Question

Use as few 3-input NOR gates as possible to design a bubble detector circuit for 8-bit thermometer code. An

n-bit thermometer code represents an integer m, with m 1s followed by (n-m) 0s. 1-bit bubble is an error in

coding when a solitary 0 (or 1) is found in between two 1s (or 0s). Implement using structural verilog. Please do not answer unless familiar with Xilinx and verilog.

Answer 1

The thermometer code is 8-bit. Therefore, it can represent integers from 0 to 8. The correct thermometer codes (without bubble error) are:

0000 0000

1000 0000

1100 0000

1110 0000

1111 0000

1111 1000

1111 1100

1111 1110

1111 1111

We divide the entire thermometer code into two equal halves. From the above, we can easily see that for each of the halves, the correct code could be one of the following four:

0000

1000

1100

1110

1111

Since we have to use NOR gates, we will make POS (Product Of Sums) expression. We make a function F(X,Y,Z,W) such that if XYZW takes one of the five values above, F(X,Y,Z,W) = 0

Note that in a POS expression, variable represents a 0, while a barred variable represents a 1. Therefore

This expression requires one NOR gate for each of the three sums inside the brackets, and one NOR gate for the product. Thus, we require four NOR gates overall for first half of the thermometer code.

If F(X,Y,Z,W) for the first half is 0, we will know that there is no bubble error in the first part.

We will need an identical circuit for the second half (requiring another four NOR gates)

The outputs of both the halves will be zero if there is no error. If these outputs are combined with a NOR gate, output will be one.

Thus, there will be no error for final output 1.

The logic circuit is as follows:

Let the 8-bit thermometer code be T = T1 T2 T3 T4 T5 T6 T7 T8

T1 represents the leftmost bit and T8 represents the rightmost bit.

The first half is T1 T2 T3 T4 and the second half is T5 T6 T7 T8

Note that ~T1 = T1 bar = inverted value of T1

We need to check for one more thing. Even if T1 T2 T3 T4 and T5 T6 T7 T8 are separately correct, we can have a situation where T1 T2 T3 T4 = 1110 and T5 T6 T7 T8 = 1000. While both these halves are correct separately, they contain bubble error when joined together. Thus, we need to check if T4 and T5 have the correct combination. From the allowed sequences, we know that T4 = 0 and T5 = 1 is the only sequence not allowed. Thus, if we use 0, T4 and ~T5 as the inputs to a 3-input NOR gate, we will get 1 if there is an error, and 0 if there is no error.

Thus, we have three sub-blocks, one each for the two halves and one for T4 and T5.

Output is Y. Y = 1 when there is no bubble error in the thermometer code. Y = 0 if bubble error is present.