Turbo Boost Increase for AKN Engine: Info

If anyone is interested!
'I purchased a so-called 'BlackBoxx' device some time ago to increase performance. Unfortunately, the performance had to be reduced due to excessive soot formation. When I posted about my problem on this forum, it turned out that the issue is due to a lack of air, which means the turbocharger pressure needs to be increased!'
The approach is now to slightly reduce the measurement voltage of the intake manifold pressure sensor, so that the control unit is given a lower boost pressure value than is actually present.
Since this intake manifold pressure sensor is essentially just an unloaded voltage divider, the necessary modifications can be easily calculated.
Please do not even consider installing a diode in the control line of the pressure sensor for this motor! There are several problems that would immediately arise:
The input impedance of the control unit is very high (not measurable with my device!), so that virtually no current flows. Therefore, a voltage drop across the diode cannot occur, not to mention that a linear characteristic curve only starts at approximately 40mA. I would not recommend connecting a resistor to ground after the diode, as this would require the current through the voltage divider to be increased to at least 48mA!
You only need to insert a 100-ohm potentiometer into the lead wire going to the voltage divider (the cable with the white marking). My calculations are very close to the actual values, so after the modification, the boost pressure can be increased by up to 0.135 bar (with a 100-ohm potentiometer). The current flowing through the voltage divider is reduced to 6.89mA!

Currently, I'm waiting for my BlackBoxx to be reprogrammed to its default settings. I'm already curious to see how much I'll ultimately need to increase the boost pressure to eliminate the black smoke!

Best regards,
Achim

Hello.

avoso wrote:

...Please don't even think about installing a diode in the control line of the pressure sensor for this motor!! There are several problems with that: The input resistance of the control unit is very high (not measurable with my measuring device!), so that virtually no current flows. Therefore, a voltage drop across the diode cannot occur, not to mention that a linear characteristic curve only starts at approximately 40mA. I would not recommend adding a resistor to ground after the diode, because the current through the voltage divider would have to be increased to at least 48mA!!

"It works perfectly for me. The Conrad datasheet function is your friend. Ge-demodulation diodes work great; just choose 2-3 types according to the datasheet and select the one that's most suitable."
If you only start to see a noticeable voltage drop at 40mA, you've probably unfortunately got the wrong type.

Best regards, Rainer.

PS: I don't quite understand your circuit .

Hello,
I used a 1N4148 diode, but there are probably better options available!
I don't understand why I should use more components than absolutely necessary, and then also increase the current flowing through the intake manifold pressure sensor?
I believe my method is significantly better because my calculations are very close to the actual values! Furthermore, a linear relationship between the measured voltage and the actual boost pressure is maintained.
No offense! / Nothing personal!

Best regards, Achim.

P.S. Information about the structure and function of unloaded voltage dividers should be easy to find (I'll check Conrad ).

Hi Achim,

I understand how a voltage divider works, but I'm not sure exactly which voltage you want to divide and where.
What is the supply voltage of the pressure sensor?
What is the initial voltage?

Best regards, Rainer.

Hello Rainer,
The pressure sensor I have has 3 connections. There's the voltage input (white marking), the 'voltage output' or ground (brown marking), and the output or measurement voltage (red marking).

Between the cable with the white and brown markings, there is a fixed resistor. The connection with the red marking can be visualized (figuratively) as a shunt, which creates a voltage divider.

Ultimately, it's up to each individual to decide which approach they want to use!

Here's a small equation that indicates the resistance required to achieve the desired increase in boost pressure:

Required resistance = 100 * desired boost pressure increase / 0.135.

Attention! Please enter the value for 'desired boost pressure increase' in the unit of bar!
The value should (in my opinion!) not exceed 0.13. I am happy to be corrected regarding this statement!!

Best regards, Achim.

Hello,

And where do you want to set the potentiometer? Please remember that the output current of the pressure sensor is limited and, as far as I recall, is less than 10mA. Did you measure the increase in boost pressure of 0.135 bar? What is the overshoot when the system is settling?
My reason for skepticism: some time ago, I introduced a voltage divider into the output signal of the boost pressure sensor, but the current load quickly became too high for the pressure sensor, rendering the control behavior unusable.

Best regards, Rainer.

Hi Rainer,
I cut the cable with the white marking and inserted the potentiometer here. Since the total resistance is now higher, the current cannot increase! The potentiometer now experiences a voltage drop, which means the current input voltage to the pressure sensor is now lower.
I calculated the 0.135 bar value, and I was also able to read it from the measurement block output (although the value had decreased accordingly).
Could you please give me some information about what 'Einregelverhalten' refers to? Unfortunately, I don't have any information about that at the moment.

Perhaps this modification can also be applied to other motor variants, as I have measured similar values in my own setup. For example, the current was originally also at 8mA!

Best regards,
Achim

Hmm, I'd be interested to see your calculation.
Because your method only works reliably if the power consumption of each pressure sensor is always constant. Have you measured the current flow in the power supply for this?

Best regards, Rainer.

Hello,
Of course, the current remains constant! The total resistance doesn't change!

Okay, so let's talk about my approach and calculations:

First, we have our voltage divider again, which is the pressure sensor, and it has 3 connections.
1. Ue - Input voltage
2. Ua - Initial or measurement voltage.
3. Mass.

For clarity: Resistor R1 is connected between Ue and Ua, and resistor R2 is connected between Ua and ground (please draw a diagram for better understanding! Again, the pressure sensor itself acts as a voltage divider!).

I disconnected the power supply (cable with a white marking) and measured the current here.
I = 8mA
The input voltage (Ue) is 5V.
Ue = 5V
UA = UR2 = 1.853V (measured).
This results in a voltage drop across R1.
UR1 = Ue - Ur2 = 3.147V
and thus the resistance values.
R1 = UR1 / I = 1.853V / 8mA = 393 Ohms.
R2 = UR2 / I = 3.147V / mA = 232 Ohms.
Dann habe ich ein Poti zwischen R2 und Masse eingebaut und somit UR2 langsam angehoben. Habe Ua = UR2 gemessen und den Ladedruck mit VAGcom ausgelesen. Engine off!
979 millibars - 1.853 volts.
1061 mbar - 2.005 V
1142 millibars - 2.149 volts.
1265 mbar - 2.383 V
From this, you can also see the linear trend!

Connecting a 100 Ohm resistor, labeled R3, in series before R1 results in a total resistance of:
Rg = R3 + R1 + R2 = 100 Ohms + 393 Ohms + 232 Ohms = 725 Ohms
And a stream of:
I = Ue / Rg = 5V / 725 Ohms = 6.89 mA
and thus a voltage drop of Ua = UR².
Ua = R2 * I = 232 Ohms * 6.89 mA = 1.598 V

Now, using the values obtained from my mbar/Ua measurements, we can create an equation:

979 millibars / 1.853 volts - previous ratio.
X mbar / 1.598V. Ratio after the process.

979 mbar / 1.853 V = X mbar / 1.598 V

x = 979 mbar * 1.598 V / 1.853 V = 844.27 mbar

Result:

Using a 100-ohm resistor results in a reduction of the measured boost pressure of:

979 mbar - 844.27 mbar = 135 mbar!

Is that clear enough?

Best regards, Achim.

Hello, yes, that makes sense. I didn't expect it to work so smoothly, and especially that the power consumption of the pressure sensor remains relatively constant...

Best regards, Rainer.