Hey everyone. I am building a keypad lock for a project using a PIC and one of those passive keypa with 8 pins, 4 input 4 output. Each of the pins corresponds to a row of numbers on the 4 by 4 keypad, and the 4 output pins correspond to the numbers in that row. so putting a 1 into input pin 1 will make the 4 output pins correspond to 1, 4, 7, dot. Putt
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ing a 1 into input pin 2 now makes 4 output pins correspond to 2, 5, 8, 0 etc. this then goes into the pic and that advances the programme etc. The problem is, the keypad keeps advancing the program without you pressing a key, the voltage into the pic floats around 1-2. 5v which I can`t get rid of. I have tried putting it through a 4050 non inverting buffer but then the output of the buffer just floats too. it should be giving out a solid 0v or 5v, not 2. 5v I`m not sure what you mean by the weak internal pull ups, the keypad is all inside a plastic housing, so I can`t get to it, but I`m pretty sure it`s just a PCB. On the pins of the Picaxe that are used as inputs from the keypad, you need to have on each of the 4 pins a resistor from the pin to the +5V supply. The value of the resistors can be from approx 3. 3K thru 10K, these will stop the pins from `floating`. The pins are not in anyway connected to the 5v rail though, I`m not sure how to add that in, the only power the keypad recieves are the 5v signals from the PIC For a Column that has been set to 5V by the program, will, if a key on that Column is closed, then Row pin will be pulled to +5V, which the program detects as a `1`. Note: In some cases depending upon the sense the Column is set active, it could be that the unselected Cols are all high and the selected Col is pulled low. On many of the `larger` PIC`s on PORTB, there is an option in the configuration part of...
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