# How do you implement a 3/8 decoder?

## How do you implement a 3/8 decoder?

From the above Boolean expressions, the implementation of 3 to 8 decoder circuit can be done with the help of three NOT gates & 8-three input AND gates. In the above circuit, the three inputs can be decoded into 8 outputs, where every output represents one of the midterms of the three input variables.

### How many 3 line to 8 line decoder are required for a 4 to 16 decoder assume that one inverter is also available?

4 to 16 decoder circuit is obtained from two 3 to 8 decoder circuits or three 2 to 4 decoder circuits. When two 3 to 8 Decoder circuits are combined the enable pin acts as the input for both the decoders.

**How many 3 to 8 line decoder with an enable input are needed to construct a 62 64 line decoder without using any other logic gates?**

How many 3-to-8 line decoders with an enable input are needed to construct a 6-to-64 line decoder without using any other logic gates? hence required decoders (from fig.) = 9 so ans is ( C) part.

**How many 3 lines to 8 line decoders are required for a 5 of 32 decoder submit detailed design?**

Altogether 5 3 to 8 decoders are required to produce 32 output lines of 5 to 32 decoders.

## What is 2×4 decoder?

The 2-to-4 line binary decoder depicted above consists of an array of four AND gates. The 2 binary inputs labelled A and B are decoded into one of 4 outputs, hence the description of 2-to-4 binary decoder. Each output represents one of the minterms of the 2 input variables, (each output = a minterm).

### How many inputs are required for a 6 to 64 decoder?

The answer is 4. Say the 8 input variables are A,B,C,D,E,F,G,H. Pass C,D,E,F,G,H as inputs to each of the four 6->64 decoders.

**How many 2 to 4 line decoders with an enable input are required to construct a 4 to 16 line decoder without using any other logic gates?**

Four 2*4 encoders are required to produce 16 outputs and 1 extra 2*4 decoder is required to enable the decoder when creating 4*16 decoders….Implement 8:1 mux using 4:1 mux.

ABCD Input Lines | Decoder Enabled | Output Lines |
---|---|---|

0101 | D2 Decoder | Y5 |

0110 | D2 Decoder | Y6 |

0111 | D2 Decoder | Y7 |

1000 | D3 Decoder | Y8 |

**How many 2×4 decoders are needed for 5×32?**

Similarly if in=101=5, the output D5 will be 1 and rest all outputs will 0. It is mandatory that en is 1 for the output to appear. If en is 0, the output is 0 no matter what the input is. So we have 4 3:8 decoders and 1 2:4 decoder, how to design a 5:32 from them?