Ladedruckanhebung mittels 10ct-Tuning - diesmal langsam

[Translating...]

Hallo,

Ich habe bislang eine 10ct-Tuning des Ladedrucks mittels einer Diode um ca. 0.1 bar realisiert.
Ich weiss nicht mehr genau, was das für eine Diode ist, auf jeden Fall erzeugt sie einen Spannungsabfall von ca. 0.3 V ab einem Strom von 5 mA.

Jetzt bin ich mitr "nur" 0.1 bar Ladedruck allerdings nicht mehr zufrieden, demnächst sollen große Düsen vom AEL rein und dann will ich auch mit 1.2 bar fahren.

Avoso hat hier https://community.dieselschrauber.org/en/viewtopic.php?t=1308&highlight=
beschrieben, wie eine einstellbare Ladedruckerhöhung zu realisieren ist. Diese habe ich noch nicht ausprobiert, möchte dies aber tun.
Im wesentlichen wird einfach ein variabler Widerstand in die Spannungsversorgung des Drucksensors eingeschleift.

Um Überschwinger so gering wie möglich zu halten, stelle ich mir aber noch eine weiter Modifikation dazu vor, verstehe aber selbst zu wenig von Elektronik, um eine Schaltung dafür selbst mit gutem Gewissen aufzubauen.

Frage:
Könnte man nicht parallel zu dem eingeschleiften Widerstand einen Kondensator schalten? Soweit ich weiss, zeigen Kondensatoren folgendes Leitfähigkeitsverhalten: Ist der Kondesator komplett entladen, fliesst Strom durch ihn durch, allerdings klingt dieser Stromfluss exponentiell mit der Zeit ab. Wenn also kein Strom mehr durch den Kondensator fließen kann muss der Stromfluss also zu 100% durch den variablen Widerstand erfolgen und somit die Ladedruckerhöhung voll eingreifen.
Natürlich soll das ganze aber erst ab einem bestimmten Ladedruck so funktionieren, sonst würde der Kondensator ja schon im Leerlauf geladen werden und stromundurchlässig sein.

Ziel ist, dass der Ladedruck bis zum Serienwert (1 bar) mit dem normalen Regelverhalten aufgebaut wird, errreicht der Ladedruck 1 bar soll allerdings langsam dem Steuergerät eine niedrigerer Ladedruck vorgetäuscht werden, so dass das STG langsam auf 1,2 bar einregelt.

Es wäre schön, wenn einer der Elektronik-Cracks hier im Forum dazu was schreiben könnte.

Have you ever heard of chip tuning?
It can be implemented much more effectively with that.
"Above all, you know that simply increasing the boost pressure alone achieves absolutely nothing."
It has a lower efficiency.

If you still want to manually adjust the boost pressure, you can get a MAP sensor.
Solder it into your control unit. You can find information about that using Google.

Hello Steffarn,

Whoever can read has a clear advantage. javascript:emoticon('')
You probably already know that an ASV (Abgas-Steuergerät) has a MAP sensor in the charge air path. Therefore, there's no need to solder anything on the ECU (Engine Control Unit). Or should I solder another additional pressure sensor into the ECU? javascript:emoticon('')

Quote:

"Above all, you know that simply increasing the boost pressure alone achieves absolutely nothing." It has a lower efficiency.

As I mentioned, nozzles from AEL are also supposed to be included.
Currently, I have an adjustable 3-potentiometer box installed, and it's sufficient for adjusting the additional fuel injection amount. It's clear that things can be even better with chip tuning.

Best regards, Georg.

Did you mean transistor?

When current flows through a capacitor, it is...
In my humble opinion, it's broken!

When you create a system that can be financed...
'Effort! Get rid of those overachievers, then write to me.'

Why would you subject a VNT15 to such hardship? Just buy one from...
'Let's get a VNT 17, or even better, a 20.'

Did you mean transistor, or?

When current flows through a capacitor, it is...
In my humble opinion, it's broken!

When you create a system that can be financed...
'Effort! Get rid of those overachievers, then write to me.'

Why would you subject a VNT15 to such hardship? Just buy one from...
#### a VNT 17, or even 20.

Hello 'ObenbeiMutti',
Alternating current always flows through a capacitor, while direct current only flows until the capacitor is charged.
Best regards,
Günther.

Inspired by the ice cream maker that I vaguely remember, it is supposed to be used in a...
The circuit seems to be working as you intended.

However, nothing actually flows through it; think about the school experiments with plate capacitors.
If the potential difference became large enough to cause a spark, it would be broken.

Hi.

Georg_G wrote:

... Could one not connect a capacitor in parallel with the shunted resistor? As far as I know, capacitors exhibit the following conductivity behavior: When a capacitor is completely discharged, current flows through it, but this current flow decreases exponentially over time. Therefore, if no more current can flow through the capacitor, the current flow must proceed 100% through the variable resistor, and thus the pressure increase due to charging will fully take effect. Of course, this should only start working above a certain boost pressure, otherwise the capacitor would start charging even at idle and become non-conductive. The goal is to have the boost pressure build up to the standard value (1 bar) using normal control behavior. However, once the boost pressure reaches 1 bar, the system should simulate a lower boost pressure to the control unit, so that the control unit gradually adjusts to a boost pressure of 1.2 bar. It would be great if one of the electronics experts here in the forum could write something about it.

It would generally work. A current flows whenever the potentials at the capacitor electrodes change. "That means that with a constant voltage of x volts, where x is less than the breakdown voltage, no current flows. -> Your circuit is called a high-pass filter. You can find good descriptions of it online, and I think there are also online calculators with nice diagrams."

If you adjust the time constant to be on the order of the settling time, the control system should perform better.
http://www.computerbase.de/lexikon/Zeitkonstante
You just need to make sure that the feedback loop remains stable. More fundamentals:
http://141.99.140.157/d/imr5/RTKurs/13seite/13seite.html

Best regards, Rainer.

Hello 'ObenbeiMutti',
You have a private message.
Best regards,
Günther.

@Rainer

How does he plan to fine-tune it so that it fits perfectly? At what point does the capacitor overload, and at what point does the resistor dissipate the excess energy? How quickly does a capacitor discharge? With short gas exchanges, it won't work. Should he use a transistor that he can switch on or off depending on the control signal?
Or am I just imagining things?

@Günther

Okay, no offense, but 'current flows through the capacitor' is syntactically incorrect. What goes in on one side *must* come out on the exact same side, otherwise it's broken, honestly.

@Rainer

How does he plan to fine-tune it so that it fits perfectly? At what point does the capacitor overload, and at what point does the resistor dissipate the excess energy? How quickly does a capacitor discharge? With short gas exchanges, it won't work. Should he use a transistor that he can switch on or off depending on the control signal?
Or am I just imagining things?

@Günther

'No offense, but 'current flowing through the capacitor' is syntactically incorrect. What goes in on one side *must* come out on exactly the same side, otherwise it's broken, honestly.'

Hello,
You can quite vividly compare a capacitor to a balloon!
Best regards,
Günther.

PS: (Of course, I meant a rather large capacitor in the private message!)

ObenbeiMutti wrote:

@Rainer How does he plan to fine-tune it so that it fits perfectly, or when will the capacitor burst...?

The capacitor doesn't explode; it's supposed to withstand 5V.

http://www.elektronik-kompendium.de/sites/grd/0205301.htm

Hi,

As long as you don't exceed the capacitor's maximum rated voltage, nothing will explode. The comparison to a balloon is, therefore, only very, very limitedly applicable.

If a capacitor were to explode due to continuous "charging," every PC power supply would fail, and virtually every electronic device would be damaged.

Best regards,

Jan.

Hello,
The balloon won't burst as long as you don't increase the pressure (which is equivalent to the voltage in a capacitor) too much!
Best regards,
Günther.

You can interpret that however you want...

No electron flow occurs THROUGH the capacitor.
However, when the electrodes ('plates') are charged, electrons are moved, which constitutes an electric current. It can also be easily measured from the outside using an ammeter.

Such transfer currents can reach drastic magnitudes. Many hobbyists prematurely choose field-effect transistors because of their 'lossless' control. This is true in the static range (no gate current flows), but not always in practice. In dynamic applications (audio, PWM), currents in the ampere range can occur here, which can then drastically overload the upstream circuitry.

A capacitor connected in parallel with a resistor initially introduces a certain time delay. However, this delay occurs in BOTH directions. The boost pressure would therefore build up more slowly, but also decrease more slowly.

However, this whole thing doesn't work if you reduce the operating voltage of the sensor. This can cause the sensor to operate in an undefined range.
Typically, the output voltage of the sensor is reduced using a voltage divider (resistor from the output to ground, do not overload the sensor output!).
Here as well, a capacitor placed in parallel would integrate the changes and provide a time delay in both directions.

Since the entire system needs to react dynamically starting from a certain pressure level (i.e., output voltage), it can no longer be achieved using passive components. More effort is needed here.

CU

Hello,
I'm from the electrical engineering field, and I intentionally didn't want to delve too deeply into that topic here (it's just ----!).
In FET circuits, the 'problem' is often simply shifted from the output stage to the input stage, because the gate capacitance can also cause issues at higher frequencies, especially when fast switching is required.
Best regards,
Günther

PS: A capacitor connected in parallel with the resistor 'accelerates' the signal; only an RC circuit integrates (averages) the signal.

PS: A capacitor in parallel with the resistor 'accelerates' the signal; only an RC circuit integrates (delays)

You are forgetting about the sensor's output impedance and the controller's input impedance!

CU