This set of Linear Integrated Circuit Questions and Answers for Aptitude test focuses on “Open-Loop Voltage Gain as a Function of Frequency – 2”.

1. Which of these statements is false?

a) The open loop gain A_{OL}(f) dB is approximately constant from 0Hz to f_{o}

b) When input signal frequency and f is equal to break frequency f_{o}, the gain frequency is called -3dB frequency

c) The open loop gain A_{OL}(f) dB is approximately constant upto break frequency f_{o}, but there after it increase 20dB each time there is a tenfold increase in frequency.

d) At unity gain crossover frequency, the open loop gain A_{OL}(f) dB is zero

View Answer

Explanation: When A

_{OL}(f) is approximately constant up to break frequency, there will be 20dB decrease each time there is a tenfold increase in frequency. Therefore it may be considered that the gain roll of at the rate of 20dB/decade.

2. What is the maximum phase shift that can occur in an op-amp with single capacitor?

a) 180^{o}

b) 60^{o}

c) 270^{o}

d) 90^{o}

View Answer

Explanation: At corner frequency, the phase angle is -45

^{o}(lagging) and at infinite frequency the phase angle is -90

^{o}. Therefore, a maximum of 90

^{o}phase change can occur in an op-amp with a single capacitor.

3. How can the gain roll off represented in dB/octave?

a) 12 dB/octave

b) 6 dB/octave

c) 10 dB/octave

d) 8 dB/octave

View Answer

Explanation: Octave represents a two fold increase in frequency. Therefore, 20 gain roll off at the rate 20 dB/decade is equivalent to 6 dB/octave.

4. Select the correct magnitude and phase for the frequency range.

List-I | List-II |

1. f< |
i. Gain is 3dB down from the value of A_{OL} in dB |

2. f=f_{1 } |
ii. Gain roll off at the rate of 20 dB/decade |

3. f>>f_{1} |
iii. Magnitude of the gain is 20logxA_{OL} in dB |

a) 1-iii, 2-i, 3-ii

b) 1-I, 2-ii, 3-iii

c) 1-iii, 2-ii, 3-i

d) 1-ii, 2-iii, 3-i

View Answer

Explanation: Properties of magnitude and phase angle characteristics equations.

5. The specific frequency at which A_{OL} (dB) is called

a) Gain bandwidth product

b) Closed loop bandwidth

c) Small signal bandwidth

d) All of the mentioned

View Answer

Explanation: A

_{OL}(dB) is zero at some specific value of input signal frequency called unity gain band width. The mentioned terms are the equivalent terms for unity gain band width.

6. Find out the expression for open loop gain magnitude with three break frequency?

a) A/[1+ j(f/f_{o1})]x[1+ j(f/f_{o2})]x[1+ j(f/f_{o3})].

b) A/[1+ j(f/f_{o})]^{3}

c) A/A/[1+ j(f/f_{o1})]+ [1+ j(f/f_{o2})]+ [1+ j(f/f_{o3})].

d) All of the mentioned.

View Answer

Explanation: The gain equation for op-amp is A

_{OL}(f) = A/[1+ j(f/f

_{o1})]x[1+ j(f/f

_{o2})]x[1+ j(f/f

_{o3})].

7. Find out the frequencies that are avoided in frequency response plot?

a) None of the mentioned

b) Single break frequency

c) Upper break frequency

d) Lower break frequency

View Answer

Explanation: Op-amps have more than one break frequency because only a few capacitors are present. Often upper break frequencies are well above the unity gain bandwidth. So, they are avoided in frequency response plot.

8. In a phase response curve of MC1556, the phase shift is -162.5^{o} about 4MHz. If the approximate value of first break frequency is 5.5Hz. Determine the approximate value of the second break frequency?

a) 945.89MHz

b) 945.89kHz

c) 945.89GHz

d) 945.89Hz

View Answer

Explanation: The phase angle equation is φ(f) = – tan

^{-1}(f/f

_{o1})- tan

^{-1}(f/f

_{o2})

=> -162.5

^{o}= – tan

^{-1}(3MHz/5.5) – tan

^{-1}(3MHz/f

_{o2})

=> tan

^{-1}(3MHz/f

_{o2}) = +162.5

^{o}-89.99

^{o}

=> 3MHz/f

_{o2}= tan(72.50

^{o})

=> f

_{o2}= 3MHz/tan(72.50

^{o})=945.89kHz.

9. Consider a practical op-amp having two break frequency due to a number of RC pole pairs. Determine the gain equation using the given specifications:

A ≅ 102.92dB; f_{o1} =6Hz ; f_{o2} =2.34MHz .

a) 140000/[1+ j(f/6)]x[1+ j(f/2.34)].

b) 140000/[1+ j(f/6)]x[1+ j(f/234)].

c) 140000/[1+ j(f/6)]x[1+ j(f/23400)].

d) 140000/[1+ j(f/6)]x[1+ j(f/2340000)].

View Answer

Explanation: Gain of op-amp at 0Hz, A = 10

^{102.92/20}

=> A = 10

^{5.146}=139958.73 ≅ 140000.

The gain equation with two break frequency value is given as

A

_{OL}(f) = A/[1+ j(f/f

_{o1})]x[1+ j(f/f

_{o2})] =140000/[1+ j(f/6)]x[1+ j(f/2.34)].

10. What is the voltage transfer function of op-amp in s-domain.

a) S×A/(s+ω_{o})

b) A/(s+ω_{o})

c) (A×ω_{o})/(s+ω_{o})

d) None of the mentioned

View Answer

Explanation: For the voltage transfer function in s-domain is expressed as A

_{OL}(f) = A/[1+ j(f/f

_{o})] =A/[1+j(ω/ω

_{o})]= A×ω

_{o}/(jω/jω

_{o}) = A×ω

_{o}/ (s+ω

_{o}).

11. Calculate the unity gain bandwidth of an op-amp. Given, A≅ 0.2×10^{6} & f_{o} =5Hz?

a) 10^{6}

b) 10^{8}

c) 10^{-5}

d) 10^{-10}

View Answer

Explanation: The break frequency, f

_{o}= UGB/A

=> =f

_{o}×A = 0.2×10

^{6}× 5 = 10

^{6}.

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