MSP-EXP430F5529LP: Delay calculation

BEERESH GUDADAR

Hi Friends,

I am using MSP430F5529LP and default DCO of 1.048576 MHz clock.I am not getting 1ms delay for 1000 counts(i.e accurate 954 counts) instead i am getting 10ms and i attached my code also,please help guys.

Thanks in advance

Beeresh

over 8 years ago

0 zrno soli over 8 years ago

Guru 34793 points

Display generated code in assembler from your C source, and than you will see how many CPU cycles loop will take.

0 Bruce McKenney over 8 years ago

Guru 101335 points

1) Do as zrno soli says, because it will be enlightening.
2) Turn on the optimizer, and repeat (1); this will also be enlightening.
3) Use __delay_cycles (or better yet a timer) so as to avoid second-guessing the compiler. __delay_cycles was specifically designed to avoid the pitfalls observed in (1) and (2).

0 BEERESH GUDADAR over 8 years ago in reply to zrno soli

Intellectual 390 points

Hi zrno soli

I attached the Disassembly code below,how many machine cycles are consumed for 1000 counts.

Thanks in advance

Beeresh

0 zrno soli over 8 years ago in reply to BEERESH GUDADAR

Guru 34793 points

L2 inc.w &i (add.w #1, &i) ; 4 cycles

cmp.w #03e8, &i ; 5 cycles

jlo L2 ; 2 cycles

As you can see, each iteration is 11 times more than you think. And because of this instead of 1ms you measured 10ms.

0 BEERESH GUDADAR over 8 years ago in reply to zrno soli

Intellectual 390 points

I got it,thanks a lot

0 Jürgen Wissenwasser over 8 years ago in reply to BEERESH GUDADAR

Intellectual 490 points

If you want to ensure a clear delay you may include a short piece of assembler code that may look like this (I do most of my time-critical work in asm):

    push.w R15 // Save R15, 2 cycles
    mov #[Value],R15 // For values 1, 2, 4, 8, 0xFFFF this takes 2 cycles, otherwise 3.
Loop: dec R15 // 1 cycle
    jnz Loop // 2 cycles
    pop.w R15 // Restore R15, 2 cycles

This means a total of (2+2+2+3*Value)=(6+3*Value) or (2+2+3+3*Value)=(7+3*Value)

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MSP-EXP430F5529LP: Delay calculation