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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
write a programme the addition two 3*3 matrix and stored in from list
#I submitted my assignment this morning and it is still processing. How long does it take?
how to find out the given number is positive or negative?
Cache Memory Caching is a technology based on the memory subsystem of any computer. The majoraim of a cache is to accelerate the computer while keeping the cost of the computer
write and run a programme using 8086 assembly language that interchange the lower four bits of AL registered with upper four bits.
How to design 4 bit signed 2s complement multiplier?
Program is written but has errors returning values from the procedure.
CISC Characteristics : The design of an instruction set for a computer might take into consideration not only machine language constraints, but also the requirements i
assempbly language routine that takes an array named A containing n bytes of postive numebrs and fills two arranys, array B containing n words and array C containing n long words
Instruction set of 8086 : The 8086/8088 instructions are categorized into the following major types. This section describes the function of each of the instructions with approp
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