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+12 votes
A priority encoder accepts three input signals $\text{(A, B and C)}$ and produces a two-bit output $(X_1, X_0 )$ corresponding to the highest priority active input signal. Assume $A$ has the highest priority followed by $B$ and $C$ has the lowest priority. If none of the inputs are active the output should be $00$, design the priority encoder using $4:1$ multiplexers as the main components.
asked in Digital Logic by Veteran (413k points)
edited by | 1k views

This might help ...

be aware, in this above lecture, he did some mistake ..!
yes @ankit, he did a mistake, 0's will be put only in 01 rows in second k-map.

5 Answers

+16 votes
Best answer

$\text{MSB} - \text{Most Significant Bit}$
$\text{LSB} - \text{Least Significant Bit}$

\rlap{\text{TRUTH TABLE}} \\
\rlap{\text{Inputs}: A,B,C}\\
\rlap{\text{Outputs}: \text{MSB},\text{LSB}}\\
A &B&C&\quad\text{MSB}&\text{LSB}\\

$\text{MSB} = A+B$
$\text{LSB} =A + \bar B C$ 


It can be implemented using two $4 \times 1$ Multiplexers.

answered by Boss (14.1k points)
edited by
Here why u select C as input of MUX?

shouldnot it be signal to see which MUX is active at a time and other deactive at that time?
no we can't do like that, we need both MSB and LSB at any time thats why both the mux should be active all the time

why u took o/p 1 for 001 i/p??


अनुराग पाण्डेय how you derive expression (K-map??) if yes then how you have taken the inputs please comment.

Is this realisation possible if "If none of the inputs are active the output should be 00" - this statement is not present in question?


A &B&C&\quad\text{MSB}&\text{LSB}\\


$LSB=A \bar{B}$

2018 Given in question "If none of the inputs are active the output should be 00", so lowest priority should start with "01" and not "00"


Assume A has the highest priority followed by Band C has the lowest priority.

and wiki says

If two or more inputs are given at the same time, the input having the highest priority will take precedence.[1] An example of a single bit 4 to 2 encoder is shown, where highest-priority inputs are to the left and "x" indicates an irrelevant value - i.e. any input value there yields the same output since it is superseded by higher-priority input. 


srestha how the expression for MSB and LSB is derived?

for final circuit it is just an assumption
+8 votes

We can implement the priority encoder using two $4:1$. Multiplexers as main components and $1$ NOT gate and $1$ OR gate
Using following truth table (Priority encoder with highest priority to $A$)$$\begin{array}{cccc|cc}  \textbf{A} & \textbf {B} &\textbf {C} & \textbf {X} &  \textbf{$X_1$ } & \textbf{$X_0$}\\ \hline 0& 0& 0  & \text{X} &0 &0\\ 0& 0& 1  & \text{X} &0 &1\\  0& 1& \text{X} & \text{X} &1 &0 \\ 1& \text{X}& \text{X}  & \text{X} &1 &1 \\ \end{array}$$

answered by Boss (11.4k points)
edited by
Can you please explain logic behind second mux? I'm getting the answer here :
i know the priority encoders but not able to understand this above concept plz clear me????
+3 votes

we know the truth table of priority encoder .

now we have a multiplexer. we need to do selection in a abnormal way. means we have to play with select lines . and we have 2 output from the above function. relize them and then give input to the select line of muxs.

y1= i3+i2

y0=i3+ i2' i1

now make y1 as s1 select line .and y0 as s0.

answered by Boss (15.8k points)
I have understood that here we have made selection i/p(s0 and s1) a bit complex in order to select any of the (x3,x2,or x1). But this encoder o/p should be 2 bits. How are we acheiving that thing?
0 votes

$X_0 = \bar{A} \bar{B} C + A$

$X_1 = \bar{A} B  + A$


answered by Boss (34.5k points)
0 votes


Suggestions are welcome.

answered by (33 points)

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