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This mock test of Digital Electronics - MCQ Test - 2 for GATE helps you for every GATE entrance exam.
This contains 20 Multiple Choice Questions for GATE Digital Electronics - MCQ Test - 2 (mcq) to study with solutions a complete question bank.
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*Answer can only contain numeric values

QUESTION: 1

In the circuit sho

Q_{o} = Q_{1} = 0 Then the values of QO and QI after 335^{th} clock pulse are _____________

Solution:

It is a MOD - 4 counter, because Johnson counter =

QUESTION: 2

An R — 2R ladder type DAC is shown in Figure if a switch status is '0', then 0 volt is applied and If a switch status is '1' then 5 volt is applied to the corresponding terminal of DAC.

What is value of output voltage V_{0} for switch status s_{o} = 0, s_{1}= 1, s_{2} = 1.

Solution:

QUESTION: 3

An R — 2R ladder type DAC is shown in Figure if a switch status is '0', then 0 volt is applied and If a switch status is '1' then 5 volt is applied to the corresponding terminal of DAC.

Q. What is the step size of DAC.

Solution:

Step size

QUESTION: 4

For the given circuit shown in figure signal generated at the output of AND gat is Y. there clock has signal frequency of 4 kHz, with duty cycle 50%

Q. What is the value of frequency of output Y

Solution:

A FF is simply MOD-2 counter.

QUESTION: 5

The following waveform pattern is for a(n) ________.

Solution:

QUESTION: 6

The input X for count sequence 0, 2, 3, 1, 0, 2 ....

Solution:

Just Solve by K-Map.

QUESTION: 7

For a Flip-flop formed from 2 NAND gates as shown in figure, the unusable state corresponds to

Solution:

when both X = Y = 0 then value of Q & will remain same so unstable.

QUESTION: 8

In the figure as long as XI = 1 and X2 = 1 the output Q remains at

Solution:

Let X_{1} = 1 & **X _{2}= 1**

IfQ=0 & X_{1}=1

The o/P of gate -2, will be 1, and which will make Q = 1

if Q = *1 & ***X _{2} = **1, then 0/P of gate - 2 will be 0 and which will make output of gate - 1 Q = 0

*Answer can only contain numeric values

QUESTION: 9

The Minimum no. of 2 inputs NAND gate required to implement Boolean function f(A, B, C, D) =

ABCD are_______

Solution:

**Correct Answer :- 3**

**Explanation : **

QUESTION: 10

The 4 variable function f is given in terms of min term as f(A, B, C, D) = E(2,3,8,10,11,12,14,15)

**Q. **If by use of K-Map function is minimized in sum of product forms then SOP is

Solution:

QUESTION: 11

The 4 variable function f is given in terms of min term as f(A, B, C, D) = E(2,3,8,10,11,12,14,15)

**Q. **Above Minimised SOP, can be implemented by how many min^{m} no. of 2 input NAND gate.

Solution:

*Answer can only contain numeric values

QUESTION: 12

Consider the following data in respect of a certain digital gate

l_{oH} = 0.2 mA I_{1H}= 40μ.A, 10_{L} = 16 mA, l_{iL} = 1.6 mA symbols have their meaning. Fan out will be

Solution:

F.0 = = 5

Smalls of both will be Fan out

F.I = =10

QUESTION: 13

If x, y and z are three Boolean variables, then F(x,y,z) = x + xy + y + yz + z + xz is equivalent to

Solution:

QUESTION: 14

How many AND gates are required for a 1-to-8 multiplexer?

Solution:

The number of AND gates required will be equal to the number of outputs in a demultiplexer.

QUESTION: 15

The logic circuit shown in figure is :

Solution:

*Answer can only contain numeric values

QUESTION: 16

In a dual slope ADC if reference voltage is 100 mV and the first integration period is set as 50 msec.

For an input voltage of 120 mV, the second integration (de-integration) period is___ ms

Solution:

t = 60 m sec

*Answer can only contain numeric values

QUESTION: 17

In a dual slope type digital voltmeter, an unknown signal voltage is integrated our 100 cycles of clock. If the signal has a 50 Hz pick up the maximum clock frequency can be__ kHz

Solution:

50 x 100

QUESTION: 18

In circuit given if both Transistors have same VT what is the approximate value of highest possible output voltage vout if v_{h}, can range from 0 to V_{DD}, it is assumed that 0 < v_{T} < V_{DD}

Solution:

QUESTION: 19

In the I.C. logic gate shown in figure.

If threshold voltage V_{BE} is o.75 volt and V_{CE} _{(sat)} = 0.2 V, calculate. Value of output voltage

**Q. If **VA = VB = 4.5 volt

Solution:

It is A NAND gate and if VA= VB = 4.5

Then o/p will be logic zero, as both are logic high

SO V_{0} = V_{CE, sat} = 0•2v

QUESTION: 20

If VA = 4.5 volt, VB = 0.2 volt

Solution:

if V_{A} = 4.5v & V_{B} = 0.2 v

so o/p will be 1, ie. V_{o} = 5 volt

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