Daisy-chaining two 74HC595 shift registers using an STM32

Question

I'm trying to daisy-chain two shift register using a STM32F407 discovery kit, but I can't seem to get it right. I use the following function to send a byte to the shift register:

void selector_SendByte(uint8_t byte)
{
  // Define timeout (ms)
  uint16_t clock_timeout = 10000;
  uint16_t latch_timeout = 1000;

  for (int8_t i = 7; i >= 0; --i)
  {
    uint8_t bit = byte & (0x1 << i);    // Read bit
    HAL_GPIO_WritePin(SL_SD_GPIO_Port, SL_SD_Pin, bit);
        
    // Toggle clock
    HAL_GPIO_WritePin(SL_SP_GPIO_Port, SL_SP_Pin, 1);
        
    while (clock_timeout--);

    HAL_GPIO_WritePin(SL_SP_GPIO_Port, SL_SP_Pin, 0);
    }

  // Toggle latch
  HAL_GPIO_WritePin(SL_SS_GPIO_Port, SL_SS_Pin, 1);
    
  while (latch_timeout--);

  HAL_GPIO_WritePin(SL_SS_GPIO_Port, SL_SS_Pin, 0);
}

Where SL_SD is the data, SL_SP is SRCLK, and SL_SS is RCLK.

I defined the following outputs:

#define Q0 0b0000000000000001
#define Q1 0b0000000000000010
#define Q2 0b0000000000000100
#define Q3 0b0000000000001000
#define Q4 0b0000000000010000
#define Q5 0b0000000000100000
#define Q6 0b0000000001000000
#define Q7 0b0000000010000000
// shift register 2
#define Q8  (Q0 << 8)
#define Q9  (Q1 << 8)
#define Q10 (Q2 << 8)
#define Q11 (Q3 << 8)
#define Q12 (Q4 << 8)
#define Q13 (Q5 << 8)
#define Q14 (Q6 << 8)
#define Q15 (Q7 << 8)

When calling selector_SendByte(Q0) and then selector_SendByte(Q8) it doesn't do what I want.

I connected a pull-up resistor of 10 kΩ to each output, and connected LEDs with 440 Ω resistors to each output. For the first shift register I can turn on the LEDs correctly, but the second shift register is not doing what I expected.

Any help on what I am missing?

Answer 1

That function is written for an 8-bit (not cascaded) shift register.

You have to send 16 bits before toggling the latch. Change the byte and bit variables to a 16-bit (or larger) data type, rename byte (because it is no longer a byte), and change the loop count.

Answer 2

what I am missing?

You call void HAL_GPIO_WritePin (GPIO_TypeDef * GPIOx, uint16_t GPIO_Pin, GPIO_PinState PinState)passing an undocumented value as PinState:

PinState: specifies the value to be written to the selected bit. This parameter can be one of the GPIO_PinState enum values:

• GPIO_PIN_RESET: to clear the port pin
• GPIO_PIN_SET: to set the port pin

from

/** @brief  GPIO Bit SET and Bit RESET enumeration */
typedef enum {
  GPIO_PIN_RESET = 0,
  GPIO_PIN_SET
}GPIO_PinState;

Prudent:
uint8_t bit = byte & (1 << i) ? GPIO_PIN_SET : GPIO_PIN_RESET; or
uint8_t bit = (byte >> i) & 1 ? GPIO_PIN_SET : GPIO_PIN_RESET;
know-it-all:
uint8_t bit = (byte >> i) & 1;

For shift-out of more than 8 bits, see CL.'s answer.