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TCAN4550: Wake Pin external components

Part Number: TCAN4550

Hi,

On page 24 (8.3.6) of the datasheet, the following sentence can be read in relation to the wake pin:
"Pin requires a 10 nF capacitor to ground for improved transient immunity in
applications that route WAKE externally. "

I understand that the 10nF MLCC is not required if the line is only routed internally on the print and is therefore not exposed to any major interference.
Can the circuitry be implemented without hesitation as shown in the appendix?
We want to be able to wake up the TCAN4550 through the microcontroller. That is the reason for the MOSFET on the wake pin.

In the data sheet on page 1 there is also a 33k resistor in front of the wake pin, the purpose of which is not entirely clear to me.
To save space, I would like to leave it out if it does not cause any problems.

Thanks in advance!

Best regards, Alen

  • Hi Allen,

    Since the WAKE pin detects a transition of either Low-to-High or High-to-Low, it is susceptible to any noise that may present itself on the pin.  The 10nF capacitor helps to filter this noise and prevent false wake up events.  As you have mentioned, if your system is designed such that the risk of noise on this pin is minimized then this capacitor is optional and can be left off.  But you are correct in that this capacitor is not necessary for the TCAN4550 to operate properly.

    I also believe your understanding of the WAKE pin control is correct and the use of a MOSFET is the recommended method for controlling the pin from a MCU.  The 33k resistor shown in the datasheet is typically included as a current limiting resistor and as a protection against possible fault conditions that can come from connections directly to the high voltage VBAT and GND connections.  In you shown schematic, the 1K resistor will offer some current limiting protection between VBAT and the WAKE pin, but you will effectively be directly shorting the WAKE pin to GND without any current limiting protection. But the WAKE pin will still detect a voltage transition if you remove the resistor.  You will just not have any external protection against faults and over-current situations that may arise.

    Regards,

    Jonathan