Test: Flip Flops - Electronics and Communication Engineering (ECE) MCQ

We can design a Modulo 2n counter with n flip-flops. We need a little more because 272 is greater than 256, i.e. 28; 29 = 512, which is greater than 272. In order to create a Modulo 272 counter, only 9 flip-flops are required. Option (B) is the right answer.

The following are correct statements: S1: An Overbeck counter has a Hamming distance of 2 while a Johnson counter has a Hamming distance of 1. S2: In the Johnson counter, output sequences of 0, 8, 12, 14, 15, 7, 3, 1, 0,... are feasible, but in the Overbeck counter, output sequences of 2, 1, 8, 4, 2, 1,... are possible. S3: A binary counter may represent 2^N states, where N is the number of bits in the code, but an Overbeck counter and a Johnson counter can only represent N states and 2N states, respectively.

We get both Q and Q' as 1 when both R and S are set to 0. This output will be permanent, and it will not be affected by the sequence in which the events occur; it will only be affected by the input values [NO RACE CONDITION]. However, if we change R=S=1 after this state, the output would be indeterminate depending on which NAND gate processes first (either Q or Q' will become 0, but we can't tell which [RACE CONDITION]), resulting in an indeterminate state.

When the Set line is set high, and the Reset line is set low, then the RS flip-flop has the state set to 1. 

A ring counter requires 'n' flip-flops for 'n' number of states. A modulus-12  has 12 states starting from 0000 to 1011. At least 12-bits are required to design a modulus-12 counter.  

RS flip flop using NAND gates. So, 00 input causes an indeterminate state which MAY lead to oscillation.

Total 4. 2 J-K flip-flops for synchronous counter + 2 J-K flip-flops to make 2 bit counter. Actually, when we have repeated again, then after three, we don't know that our Synchronous counter will go to which zero. To make it work right, we need to move it to 1st zero after 3 2nd zero after 1 3rd zero after 2 i.e. 0 -> 1 -> 0 -> 2 -> 0 -> 3 (from here it again go to 1st zero). In order to decide which zero to move on, we use counters from 1 to 3; I have attached the truth table of a normal 2 bit synchronous counter using JK flip flop.

Count sequence is 00, 00, 01, 01, 10, 10, 11, 11 the  repeated sequence is above present. As two bits will not be sufficient, we need at least three flip-flops. So, the answer is 3.

The 4 bit Johnson counter connects the complement of the output of the last shift register with the input of the first register with shift distance=1, i.e., 1 bit will shift/cycle. It work as follows:- 0000 //The last 0 complemented & fed as input to first register = ( 1000 1100 1110 1111 )//The last one complimented & fed as input to first register = ( 0111 0011 0001 0000).

At first, the Q output of D - FF = 1 in starting Q output of J K - F F = 0. Now with the help of the present state and next state, the circuit.

• Toggle: J = K = 1
• Hold : J = K = 0

Make table Q output of D-FF is going to next state input of J K-F F and the bits sequence produced is like 110110…..Including initial condition (0), we get output as 0110110110. Hence answer is (A) part.