1. Lab 9: Matrix Keypad : ”No Key Press” Analysis Slide #2 Slide #3 ”Press and Hold Key 5” Analysis

2. The AND gate will pass the 1 K PPS pulse waveform to the clock input of the MOD 16 counter because the control input is logic 1. We will analyze the system with no key being pressed down. All switches are open circuits. The 4 resistors pull up the inputs of the MUX to 5V. Thus MUX Z=1 regardless of the logic levels at S0 and S1. This symbol indicates counting…changing 1/0. The output of the counter is cycling (counting) very quickly because of the 1 K PPS clock signal. Each count state lasts only 1 millisecond. It only takes 16 milliseconds for the counter to complete cycle through its count states from 0 to 15. The 1 of 4 decoder receives the quick changing signals from the counter. The decoder is driven by the mod 4 section. A1 A0 O0 O1 O2 O3 0 0 0 1 1 1 0 1 1 0 1 1 1 0 1 1 0 1 1 1 1 1 1 0 Continue and you will see a slow motion animation of the decoder output cycling. Use the backspace key to re-run the animation a few times and you will see that net result is a logic 0 moving across the columns of the keypad from left too right. The actual speed is much faster the logic 0 moves right every millisecond. The rotating logic 0 from the decoder does not get transferred to the MUX inputs because all keys are open circuit (not pressed). The MUX inputs are selected one at a time and transferred to Z. The MUX select inputs receive the quick changing signals from the counter (Q2, Q3). A different channel is selected every 4 milliseconds S1 S0 Z 0 0 I0 0 1 I1 1 0 I2 1 1 I3 Continue and you will see a slow motion animation of the MUX cycling. Use the backspace key to re-run the animation a few times and you will see that net result is each keypad row’s logic level is transferred to Z every 4 milliseconds. The data register receives the quick changing signals from the counter. With Clk =1 the 4 bit number at Da … Dd is ignored by the data register. The cycling from the counter is thus ignored. Let’s assume that the initial number stored into the data register was 0. The end result is: The 0 gets displayed and the changing signals from the counter are ignored. Lab 9 : “No key pressed” Analysis : A matrix keypad has it’s keys arranged in rows and columns. When a key is pressed it connects a column to a row. Each key pressed is identified by a unique column number row number connection. Note: The effects of switch bounce are ignored. 1 1 1 1 1 1 0 0 0111 1 0 1011 1 1 1110 0 1 1101 0 0 0111 1 0 1011 1 1 1110 0 1 1101 0 0 0 0 0 1 0 0 0 0 1 0 0 1 1 0 0 0 0 1 0 1 0 1 0 0 1 1 0 1 1 1 0 0 0 0 1 1 0 0 1 0 1 0 1 1 1 0 1 0 0 1 1 1 0 1 1 0 1 1 1 1 1 1 1 The 0 cycling on the keypad columns and the MUX transferring the keypad row data is called scanning the keypad. This scanning will allow keys to be encoded. Slide #2 Column #0Column #1Column #2Column #3 Row #0 Row #1 Row #2 Row #3

3. 01111011 0 0 0 0 1 0 0 0 0 1 0 0 The 1 of 4 decoder receives 0,0 from the counter and grounds O0 (active low). The other 3 outputs remain at 1 (inactive). The logic 1 at O1 is transferred through the pressed key to I1 of the MUX. Lab 9: “Press and hold key 5”: The demonstration will assume the counter starts at 0 when the user presses and holds down key 5 of the matrix keypad. The actual time it takes the system to respond to this event is 6 milliseconds. Let’s also assume the data register has 0 stored. 1 1 1 1 1 0 The MUX inputs I0, I2, I3 are all at logic 1 because the keys on those rows are open and pull up resistors make the input voltage 5V. The result is MUX Z=1. With MUX Z=1. Data reg. Clk = 1. Data reg. is disabled. It ignores inputs Da…Dd and continues to display the number 0 which is stored in the data register. The 1 K PPS pulse is passed by the AND gate and the counter rolls forward to 1. The 1 of 4 decoder receives 0,1 from the counter and grounds O1 (active low). The other 3 outputs remain at 1 (inactive). The logic 0 at O1 is transferred through the pressed key to I1 of the MUX. The MUX inputs I0, I2, I3 are all at logic 1 because the keys on those rows are open and pull up resistors make the input voltage 5V. Even though I1=0 the MUX output Z stays at logic 1 because the logic level at I0 is selected (counter Q3, Q2 = 0,0) With MUX Z=1. Data reg. Clk = 1.Data reg. is disabled. It ignores inputs Da…Dd and continues to display the number 0 which is stored in the data register. The 1 K PPS pulse is passed by the AND gate and the counter rolls forward to 2. The 1 of 4 decoder receives 1,0 from the counter and grounds O2 (active low). The other 3 outputs remain at 1 (inactive). The logic 1 at O1 is transferred through the pressed key to I1 of the MUX. 011 1 1 With MUX Z=1. Data reg. Clk = 1. Data register is disabled. It ignores inputs Da…Dd and continues to display the number 0 which is stored in the data register. The 1 K PPS pulse is passed by the AND gate and the counter rolls forward to 3. The counter continues to roll forward from 3 to 4 because the MUX Z=1. Let’s look ahead to when the counter reaches 5 (the key number!). 1 0 0 1 1 0 1 0 The 1 of 4 decoder receives 0,1 from the counter and grounds O1 (active low). The other 3 outputs remain at 1 (inactive). The logic 0 at O1 is transferred through the pressed key to I1 of the MUX. 101 1 0 The MUX select inputs S1, S0 = 0,1. Channel I1 is selected and the logic 0 is passed to output Z. 0 The AND gate will block the clock because Z=0. It stops the counter at the number 5. Data reg. Clk changes to 0 because Z=0. The data register is clocked and the number 5 from the counter is stored and displayed. Slide #3