This set of Electronic Devices and Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Difference Amplifiers”.

1. For the difference amplifier which of the following is true?

a) It responds to the difference between the two signals and rejects the signal that are common to both the signal

b) It responds to the signal that are common to the two inputs only

c) It has a low value of input resistance

d) The efficacy of the amplifier is measured by the degree of its differential signal to the preference of the common mode signal

View Answer

Explanation: All the statements are not true except for the fact that it responds only when there is difference between two signals only.

2. If for an amplifier the common mode input signal is v_{c}, the differential signal id v_{d} and A_{c} and A_{d} represent common mode gain and differential gain respectively, then the output voltage v_{0} is given by

a) v_{0} = A_{d} v_{d} – A_{c} v_{c}

b) v_{0} = – A_{d} v_{d} + A_{c} v_{c}

c) v_{0} = A_{d} v_{d} + A_{c} v_{c}

d) v_{0} = – A_{d} v_{d} – A_{c} v_{c}

View Answer

Explanation: It is a standard mathematical expression.

3. If for an amplifier v_{1} and v_{2} are the input signals, v_{c} and v_{d} represent the common mode and differential signals respectively, then the expression for CMRR (Common Mode Rejection Ratio) is

a) 20 log (|A_{d}| / |A_{c}|)

b) -10 log (|A_{c}| / |Ad|)^{2}

c) 20 log (v_{2} – v_{1} / 0.5(v_{2} + v_{1}))

d) All of the mentioned

View Answer

Explanation: Note that all the expressions are identical.

4. The problem with the single operational difference amplifier is its

a) High input resistance

b) Low input resistance

c) Low output resistance

d) None of the mentioned

View Answer

Explanation: Due to low input resistance a large part of the signal is lost to the source’s internal resistance.

5. For the difference amplifier as shown in the figure show that if each resistor has a tolerance of ±100 ε % (i.e., for, say, a 5% resistor, ε = 0.05) then the worst-case CMRR is given approximately by (given K = R_{2}/R_{1} = R_{4}/R_{3})

a) 20 log [K+1/4ε].

b) 20 log [K+1/2ε].

c) 20 log [K+1/ε].

d) 20 log [2K+2/ε].

View Answer

Explanation: None.

6. For the circuit given below determine the input common mode resistance.

a) (R_{1} + R_{3}) || (R_{2}) || + (R_{4})

b) (R_{1} + R_{4}) || (R_{2} + R_{3})

c) (R_{1} + R_{2}) || (R3 + R_{4})

d) (R_{1} + R_{3}) || (R_{2} + R_{4})

View Answer

Explanation: Parallel combination of series combination of R

_{1}& R

_{3}with the series combination of R

_{3}and R

_{4}is the required answer as is visible by the circuit.

7. For the circuit shown below express v_{0} as a function of v_{1} and v_{2}.

a) v_{0} = v_{1} + v_{2}

b) v_{0} = v_{2} – v_{1}

c) v_{0} = v_{1} – v_{2}

d) v_{0} = -v_{1} – v_{2}

View Answer

Explanation: Considering the fact that the potential at the input terminals are identical and proceeding we obtain the given result.

8. For the difference amplifier shown below, let all the resistors be 10kΩ ± x%. The expression for the worst-case common-mode gain is

a) x / 50

b) x / 100

c) 2x / (100 – x)

d) 2x / (100 + x)

View Answer

Explanation: None.

9. Determine A_{d} and A_{c} for the given circuit.

a) A_{c} = 0 and A_{d} = 1

b) A_{c} ≠ 0 and A_{d} = 1

c) A_{c} = 0 and A_{d} ≠ 1

d) A_{c} ≠ 0 and A_{d} ≠ 1

View Answer

Explanation: Consider the fact that the potential at the input terminals are identical and obtain the values of V

_{1}and V

_{2}. Thus obtain the value of V

_{d}and V

_{c}.

10. Determine the voltage gain for the given circuit known that R_{1} = R_{3} = 10kΩ abd R_{2} = R_{4} = 100kΩ.

a) 1

b) 10

c) 100

d) 1000

View Answer

Explanation: Voltage gain is 100/10.

**Sanfoundry Global Education & Learning Series – Electronic Devices and Circuits.**

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