NHD-2.4-240320SF-CTXI#-FT1
Question about: NHD-2.4-240320SF-CTXI#-FT1
In what way a 16 bit command/data is mapped onto DB15 - DB0 from the datasheet?
void TFT_24S_Write_DB(uint16_t commanddata){
LATDbits.LATD2 = (0x0001&(commanddata>>15)); //TFT DB15
LATDbits.LATD3 = (0x0001&(commanddata>>14)); //TFT DB14
LATDbits.LATD12 = (0x0001&(commanddata>>13)); //TFT DB13
LATDbits.LATD13 = (0x0001&(commanddata>>12)); //TFT DB12
LATDbits.LATD4 = (0x0001&(commanddata>>11)); //TFT DB11
LATDbits.LATD5 = (0x0001&(commanddata>>10)); //TFT DB10
LATDbits.LATD6 = (0x0001&(commanddata>>9)); //TFT DB9
LATDbits.LATD7 = (0x0001&(commanddata>>); //TFT DB8
LATFbits.LATF0 = (0x0001&(commanddata>>7)); //TFT DB7
LATFbits.LATF1 = (0x0001&(commanddata>>6)); //TFT DB6
LATGbits.LATG1 = (0x0001&(commanddata>>5)); //TFT DB5
LATGbits.LATG0 = (0x0001&(commanddata>>4)); //TFT DB4
LATAbits.LATA6 = (0x0001&(commanddata>>3)); //TFT DB3
LATAbits.LATA7 = (0x0001&(commanddata>>2)); //TFT DB2
LATEbits.LATE0 = (0x0001&(commanddata>>1)); //TFT DB1
LATEbits.LATE1 = (0x0001&commanddata); //TFT DB0
/*
LATDbits.LATD2 = (0x0001&(commanddata>>0)); //TFT DB15
LATDbits.LATD3 = (0x0001&(commanddata>>1)); //TFT DB14
LATDbits.LATD12 = (0x0001&(commanddata>>2)); //TFT DB13
LATDbits.LATD13 = (0x0001&(commanddata>>3)); //TFT DB12
LATDbits.LATD4 = (0x0001&(commanddata>>4)); //TFT DB11
LATDbits.LATD5 = (0x0001&(commanddata>>5)); //TFT DB10
LATDbits.LATD6 = (0x0001&(commanddata>>6)); //TFT DB9
LATDbits.LATD7 = (0x0001&(commanddata>>7)); //TFT DB8
LATFbits.LATF0 =(0x0001&(commanddata>>); //TFT DB7
LATFbits.LATF1 =(0x0001&(commanddata>>9)); //TFT DB6
LATGbits.LATG1 =(0x0001&(commanddata>>10)); //TFT DB5
LATGbits.LATG0 =(0x0001&(commanddata>>11)); //TFT DB4
LATAbits.LATA6 =(0x0001&(commanddata>>12)); //TFT DB3
LATAbits.LATA7 =(0x0001&(commanddata>>13)); //TFT DB2
LATEbits.LATE0 =(0x0001&(commanddata>>14)); //TFT DB1
LATEbits.LATE1 =(0x0001&(commanddata>>15)); //TFT DB0
*/
}
-
Hello,
Pin 14 (DB0) is the LSB (least significant bit) of the data byte.
Pin 29 (DB15) is the MSB (most significant bit) of the data byte.
Hope this helps!0
Please sign in to leave a comment.
Comments
1 comment