Let's look at interfacing a Hex keypad, the one with 16 switches; arranged in 4 columns and 4 rows. The interesting challenge is to interface 16 switches on 8 pins. So how does this work? The simple answer is the MCU is faster than the your fingers.
Basics
Code
#include "keypad.h" #include "lcd.h" int main() { uint8_t key; /*Connect RS->PB0, RW->PB1, EN->PB2 and data bus PORTB.4 to PORTB.7*/ LCD_SetUp(PB_0,PB_1,PB_2,P_NC,P_NC,P_NC,P_NC,PB_4,PB_5,PB_6,PB_7); LCD_Init(2,16); /*Connect R1->PD0, R2->PD1, R3->PD2 R4->PD3, C1->PD4, C2->PD5 C3->PD6, C4->PD7 */ KEYPAD_Init(PD_0,PD_1,PD_2,PD_3,PD_4,PD_5,PD_6,PD_7); LCD_Printf("Key Pressed:"); while (1) { key = KEYPAD_GetKey(); LCD_GoToLine(1); LCD_DisplayChar(key); } return (0); }
Let's look at the most important function in the code in a little more details
key = KEYPAD_GetKey();
As you can guess, the function return the ASCII value of the key being pressed. It follows the following sequences to decode the key pressed:
- Wait till the previous key is released.
- Wait for the new key press.
- Scan all the rows one at a time for the pressed key.
- Decodes the key pressed depending on ROW-COL combination and returns its ASCII value.
It is defined in keypad.c as below:
uint8_t KEYPAD_GetKey(void) { uint8_t i,j,v_KeyPressed_u8 = 0; keypad_WaitForKeyRelease(); keypad_WaitForKeyPress(); for (i=0;i<C_MaxRows_U8;i++) { GPIO_PinWrite(A_RowsPins_U8[i],HIGH); } for (i=0;(i<C_MaxRows_U8);i++) { GPIO_PinWrite(A_RowsPins_U8[i],LOW); for(j=0; (j<C_MaxCols_U8); j++) { if(GPIO_PinRead(A_ColsPins_U8[j]) == 0) { v_KeyPressed_u8 = 1; break; } } if(v_KeyPressed_u8 ==1) { break; } GPIO_PinWrite(A_RowsPins_U8[i],HIGH); } if(i<C_MaxRows_U8) v_KeyPressed_u8 = A_KeyLookUptable_U8[i][j]; else v_KeyPressed_u8 = C_DefaultKey_U8; return v_KeyPressed_u8; }