encoders truth table

[Disha Karnataki]

they have taught recently about encoders to us & the truth table is something like her:https://www.google.co.in/imgres?img...a=X&ei=sOxpUu6JCMj-rAfi6YGgBw&ved=0CCwQ9QEwAA
i am unable to understand why the lsb bit x0 is written in the msb bit position & whereas the o/p is written as decimal 0
it could be written other way round also no?? then why is it written like that for all encoders???
whereas see this https://www.google.co.in/imgres?img...a=X&ei=IvBpUtueGI6Nrge1kYG4AQ&ved=0CDkQ9QEwAw it's usual way we write the truth table

 

[nimai]

The truth table from first link is the correct decimal to binary encoder.You can easily note that the Y output gives the binary code of the input signal which is active (here high)...Eg: For the input X0 or decimal zero you get binary output as 000..Similarly for all other input...

Second link shows the condition if any two of inputs are high,it gives high output...This another combinational circuits..

Encoder can be designed for generating many codes.The first one is for generating binary codes.
You can also design a decimal to gray code also.The only difference is the internal combinational circuit will be different..

 

[Disha Karnataki]

Encoder can be designed for generating many codes.The first one is for generating binary codes.
You can also design a decimal to gray code also.The only difference is the internal combinational circuit will be different..

yes i got it,
but you see the truth table for just decimal to binary priority encoder or may be even decimal to gray code convertor. Do you draw the k-map or map entered variable technique inorder to get the internal combinational circuit or how is it?? because drawing a k-map for 9 inputs(in case of decimal to bcd) is simply time consuming. so, how to we actually design the internal circuitry is my question, other-wise i am very comfortable with decoders,muxes,demuxes i.e with their internal circuitry.but,just unable to understand the same for encoders.

 

[nimai]

I will discuss the design of binary encoder .we must consider three outputs Y0,Y1 and Y2 for the inputs X0 to X7. From the truth table it can be noticed that Y2 becomes '1 or high' either X0 or X1 or X2 or X3 becomes high,implies the expression becomes Y2= X4 + X5 +X6+X7. Proceeding the same way ,you could obtain Y1=X2+X3+X6+X7 and Y0=X1+X3+X5+X7.

 

[Disha Karnataki]

thanks,
see the page number:"""""5-61"""""" of the truth table given there in the link given below:::
**broken link removed**

could u see that truth tabel???
there are dont'care conditions also it's basically a decimal to bcd """priority encoder"""" so that is why we have that kind of thruth table.
now i don't think so we can analyse it as it is as we did it previously, now we require k-map to be written right????
i actually have the internal design part of this in my book but unable to trace it by the methods i know & i am forced to byheart it ..
so if u know any way how to recognize his & built it up please do tell..

 

[nimai]

The truth table is for decimal to binary.Since the inputs and outputs are active low.You will get the binary code for X2, when it becomes '0' .(Since it is high prior to X1,the value of X1 can be 0 or 1 means don't care and since X1 is low prior to all other inputs ,they must be '1' .they must not be enabled.)Since the outputs are also active low the output '1101' is actually read or understood as '0010'.

Once if clear with truth table the internal design can be as follows:First you replace all the don't cares with one.Now You can see that D becomes zero if either X9 or X8 is zero.So simply connect them to a two input AND gate and the expression becomes D=X9.X8...Proceeding the same way C=X4.X5.X6.X7 ,B=X2.X3.X6.X7. and A=X1.X3.X7.X9.

Enjoy..

 

[Disha Karnataki]

yes correct now i am understanding the circuit if any more doubts i will soon approach u ...
another doubt is :
muxes,demuxes,decoders there are no active low o/p's whatever is input that only is the o/p.... but i don't know why unnecessarily the task of understanding the encoders is made difficult by using active low o/ps???
when i tried asking sir they gave me an answer saying that in order to protect the internal circuuitry from excess current which may damage the further circuitry.. extra not gate are used. but they would have also used two not gates & there would not have been active active low o/p as such.... i am not convinced with sir's ans..
what is your justification?????????