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How to reduce watchdog timeout interval to 1.5 mSec for TMS320F28335?

Sejal Gandhi
Intellectual 960 points

Other Parts Discussed in Thread: TMS320F28335

Hi Champs,

My query is regarding TMS320F28335.

To set watchdog Timeout interval, we use the following sequence of steps :

1) 24MHz/512/x = y Hz (OSCCLK/512/x) (where x = 1,2,4,8,16,32,64 => lesser the value of x, less Timeout interval) 

2) Inverse y so that frequency is expressed in  Seconds

 (T = 1 / f )

3) Multiply inversed y with watchdog register size, i.e., if it's a 8 bits register, multiply with 256.

4) Minimally i will get 5.46 mSeconds

I want this timeout interval reduce to 1.5 mSeconds

Please suggest something on this.

Regards,

Sejal

  • 0
    TI__Guru 61235 points

    Sejal,

    As you computed, the minimum WD timeout is:

    WD_timeout_min = (1/f_input)*512*256

    For a 30 MHz input clock (or crystal), WD_timeout_min = 4.37 ms.

    You seem to be using a 24 MHz input clock, so you get WD_timeout_min = 5.46 ms.

    You cannot decrease this without using a slower input clock, but then that limits the maximum frequency you can operate the device at.

    Regards,

    David