S12 / MagniV Microcontrollers中的主题 Re: S12 8MHz - NOP processing time

S12DP512 8MHz Bus - NOP processing time

pon — Thu, 20 Sep 2007 17:14:42 GMT

i am using HCS12 16 bit 8MHz microcontrller. can anyone give me the porcessing time of the asm "nop" instruction?

--
Alban Edit: Please always include FSL Part Number in Message Subject line.

Message Edited by Alban on 2007-09-20 11:20 AM

Message Edited by Alban on 2007-09-20 12:15 PM

Re: S12 8MHz - NOP processing time

Alban — Thu, 20 Sep 2007 17:25:21 GMT

Hello,

It depends on MANY factors.

With a PLL you can get 33MHz bus (or 25MHz, see datasheet). So NOP = 30.3ns.
If XTAL is 8MHz and no PLL, then bus = 4MHz. So NOP = 250ns.
if bus = 8MHz. NOP = 125ns

With PLL to get 1MHz bus, NOP = 1µs.

With amount details you give, answer is 30.3ns < tNOP < 1µs.

Cheers,
Alban.

Re: S12 8MHz - NOP processing time

pon — Thu, 20 Sep 2007 18:02:32 GMT

The controller i use is MC9S12DP512. i guess that PLL is 16MHz and the bus is 8MHz. so NOP is 125 ns. am i right? when i make delay in program using NOP its giving me wrong delay. i am using NOP within loop in C to make dely.. should i calculate loop timing also?

Re: S12DP512 8MHz Bus - NOP processing time

Alban — Thu, 20 Sep 2007 18:21:05 GMT

Hey Pon,

Yes, 125ns is right. You also need to count the time of the Branch (BNE)

TotalDelay = (tNOP + tBranch) * NbLoop

When you compile, in the LIST file, you can have within square brackets the number of clock cycles for each instruction.

Alban.

Re: S12DP512 8MHz Bus - NOP processing time

pon — Thu, 20 Sep 2007 19:10:33 GMT

thanks a lot!! anyhow i could not find the clock cycle deatils in LIST. here are part of the code form list, is A7 is clock cycle needed for NOP?

TIMER_EXTERN void delay10us (void)
0022           ; {
0022           ;
0022           ;     asm("nop");asm("nop");asm("nop");asm("nop");
0022 A7                nop
0023 A7                nop
0024 A7                nop
0025 A7                nop
0026           ;       asm("nop");asm("nop");asm("nop");asm("nop");
0026 A7                nop
0027 A7                nop
0028 A7                nop
0029 A7                nop
002A           ;       asm("nop");asm("nop");asm("nop");asm("nop");
002A A7                nop
002B A7                nop
002C A7                nop
002D A7                nop
002E           ;       asm("nop");asm("nop");asm("nop");asm("nop");
002E A7                nop
002F A7                nop
0030 A7                nop
0031 A7                nop
0032           ;       asm("nop");asm("nop");asm("nop");asm("nop");
0032 A7                nop
0033 A7                nop
0034 A7                nop
0035 A7                nop
0036           ;       asm("nop");asm("nop");asm("nop");asm("nop");
0036 A7                nop
0037 A7                nop
0038 A7                nop
0039 A7                nop
003A           L2:
003A                   .dbline 0 ; func end
003A 3D                rts
003B           ;              i -> 0,SP
003B           ;           time -> 2,SP
003B           _smallDelay::
003B 3B                pshd
003C 1B9E              leas -2,S
003E           ;
003E           ;   }

Re: S12DP512 8MHz Bus - NOP processing time

Alban — Thu, 20 Sep 2007 19:35:27 GMT

Hey Pon,

No A7 is the OPCODE in HEXADECIMAL.
The 0022 is the ADDRESS.

In a nutshell, A7 is the data located at $0022.

You have 6 times 4 NOP = 24 NOP = 125ns * 4 = 0.5µs.

I don't understand why there is an RTS in the delay10us() function.

Alban.

Re: S12DP512 8MHz Bus - NOP processing time

pon — Thu, 20 Sep 2007 20:20:35 GMT

i confused by the sentence "You have 6 times 4 NOP = 24 NOP = 125ns * 4 = 0.5µs." ,
i guess 25ns * 24 = 3µs.

RTS is comming after end of the delay10us() function. i coppied little remainig parts too.

Re: S12DP512 8MHz Bus - NOP processing time

Alban — Thu, 20 Sep 2007 20:36:06 GMT

Ok, yes 3µs, I dropped 20

So either you do a loop, either you extend the number of NOP to reach 10µs.