How do you design a decimal counter using j-k flip-flop?

Answer 1

Since it is a 3 bit counter( max state here is 6<8 i.e. 2^3) so 3 FF would be used here.Name these three FF A, B,C (u can assume ny other name).

Prepare a state table- here we have 2 count from state 0 to 1, 1(001 in binary) to next state(010), then from 2(010) to 4(100) , then from 4(100) to 5(101), then from 5(101) to 6(110).

After u prepare the state table next step is to prepare the excitation table from this state table.Generally RS FF is used to design the sequence with even no. in the end, for odd we gen' use JK FF's.

Using the present state as o/p transitions, and the basic excitation table for rs ff prepare the table.( this can be done here like ABC changes value as 000 to 001 write the value for Ja as X and Ka as 0, same can be done for Jb , Kb, as well as Jc, Kc.)

Using excitation table prepare K-maps for Ja , Ka , Jb ,Kb , Jc , Kc .From these Kmaps obtain simplified Boolean expressions.

And these boolean expressions can be used to prepare complete Logic.

Answer 2

step 1) make a table for the present state ,future state and the excitation state of jk flip-flop.

step 2)draw the k-map of perticular j's and k's.eg-j1,j2,j3 all will have different k maps and same will happen with k's

step3) check the output from the k-map and draw the circuit according to it.

Answer 3

better use a D flip flop to design. use the formula N<=(2^n)-1 to find the number of flip flops required.

Answer 4

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