Yet another O2 sensor fakeout trick

[Don Holtz]

Even the voltage drop across a diode is a function of current:

i=Io * EXP (-v/Vo)

It is true that for forward currents around the designed operating level,
most diodes have a voltage drop about .7volts, but this is only a very very
crude first approximation.  But for a very small forward currents the diode
will have a very small voltage drop, certainly much less than .7volts.

I suspect that, since the O2 sensor is a high-impedance voltage source,
putting a diode in series with it will not cause a nice predictable voltage
drop.

If you want to test your idea you could try this instead, and use a more
refined technique once it all works:

Use a 1.5V AA battery, and a high impedance voltage divider:


                           ------------------------------> TO ECM.
                           | 
                  100kohm  |     100kohm
    ---------o---/\/\/\/\--o---/\/\/\/\--o
    |        |                           |
    |        |            |              |              
--------     |            | |            |
|      |     -------------| |-------------
|  O2  |                + | | -
|      |                  |
--------                  1.5V (AA)
   |
   |
  GND

The 1.5V battery and 100kOhm divider combination create a floating .75volt
voltage drop.  The polarity of the voltage drop can be changed by reversing
the battery connection.  The voltage drop that is created has an impedance
of 50kOhm which is much lower than the O2 sensore impedance (approx
10Mohm), and the 100kohm resistors will not drain the battery too quickly.

Just an idea!

Cheers,
Don





   

At 08:31 AM 5/15/99 EDT, you wrote:
>>Thinking about trying yet another trick. I just can't leave this one thing
>>alone. I would really like to get rid of my 5th injector. The fuel line is
>>touching several items on the motor and I am afraid it might chap and
>>spring a leak. Also it doesn't look very professional anyway. And the fact
>I> am sure the injectors will provide enough fuel if I could just boost the
>>pulse rate up a bit.
>
>>Here's my plan.
>
>>Since my pressure switch has three poles, I can wire it up with a resistor
>>in parallel.
>
>>When I hit the boost zone, the O2 sensor's output which is either .8 volts
>>or .9 volts, however if I place a resistor between this circuit which
>>brings the voltage down to .1 volts the computer will think the engine is
>>running lean and increase the pulse rate trying to catch up.
>
>>The trick will be to find the right resistance to create such a voltage 
>drop. 
>
>>I know there is a formula somewhere. What size resistor do I need to drop
>>the voltage from .9 to .1 volts?
>
>>I am also assuming if I need a little more fuel, I can increase the
>>resistance some more.
>
>  James
>  I'm no wizard in solid state , but I know that almost all silicon diodes 
>have a forward voltage drop of about 3/4 volt. So, if you put one in series 
>with a circuit drawing current, it will "drop" that voltage. That way you 
>wont need to contend with the voltage divider, which will be current 
>dependent.
>
>    Jerry
>
>