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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
Mov ax, [1234h: 4336h + 100]
Assembly Language: Inside the 8085, instructions are really stored like binary numbers, not a very good manner to look at them and very difficult to decipher. An assembler is
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
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Cache controller The cache controller is the mind of the cache. Its responsibilities include: performing the snarfs and snoops, updating the TRAM and SRAM and implementing
do you have experts that know 4 digit 7_Seg dispaly
write a program to divide 2 numbers
The main objective of the assignment is to explore the knowledge regarding parallel ports of a computer system. You can read and write datato/from the parallel port using IN and OU
DAS: Decimal Adjust after Subtraction:- This instruction converts the result of subtraction operation of 2 packed BCD numbers to a valid BCD number. The subtraction operation has
8254 Programmable Timer A diagram of Intel's 8254 interval event/timer counter is given in Figure. The 8254 consists of 3 identical counting circuits, per of which has GATE and
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