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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
program to accept 23 students name using while loop let your variable control the value negative 4
using 8086 assembly language that interchange upper four bits to lower four bits. assume that data store in byte memory and it written back to same location. and assume the data as
NEG: Negate:- The negate instruction forms the 2's complement of the particular destination in the instruction. For obtaining 2's complement, it subtracts the contents of destinat
IRET : Return from ISR:- When an interrupt service routine is called, before transferring control to it, the IP, CS register and flag registers are stored in the stack to ment
EVOLUTION OF MICROPROCESSOR : The digital circuits and systems may be broken into two part: 1) Sequential Circuit and 2) Combinational Circuits Norm
INT N : Interrupt Type N:- In the interrupt structure of 8086/8088, 256 interrupts are distinct equivalent to the types from OOH to FFH. When an instruction INT N is executed,
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
Computes the integral square root: Problem: Square Root: For this problem you will write a short assembly program that computes the integral square root of an input numb
Assembler Directives and Operators The major advantage of machine language programming is directly that the memory control is in the hands of the programmer, so that, he can be
Assembly Language Example Programs We studied the entire instruction set of 8086/88, pseudo-ops and assembler directives. We have explained the process of entering an assembly
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