Analog and Digital Electronics - Online Test
Q1. The following FIVE instructions were executed on an 8085 microprocessor.
MVI A, 33H
MVI B, 78H
ADD B
CMA
ANI 32H
The Accumulator value immediately after the execution of the fifth instruction is
Answer : Option B
Explaination / Solution:
MVI A, 33H A⟵33H
MVI B, 78H B⟵78H
ADD B B⟵ABH
CMA A⟵54H
ANI 32H A⟵10H
A⟶0011 0011 A⟶1010 1011 0101 0100
B⟶0111 1000 B⟶0101 0100 0011 0010
1010 1011 0001 0000
Q2.
Which one of the following gives the simplified sum of products expression for the Boolean function F =
m0 + m2 + m3 + m5, where m0,
m2, m3
and m5 are minterms corresponding to the inputs A, B and C with A as the MSB and C as the LSB?
and m5 are minterms corresponding to the inputs A, B and C with A as the MSB and C as the LSB?
Answer : Option B
Explaination / Solution:
F =
m0 + m2 + m3 + m5, ⟶ minterm
Q3. In an 8085 microprocessor, the shift registers which store the result of an addition and the overflow bit are, respectively
Answer : Option B
Explaination / Solution:
The shift registers A and F store the result of an addition and the overflow bit.
Q4. 
In the circuit shown below, the op-amp is ideal, the transistor has VBE = 0.6 V and β= 150. Decide whether the feedback in the circuit is positive or negative and determine the voltage V at the output of the op-amp.
Answer : Option D
Explaination / Solution:
The circuit is shown in fig below
Q5. A 3-input majority gate is defined by the logic function M(a,b,c) = ab + bc + ca. Which one of the following gates is represented by the function 
Answer : Option B
Explaination / Solution:
3 input majority gate is given as
M(a,b,c) = ab + bc + ca
We have to obtain 
We obtain truth table for the function as
So, the function is odd number of 1’s detector. This function represent the 3-input XOR gate.
Q6. The Boolean expression 
converted into the canonical product of sum (POS) form is
converted into the canonical product of sum (POS) form is
Answer : Option A
Explaination / Solution:
We have the SOP Boolean form,
Hence, in POS form, we have
Q7. If X = 1 in logic equation 
Then
Then
Answer : Option D
Explaination / Solution:
Substituting X = 1 and
we get
1 + A = 1 and 0 + A = A
Substituting X = 1 and
we get 1 + A = 1 and 0 + A = A
Q8. What are the minimum number of 2- to -1 multiplexers required to generate a 2- input AND gate and a 2- input Ex-OR gate
Answer : Option A
Explaination / Solution:
The AND gate implementation by 2:1 mux is as follows
Q9. The Boolean expression 
simplifies to
simplifies to
Answer : Option C
Explaination / Solution:
Given the Boolean expression
We simplify the expression as
Q10. What are the counting states (Q1,Q2) for the counter shown in the figure below
Answer : Option A
Explaination / Solution:
The given circuit is as follows.
The truth table is as shown below. Sequence is 00, 11, 10, 00 ...
The truth table is as shown below. Sequence is 00, 11, 10, 00 ...
