I want to dedicate a thread for those who want to experiment with BMW ECUs to do the SAS Delete mod. The following is just a theory based on the Volvo, but I am hopeful that some BMW gurus will use this info to do a BMW SAS Delete.
So, if you use the info here and do a successful E39 (or other BMW models) SAS Delete, please post your success stories.
Remember that virtually all European cars, whether VW, Audi, Volvo, BMW, Mercedes, Porsche etc., use similar systems:
- ECU made by Bosch or Siemens.
- Air Valves (small electrical air valve, mechanical air valve, and air pump) made by Pierburg.
- Similar principles of operations of the SAS system.
---------------
For those of you who also own Volvo, you would be familiar with "SAS Delete". Remember that virtually all European cars (Volvo, VW, Audi, BMW, Mercedes, Porsche etc.) use the same setup for SAS, i.e., same Pierburg air pump, same air valve, similar ECU monitoring system.
Basically in the Volvo world, Radio Shack Diode 1N4003 (90-cent diode) bridged between FRONT O2 sensor and Aux Air Valve Solenoid circuit does the trick in fooling the ECU that the Air Pump is working. I have done this diode thingy in my Volvo and basically forget about the Air Pump. Zero CEL.
How does this work? The FRONT O2 sensor monitors exhaust condition for LEAN mixture during cold start. This happens b/c the Air Pump injects extra air during cold start. Let's say the FRONT O2 sensor sends a voltage of let's say 0.2V, the ECU knows it is lean and therefore "happy".
On the other hand, if the FRONT O2 sensor sends a voltage of let's say 0.8V, the ECU knows it is rich and therefore "unhappy", setting a SAS error code.
Just read a bit on how O2 sensor works on the web and you will see.
http://en.wikipedia.org/wiki/Oxygen_sensor
So if you know the Pinout of the E39 ECU, you can do this diode thing for 90 cents.
Below are photos of the 1998 Volvo S70 circuit for your use.
---> First photo shows the circuit diagram and how the diode is installed.
---> Second photo shows the actual mod...
--------
During cold start, the ECU sends a signal lasting about 100 seconds to 2 devices:
1. The Small Electric Air Valve (under intake manifold) to open its solenoid, allowing vacuum to feed the Mechanical Air Valve sitting on the exhaust manifold. This opens the port for the Air Pump to inject extra air into the Exhaust.
2. The Air Pump to run and inject extra air into the Exhaust.
During this time, if the SAS system works as designed, the O2 sensor will send different signal to the ECU:
- At first start, no extra air in Exhaust, i.e., rich condition, a ~ 0.8V[/] signal is sent to the ECU.
- Now the Air Pump injects extra air, the O2 sensor senses this, and sends a ~ 0.2V signal to the ECU, telling the ECU that the SAS system works as designed.
- In other words, the ECU is looking for voltage drop signal, in this case, about 0.6V or so.
- When your SAS system fails (either the small electrical air valve, mechanical air valve, or air pump malfunctions), the O2 sensor sends the same signal of 0.8V to the ECU, no voltage drop signal. The ECU will set an error code of SAS failure.
- The way to do a SAS Delete is to somehow fool the ECU into thinking there is a voltage drop signal.
I was helping my kids with AP Physics (College Physics), so here you go***8230;.Ohm's law states the following; where R = resistance (in Ohms) , V = voltage, and I = current (in Amperes).
This is for single resistor circuit, when you add a diode in series (as in the Volvo SAS Delete), you need to calculate voltage drop across the resistor (recall that the O2 sensor is essentially a resistor) with and without the diode.
I can go into complex calculation (Calculus, and differentiation etc.) but no need to do that. In a series circuit:
R (total) = R(1) + R(2)
And your college professor will tell you that when you add R(2) to the existing R(1) circuit, there is no change in current, but there is a change in voltage drop across R(1) because this is a series circuit. When you add a diode (a diode is basically a 1-way resistor) to a resistor, there should be a voltage drop across the resistor: I verified by a real experiment...
Think about the O2 sensor as a resistor, actually it is but its resistance varies with O2 in the exhaust.
1. Reference voltage is 8.92V. The diode is the 90-cent item from Radio Shack 1N4003.
2. With resistor only, Voltage = 8.88V.
3. With diode connector in series (note the direction of the diode), Voltage = 8.21V.
Or Voltage drop is 0.67V.
Now you know why when you install the diode into the Volvo ECU, for the first 100 seconds or so, it causes a Voltage Drop to pin #32, fooling the ECU!
That is it boys and girls, a multiple-choice exams will follow LOL.
So, if you use the info here and do a successful E39 (or other BMW models) SAS Delete, please post your success stories.
Remember that virtually all European cars, whether VW, Audi, Volvo, BMW, Mercedes, Porsche etc., use similar systems:
- ECU made by Bosch or Siemens.
- Air Valves (small electrical air valve, mechanical air valve, and air pump) made by Pierburg.
- Similar principles of operations of the SAS system.
---------------
For those of you who also own Volvo, you would be familiar with "SAS Delete". Remember that virtually all European cars (Volvo, VW, Audi, BMW, Mercedes, Porsche etc.) use the same setup for SAS, i.e., same Pierburg air pump, same air valve, similar ECU monitoring system.
Basically in the Volvo world, Radio Shack Diode 1N4003 (90-cent diode) bridged between FRONT O2 sensor and Aux Air Valve Solenoid circuit does the trick in fooling the ECU that the Air Pump is working. I have done this diode thingy in my Volvo and basically forget about the Air Pump. Zero CEL.
How does this work? The FRONT O2 sensor monitors exhaust condition for LEAN mixture during cold start. This happens b/c the Air Pump injects extra air during cold start. Let's say the FRONT O2 sensor sends a voltage of let's say 0.2V, the ECU knows it is lean and therefore "happy".
On the other hand, if the FRONT O2 sensor sends a voltage of let's say 0.8V, the ECU knows it is rich and therefore "unhappy", setting a SAS error code.
Just read a bit on how O2 sensor works on the web and you will see.
http://en.wikipedia.org/wiki/Oxygen_sensor
So if you know the Pinout of the E39 ECU, you can do this diode thing for 90 cents.
Below are photos of the 1998 Volvo S70 circuit for your use.
---> First photo shows the circuit diagram and how the diode is installed.
---> Second photo shows the actual mod...
--------
During cold start, the ECU sends a signal lasting about 100 seconds to 2 devices:
1. The Small Electric Air Valve (under intake manifold) to open its solenoid, allowing vacuum to feed the Mechanical Air Valve sitting on the exhaust manifold. This opens the port for the Air Pump to inject extra air into the Exhaust.
2. The Air Pump to run and inject extra air into the Exhaust.
During this time, if the SAS system works as designed, the O2 sensor will send different signal to the ECU:
- At first start, no extra air in Exhaust, i.e., rich condition, a ~ 0.8V[/] signal is sent to the ECU.
- Now the Air Pump injects extra air, the O2 sensor senses this, and sends a ~ 0.2V signal to the ECU, telling the ECU that the SAS system works as designed.
- In other words, the ECU is looking for voltage drop signal, in this case, about 0.6V or so.
- When your SAS system fails (either the small electrical air valve, mechanical air valve, or air pump malfunctions), the O2 sensor sends the same signal of 0.8V to the ECU, no voltage drop signal. The ECU will set an error code of SAS failure.
- The way to do a SAS Delete is to somehow fool the ECU into thinking there is a voltage drop signal.
I was helping my kids with AP Physics (College Physics), so here you go***8230;.Ohm's law states the following; where R = resistance (in Ohms) , V = voltage, and I = current (in Amperes).
This is for single resistor circuit, when you add a diode in series (as in the Volvo SAS Delete), you need to calculate voltage drop across the resistor (recall that the O2 sensor is essentially a resistor) with and without the diode.
I can go into complex calculation (Calculus, and differentiation etc.) but no need to do that. In a series circuit:
R (total) = R(1) + R(2)
And your college professor will tell you that when you add R(2) to the existing R(1) circuit, there is no change in current, but there is a change in voltage drop across R(1) because this is a series circuit. When you add a diode (a diode is basically a 1-way resistor) to a resistor, there should be a voltage drop across the resistor: I verified by a real experiment...
Think about the O2 sensor as a resistor, actually it is but its resistance varies with O2 in the exhaust.
1. Reference voltage is 8.92V. The diode is the 90-cent item from Radio Shack 1N4003.
2. With resistor only, Voltage = 8.88V.
3. With diode connector in series (note the direction of the diode), Voltage = 8.21V.
Or Voltage drop is 0.67V.
Now you know why when you install the diode into the Volvo ECU, for the first 100 seconds or so, it causes a Voltage Drop to pin #32, fooling the ECU!
That is it boys and girls, a multiple-choice exams will follow LOL.