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De-Multiplexer is a combinational circuit that performs the reverse operation of Multiplexer. It has single input, ‘n’ selection lines and maximum of 2n outputs. The input will be connected to one of these outputs based on the values of selection lines.
Since there are ‘n’ selection lines, there will be 2n possible combinations of zeros and ones. So, each combination can select only one output. De-Multiplexer is also called as De-Mux.
1×4 De-Multiplexer
1×4 De-Multiplexer has one input I, two selection lines, s1 & s0 and four outputs Y3, Y2, Y1 &Y0. The block diagram of 1×4 De-Multiplexer is shown in the following figure.
The single input ‘I’ will be connected to one of the four outputs, Y3 to Y0 based on the values of selection lines s1 & s0. The Truth table of 1×4 De-Multiplexer is shown below.
| Selection Inputs | Outputs | ||||
|---|---|---|---|---|---|
| S1 | S0 | Y3 | Y2 | Y1 | Y0 |
| 0 | 0 | 0 | 0 | 0 | I |
| 0 | 1 | 0 | 0 | I | 0 |
| 1 | 0 | 0 | I | 0 | 0 |
| 1 | 1 | I | 0 | 0 | 0 |
From the above Truth table, we can directly write the Boolean functions for each output as
$$Y_{3}=s_{1}s_{0}I$$
$$Y_{2}=s_{1}{s_{0}}”I$$
$$Y_{1}={s_{1}}”s_{0}I$$
$$Y_{0}={s_1}”{s_{0}}”I$$
We can implement these Boolean functions using Inverters & 3-input AND gates. The circuit diagram of 1×4 De-Multiplexer is shown in the following figure.
We can easily understand the operation of the above circuit. Similarly, you can implement 1×8 De-Multiplexer and 1×16 De-Multiplexer by following the same procedure.
Implementation of Higher-order De-Multiplexers
Now, let us implement the following two higher-order De-Multiplexers using lower-order De-Multiplexers.
- 1×8 De-Multiplexer
- 1×16 De-Multiplexer
1×8 De-Multiplexer
In this section, let us implement 1×8 De-Multiplexer using 1×4 De-Multiplexers and 1×2 De-Multiplexer. We know that 1×4 De-Multiplexer has single input, two selection lines and four outputs. Whereas, 1×8 De-Multiplexer has single input, three selection lines and eight outputs.
So, we require two 1×4 De-Multiplexers in second stage in order to get the final eight outputs. Since, the number of inputs in second stage is two, we require 1×2 DeMultiplexer in first stage so that the outputs of first stage will be the inputs of second stage. Input of this 1×2 De-Multiplexer will be the overall input of 1×8 De-Multiplexer.
Let the 1×8 De-Multiplexer has one input I, three selection lines s2, s1 & s0 and outputs Y7 to Y0. The Truth table of 1×8 De-Multiplexer is shown below.
| Selection Inputs | Outputs | |||||||||
|---|---|---|---|---|---|---|---|---|---|---|
| s2 | s1 | s0 | Y7 | Y6 | Y5 | Y4 | Y3 | Y2 | Y1 | Y0 |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | I |
| 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | I | 0 |
| 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | I | 0 | 0 |
| 0 | 1 | 1 | 0 | 0 | 0 | 0 | I | 0 | 0 | 0 |
| 1 | 0 | 0 | 0 | 0 | 0 | I | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 | 0 | I | 0 | 0 | 0 | 0 | 0 |
| 1 | 1 | 0 | 0 | I | 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 1 | 1 | I | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
We can implement 1×8 De-Multiplexer using lower order Multiplexers easily by considering the above Truth table. The block diagram of 1×8 De-Multiplexer is shown in the following figure.
The common selection lines, s1 & s0 are applied to both 1×4 De-Multiplexers. The outputs of upper 1×4 De-Multiplexer are Y7 to Y4 and the outputs of lower 1×4 De-Multiplexer are Y3 to Y0.
The other selection line, s2 is applied to 1×2 De-Multiplexer. If s2 is zero, then one of the four outputs of lower 1×4 De-Multiplexer will be equal to input, I based on the values of selection lines s1 & s0. Similarly, if s2 is one, then one of the four outputs of upper 1×4 DeMultiplexer will be equal to input, I based on the values of selection lines s1 & s0.
1×16 De-Multiplexer
In this section, let us implement 1×16 De-Multiplexer using 1×8 De-Multiplexers and 1×2 De-Multiplexer. We know that 1×8 De-Multiplexer has single input, three selection lines and eight outputs. Whereas, 1×16 De-Multiplexer has single input, four selection lines and sixteen outputs.
So, we require two 1×8 De-Multiplexers in second stage in order to get the final sixteen outputs. Since, the number of inputs in second stage is two, we require 1×2 DeMultiplexer in first stage so that the outputs of first stage will be the inputs of second stage. Input of this 1×2 De-Multiplexer will be the overall input of 1×16 De-Multiplexer.
Let the 1×16 De-Multiplexer has one input I, four selection lines s3, s2, s1 & s0 and outputs Y15 to Y0. The block diagram of 1×16 De-Multiplexer using lower order Multiplexers is shown in the following figure.
The common selection lines s2, s1 & s0 are applied to both 1×8 De-Multiplexers. The outputs of upper 1×8 De-Multiplexer are Y15 to Y8 and the outputs of lower 1×8 DeMultiplexer are Y7 to Y0.
The other selection line, s3 is applied to 1×2 De-Multiplexer. If s3 is zero, then one of the eight outputs of lower 1×8 De-Multiplexer will be equal to input, I based on the values of selection lines s2, s1 & s0. Similarly, if s3 is one, then one of the 8 outputs of upper 1×8 De-Multiplexer will be equal to input, I based on the values of selection lines s2, s1 & s0.
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