14V the resitor will dissipate roughly 4.2W of power. It will get hot. Using a 5W rated resistor is this case is not a good engineering practice. A resistor should be derated by 50%, so use 10W ceramic resistor. It won't cost more than 5W but live a long happy life. Electronic components needs to be derated. This is how we design all of our electronics.
BTW you can also connect two 22 ohm 5W resistors in series. The wattage rating will double and 44 ohm load will keep the ECU CEL happy.
Your link didn't work for me so I figured I'd quote it for others
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Old Jun 28, 2008 | 04:35 PM
  #323  
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Should work now.
Thank you J of T
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Old Jun 30, 2008 | 12:02 AM
  #324  
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Originally Posted by Jack_of_Trades
Current draw at 14 volts:
Stock sensor draws = 1.2 amps@12 ohms
two 50 ohm resistors in parallel draw = 0.56 amps
one 50 ohm resistor draws = .28 amps
just to be clear - Temperature is a function of power. power splits along with current in a parallel circuit.

If you halve the total resistance, you're doubling total power - as shown by your own calculations.

so whats the difference between splitting twice the power between 2 50 Ohm resistors, or just having one 50 resistor? the difference is you're actually creating more heat, by having 2 resistors dissipating the same power, then you would have if you would have just used one.

heat per resistor doesn't change, its exactly the same in both configurations - except now your creating twice the total heat because you have two resistors generating heat as opposed to one.



Bottom line - If you're using 50 Ohm 10W resistors, using 2 of them in parallel will create more heat, than using a single 50 Ohm.





Originally Posted by erioshi
I've been using one 50 ohm 10 watt for a few months now without problems. I had never considered running parallel resistors though, lol.

For my set-up I spliced a new plug set-up off of the original O2 heater leads (leaving them intact) and enclosed the resistors in a project box being careful to keep them away from the plastic & project board. That leaves the whole assembly something I can unplug & remove (to reinstall the stock O2) without needing to make any changes.

After reading about the dual resistor idea I did just added the second resistor to my set-up. They come as a 2-pack (Radio Shack pn: 271-133) so I had a spare.
erioshi - see above.

Last edited by EvoBroMA; Jun 30, 2008 at 12:54 AM.
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Old Jun 30, 2008 | 12:09 AM
  #325  
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From: MA
double post
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Old Jun 30, 2008 | 04:06 AM
  #326  
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From: Opelika,AL
Originally Posted by dexmix

Bottom line - If you're using 50 Ohm 10W resistors, using 2 of them in parallel will create more heat, than using a single 50 Ohm.
Dude, you're missing the point. I was never suggesting placing a 50 ohm load on the heater circuit because it was too far from the original 12 ohm load the stock sensor presented. My arguement was between a SINGLE 25 ohm,10 watt resistor opposed to 2 50 ohm, 10 watt resistors in parallel. the dual resistors will in FACT create less heat and be more stable since heat dissipation isn't a constant like you think when you start to push more current through a resistor.

Obvious a 50 ohm load draws less than 25 but that was never what I suggested...just read any post of mine above
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Old Jun 30, 2008 | 08:36 AM
  #327  
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From: MA
Originally Posted by Jack_of_Trades
Dude, you're missing the point. I was never suggesting placing a 50 ohm load on the heater circuit because it was too far from the original 12 ohm load the stock sensor presented. My arguement was between a SINGLE 25 ohm,10 watt resistor opposed to 2 50 ohm, 10 watt resistors in parallel. the dual resistors will in FACT create less heat and be more stable since heat dissipation isn't a constant like you think when you start to push more current through a resistor.

Obvious a 50 ohm load draws less than 25 but that was never what I suggested...just read any post of mine above

No i'm definitely not missing the point - and i'm not arguing with you, just trying to clear it up for everyone else. Its definitely not obvious.

You're posts were confusing people - which is why erioshi went and plopped in his second 50 Ohm, when its actually not doing anything for him, except doubling his heat output, when he intended to lessen it. Not everyone has an understanding of electronics.
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Old Jun 30, 2008 | 10:35 AM
  #328  
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From: Opelika,AL
Originally Posted by dexmix
No i'm definitely not missing the point - and i'm not arguing with you, just trying to clear it up for everyone else. Its definitely not obvious.

You're posts were confusing people - which is why erioshi went and plopped in his second 50 Ohm, when its actually not doing anything for him, except doubling his heat output, when he intended to lessen it. Not everyone has an understanding of electronics.
Ah gotcha. Well, what we really need to find out is what load actually triggers the CEL for a failed heater circuit. The higher the resistance value can be...the better. I doubt there is any way in the code to find out the trigger point for that CEL???
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Old Jun 30, 2008 | 10:44 AM
  #329  
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Originally Posted by Jack_of_Trades
Ah gotcha. Well, what we really need to find out is what load actually triggers the CEL for a failed heater circuit. The higher the resistance value can be...the better. I doubt there is any way in the code to find out the trigger point for that CEL???
Don't you guys read what I post? I posted up the values almost a month ago, and no one seemed to notice. I reposted the values a few days ago (see first post on this page), and even determined a suitable resistance value.
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Old Jun 30, 2008 | 10:53 AM
  #330  
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From: Opelika,AL
Originally Posted by mrfred
Don't you guys read what I post? I posted up the values almost a month ago, and no one seemed to notice. I reposted the values a few days ago (see first post on this page), and even determined a suitable resistance value.
I totally missed that post, my apologies.

Originally Posted by mrfred
Assuming the heater power is 12-14V, then the resistor needs to be at least 1.1 ohm and less than 80 ohms. dexmix's 50 ohm, 10 watt resistor should be perfect because the current will be in the right range, and the total power that has to be dissipated is only 4 watts.
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