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Pelatihan Dasar Mikrokontroller
14-15 April 2007
Surabaya
Rp. 300.000
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Microcontroller Kits
Programmer and Target 89s51
Rp.100.000
(USD $10)
Simple Mikrokontroller 89s51 Trainer
Rp, 350.000
(USD $35)
Standart
Mikrokontroller 89s51 Trainer
Rp. 650.000
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Super Mikrokontroller Trainer 89s51
Rp.1.250.000
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32bit Floating point package for the 8051
; 32bit Floating point package for the 8051; Created by Jerson Fernandes for the C16 meta assembler
; floating point format ; exponent, 24 bits mantissa ; xxxxxxxx, Sxxxxxxx xxxxxxxx xxxxxxxx ; stored internally from right to left
CPU "8051.tbl" hof "INT8" incl "8051sfr.asm"
ORG 8 ;internal memory float_arg: equ 3 ;3 bytes mantissa float_exp: equ 1 ;1 byte exponent expbias: equ 128 ;exponent bias
Quo: dfs float_arg ;holds quotient of divide Aarg: dfs float_arg Aexp: dfs float_exp Barg: dfs float_arg Bexp: dfs float_exp Aext: dfs 1 ;extension for Aarg Aext2: dfs 1 ;for holding the aligned bit TOS:
ORG 0
;This segment is to test the functionality of the librarystart: mov sp,#TOS ;point to the top of stack mov Aarg,#low 1 mov Aarg+1,#high 1 mov Aarg+2,#0 ;high -10 acall int2float ; acall copy_ab ; mov Aarg,#low 1000 mov Aarg+1,#high 1000 mov Aarg+2,#0 acall int2float ; acall fpmul32 mov a,aarg push acc mov a,Aarg+1 push acc mov a,Aarg+2 push acc mov a,Aexp push acc mov Aarg,#low 2000 mov Aarg+1,#high 2000 mov Aarg+2,#0 acall int2float acall copy_ab
pop acc mov Aexp,a pop acc mov Aarg+2,a pop acc mov Aarg+1,a pop acc mov Aarg,a
lcall fpdiv32 acall float2int sjmp start
; This is where the library begins
;Copy ArgA to ArgBCopy_AB:mov r7,#float_arg mov r0,#Aarg mov r1,#Bargcopy_AB_5: mov a,@r0 mov @r1,a inc r0 inc r1 djnz r7,copy_AB_5 mov Bexp,Aexp ret; Convert a integer to a floating point number; Input - 24 bit signed integer @ Aarg; Output- 32 bit floating point number @ Aargint2float: acall Normalize ret; Normalize the 24 bit signed integer in AargNormalize: mov r6,#24 ;exponent counter mov a,Aarg+2 ;extract sign anl a,#80h mov r5,a ;keep in R5 jz norm5;negative number make +ve as we have the sign mov a,Aarg cpl a add a,#1 ;2s complement mov Aarg,a mov a,Aarg+1 cpl a addc a,#0 mov Aarg+1,a mov a,Aarg+2 cpl a addc a,#0 mov Aarg+2,a Norm5: mov r0,#Aarg+2 mov a,@r0 orl a,#0 jnz Norm10 ;the highest 8 are 0, shift 8 mov a,#0 xch a,Aarg xch a,Aarg+1 xch a,Aarg+2 clr c mov a,r6 subb a,#8 mov r6,a ;update the exponent ; jnc Norm5 Norm10:mov a,@r0 ;read the msb jb ACC+7,Norm_end ;shift one bit left mov a,Aarg clr c rlc a mov Aarg,a mov a,Aarg+1 rlc a mov Aarg+1,a mov a,Aarg+2 rlc a mov Aarg+2,a djnz r6,Norm10 ;note the shift in the exponent too Norm_end: mov a,#expbias add a,r6 mov Aexp,a;now make the msbit implicit and put in the sign there mov a,r5 orl a,#7fh anl a,Aarg+2 mov Aarg+2,a ;sign is always for int2float ret
; Float to integer conversion routine ; Input - floating point number in Aarg,Aexp ; Output- 24 bit signed integer in Aarg float2int: mov a,Aexp clr c subb a,#expbias jc res032 ;result is 0 cjne a,#24,$+3 ;overflow,return a 0 int jnc res032 mov r6,a ;keep the exponent here;find out how much to shift the float to get our int mov a,#24 clr c subb a,r6 mov r6,a ;take out the sign and make msb explicit mov a,Aarg+2 ;keep in R5 mov r5,a orl a,#80h mov Aarg+2,a ;explicit msb ;now shift the float to form the int flint10: mov a,r6 ;if exp >= 8, shift bytes clr c subb a,#9 jnc flint20 ;now do bit shifts flint15: clr c mov a,Aarg+2 rrc a mov Aarg+2,a mov a,Aarg+1 rrc a mov Aarg+1,a mov a,Aarg+0 rrc a mov Aarg+0,a djnz r6,flint15 ;we're done getting the int,restore the sign mov a,r5 anl a,#80h jz flint30 ;restore the negative number mov a,Aarg cpl a add a,#1 mov Aarg,a mov a,Aarg+1 cpl a addc a,#0 mov Aarg+1,a mov a,Aarg+2 cpl a addc a,#0 mov Aarg+2,a flint30:ret flint20: ;shift right 8 bits clr a xch a,Aarg+2 xch a,Aarg+1 xch a,Aarg+0 ;and adjust the exp counter in r6 mov a,r6 subb a,#8 mov r6,a sjmp flint10 res032: clr a mov Aarg+0,a mov Aarg+1,a mov Aarg+2,a ret
;Floating point divide ; Input - Arg 1 in Aarg, Aexp ; - Arg 2 in Barg, Bexp ; Output- Divide arg1 by arg2 ; - result in Aarg ; Temp - R5 resulting sign ; acc for dividend is ; Aarg[2,1,0],Quo[2,1,0] msb first fpdiv32: ;check if Bexp == 0 mov a,Bexp jnz fpd10 ajmp fpdivz fpd10: mov a,Aarg+2 ;extract resulting sign xrl a,Barg+2 anl a,#80h ;keep the sign bit here mov r5,a ;make the msb explicit in args mov a,#80h orl a,Aarg+2 mov Aarg+2,a mov a,#80h orl a,Barg+2 mov Barg+2,a ;if A<B mantissa, divide, else align clr a mov Quo,a mov Quo+1,a mov Quo+2,a clr c mov a,Aarg subb a,Barg mov a,Aarg+1 subb a,Barg+1 mov a,Aarg+2 subb a,Barg+2 jc fpdok ;make A mantissa < B mantissa clr c mov a,Aarg+2 ;msb rrc a mov Aarg+2,a mov a,Aarg+1 rrc a mov Aarg+1,a mov a,Aarg rrc a mov Aarg,a ;lsb mov a,Aext2 ;msb rrc a mov Aext2,a ;move into quotient ;adjust the Aexp accordingly mov a,Aexp inc a mov Aexp,a ;get the resulting exponent in Aexp fpdok: clr c mov a,Aexp subb a,Bexp add a,#expbias mov Aexp,a ;resulting exponent here fpdargok: clr a mov Aext,a mov r7,#24 ;initialize bit counter for divide fpdloop32: ;shift left 1 bit clr c mov a,Aext2 rlc a mov Aext2,a mov a,Aarg+0 rlc a mov Aarg+0,a mov a,Aarg+1 rlc a mov Aarg+1,a mov a,Aarg+2 rlc a mov Aarg+2,a ;move the bit into our accumulator mov a,Aext rlc a mov Aext,a ;subtract clr c mov a,Aarg+0 subb a,Barg+0 mov a,Aarg+1 subb a,Barg+1 mov a,Aarg+2 subb a,Barg+2 mov a,Aext subb a,#0 jc nosub ;cannot subtract clr c mov a,Aarg+0 ;subtract subb a,Barg+0 mov Aarg+0,a mov a,Aarg+1 subb a,Barg+1 mov Aarg+1,a mov a,Aarg+2 subb a,Barg+2 mov Aarg+2,a mov a,Aext subb a,#0 mov Aext,a nosub: cpl c ;shift in the complement of C mov a,Quo ;into the result rlc a mov Quo,a mov a,Quo+1 rlc a mov Quo+1,a mov a,Quo+2 rlc a mov Quo+2,a djnz r7,fpdloop32 ;move the quotient to Aarg and make msb implicit mov a,Quo+1 mov Aarg+1,a mov a,Quo+0 mov Aarg+0,a mov a,r5 jnz minusdiv mov a,Quo+2 anl a,#7fh ;make msb implicit if +ve number mov Aarg+2,a ret minusdiv: mov a,Quo+2 mov Aarg+2,a ret fpdivz: ;return divide by zero error ret fpov: ;return divide overflow ret
;Floating point multiplication ; Input - Aarg,Aexp ; - Barg,Bexp ; Output- Aarg,Aexp ; partial product = Quo[2..0],Aarg[2..0] fpmul32: ;if A == 0 || B == 0 result = 0 mov a,Aexp jz fpm10 mov a,Bexp jnz fpm20 fpm10: ljmp res032 fpm20: ; get the resulting exponent clr c mov a,Aexp subb a,#expbias add a,Bexp mov Aexp,a ;store resulting exponent here jnc fpm30 ljmp fpov fpm30: ;make the msbs explicit mov a,Aarg+2 orl a,#80h mov Aarg+2,a mov a,Barg+2 orl a,#80h mov Barg+2,a ;now multiply using Quo as the partial product clr a mov Quo,a mov Quo+1,a mov Quo+2,a mov r7,#24 ;bits fpmloop: ; mov a,Aarg rrc a jnc fpmnoadd ;add the multiplier to the product mov a,Quo add a,Barg mov Quo,a mov a,Quo+1 addc a,Barg+1 mov Quo+1,a mov a,Quo+2 addc a,Barg+2 mov Quo+2,a fpmnoadd: ;right shift the product mov a,Quo+2 rrc a mov Quo+2,a mov a,Quo+1 rrc a mov Quo+1,a mov a,Quo rrc a mov Quo,a ;shift right the multiplicand clr c mov a,Aarg+2 rrc a mov Aarg+2,a mov a,Aarg+1 rrc a mov Aarg+1,a mov a,Aarg rrc a mov Aarg,a fpm40: djnz r7,fpmloop ;multiplication is over ;check for normalisation now mov a,Quo+2 jb acc+7,fpm50 clr c mov a,Quo rlc a mov Quo,a mov a,Quo+1 rlc a mov Quo+1,a mov a,Quo+2 rlc a mov Quo+2,a dec Aexp fpm50: mov Aarg,Quo mov Aarg+1,Quo+1 mov a,r5 anl a,#80h ;check sign of result jz fpmplus mov Aarg+2,Quo+2 ret fpmplus: mov a,Quo+2 anl a,#7fh mov Aarg+2,a ret
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Programmer
ISP 89s
Free Software
a. Edsim 51
b. MIDE-51
c. ATMEL ISP
Lesson 1:
Architecture
1.1.Memory
1.2.SFR
1.3.Addressing
1.4.Instruction Set
1.5.Assignment
Lesson 2:
Input Output
2.1.LED
2.2.Swicht
2.3.7 Segmen
2.4.LCD Character
2.5.ADC
2.6.DAC
2.7.Motor Stepper
2.8.Keypad
2.9.Assignment
Lesson 3:
Timer Counter
3.1.Basic
3.2.Mode 0
3.3.Mode 1
3.4.Mode 2
3.5.Mode 3
3.5.Assignment
Lesson 4:
Serial Comm.
4.1.Basic
4.2.LED
4.3.Rotate LED
4.2 ADC
4.3.LCD
4.4.Assignment
Lesson 5:
Interuption
5.1.Basic
5.2.Timer
5.2.External
5.3.Assignment
