How long does a nop instruction take at max. HSI (16MHz) ?

[jeffrey23]

Posted on April 29, 2010 at 01:51

How long does a nop instruction take at max. HSI (16MHz) ?

[fggnrc2]

Posted on May 17, 2011 at 15:08

Boltnut,

an empty ISR doesn't takes as much clock cycles as a nop.

The STM8 CPU Programming Manual shows what happens when an ISR is serviced:

the STM8 core pushes PCL, PCH, PCE, Y, X, A and CC registers onto the stack before fetching the ISR interrupt vector. This takes 9 clock cycles.

If your ISR is empty, i.e. it consists only in an IRET, another 9 clock cycles are needed to restore those registers.

EtaPhi

[jeffrey23]

Posted on May 17, 2011 at 15:08

EtaPhi,

Do you mean that a NOP instruction takes 1/16MHz = 62.5 ns

if fMASTER = 16MHz ?

[fggnrc2]

Posted on May 17, 2011 at 15:08

Boltnut,

the STM8 core has got a three stage pipeline.

If an instruction stream contains no jump instructions, each NOP instruction takes one clock cycle, i.e. 1/16MHz = 62.5 ns.

Your conclusion is true, but only if the instruction pipeline is full.

Regards

[jeffrey23]

Posted on May 17, 2011 at 15:08

Hi EtaPhi,

I need a shorter delay time for I2C timing and

implement the following function:

void I2C_Delay (UB8 bTime)

{

    bTime /= 3;             // 3.5 cycles each loop !

//69               ; 97     while (bTime--)

//71  0005 7b01             ld  a,(OFST+1,sp)       ; 1.0 cycle

//72  0007 0a01             dec (OFST+1,sp)         ; 1.0 cycle

//73  0009 4d               tnz a                   ; 1.0 cycle

//74  000a 26f9             jrne    L13             ; 0.5 cycle

    while (bTime--)

        ;

} /* I2C_Delay */

Is it correct ?

How to estimate the delay time correctly ?

Regards,

Boltnut

[fggnrc2]

Posted on May 17, 2011 at 15:08

Hello Boltnut.

If you mean ''smaller delay granularity'' in your request, you can use:

while ( --bTime )

   ;

which your compiler should translate in:

      ld  a,(OFST+1,sp)

L13:

      dec a

      jrne L13

This code should take 3 x bTime + 1 clock cycles to execute (if interrupts aren't allowed)

There is a simple way to measure how long a delay routine takes: add an instruction which toggles an output pin before and after the delay routine.

Then use an oscilloscope to see and measure the waveform.

Regards

EtaPhi

[jeffrey23]

Posted on May 17, 2011 at 15:08

Hi EtaPhi,

Thanks for your advice.

I tried to revise it as below:

void I2C_Delay (UB8 bTime)

{

//  59  0000 88             push a

//  75            ; 106     while (--bTime)

//  77  000d 0a01           dec (OFST+1,sp)         ; 1.0 cycle

//  78  000f 26fc           jrne    L13                      ; 0.5 cycle

    while (--bTime)

        ;

//  71  0005 84             pop a

//  72  0006 81             ret

} /* I2C_Delay */

It is good enogh to my application.