Lesson : Combinational Logic Circuit Example 2

We recommend having the cheat sheet for this chapter to help you understand the example. In this second combinational logic circuit example, we will have two logic gates seen in this chapter : “AND” and “OR”. We will analyze one possible inputs combination of this circuit and then complete the truth table of this combinational logic circuit. Below, we have the example:

Figure 1 : Combinational Logic Circuit Example 2

We will take the combinational circuit from Figure 1 but we are going to set A, B, C and D inputs to specific values (0 or 1). We will then see how to analyze the circuit. For our example : A = 1, B = 0, C = 1 and D = 0 as shown on Figure 2. We want to find the resulting output (0 or 1) of this combinational logic circuit. We could start with the “AND” or “OR” gates on the left side since we know the inputs value of these two gates. We are going to start with the “AND” gate in the top left.

Figure 2 : Combinational Logic Circuit Example 2 Step 1

The “AND” gate produces a “High” or “1” at its output if all of its inputs are “1s”. In our case, one of its inputs is a “0” which means the output of the top left “AND” gate is going to be “0”.

Figure 3 : Combinational Logic Circuit Example 2 Step 2

We are going to continue with the “OR” gate in the bottom left since we don’t have the value of the bottom input of the right “OR” gate. We need to have all inputs value to be able to find the resulting output. The “OR” logical operation produces a “Low” output if all of its input are “0s”. If one or multiples of its inputs are “1s”, the output of the “OR” logic gate will be a “1”. In our case, one of its inputs is a “1” which means the output of the left “OR” gate is going to be “1”.

Figure 4 : Combinational Logic Circuit Example 2 Step 3

We are going to continue with the last gate which is the “OR” gate on the right. The “OR” logical operation produces a “Low” output if all of its input are “0s”. If one or multiples of its inputs are “1s”, the output of the “OR” logic gate will be a “1”. In our case, one of its inputs is a “1” which means the output of the right “OR” gate is going to be “1”. Below you can find the final figure after analyzing this specific inputs combination with its resulting output:

Figure 5 : Combinational Logic Circuit Example 2 Step 4

We will now complete the truth table of this circuit.

Figure 6 : Combinational Logic Circuit Example 2

We will first calculate how many possible inputs combination there is. We have four inputs which means we have 2 exponent 4 possibilities for this circuit.

$2^{number\ of\ inputs}$

$2^4 = 2 \cdot 2 \cdot 2 \cdot 2 = 16$

We have sixteen possible inputs combinations for this circuit. Starting from 0 to 15 in the binary system. I have prefill the inputs combinations row for you since we haven’t seen yet how to count in binary yet. I also added the result for the “1010” combination since we already analyzed this inputs combination. This is the table we have to fill :

$\begin{array}{c c c c|c} A & B & C & D & Output\\ \hline 0 & 0 & 0 & 0 & ? \\ 0 & 0 & 0 & 1 & ? \\ 0 & 0 & 1 & 0 & ? \\ 0 & 0 & 1 & 1 & ? \\ 0 & 1 & 0 & 0 & ? \\ 0 & 1 & 0 & 1 & ? \\ 0 & 1 & 1 & 0 & ? \\ 0 & 1 & 1 & 1 & ? \\ 1 & 0 & 0 & 0 & ? \\ 1 & 0 & 0 & 1 & ? \\ 1 & 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 1 & ? \\ 1 & 1 & 0 & 0 & ? \\ 1 & 1 & 0 & 1 & ? \\ 1 & 1 & 1 & 0 & ? \\ 1 & 1 & 1 & 1 & ? \end{array}$

We can start filling the truth table of this circuit. There are some shortcuts we can do to fill this table. If we look at the logic gate that drives the output, we have an “OR” gate. The “OR” gate produces a “Low” output if all of its input are “0s”. If one or multiples of its inputs are “1s”, the output of the “OR” logic gate will be a “1”. We need either a “1” on the top or bottom inputs of the “OR” gate to have an output “High” or “1”. The top input of the “OR” logic gate will be “1” if the inputs “A” and “B” of the “AND” gate are “1s”. The “AND” gate produces a “High” or “1” at its output if all of its inputs are “1s”. For every row in our truth table where “A” and “B” are both “1”, we will have an output “High” or “1” at the “AND” gate which result in an output “High” at the “OR” gate on the right. This gives us the following truth table :

$\begin{array}{c c c c|c} A & B & C & D & Output\\ \hline 0 & 0 & 0 & 0 & ? \\ 0 & 0 & 0 & 1 & ? \\ 0 & 0 & 1 & 0 & ? \\ 0 & 0 & 1 & 1 & ? \\ 0 & 1 & 0 & 0 & ? \\ 0 & 1 & 0 & 1 & ? \\ 0 & 1 & 1 & 0 & ? \\ 0 & 1 & 1 & 1 & ? \\ 1 & 0 & 0 & 0 & ? \\ 1 & 0 & 0 & 1 & ? \\ 1 & 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 1 & ? \\ 1 & 1 & 0 & 0 & 1 \\ 1 & 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 1 & 1 \end{array}$

We need either a “1” on the top or bottom inputs of the “OR” gate on the right to have an output “High” or “1”. We already analyzed the top input so we will now look at the bottom input. The bottom input of the “OR” logic gate will be “1” if the input “C” or “D” of the “OR” gate on the left side is a “1”. The “OR” gate produces a “Low” output if all of its input are “0s”. If one or multiples of its inputs are “1s”, the output of the “OR” logic gate will be a “1”. For every row in our truth table where “C” or “D” is a “1”, we will have an output “High” or “1” at the left “OR” gate which result in an output “High” at the right “OR” gate. This gives us the following truth table :

$\begin{array}{c c c c|c} A & B & C & D & Output\\ \hline 0 & 0 & 0 & 0 & ? \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 1 & 0 & 0 & ? \\ 0 & 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & 0 & 0 & ? \\ 1 & 0 & 0 & 1 & 1 \\ 1 & 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 0 & 1 \\ 1 & 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 1 & 1 \end{array}$

We know in the 3 rows we haven’t completed that the bottom and top inputs of the right “OR” gate are “0s” since we have solved every case where either the top or bottom input is a “1”. We can complete the table with this information :

$\begin{array}{c c c c|c} A & B & C & D & Output\\ \hline 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 1 \\ 1 & 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 0 & 1 \\ 1 & 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 1 & 1 \end{array}$

We are done with this example on combinational circuit. The last lesson is going to be exercises similar to the examples. You will have to solve these exercises using what you learned in this chapter.