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G30 rlp spi


#42

@Brett The display cursor position resets itself after sending 8 characters on its own. I was taking advantage of that to not do an additional write to the display but I will give it a shot. The ToString() barrage was just testing the positioning on the display / making sure a shorter string was not getting passed.

I am actually not sure what the command would be in 4 bit mode (DB0-DB3 unused), clear display and return home look the same to me.


#43

I actually found this a bit confusing for a minute. The information is there but not at first obvious.

4-bit mode:

Clear Display: command(0x01);

Return Home: command(0x02); OR command(0x03)

Imagine 0x03 written as 0b00000011 instead. It would be like this in the 8-bit mode

DB7 = 0, DB6 = 0, DB5 = 0, DB4 = 0, DB3 = 0, DB2 = 0, DB1 = 1, DB0 = 1

This assumes you implement the command(byte data) function just like in the datasheet for the 4-bit example


#44

Um, no, clearly not. Read the description. One clears DDRAM, one does not.

ok, so apparently you don’t understand the code? 4-bit mode simply uses two writes to write the 8-bit value.

I am not sure why you’d think that’s a smart test. Create a single string that you know doesn’t change, because who knows how the value you want to display could change while your code is executing. Plus, even if this was how you were testing it, the smart thing to do would have been to admit it made no difference, and go back to the streamlined version you really want to use…


#45

I am lost in all the conversions.

In the driver I adapted my code from, the have CUR_HOME=2 and do SendCommand(CUR_HOME);

 private void SendCommand(byte command)
{
    lcdRS.Write(false);                         //set LCD to data mode

    WriteNibble((byte)(command >> 4));
    WriteNibble(command);

    Thread.Sleep(2);

    lcdRS.Write(true);                          //set LCD to data mode  
}

How do they get a command of 2 from all that? It seems to work for me.


#46

You are just writing the top four bits and then the lower four bits since you only have four wires.

CUR_HOME = 0x02 aka 0b 0000 0010

So first we set up :

DB7 = LOW, DB6 = LOW DB5 = LOW, DB4 = LOW (The High Nibble: 0000)

Then once we are sure that it is clocked in we can set up the Low Nibble:

DB7 = LOW, DB6 = LOW, DB5 = HIGH, DB4 = LOW (The Low Nibble: 0010)

So once that is clocked in the device knows that since it got two Nibbles and it is configured in 4-bit mode that it should reconstruct an 8 bit command by putting them together.

If it is the shift operator causing you trouble then let me try to explain what to expect given certain input:

The operators << and >> are bit-shift operators. They move the bits either left ( << ) or right ( >> ).

If a bit is “shifted out” of the value… the information is lost. Bits “shifted in” will be 0.

byte B = 0x20; // 0010 0000

byte ShiftedRightBy4 = B >> 4; // 0000 0010  - the one was shifted right 4 spaces. zeros filled in from the left.

byte C = 0x4F; // 0100 1111

byte ShiftedLeftBy5 = C << 5; // 1110 0000  - information that was shifted out is lost, zeros filled in from the right.

#47

@jwizard93 thank you that is much more clear. What does the 0b mean though? That is the last part I am missing


#48

0xFF is an example of a hexadecimal value (notice the 0x in front)

0b00100110 is an example of a binary value (notice the 0b in front)

The prefixes are just ways of telling the reader what base the value is in. Sometimes they are set up to convey this information to the compiler as well… like we can use " 0x34 " as a value in our netmf apps.


#49

Thank you! I knew about the 0x but did not realize (or remember) that 0b was binary