3bit cmos encoder cicuit

i need a 3 bit cmos encoder circuit with all the 8 inputs
i have created a 7 input encoder using rom design bt i need to have the '000' condition also
please help

What about a 3-input NOR ?

i need a encoder
why would i need to think about a nor??
its very simple
just add 2 nor with one another in series

Aritra17 said:

just add 2 nor with one another in series

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... and then think about why it won't work, Baby! :roll:

i can't understand what you r trying to say
how is that possible to create a 3bit encoder using 3input nor??
2 nor joined in series means the relation is like this:
((a+b)bar +c)bar
if i consider a=1 & rest 0 then the result is 1
same thing happens for b
but in case of c when it is 1 and rest 0 then the result is 0
which is incorrect in terms of encoder

and secondly this happens for 3 inputs & one out put
i need 8 inputs & 3 outputs
how am i supposed to join these nor gates??

if you can explain in details with a circuit diagram that would be helpful
i can't understand this way

Aritra17 said:

i can't understand what you r trying to say
how is that possible to create a 3bit encoder using 3input nor??

Click to expand...

Hi Paul,
in your original question (above) you asked

Aritra17 said:

... bt i need to have the '000' condition also
please help

Click to expand...

... so I just wanted to give you a hint how to master the '000' condition, but erroneously I thought of decoding (3 to 8 ) instead of encoding (8 to 3), sorry!

Aritra17 said:

2 nor joined in series means the relation is like this:
((a+b)bar +c)bar
if i consider a=1 & rest 0 then the result is 1
same thing happens for b
but in case of c when it is 1 and rest 0 then the result is 0
which is incorrect in terms of encoder

Click to expand...

So you found out that this wouldn't work! That's what I wanted to tell you above. Still thinking of decoding.

Aritra17 said:

... and secondly this happens for 3 inputs & one out put

Click to expand...

No: for decoding the '000' condition a 3-input NOR would work (output high in this case).

Aritra17 said:

i need 8 inputs & 3 outputs
how am i supposed to join these nor gates??

Click to expand...

For encoding the '000' condition (only) you'd need an 8-input NOR gate e.g. like MC14078B or 6x2-input OR + 1x2-input NOR (for active high output).

Schematics for a full 3 to 8 encoder can easily be found via G00gle images search.

erikl said:

For encoding the '000' condition (only) you'd need an 8-input NOR gate e.g. like MC14078B or 6x2-input OR + 1x2-input NOR (for active high output).

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couldn't understand what you tried to explain
if i use 6 two input or gates the number of inputs will be 8 i agree
but how will it end up having only 3 outputs????

and i need the full encoder with all the 8 input conditions
i.e '000' condition to '111' condition

google image search does not give 8 to 3 bit encoder
not in cmos schematic level

actually im making a 3 bit adc circuit in tanner eda software
i have created all of the basic components
i have also created a encoder for the circuit but its giving some abnormality
you can see that abnormal behavior of the output waveform in another thread of mine
named as "output waveform of a 3bit flash adc in tanner eda software"

the encoder i built using cmos schematic has 7 inputs & has 7 high-low conditions in the output waveform
but i don't think think that's right
that's why i need help

please understand the full condition
and please help

Aritra17 said:

google image search does not give 8 to 3 bit encoder
not in cmos schematic level

Click to expand...

Try this G00gle request. Already the first row shows appropriate schematics:

erikl [img said:

https://obrazki.elektroda.pl/79_1300713497.gif[/img]

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where does the I0 input goes to?

Aritra17 said:

where does the I0 input goes to?

Click to expand...

Nowhere, as you can see. It's unused.

I<7:0>, representing the numbers 0..7 , are encoded to a binary (resp. BCD) representation of the respective number.

The '000' condition is already covered by the 3 OR gates: if I<7:1> = 0 --> Y=0 , no matter if I<0> is set (which it should be, if the input is non-ambiguous) - or not.

see i know this condition
i told you before also
i need all the 8 input conditions '000' up to '111'
my encoder has 7 input terminals and also shows 7 high low conditions but i need to have 8 high low conditions using cmos technology

your given circuit seems to be made through vldl programing

i am using tanner eda software

please try to give me a 8 input 3 output cmos encoder

see
this is the circuit and i want to make this using cmos technology

The above circuit exactly fulfills the function of your "8-line to 3-line encoder" box.

A 3-bit binary output can only represent the numbers 0..7, and not 1..8 . In your above schematic, if your Vin is either below the ref. voltage of the bottom-most comparator (then its output is low), or if Vin is between the ref. voltages of the 2 bottom-most comparators (in this case the output of the bottom-most comparator is high), in both cases the '000' condition is encoded, hence the output state of this comparator (I<0>) is irrelevant, and both the bottom-most comparator and EXOR are dispensable - like the top-most EXOR, BTW.

If you don't want to loose the bottom-most half-bit, I'd suggest to change your circuit to the following one.
With this, you can save one comparator and 2 EXORs:

this is the encode circuit



and this is my flash adc 3 bit output waveform



please tell what should be the desired output