Hi Ashgar!
This is a 24bit BIN to 32bit BCD-Routine i wrote a few weeks ago, it´s tested ok.
;************************************************************************************************************
;
; 24BIT_BIN_to_BCD
;
;Eingabe: r2=High r1=middle r0=Low
;
;Ausgabe: r3=High im BCD-Format
; r2=middle2
; r1=middle1
; r0=Low im BCD-Format
;
; r4=Schleifenzähler
; r5=buffer
; 25h= low buffer
; 26h= middle buffer
; 27h= high buffer
;
;
I24BIT_BIN_to_BCD:
mov 25h,r0
mov 26h,r1
mov 27h,r2
mov r0,#00
mov r1,#00
mov r2,#00
mov r3,#00
mov r4,#24
BIN2BCD1:
clr c
mov a,25h
rlc a
mov 25h,a
mov a,26h
rlc a
mov 26h,a
mov a,27h
rlc a
mov 27h,a
mov a,r0
rlc a
mov r0,a
mov a,r1
rlc a
mov r1,a
mov a,r2
rlc a
mov r2,a
mov a,r3
rlc a
mov r3,a
dec r4
cjne r4,#00,BIN2BCD2
ret

BIN2BCD2:
mov a,r0
add a,#3h
clr c
jnb acc.3,BIN2BCD3
;ist bit 3gesetzt?
;Falls nein zur
;Stelle BIN2BCD3
;springen
mov r0,a
BIN2BCD3:
mov a,r0
add a,#30h
clr p
jnbacc.7,BIN2BCD4
;ist bit 7 gesetzt?
;falls nicht zu BIN2BCD4 springen
mov r0,a
BIN2BCD4:
mov a,r1
add a,#3h
clr p
jnb acc.3,BIN2BCD5
mov r1,a
BIN2BCD5:
mov a,r1
add a,#30h
clr p
jnb acc.7,BIN2BCD6
mov r1,a
BIN2BCD6:
mov a,r2
add a,#3h
clr p
jnb acc.3,BIN2BCD7
mov r2,a
BIN2BCD7:
mov a,r2
add a,#30h
clr p
jnb acc.7,BIN2BCD8
mov r2,a
BIN2BCD8:
mov a,r3
add a,#3h
clr p
jnb acc.3,BIN2BCD9
mov r3,a
BIN2BCD9:
mov a,r3
add a,#30h
clr p
jnb acc.7,BIN2BCD10
mov r3,a
BIN2BCD10:
sjmp BIN2BCD1