Arduino Uno - combined LCD and keypad interfacing problem

Question

I have a problem with the Arduino Uno.

I need to use a 4x3 keypad and 16x2 LCD together with an Arduino Uno. As 4x3 keypad requires 7 pins, I have defined pins 2, 3, 4, 5, 6, 7 and 8 to it. A 16x2 LCD requires 6 pins but only 5 pins remain (9, 10, 11, 12 and 13). Pins 0 and 1 are used for serial.

Please suggest a probable solution.

Answer 1

You could use a shift register to read the switches as outlined by the excellent tutorial Parallel to Serial Shifting-In with a CD4021BE. Quoting the summary:

Sometimes you'll end up needing more digital input than the 13 pins on your Arduino board can readily handle. Using a parallel to serial shift register allows you collect information from 8 or more switches while only using 3 of the pins on your Arduino.

So you would only use 3 pins to read the seven switches, which would leave you with 9 pins to play with, for the LCD.

In addition, you could use a similar method for the LCD, using a shift-out register, and use just 3 pins for the 6 inputs of the LCD. A tutorial for that is Alphanumeric LCD with Shift Register on Arduino – part 1, using the AlphaLCD library or 3-Wire Serial LCD using a Shift Register, which has no library, but provides code.

This would leave you with 6 pins free (out of the total 12 pins, if you exclude pins 0 and 1 for serial), after both the switches and LCD are wired up.

Of course, this solution requires you to purchase one or two shift (in and/or out) registers.

The Arduino IDE may come with a built-in library that use shift registers, but I am not sure about that. Maybe someone can confirm, or refute, this?

I hope that this helps.

Answer 2

Am new at this but I am working on a similar work, it is possible to use the pin 0 for your keypad, the pin is a RX pin by default making the pins(0,2,3,4,5,6,7).

As for the LCD I would suggest you use an i2c backpack, it is a parallel to serial converter and means you would only need to use 2 pins (SDA, SCL which would be connected to pin A4, A5)

Try the code below for the keypad (The code is if you have an i2c backpack)

#include <Wire.h>
#include <Keypad.h> //keypad library
#include <LiquidCrystal_I2C.h> //library for the backpack

LiquidCrystal_I2C lcd(0x27, 2, 1, 0, 4, 5, 6, 7, 3, POSITIVE); 

const byte ROWS = 4; //four rows
const byte COLS = 3; //three columns

//define the Symbols on the buttons of the keypads
  char hexaKeys[COLS][ROWS] = {
  {'1','2','3'},
  {'4','5','6'},
  {'7','8','9'},
  {'*','0','#'}
};

byte rowPins[ROWS] = {0, 2, 3, 4}; //connect to the row pinouts of the keypad
byte colPins[COLS] = {5, 6, 7, 8}; //connect to the column pinouts of the keypad

//initialize an instance of class NewKeypad
Keypad customKeypad = Keypad( makeKeymap(hexaKeys), rowPins, colPins, ROWS, COLS); 

void setup(){
  Serial.begin(9600);
  lcd.begin(16,2);
  for(int i = 0; i< 3; i++)
  {
    lcd.backlight();  delay(250);
    lcd.noBacklight();  delay(250);
  }
  lcd.backlight();
}

void loop(){

  char customKey = customKeypad.getKey();

  if (customKey){
    lcd.print(customKey);
  }
}