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I am working on a project with an Arduino Nano that includes a keypad, screen, and the NRF4001 transceiver. The keypad requires 7 digital pins to operate, the wireless transceiver uses pins D9 - D13 with the RF24 library I am using. Now, that leaves me with 1 Digital Pin left for a screen unless there is an I2C screen I could use.

That would be easy enough to find except that I have some limitations. If at all possible I would like the screen to be under 5 dollars and about the size of a Nokia 5110. The Nokia uses to many pins and does not tolerate 5v logic levels I've heard. An OLED 0.96 inch is only a 4 bucks and is I2C, but it is too small. I could make a board for the keypad that used different values of resistors and one analog pin to detect different buttons but would rather not go that route if there is any easier way to get around this.

Are there any options out there that would suffice my pin, cost and size conditions or am I being too picky?

  • Has anyone seen a backpack for Nokia's that make them I2c capable for example? I know its been done with the typical 16x2s. – NULL Aug 28 '15 at 16:22
  • You can also use the analog pins as regular digital pins. So you have 6 pins left. The Nokia 5110 seems to run fine on 5v, from what I've heard. But you could add a buffer chip to change the voltage levels (example). – Gerben Aug 28 '15 at 18:39
  • How would you do that....use analog as digital? – NULL Aug 28 '15 at 18:41
  • Ok...I saw how they could be used as inputs 14 - 19 but can they be used as digital outputs too in the same way or do I have to say analogWrite(pin, 1023) to get the same result as a digital HIGH – NULL Aug 28 '15 at 18:57
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    Just use pinMode(A3); and digitalWrite(A3, LOW); e.g.. 5v on the 5110 is out of spec, but appears to work with no problem. But use at own risk. – Gerben Aug 29 '15 at 11:31
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You can use a single analog pin to read an entire keypad.

First, read up on voltage dividers - they let you convert a resistance into a voltage. You will need to know this later on.

A keypad typically has 4 inputs, and 3 outputs (or vice-versa). Label the 4 pins A,B,C,D and the 3 pins E, F, G. You will be wiring A up to a voltage supply, and G up to an analog pin, and via a suitable resistor to ground. This becomes your voltage divider.

Then, put a resistor between A+B, B+C, C+D, and between E+F, F+G. So, if you press the button that connects A to G, there will be no resistance (=+5V). If you press the button that connects A to F, then the resistance will be that of the resistance F->G. If you press the button that connect D and E, the circuit goes: VCC-A-resistor-B-resistor-C-resistor-D-pushbutton-E-resistor-F-resistor-G - Arduino analog pin. With no button pressed, the voltage will be 0.

You will need to work out different resistors in advance, to give you as even a range as possible between 0 and 5 volts (the critical thing being the smallest difference). You can calculate the resistance of any combination here: http://www.calculator.net/resistor-calculator.html and you can use Ohm's law to calculate the final voltage. This depends on what you have in your kit. I ended up making a spreadsheet, and then trying random resistor values until I got a good spread.

DO NOT use any resistors under 1kOhm.

Once you have the right spread, make it up on a solderless breadboard. Measure the voltage with your arduino (use the "readanalogvoltage" sketch, take out * (5.0 / 1023.0) , and make up a table - press button 1, you get a reading if A, press button 2, get a reading of B. Include the reading when no button is pressed, which should be 0.

Then find the numbers exactly in between.

For example, if the lowest three readings are 0,20,37 for no button, A, and B pressed, respectively, then in your code, you will say: if the pin measures less than (half way between 0 and 20=) 10, then no button is pressed; else if the pin measures less than (half way between 20 and 37=) 29, then button A is pressed... and so forth.

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