The objective of this lab is to build circuits to verify the DeMorgan’s Laws.
Chips: 74HCT08 Quad 2-input AND gate, 74HCT32 Quad 2-input OR gate, 74HCT04 inverter
Procedure
Design a circuit with AND, OR and NOT gates that will prove the DeMorgan's Laws:
(X+Y)’ = X’ Y’ (XY)’ = X’ + Y’
Derive the truth table:
X Y
F1=(X+Y)’
F2=X’Y’
F3=(XY)’
F4=X’+Y’
0 0
1
1
1
1
0 1
0
0
1
1
1 0
0
0
1
1
1 1
0
0
0
0
Draw the logic diagram
Circuit Construction on Tinkercad
The circuit can be built with one 7408 AND chip, one 7432 OR chip, and one 7404 inverter chip. DIP switches will be used to supply the two input logic levels X and Y, and LEDs will be used to provide readout of the four output bits F1 - F4.
After you log into the Tinkercad, copy and paste the following link to your web browser https://www.tinkercad.com/things/19XqriP2dMF. Click on the button of “Copy and Tinker” on the right to make a copy of the CSC 347 starter circuit and rename as “Lab 2”.
For each chip (08, 32, 04), connect its pin 14 to the power, and pin 7 to the ground.
Connect the chips together according to the schematic diagram obtained in Step 2.
Connect the inputs (X, Y) to two switches, and the outputs (F1, F2, F3, F4) to four LEDs. (Remember that you can only use the holes in the lower half of the breadboard for the LEDs and Switches)
Press “Start Simulation”
https://www.tinkercad.com/things/eJIg8HTydH1-copy-of-csc-347-starter-kit/editel?sharecode=Co85fg01GpKUbt3fGqaLRBm_9lkt9Elwa0dPVoz0pxE
Experiment:
Apply all four combinations of logic LOW and HIGH levels to the two gate inputs using the DIP switches. Observe the outputs, and record the results in the following table, in the form of 0s and 1s. Verify if your table matches with the truth table shown obtained in the first page.
X Y
F1=(X+Y)’
F2=X’Y’
F3=(XY)’
F4=X’+Y’
0 0
1
1
1
1
0 1
0
0
1
1
1 0
0
0
1
1
1 1
0
0
0
0
Homework: Design a circuit to prove the Distributive Laws using TinkerCad and include it (equation, truth table, logic diagram, screenshot of circuit) in your lab report.
A(B + C) = AB + AC A + BC = (A + B)(A + C)
