system timer

[xxxx]

how do i find out the CLK input of the system timer?
my motherboard is 81845gvm-rz with intel 845gv chipset.the CPU is pentium 4 533/400mhz fsb

in the user's manual there's a block diagram for the clock generator and it has 10 output lines:

1.PCICLK(33MHZ)
2.USBCLK(48MHZ)
3.14.318MHZ
4. 33MHZ
5. 24MHZ
6.MCHCLK +/-(100/133MHZ)
7.CPUCLK +/-(100/133MHZ)
8.AGPCLK(66MHZ)
9.GMCHCLK(66MHZ)
10.ICH3V66(66MHZ)

i don't know which out is for the system timer.plz help

thanks

[MichaelW]

The standard clock for the system timer runs at 1,193,182 Hz, derived for the original IBM PC as the 4.77 MHz processor clock divided by 4. I doubt that your system uses a different clock speed, but you can test it, under Windows 9x or under MS-DOS only, with this DOS program. Note that system timer 2 is not available under Windows 2000 or XP, so this program will not delay.

Code Select Expand


;--------------------------------------------------------------
; If the system timer clock is somewhere close to 1,193,182Hz,
; This program should delay 10 seconds between prompts.
;--------------------------------------------------------------
    .model small
    .386
.stack
.data
    tobegin   db "Press any key to begin",13,10,"$"
    finished  db "Finished, press any key to exit",13,10,"$"
.code
.startup

    ; Set the gate for timer 2 (bit 0 at I/O port 61h) to off.
    ; To avoid changing other bits in the register, read the
    ; current value, set bit 0, and write the altered value
    ; back.
    mov   dx,61h
    in    al,dx
    and   al,NOT 1
    out   dx,al

    ; Program timer 2 for LSB then MSB, mode 0, binary.
    ; The system timer control word register is at I/O
    ; port 43h. The system timer control word is set as
    ; follows:
    ;   bit 7-6:  10  = timer 2
    ;   bit 5-4:  11  = R/W LSB then MSB
    ;   bit 3-1:  000 = single timeout
    ;   bit 0:    0   = binary
    mov   dx,43h
    mov   al,0b0h
    out   dx,al

    ; The system timer normally has a 1,192,182Hz clock.
    ; For a full timer cycle you load an initial count
    ; of zero, which, because the count is decremented
    ; before it is checked for zero, causes the timer
    ; to count 65,536 clock cycles before it times out.
    ; Each cycle with take 65536/1192183 ~ 55ms, so
    ; 182 cycles should take ~10 seconds.

    mov   ah,9
    mov   dx,OFFSET tobegin
    int   21h 
    mov   ah,0
    int   16h

    mov   cx,182
  looper:
    ; Load the starting value, LSB then MSB.
    mov   dx,42h
    mov   al,0
    out   dx,al
    out   dx,al

    ; Set the gate for timer 2 (bit 0 at I/O port 61h) to on.
    mov   dx,61h
    in    al,dx
    or    al,1
    out   dx,al
   
    ; Wait until the output bit (bit 5 at I/O port 61h) is set.
  @@:
    in    al,dx
    and   al,20h
    jz    @B 

    loop looper

    mov   ah,9
    mov   dx,OFFSET finished
    int   21h
    mov   ah,0
    int   16h

    .exit
end

MAKEIT.BAT:

ML /c timer.asm
PAUSE
LINK16 timer.obj;


[xxxx]

my os is windows XP and i ran it from the command prompt and it did not delay.so i guess this program won't delay even if it was run ifrom the command prompt.

anyway:

1." ; Wait until the output bit (bit 5 at I/O port 61h) is set.
  @@:
    in    al,dx
    and   al,20h
    jz    @B  "

   what's the reason for waiting bit 5?is it because it changes it's state every 15.085microseconds?according to my book,it is bit 4 that changes it state every 15.085 microseconds.what is ur comment?

2.accoding to my book,this is the code to generate a 10 second  time  time delay.it only uses port 61h.any comment?
.MODEL SMALL
    .STACK 64
    .386
    .DATA
   
   
    tobegin   db "Press any key to begin",13,10,"$"
    finished  db "Finished, press any key to exit",13,10,"$"
   
   
   
   
    .CODE
    MAIN PROC FAR
    MOV AX,@DATA
    MOV DS,AX
   
   
   mov   ah,9
    mov   dx,OFFSET tobegin
    int   21h
   
    mov   ah,0
    int 16h
   
    sub bx,bx
    mov bl,48
again:   mov cx,55260
   
    call delay
    dec bl
    jnz again
   
   
   
   
   mov   ah,9
    mov   dx,OFFSET finished
    int   21h
     mov   ah,0
    int   16h
   
   
   MOV AH,4CH
   INT 21H
   
   
   MAIN ENDP
   
   DELAY PROC
   
   PUSH AX
   WAITF1: IN AL,61H
           AND AL,10H
           JE WAITF1
           MOV AH,AL
           LOOP WAITF1
   
   
   POP AX
   Ret
DELAY EndP
END MAIN

3.why did you load counter 2 with mode 0?






thnak you

[MichaelW]

Bit 5 of I/O port 61h is the output for timer 2, and bit 4 is the output for timer 1.

I used mode 0, Single Timeout, because it is suitable for creating a short programmable delay, which is what I originally used the code for. I simply adapted existing code as a crude method of checking the system timer clock speed.

I recall trying to use timer 1 to create a programmable delay for DOS programs running under Windows 2000/XP, but I ended up using a different method because the output of timer 1 is far too erratic for short delays under Windows 2000/XP. I suspect that the output you see is just some crude emulation that Microsoft provided to support some common application that depended on it. The code below will delay for 10 seconds when run from a Windows 9x boot disk, but the delay is more like 8 seconds under Windows 2000.

Code Select Expand


;--------------------------------------------------------------
; If the system timer clock is somewhere close to 1,193,182Hz,
; this program should delay 10 seconds between prompts.
;
; System timer 1 was originally used to generate periodic DRAM
; refresh requests. It is normally programmed to operate in
; mode 2 and loaded with an initial count of 18. This causes
; it to cycle once each 18/1193182 = 15.086 microsecond and
; drive bit 4 of I/O port 61h low for a period of 1 count
; (1/1193182 = 838 nanosecond).
;--------------------------------------------------------------
    .model small
    .386
.stack
.data
    tobegin   db "Press any key to begin",13,10,"$"
    finished  db "Finished, press any key to exit",13,10,"$"
.code
.startup   

    mov   ah,9
    mov   dx,OFFSET tobegin
    int   21h 
    mov   ah,0
    int   16h

    mov   ecx,10*1193182/18
    mov   dx,61h
  looper:   
    in    al,dx
    and   al,10h
    mov   ah,al
  @@:
    in    al,dx
    and   al,10h
    cmp   ah,al
    je    @B   
    loopd looper

    mov   ah,9
    mov   dx,OFFSET finished
    int   21h
    mov   ah,0
    int   16h

    .exit
end

MAKEIT.BAT:

ML /c timer.asm
PAUSE
LINK16 timer.obj;


[xxxx]

thanks alot.very helpful info