An introduction to AND gates - Answers
Q1. When we have just 2 inputs, there are a total of 4 combinations of 1s and 0s for the inputs (2 to the power 2). If you had 3 inputs, there are 8 possible combinations. (2 to the power 3).
Q2. If you had 4 inputs, there would be 16 possible combinations (2 to the power 4).
Q3. The truth table for this diagram ....
| A | B | D | C | E |
| 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 |
| 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 | 0 |
| 1 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 | 0 |
| 1 | 1 | 0 | 1 | 0 |
| 1 | 1 | 1 | 1 | 1 |
Something very interesting is happening in columns A, B and D. We are counting in binary to get the different permutations.
000 - 0
001 - 1
010 - 2
011 - 3
100 - 4
101 - 5
110 - 6
111 - 7
Q4. There are 4 inputs in this diagram, A, B, D and F.
Q5. There 2 to the power of 4 = 16 different combinations of inputs.
Q6. There is only one output to the whole diagram, G. There are outputs from other logic gates though. These are C and E.
Q7. The truth table is as follows:
| A | B | D | F | C | E | G |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 0 | 1 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 | 0 | 0 | 0 |
| 0 | 1 | 0 | 0 | 0 | 0 | 0 |
| 0 | 1 | 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 1 | 0 | 0 | 0 |
| 1 | 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 | 0 | 0 | 0 |
| 1 | 1 | 0 | 0 | 1 | 0 | 0 |
| 1 | 1 | 0 | 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 0 | 1 | 1 | 0 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 |
We are counting in binary in columns A, B, D and F, again! This will give us all of the permutations we need to try out!
0000 - 0
0001 - 1
0010 - 2
0011 - 3
0100 - 4
0101 - 5
0110 - 6
0111 - 7
1000 - 8
1001 - 9
1010 - 10
1011 - 11
1100 - 12
1101 - 13
1110 - 14
1111 - 15
Q8. This diagram also has 4 inputs.
Q9. That means there are 2 to the power of 4 = 16 different permutations of inputs.
Q10. Here is the truth table. N0te that you could layout the 1s and 0s in columns D and E in the same way as A and B, but if you understand counting in binary, it makes getting all the permutations down in the table quick and easy.
| A | B | D | E | C | F | G |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 0 | 1 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 | 0 | 0 | 0 |
| 0 | 1 | 0 | 1 | 0 | 0 | 0 |
| 0 | 1 | 1 | 0 | 0 | 0 | 0 |
| 0 | 1 | 1 | 1 | 0 | 1 | 0 |
| 1 | 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 0 | 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 0 | 1 | 0 | 0 |
| 1 | 1 | 0 | 1 | 1 | 0 | 0 |
| 1 | 1 | 1 | 0 | 1 | 0 | 0 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 |
It should be clear to students now that all the inputs must be a 1 if the output is to be a one, in these examples using just AND gates. It's good practice to get them to draw out the truth tables and start thinking in binary as well as logic.