digital clock help required URGENT
Sun Oct 07 2007, 07:55 pm
well sorry.. i mis interpreted the data you gave before...
LCD data bits are connected to the higher nibble of P0 and control pins are connected as mentioned.. in the #define context.
is there anything else you want to know?
LCD data bits are connected to the higher nibble of P0 and control pins are connected as mentioned.. in the #define context.
is there anything else you want to know?
Mon Oct 08 2007, 07:57 am
if i have to make changes in the LCH header file to incorporate my design ie in the 8 bit mode then what all changes will i have to make??
or can u plz highlight the instructions that will need some change! thanks
or can u plz highlight the instructions that will need some change! thanks
Mon Oct 08 2007, 08:07 am
one more thnig, if i have to send nos to the LCD to display em in decimal, will 4 bit mode help or do i make it 8 bit??? and in case of 4 bit, do i connect only 4 pins i.e. 0.4 to 0.7 with 0.7 as HSB
Mon Oct 08 2007, 01:19 pm
you have to make a lot of changes in the code..
4-bit and 8-bit modes are totally different from each other..
also regarding pins..
4 bit mode need a max of 6 pins where as 8-bit more need 11 pins to connect LCD to controller.
if you want to program using 8-bit mode.. see the lcd tutorials in tutorial section.
4-bit and 8-bit modes are totally different from each other..
also regarding pins..
4 bit mode need a max of 6 pins where as 8-bit more need 11 pins to connect LCD to controller.
if you want to program using 8-bit mode.. see the lcd tutorials in tutorial section.
darshini like this.
Mon Oct 08 2007, 10:02 pm
P0=((cmd&0xF0))|(0x08); // Higher nibble first
PULSE();
P0=(((cmd&(0x0F))*16))|(0x08); // Lower nibble next
PULSE();
where CMD is an unsigned char
can u plz telll me why is it bitwise ORed with 08??
PULSE();
P0=(((cmd&(0x0F))*16))|(0x08); // Lower nibble next
PULSE();
where CMD is an unsigned char
can u plz telll me why is it bitwise ORed with 08??
Mon Oct 08 2007, 10:05 pm
they are setting the bit P0.3 of Port 0 check where it is connected in the original circuit. only then i can tell why they are doing so...
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