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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
General terms for Cache : Cache Hits : When the cache consisted the information requested, the transaction is said to be a cache hit. Cache Miss : When the cache does n
) What is the difference between re-locatable program and re-locatable data?
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
DMA Hardware (8237 DMAC) : 1)Processor contain HOLD/HOLD Acknowledge lines to interact with 8237 o DMAC can achieve control of ISA bus by asserting HOLD o P
You have to write a subroutine (assembly language code using NASM) for the following equation.
Programming with an assembler The procedure of hand-coding 8086 programs is somewhat tiresome; hence generally a programmer may find it hard to get a correct listing of the mac
TEST : Logical Compare Instruction: The TEST instruction performs bit by bit logical AND operation on the 2 operands. Each bit of the result is then set to value I, if the equival
Hello
1. Write an assembly program that adds the elements in the odd indices of the following array. Use LOOP. What is the final value in the register? array1 DWORD 10, 20, 30, 40, 50, 6
calculate the number of one bits in bx and complement an equal number of least significant bits in ax hint use the xor instruction
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