Flowcode Support Board

Pages 13 and 14 of the Locktronics Curriculum CP8408 (Automotive ECU Architecture and Logic) show two 5.6kΩ resistors in series between the input I1 and the 12V power rail to create a pull-up. Is this correct? I believe the voltage on I1 will be 12V when the input is floating, which exceeds the maximum voltage rating for the microcontroller's input pin. Instead, I would use a voltage divider with the two resistors between the input and the 12V rail, with the junction between the resistors connected to the microcontroller's input pin. Or am I overlooking something?

What is the purpose of the 220ohm resistors connected beween the terminals of the Micro ECU and the pins of the microcontroller?

Not sure if we are looking at the same thing, but I see the two resistors are arranged as a potential divider.
The 220 ohm resistor will give the MCU pin some protection by limiting the current out, or into the catch diodes.

Hi

Further to LeighM, I don't think 12V will be an issue as the ECU module is designed for 12V operation within an automobile environment. I sincerely doubt that Matrix would offer up a product incompatible with their course. Without seeing the circuit diagram of the module I can't be 100% sure but I would expect any voltage division is taken care of within the module itself.

Regards

Hi
I attached the schematic from page 14, which shows the two series resistors used to create a pull-up for input I1 and the voltage divider used to create a pull-down for input I2.

Kind regards.

Hi

The diagram only shows a Block Diagram of the ECU module, not the internal components that form the ECU, and you can see that the ECU has direct connections to both +12v and 0v. Therefore I would again suggest that any voltage division necessary for the ECU is taken care of within the ECU module itself.

However I'm sure Matrix themselves will respond once the New Year break is over.

Regards

Looks like CP8408 might have had revisions over time, as that schematic is different to what I have.
As chipfryer says, Matrix will no doubt give a definitive answer in the new year.

Hello.
The ECU has a 5.1 Zener diodes to protect the inputs of a any voltage greater than 5.1V.
Therefore no damage will occur.
You can't connect 12V directly to the inputs, you will need current limiting resistors.
Suppose the Zener diode was omitted (its not), there will be damage if a 12v low impedance is connected directly across the input.
However if 12V is connected via two 5K6 resistors then the current will be limited (12-5.8)/11.2E3 = 554uA
Since the maximum clamp current is 20mA then the input is protected.

The conclusion is it's perfectly safe to connect the 12V via the two x 5K6 resistors due to the external Zener diode and the internal clamp diode.

I hope this puts your mind to rest?

Hi Martin

Thank you for your explanation. It has clarified things and put my mind at ease. However, I still have a didactic concern. Without explaining the internal structure of the Micro ECU (e.g., the Zener diode) to the students, the pull-up configuration with two resistors to the 12V power supply could be confusing. Is a (principle) schematic of the Micro ECU available?
Additionally, I assume that a voltage divider, with two resistors between the input and the 12V rail and the junction connected to the microcontroller's input pin, would also function as a pull-up for the input. Is this correct?

Kind regards

Patrick