Digital Electronics - Counters - Discussion
Discussion Forum : Counters - General Questions (Q.No. 16)
16.
How many AND gates would be required to completely decode ALL the states of a MOD-64 counter, and how many inputs must each AND gate have?
Discussion:
6 comments Page 1 of 1.
Harini said:
8 years ago
i.e we need 2^n:n Decoder.
Here n=6,
To decode mod64 or 6-bit counter o/p we need 64:6 decoder.
so it consists 64 AND gates.
Here n=6,
To decode mod64 or 6-bit counter o/p we need 64:6 decoder.
so it consists 64 AND gates.
(1)
Vaibhav Pawar said:
9 years ago
@Satyendra Singh.
Good explanation, thanks.
Good explanation, thanks.
(1)
Satyendra singh said:
1 decade ago
MOD = 64;
MOD = 2^6;
Because,
STARTING MOD = 000000.
.
.
.
ENDING MOD = 111111 FOR UNDERSTANDING(32 16 8 4 2 1).
Total state = 64(0 to 63).
For every state need of 6 input gate then,
Total no.of AND gate = 64;(every 6 input AND GATE).
MOD = 2^6;
Because,
STARTING MOD = 000000.
.
.
.
ENDING MOD = 111111 FOR UNDERSTANDING(32 16 8 4 2 1).
Total state = 64(0 to 63).
For every state need of 6 input gate then,
Total no.of AND gate = 64;(every 6 input AND GATE).
Santosh said:
1 decade ago
Where we use 64 and gates? Please explain.
Prakash said:
1 decade ago
Total 64 states, so we need 6 filp-flops (since we are using synchronous counter).
In this, 1st and 2nd ff does not require AND gate. Starting from the 3rd ff to the 6th we need 4 ff and the final one will have 5 inputs.
CORRECT ME IF I AM WRONG.
In this, 1st and 2nd ff does not require AND gate. Starting from the 3rd ff to the 6th we need 4 ff and the final one will have 5 inputs.
CORRECT ME IF I AM WRONG.
Sreekala.k.s said:
1 decade ago
Total 64 states ie input combination so 64 gates.
For 64 input combinations need 6 variables so 6 input for each gates.
For 64 input combinations need 6 variables so 6 input for each gates.
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