SAS Delete: the Math and Physics behind it

[cn90]

For those who did the SAS Delete...

Have you ever wondered how it works?


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 b/c it could find the expected "Voltage Drop" signal of ~ 0.6V.

Just read a bit on how O2 sensor works on the web and you will see.

http://en.wikipedia.org/wiki/Oxygen_sensor

Below are photos of the Volvo circuit for your use.

---> First photo shows the circuit diagram and how the diode is installed.
---> Second photo shows the actual mod...

SASDelete1.JPG (120.18 KiB) Viewed 3548 times

SASDelete2.JPG (89.83 KiB) Viewed 3548 times

--------
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…Ohm’s law states the following; where R = resistance (in Ohms) , V = voltage, and I = current (in Amperes).

Ohmlaw.JPG (2.89 KiB) Viewed 3548 times

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 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 change in voltage drop across the resistor...I found out, 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.

SASDelete3.JPG (56.45 KiB) Viewed 3548 times

SASDelete4.JPG (21.87 KiB) Viewed 3548 times

SASDelete5.JPG (91.6 KiB) Viewed 3548 times

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.

[mikealder]

0.6V to 0.7V drop across a silicon diodes junction is quite normal, if the diode were made using Germanium then you would get somewhere between 0.25 and 0.3V drop - Mike

[cn90]

Actually this is the voltage drop across the resistor (in this case the O2 sensor).

[erikv11]

Nice post. It sounds like you are not clear on what "voltage drop" means.

...
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 voltage drop across R(1) because this is a series circuit. ...

Your professor won't tell you that. At least not if he wants to keep his job! He'll tell you that when two resistors are in series, the voltage drop across the entire circuit is unchanged, and the drop across each resistor is proportional to its fraction of the total resistance.

A voltmeter reports the voltage drop across whatever component is between the leads. In this case you have the leads across the same resistor, in both circuits. So in the circuit on the left the drop across the resistor is 8.88. In the right picture the drop across the resistor is 8.21, because the drop across the diode is 0.67; total drop across the circuit is unchanged, drop across the individual elements is proportional to the fraction of the total resistance. That's what Mike is talking about - he's right.

For example, if you used an identical resistor instead of the diode in that series circuit, then the meter would report 4.44 Volts. The voltage drop would be 4.44 across each resistor.

[cn90]

In the last photo above, think about the black probe as pin #32 of the ECU.
This is why I put the black probe there. Pin #32 is exactly how the ECU monitors the O2 sensor output.

All I want to illustrate is the effect of a "diode in series" on the "voltage drop across a resistor".

[cn90]

Thanks erik,

I just edited the original post for typo error.

I should verify 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 because of the added diode."

[erikv11]

... "...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 because of the added diode."

I just came across this old post and saw your edit. Again, you may wish to look up what "voltage drop" is. In a live circuit there is always a voltage drop across a resistor, VD isn't newly created because of the diode. So even after editing there is an error in that section of text. The same mistake occurs in several other places in the post.

e.g. the section you started to edit could be fixed like this

... "...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. So when you add a diode D (a diode is basically a 1-way resistor) in series with a resistor R, there will be less voltage drop across the resistor R."

Don't get me wrong this post that is mostly spot on and helpful, the main point gets across if you already know what is going on, but the physics is garbled because terminology is used improperly. And you wanted it to be about the physics so ...

[pjco217]

looks like volvospeed is down that has another DIY for this mod. I would like to add that I performed this SAS delete and the A37 and A32 are the correct pins for the diode within the ECU. Note that you have to keep the solenoid/wiring in addition to the secondary air pump relay wiring as well. I was going crazy thinking I fried the diode but it ended up being a bad relay/corroded relay contact.

A new relay (p/n 9442932) and new wire contacts for the control wire pin B38 (p/n 3523813)

You are fooling the ECU in thinking the relay is powered B38 (air pump is powered) and the solenoid A37 (solenoid is opened and vacuum is pulled from the tree which opens the mechanical valve where air should be traveling into the exhaust from the pump) to know the system is working in addition to seeing a voltage drop from the oxygen sensor.

Nice mod (13 years late to the fix)

sasdeleterelay.PNG (190.21 KiB) Viewed 1129 times

[jreed]

Welcome to MVS! Never too late to join the party -- lots of great information here and helpful gurus.

[MrAl]

Hi,

Just to clear up one thing i read in the first post.
The statement was that if you have a resistor in a circuit and it has a voltage drop of say 2 volts and introduce a second resistor in series with the first resistor that there will be another 2 volts drop, and that is almost correct in many cases (but not all), but then it also says that the current will not change, which is not true in the usual case. This is because in order to see the current remain the same with the second resistor too, the first (and now the second) resistor would have to be powered by a constant current.
Now that could be the case, but the more usual case is that the whole thing is powered by a (nearly) constant voltage like the car battery. That's the more usual case, and in that case both the voltage drops increase and also the current decreases.
So say we have a 1k resistor in series with a 12 volt battery, the current is 12ma and the voltage is 12 volts across that resistor.
Now place a second 1k resistor in series with that first one. The current drops to 6ma, and now the voltage drop across that first resistor is 6 volt. So both the voltage drop across the first resistor changed, and the current changed too, and it changed a lot.

With a diode however it's a little different. The diode has a nearly constant voltage drop, as compared to a resistor. If we have a resistor of 1k in series with a diode and the diode is connected with the right polarity, the diode drops roughly 0.6 to 0.8 volts, and so now the resistor would have 11.2 to 11.4 volts across it. In this case the current does not matter as much for the diode, although it still does for the resistor. The diode will always drop about 0.6 to 0.8 volts (standard silicon diode) but the resistor will drop whatever it drops with the current through it, which can change. If you use a Schottky diode it will drop less, from 0.2 to 0.5 volts (roughly), so the results would be different. A typical common Schottky diode is the 1N5817 and a common regular silicon diode is 1N4001 and similar.
A Schottky diode comes in handy if you want to boost your voltage regulator output voltage to charge the battery better on short trips and you dont want it to go too much higher. In my case my regulator puts out 14.0 volts to start, and if i were to add a Schottky diode to the regulator it would go up to around 14.4 volts which would be really nice, but with a regular silicon diode it would go up to around 14.7 volts and not sure i would want it that high. The mod is simple too but you have to be able to get your regulator off of the alternator, and may have to get the alternator out of the car to do this, something that is not that easy on my 1998 v70.
The simple trick is to place the diode in series with the voltage regulator sense wire (or connection) so that it tricks the alternator into thinking that the battery voltage is 0.4 volts lower than it really is (or 0.7v lower with a regular Si diode). That means it keeps putting out current into the battery until the actual battery voltage goes up to about 14.4 volts instead of 14.0 volts, which gives the battery a better charge over short trips. I wouldnt mind having a switch across the diode too so i could short it out for long trips, which would also be easy to do.

Good luck with it.