CC1110-CC1111: Instruction Cycles

Rene Koenig

Part Number: CC1110-CC1111

Hi,

the instruction set reference for the CC1111 says that the "MOVX A,@DPTR" instruction takes 3-10 cycles. The "MOVX @DPTR,A" is similar, takes 4-11 cycles. What does it depend on? Is there way to tell in advance how many cycles these instructions take take? And why does the same opcode have different execution speeds?

I'm reading the Rev. H datasheet, table 37, page 58.

Thanks,

Rene

over 2 years ago

0 TER over 2 years ago

TI__Guru**** 317180 points

I believe the difference depends on the type of memory for the XDATA address you want to access since a register access will be faster than a flash access.

0 Rene Koenig over 2 years ago in reply to TER

Intellectual 390 points

So how long does it take to access the memory beginning at 0xF000, which is not register and not flash?

0 TER over 2 years ago in reply to Rene Koenig

TI__Guru**** 317180 points

I'm not able to easily find a defined answer to this but I would assume 1 -2 cycles more than accessing an register. (4-5 cycles).

0 Rene Koenig over 2 years ago in reply to TER

Intellectual 390 points

Doing development based on assumptions is not what I'm usually doing. However. you're basically saying that it is (probably) faster to use CLR A followed by MOVC A,@A+DPTR, because it is always 4 cycles in total compared to maybe 4-5 or even more cycles by using a simple MOVX A,@DPTR. Did I get that correct?

0 TER over 2 years ago in reply to Rene Koenig

TI__Guru**** 317180 points

I looked through the old project folder and I think I found the documentation for the IP. The timing there is the same as in the datasheet for CC1110/ CC1111. I'm therefor not able to provide you with more information. If the timing is very important I would do some tests accessing the different memory regions and measure the access time.

Sub-1 GHz

Sub-1 GHz forum

CC1110-CC1111: Instruction Cycles