GATE 2019 ECE (Electronics and Communication) Digital circuits questions

Digital Circuits Questions and Answers: GATE 2019 ECE (Electronics and Communication)

Ques1. In the circuit shown, what are the values of F for EN = 0 and EN =1, respectively.

A) 0 and D

B) Hi-z and D

C) 0 and 1

D) Hi-z and D’

sol: Let’s consider the NAND gate output is x and this is the input for PMOS transistor and NOR gate output is y and this is the input for the NMOS transistor.

So x = (EN.D)’ and y = (EN’ + D)’

So there are two inputs EN and D then 4 combinations are present. We all know that PMOS will conduct at logic 0 and NMOS will conduct at logic 1. The truth table is 

From the truth table when EN=0, both the transistors are OFF and not connected to ground and supply (VDD) so the transistor is in high impedance state.

From the truth table when EN=1, the output is equal to D value.

Option (B)

Ques2. In the circuit shown, A & B are the inputs, and F is the output. What is the functionality of the circuit?

A) Latch

B) XNOR

C) SRAM Cell

D) XOR

Sol: We all know that PMOS will conduct at logic 0 and NMOS will conduct at logic 1.

Here also two inputs A and B so 4 combinations of input will present.
Rearranging the circuit, for the ease of convenience. 

The truth table is:

Case1: when A = 0, B = 0

Both PMOS transistors are ON and NMOS transistors are OFF so F is directly connected to the VDD. In this case output is 1.

Case2: when A = 0, B = 1

Here, the transistor PMOS1 is OFF, PMOS2 is ON, the transistor NMOS1 is ON and NMOS2 is OFF so from the figure we can see that F is directly connected to the A and the value of A is 0 so F is also 0.

Case3: when A = 1, B = 0

Here, the transistor PMOS1 is ON, PMOS2 is OFF , the transistor NMOS1 is OFF and NMOS2 is ON so from the figure we can see that F is directly connected to the B and the value of B is 0 so F is also 0.

Case4: when A = 1, B = 1

Both PMOS transistor are OFF and Both NMOS transistors are ON So F is directly connected to the AB and A & B values are 1 and behave as AND gate so the output is 1

So from the truth table we conclude that this is an XNOR gate truth table.

Option (B)

Ques3. In the circuit shown, the clock frequency, i.e. the frequency of clk signal, is 12 kHz. The frequency of the signal at Q2 is …….kHz.

Sol: from the fig.

D1 = Q1’Q2’

D2 = Q1

Q1 = D1

Q2 = D2

State table:

Initially assume FF in reset mode i.e. Q2 and Q1 are zero.

So this is the MOD-3 counter.

NOTE: in MOD-N counter, if the applied frequency is “f” then output frequency is f/N.

“MOD N” indicates the number of states in the counting sequence.

fout = fclk/3

        = 12/3

        = 4 kHz

Ques4. The state transition diagram for the circuit shown is.

Sol: let’s take mux output is Y.

Y = A’Q_n’ + AQ_n …………. (i)

And D = (Q_nY)’…………. (ii)

Put Y value in equation (ii),

D = (Q_n. (A’Q_n’ + AQ_n))’

On solving D = (AQ_n)’

And we know the next state equation of D FF is Q_n+1 = D

From the table, it is clear that it is NAND gate so state diagram is:

When Q=0 and A=0 then it goes to the Q=1

When Q=0 and A=1 then it goes to the Q=1

When Q=1 and A=0 then it goes to the Q=1 (same state)

When Q=1 and A=1 then it goes to the Q=0
option (c)