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asked in Digital Logic by Junior (965 points) | 184 views

3 Answers

+3 votes
Best answer

Consider a state 100 representing 001 as Q2 is the rightmost in figure. The FFs are triggered on rising edge (0-1). Ignoring the flip flop delays as it is asynchronous counter,

111 - next state 011, CLK1 changed from 1-0
011 - next state 101, CLK1 changed from 0-1, CLK2 changed from 1-0
101 - next state 001, CL1 changed from 1-0, 
001 - next state 110, CLK1 changed from 0-1, CLK2 changed from 0-1
110 - next state 010, CLK1 changed from 1-0
010 - next state 100, CLK1 changed from 0-1, CLK2 changed from 1-0
100 - next state 000, CLK1 changed from 1-0
000 - next state 111, CLK1 changed from 0-1, CLK2 changed from 0-1

So, cycle repeats 7-6-5-4-3-2-1-0-7 mod 3 down counter. 

So, A is the answer.

answered by Veteran (326k points)
edited by
Q2 is MSB given in questn then how is it pssbl??
ans should be (a) always down counter in fact it is mod 8 down counter..
yes its a down counter .
Yes. I missed "negative" triggering. Corrected now.
its positive edge triggered and count will be like this  

000 > 111 > 110 > 101 > 100 > 011 > 010 > 001
+2 votes

sir , plzz check it in above explanation i hav done this question for negative edge triggering so it will give up counter but if i will do it by positive edge triggering then we will get down counter 

so answer will be always down couter from any initial state 

answered by Loyal (4.5k points)
so option will be (a)

two another way to draw a down counter

Yes, you are correct. I had missed the "rising edge" part.
0 votes
always down bcz Q of previous is connected to clk of next which is +edge triggrd..
answered by (327 points)

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