How to output a watchdog signal

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SUMMARY

The discussion focuses on interfacing the Infineon TLE4263-2ES voltage regulator with a watchdog signal using a Freescale S12G microcontroller. It is established that the pulse signal sent to the voltage regulator must have a frequency that exceeds the timeout period, which is determined by the capacitor value (Cd). For a capacitor value of 100nF, the shortest timeout is 16ms, while 200nF extends it to 32ms. The correct implementation involves sending a positive edge to the watchdog input followed by a LOW signal until the next positive edge is required.

PREREQUISITES
  • Understanding of Infineon TLE4263-2ES voltage regulator specifications
  • Familiarity with Freescale S12G microcontroller architecture
  • Knowledge of timer-based interrupts in embedded systems
  • Basic electronics concepts, particularly capacitive timing
NEXT STEPS
  • Research the timing characteristics of the Infineon TLE4263-2ES voltage regulator
  • Learn about configuring timer-based interrupts in Freescale S12G microcontrollers
  • Study the effects of different capacitor values on timeout periods for watchdog signals
  • Explore best practices for implementing watchdog timers in embedded systems
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Embedded systems engineers, microcontroller developers, and anyone involved in designing reliable power management solutions with watchdog timers.

ashifulk
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Hello, I'm trying to interface an Infineon TLE4263-2ES voltage regulator with a external watchdog figure. The microcontroller is from the Freescale S12G family. My question is kind of gereic, in a way. When I'm sending a pulse signal to the voltage regulator from the micro, at what frequency should I send the signal in? I will be using a timer based interrupt to do this.

Attached are the watchdog and reset information from the voltage regulator datasheet. Any help is appreciated.
 

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If you look at figure 5 at the very end of the data sheet, it shows that you need a positive edge on the watchdog input to reset the timer. The rate of positive edges has to be faster than the timeout time, obviously. The equations under the picture explain how to determine the timeout time.

With Cd = 100nF the shortest timeout time (given part to part variation) is 16ms. Change Cd to 200nf for 32ms, etc (per the formula)
 
Thank you for the reply. I do have one more question. After sending a positive edge on the watchdog input, I just bring that pin to "LOW" until the next time I send out a positive edge, correct? Sorry if this question is very simple. For some reason I'm having a hard time understanding this.
 
That will work fine.
 

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