T02 - LED display driver

4-digit LED display is connected to microcontroler by the use of two serial-in shift registers with output latches SN74LS595. To display any information on the display it is necessary to send two bytes:

• 1st byte represents individual segments of single LED display (D7..D0)
• 2nd byte represents address of digit (A7..A0)

Used pins of microcontroller:

• GPIOA9 - serial data
• GPIOA8 - clock
• GPIOB5 - latch

Figure 1 shows time diagram of communication between microcontroller and LED display. Bits are sent starting from MSB.

Example of serial data sequence (two bytes) to show number 0 on 1st digit (from the left): 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1

shift register (1B):

     output (8 bits)
  ^   ^   ^   ^   ^   ^   ^   ^
| x | x | x | x | x | x | x | x | < input data  (1 bit) GPIOA9
                                  < input clock (1 bit) GPIOA8
                                  < input latch (1 bit) GPIOB5

00110011 -> store it in register memory
   
series of bits on data input (GPIOA9)
clock (change from 0->1 on GPIOA8) at the moment, when value on the data input is set
   
set 0 to data input  | x | x | x | x | x | x | x | x | < 0
clock                | x | x | x | x | x | x | x | 0 |
set 0 to data input  | x | x | x | x | x | x | x | 0 | < 0
clock                | x | x | x | x | x | x | 0 | 0 |
set 1 to data input  | x | x | x | x | x | x | 0 | 0 | < 1
clock                | x | x | x | x | x | 0 | 0 | 1 |
set 1 to data input  | x | x | x | x | x | 0 | 0 | 1 | < 1
clock                | x | x | x | x | 0 | 0 | 1 | 1 |
 ...
set 1 to data input  | x | 0 | 0 | 1 | 1 | 0 | 0 | 1 | < 1
clock                | 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 |
latch - change from 0->1 propagates 1B word stored in memory to the output of register
   
1 byte representation of 0/1
0: 00000000
1: 00000001
byte to serialize: X
bit to be set on the GPIOA9: bit
   
for (int i = 7; i >= 0; i--)
{
	bit = (X >> i) & 0x01;	// -> 00010001 & 0x01 -> 00000001
}
   
LED segments LOW - LED segment in on, HIGH - LED segment is off
0: 11000000
1: 11111001
   
algorithm: 
   
unsigned char segments[10] = {192, 249, ....};
unsigned char digits[3] = {1, 2, 4, 8};

we want to show 0 on the 2 position from the left

select 0 from segments -> A
select 1 from digits   -> B
   
setup of registers: most significan bit of B is connected to least significant bit of A 
A: | x | x | x | x | x | x | x | x |    B: | x | x | x | x | x | x | x | x | 
                                 ^           ^
                                  \---------/
latch = 0
loop for A:
7..0
	clock = 0               (8th bit in ODR)
   	bit = (A >> i) & 0x01;  (9th bit in ODR)
   	clock = 1
loop for B
7..0
	clock = 0
   	bit = (B >> i) & 0x01;  (9th bit in ODR)
   	clock = 1
latch = 1

1s precission stop watch
this display can only show one digit of four in one time (have no memory)

long T = 1;	// global
   
while(1)
{
	// update display without any delay
	display(T);
}
   
// systick interrupt
void systick_handle(void)
{
	T++;
}