Binary counter for large numbers
For high frequencies, i.e. short times, the controllers spoil us with timers.
But Software is necessary for very long periods of time. So, I developed a small counting loop with a few program lines with my S08 BDM board. A few RAM bytes are required as a counter. The simple algorithm is independent of the controller and programming language.
Algorithms for the 3-byte counter cycle (Waiting loop for a single run)
- Binary counter means that all bits of the bytes used are counted with increasing significance.
- Counter for large numbers means: The maximum count number increases with 2n of n bits. Three bytes of 8 bits each, can also count up to 224 = 16.777.216, Four Bytes can count to 4.294.967.296 and so on.
Time counters need a clock. For this I used a TPM channel xxxxx with a toggle output. This enabled me to measure the clock on port PTA0.
Here is the algorithm for the TPM-ISR controlled step-by-step operation of the counting loop:
Algorithms for step-by-step operation
The assembler program:
; -- 3 Byte with TPM --
; TPMCH0V, Bus cycle without divider
XDEF Entry,main,T_ISR
Include 'derivative.inc'
Z1 EQU $67 ; Define 3 byte RAM places
Z2 EQU $68
Z3 EQU $69
Init_Timer:
mov #$4D,TPMC0VH ; H,L Comparison values for 10 ms
mov #$E6,TPMC0VL ;
mov #$54,TPMC0SC ; Channel 0 settings for toggle output on compare
bset 3, TPMSC ; Switch on timer start by bus
rts
; ---------------------------------------------------------------------
Load:
mov #$0F, Z1
mov #$27, Z2
mov #$00, Z3
rts
; ---------------------------------------------------------------------
main:
Entry:
Main_Init:
bsr Init_Timer ; Initialize Timer
bsr Load ; Load 3 byte counter values
lda #$12
sta SOPT1 ; disable watchdog
cli ; Allow interrupts to happen
lda #$FF
sta PTBDD ; Port B Pins are output and low
; ----------------------------------------------------------------------
loop:
bra loop
; ---------------------------------------------------------------------
T_ISR:
bclr 7, TPMC0SC ; Clear overflow flag
bset 7, TPMCNTH ; setting of any bit clears the counter
; -----------------------------------------------------------------------------
; Counting the Interupts with 3 bytes,
; ----------------------------------------------------------------------------
tst Z1 ; see figure with the algorithm
E: beq Z
Ee: dec Z1
rti
Z: tst Z2
beq Ha
Zz: dec Z2
bra Ee
Ha: tst Z3
beq La
dec Z3
M01: bra Zz
La: bsr Load
; --------------------------------------------------------------------
; Switch PTB6
; ---------------------------------------------------------------------
brset 6,2,L1
bset 6,2
rti
L1: bclr 6,2
rti
nop
END
I hope my thoughts are understandable and maybe of use to someone
Gerd Meinert
