Decoder

 

Implementation of Higher-order Decoders

Now, let us implement the following two higher-order decoders using lower-order decoders.

• 3 to 8 decoder
• 4 to 16 decoder

3 to 8 Decoder

In this section, let us implement 3 to 8 decoder using 2 to 4 decoders. We know that 2 to 4 Decoder has two inputs, A1 & A0 and four outputs, Y3 to Y0. Whereas, 3 to 8 Decoder has three inputs A2, A1 & A0 and eight outputs, Y7 to Y0.

We can find the number of lower order decoders required for implementing higher order decoder using the following formula.

Required number of lower order decoders= m2/m1

Where,

1. m1 is the number of outputs of lower order decoder.
2. m2 is the number of outputs of higher order decoder. 
3. m1 = 4 and m2 = 8. Substitute, these two values in the above formula.
4. Required number of 2 to 4 decoders=8/4=2

Therefore, we require two 2 to 4 decoders for implementing one 3 to 8 decoder. The block diagram of 3 to 8 decoder using 2 to 4 decoders is shown in the following figure.

The parallel inputs A1 & A0 are applied to each 2 to 4 decoder. The complement of input A2 is connected to Enable, E of lower 2 to 4 decoder in order to get the outputs, Y3 to Y0. These are the lower four min terms. The input, A2 is directly connected to Enable, E of upper 2 to 4 decoder in order to get the outputs, Y7 to Y4. These are the higher four min terms.