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init_lcd ;(this initialises a 2 row lcd) bcf TRISA,0 ;PORTA bit 0 as an output (lcd RS pin) bcf TRISA,1 ;PORTA bit 1 as an output (lcd R/W pin) bcf TRISA,2 ;PORTA bit 2 as an output (lcd E pin) bcf lcd_rs bcf lcd_rw bcf lcd_en movlw b'00110000' ;Instruction - function set. First of 3 call lcd_write call Delay5 ;wait MORE THAN 4.1mS movlw b'00110000' ;Instruction - function set. Second of 3 call lcd_write movlw .33 ;wait MORE THAN 100uS [(33 x 3) +1 +1 + (1 x 2)] - measured 102uS on scope movwf counta ;1 cycle decfsz counta ;1 cycle until true then a 2 cycle goto $-2 ;2 cycle instruction movlw b'00110000' ;Instruction - function set. Third of 3 call lcd_write movlw b'00111000' ;8 bit data / 2 line display / 5x8 character font call lcd_write call busy movlw b'00001111' ;instruction bit / display on / cursor position on / blinking cursor on call lcd_write call busy movlw b'00000001' call lcd_write call busy movlw b'00000110' ;increment cursor position by 1 / don't shift display call lcd_write call busy return ;return from initialization routinelcd_write movwf lcd_data ;data to port bsf lcd_en ;sets LCD enable pin high bcf lcd_en ;data is latched into lcd on high to low transition of E returnbusy bsf TRISC,7 ;was output when sending init data, now input to read busy flag bsf lcd_rw ;set high to read data from lcd bsf lcd_en ;enable btfsc lcd_busy ;is the busy flag low? goto $-2 ;no, go back and keep checking until it is bcf lcd_en ;busy flag now clear bcf lcd_rw ;set lcd back to write mode bcf TRISC,7 ;port bit as an output again returnDelay30 movlw .30 ;delay 30mS goto d0Delay5 movlw .5 ;delay 5msd0 movwf count1d1 movlw 0xC7 movwf counta movlw 0x01 movwf countbDelay_0 decfsz counta goto $+6 decfsz countb goto Delay_0 decfsz count1 goto d1 retlw .0 ;return from the delay routine
i have trying to do the homework but there is a mistake. (Counting positive and negative numbers and computing the average of numbers) write a program that reads an unspecified nu
move a byte string ,16 bytes long from the offset 0200H to 0300H in the segment 7000H
Segment Registers The 8086 addresses a segmented memory unlike 8085. The complete 1 megabyte memory, which 8086 is capable to address is divided into 16 logical segments.Thusea
errorlevel -302 ;prevents error code for this chipset __config 0x373A ;chip config PIC spec page 146 processor 16F877A ;chipset reset code
DEC : A powerful new Alpha 64 bit RISC computer chip was introduced in the year 1977, as new VAX (Virtual Address Extension) Computer. The VAX was 32 bit computer line based on
Write a program to calculate the first 20 numbers of Fibonacci series. Use the stack (memory) to store the calculated series. Your debugger output should look like the following sc
segmentation and overlapping in assemble language
I have two homework assignments due in 10 hours for the x86 processor assembly language
Need help with 2 homework assignments
When Seen in the choir, Terry was the picture of an angelic devil. I have to underline the predicate twice
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