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## Linear Integrated MCQ Set 1

1. How is the higher order filters formed?
a) By increasing resistors and capacitors in low pass filter
b) By decreasing resistors and capacitors in low pass filter
c) By inter changing resistors and capacitors in low pass filter
d) All of the mentioned

Answer: c [Reason:] High pass filter are often formed by interchanging frequency determining resistors and capacitors in low pass filters. For example, a first order high pass filter is formed from a first order low pass filter by inter changing components Rand C.

2. In a first order high pass filter, frequencies higher than low cut-off frequencies are called
a) Stop band frequency
b) Pass band frequency
c) Centre band frequency
d) None of the mentioned

Answer: b [Reason:] Low cut-off frequency, fL is 0.707 times the pass band gain voltage. Therefore, frequencies above fL are pass band frequencies.

3. Compute the voltage gain for the following circuit with input frequency 1.5kHz. a) 4dB
b) 15dB
c) 6dB
d) 12dB

Answer: d [Reason:] |VO/Vin|= [AF×(f/fL)]/ [√1+(f/fL)2] = [4×(1.5kHz/225.86)] / √[1+(1.5kHz/225.86)2] =26.56/6.716=3.955 =20log(3.955)=11.9. |VO/Vin|≅12 dB AF= 1+(RF /R1)= 1+(12kΩ/4kΩ) =4. fL= 1/(2πRC) = 1/2π×15kΩ×0.047µF= 1/4.427×10-3 =225.86Hz.

4. Determine the expression for output voltage of first order high pass filter?
a) VO = [1+(RF /R1)]× [(j2πfRC/(1+j2πfRC)] × Vin
b) VO = [-(RF /R1)]× [(j2πfRC/(1+j2πfRC)] × Vin
c) VO = {[1+(RF /R1)]× /[1+j2πfRC] }× Vin
d) None of the mentioned

Answer: a [Reason:] The first order high pass filter uses non-inverting amplifier. So, AF= 1+(RF /R1) and the output voltage, VO = [1+(RF /R1)]× [(j2πfRC/(1+j2πfRC)]× Vin.

5. The internal resistor of the second order high pass filter is equal to 10kΩ. Find the value of feedback resistor?
a) 6.9kΩ
b) 5.86kΩ
c) 10kΩ
d) 12.56kΩ

Answer: b [Reason:] Pass band gain for second order butterworth response, AF =1.586. => AF= [1+(RF/R1)] => RF= (AF-1)×R1 =(1.586-1)×10kΩ =5860 =5.86kΩ.

6. Consider the following circuit and calculate the low cut-off frequency value? a) 178.7Hz
b) 89.3Hz
c) 127.65Hz
d) 255.38Hz

Answer: a [Reason:] The low cut-off frequency for the given filter is fL =1/√[2π√(R2×R3×C2×C3)]=178.7Hz.

7. Determine voltage gain of second order high pass butterworth filter.
Specifications R3 =R2=33Ω, f=250hz and fL=1khz.
a) -11.78dB
b) -26.51dB
c) -44.19dB
d) None of the mentioned

Answer: c [Reason:] Since R3 =R2 => C2 = 1/(2π ×fL×R2) = 1/(2π ×1kHz×33Ω) => C3 =C2= 4.82µF. Voltage gain of filter |VO/Vin|=AF / [√ 1+(fL/f)4] = 1.586/[1+(1kHz/250kz)4] =1.586/252=6.17×10-3 =20log(6.17×10-3)= -44.19dB.

8. From the given specifications, determine the value of voltage gain magnitude of first order and second order high pass butterworth filter?
Pass band voltage gain=2;
Low cut-off frequency= 1kHz;
Input frequency=500Hz.
a) First order high pass filter =-4.22dB , Second order high pass filter=-0.011dB
b) First order high pass filter =-0.9688dB , Second order high pass filter=-6.28dB
c) First order high pass filter =-11.3194dB , Second order high pass filter=-9.3257dB
d) First order high pass filter =-7.511dB , Second order high pass filter=-5.8999dB

Answer: b [Reason:] For first order high pass filter, |VO/Vin|=AF ×(f/fL) / [ √1+(f/fL)2] =(2×(500Hz/1kHz)) /√[1+(500Hz/1kHz)2] => |VO/Vin| = 1/1.118= 0.8944 =20log(0.8944) =-0.9686dB. For second order high pass filter, |VO/Vin|=AF / [ √ 1 +(fL/f)4] =2/√[1+ (1kHz/500Hz)2] =>|VO/Vin|=2/4.123= =0.4851 = 20log(0.4851) = -6.28dB.

9. How is the higher order filters formed?
a) Using first order filter
b) Using second order filter
c) Connecting first and second order filter in series
d) Connecting first and second order filter in parallel

Answer: c [Reason:] Higher filters are formed by using the first and second order filters. For example, a third order low pass filter is formed by cascading first and second order low pass filter.

10. State the disadvantage of using higher order filters?
a) Complexity
b) Requires more space
c) Expensive
d) All of the mentioned

Answer: d [Reason:] Although higher order filter than necessary gives a better stop band response, the higher order type is more complex, occupies more space and is more expensive.

11. The overall gain of higher order filter is
a) Varying
b) Fixed
c) Random
d) None of the mentioned

Answer: b [Reason:] The overall gain of higher order filter is fixed because all the frequency determining resistor and capacitors are equal.

12. Find the roll-off rate for 8th order filter

Answer: a [Reason:] For nth order filter the roll-off rate will be -n×20dB/decade. =>∴ for 8th order filter= 8×20=160dB/decade.

## Linear Integrated MCQ Set 2

1. Variation in the operating frequency of op-amp causes
a) Variation in gain amplifier
b) Variation in gain phase angle
c) Variation in gain amplitude and its phase angle
d) None of the mentioned

Answer: c [Reason:] The gain of the op-amp is a function of frequency. It will have a specific magnitude as well as a phase angle.

2. A graph of the magnitude of the gain versus frequency is called
a) Break frequency
b) Frequency response plot
c) Frequency stability plot
d) Transient response plot

Answer: b [Reason:] A frequency response plot is obtained by plotting the gain of the op-amp responding to different frequencies.

3. In the frequency response plot, the frequency is expressed in
a) Anti-logarithmic scale
b) Logarithmic scale
c) Linear scale
d) Exponential scale

Answer: b [Reason:] To accommodate large frequency ranges the frequency is assigned to a logarithmic scale.

4. Why the gain magnitude in frequency response plot is expressed in decibels (dB)
a) To obtain gain > 105
b) To obtain gain < 105
c) To obtain gain = 0
d) To obtain gain = ∞

Answer: a [Reason:] In frequency response plot, gain magnitude is assigned a linear scale and is expressed in decibels to accommodate very high gain ( ≅ of the order 105 or higher).

5. Which technique is used to determine the stability of op-amp?
a) Frequency response plot
b) Transient response plot
c) Bode plot
d) All of the mentioned

Answer: c [Reason:] Although frequency response and bode plots indicate the effect of frequency variation on gain, the Bode plot is generally used for stability determination and network design.

6. How many types of plots can be obtained in the AC analysis of network using Bode plot?
a) Five
b) Four
c) Three
d) Two

Answer: d [Reason:] Two types of plots can be obtained using Bode plot. They are magnitude versus frequency and phase angle versus frequency plots.

7. What happens when the operating frequency of an op-amp increase?
a) Gain of the amplifier decrease
b) Phase shift between output and input signal decrease
c) Gain and phase shift of amplifier decreases
d) None of the mentioned

Answer: a [Reason:] When the operating frequency is increased the gain of the amplifier decrease. As it is linearly related to frequency, the phase shift is logarithmically related to frequency.

8. Which of the following causes change in gain and phase shift?
a) Internally integrated Resistor
b) Internally integrated inductors
c) Internally integrated Capacitor
d) All of the mentioned

Answer: c [Reason:] The change in function of frequency is attributed to the internally integrated capacitor as well as stray capacitor. These capacitors are due to the physical characteristic of semiconductor device.

9. Which plot is not provide by the manufactures?
a) Magnitude plot
b) Phase angle plot
c) Frequency response plot
d) None of the mentioned

Answer: b [Reason:] Phase angle plot are not generally provided because phase shift of later generation op-amp are less than 90o even at cross over frequency.

10. Find out the non-compensating op-amp from the given circuit Answer: c [Reason:] Non-compensating op-amp has external compensating components, that is , resistors and / or capacitors, are added at designated terminals. The mentioned op-amp has three compensating components: a resistor and two capacitors.

## Linear Integrated MCQ Set 3

1. Open loop bandwidth of an op-amp extend its bandwidth from
a) 0Hz to fo
b) 20dB to fo
c) 3dB to fo
d) 0.704dB to fo

Answer: a [Reason:] The gain of the op-amp remains essentially constant from 0 to the break frequency fo and therefore rolls off at a constant rate of 20dB per decade. Thus, the open-loop bandwidth is the frequency band extending from 0Hz to fo.

2. What happens if 741 op-amp is configured as a closed loop inverting amplifier?
a) Gain increases
b) Gain roll-off at a rate 20dB/decade
c) No gain roll-off takes place
d) Gain decreases

Answer: b [Reason:] Whether the op-amp is inverting / non-inverting the gain will always roll-off at a rate of 20dB/decade, using only resistive components regardless of the value of its closed loop gain.

3. Op-amp requiring external compensating components is called as
a) Tailored frequency response op-amp
b) Compensating op-amp
c) Transient op-amp
d) High frequency op-amp

Answer: a [Reason:] Op-amp using external components like resistor and capacitor to form the compensating network are sometimes called tailored frequency response op-amps because the user has to provide the compensation if it is needed to tailor the response.

4. In the first generation op-amp 709c, the open loop bandwidth of gain versus frequency curve
a) Increases from the innermost compensated curve to the outermost
b) Decrease from the innermost compensated curve to the outermost
c) Increases from the outermost compensated curve to the innermost
d) Decreases from the outermost compensated curve to the innermost

Answer: d [Reason:] The gain versus frequency curve of 709c decreases from the outermost compensated curve to the innermost. For example, if C1 =10pF, R1 = 0Ω and C2 = 3pF, the bandwidth ≅ 5kHz. While if C1 =5000pF, R1 =1.5Ω and C2= 200pF, the bandwidth will be 100Hz.

5. Which type of op-amp offer relatively broader open-loop bandwidth?
a) Compensated op-amp
b) Uncompensated op-amp
c) Tailored frequency response op-amp
d) Non-compensated op-amp

Answer: b [Reason:] The uncompensated op-amps offer broader open loop bandwidth whereas; the internally compensated op-amps have very small open-loop bandwidth.

6. How the performance of an op-amp circuit can be improved?
a) By using non-compensating network
b) By using frequency network
c) By using compensating network
d) None of the mentioned

Answer: c [Reason:] The compensating networks are used to improve /modify the performance of an op-amp circuit over the desired frequency range by controlling it gain and phase shift.

7. Which op-amp require external compensating network?
a) Op-amp 771
b) Op-amp 351
c) Op-amp 709
d) Op-amp 741

Answer: c [Reason:] Op-amp 709 is a first generation op-amp. Generally first generation op-amp are required for external compensating network.

8. IC 741c op-amp belongs to
a) Compensated op-amp
b) Uncompensated op-amp
c) Non-compensated op-amp
d) None of the mentioned

Answer: a [Reason:] 741c belongs to later generation op-amp and it has internal compensating network. In internal compensated op-amp, the compensating network is designed into the circuit to control the gain and phase shift of the op-amp and they are called as compensating op-amp.

## Linear Integrated MCQ Set 4

1. Which factor is responsible for the gain of the op-amp to roll-off after a certain frequency is reached
a) Capacitive effect
b) Resistive effect
c) Inductor effect
d) None of the mentioned

Answer: a [Reason:] The gain of the op-amp roll-off due to the presence of capacitive component, in the equivalent circuit of op-amp. The frequency increases as the reactance of the component decreases.

2. What remedy can be followed to maintain the operating frequency in the op-amps?
a) Use LC circuit
b) Use resistor
c) Use capacitor
d) Use transistor

Answer: c [Reason:] Capacitors are used in the high frequency model of the op-amp at the output terminal to maintain or control the frequency.

3. How does the physical characteristic of semiconductor account for the increase in frequency of op-amps?
a) Transistor values
b) Junction capacitance
c) Dopant concentration
d) None of the mentioned

Answer: b [Reason:] Op-amps are composed of BJTs and FETs which contain junction capacitors. These junction capacitors are very small and take finite values at higher frequency. So, when the reactances of these capacitors decrease, the frequency increases.

4. How are op-amp with three break frequency are represented
a) Using two capacitors
b) Using three capacitors
c) Using one capacitor
d) All of the mentioned

Answer: b [Reason:] Op-amps with more than one break frequency may be represented by using as many capacitors as the break frequency they have.

5. In what way the internal construction of the op-amp contribute to capacitive effect in op-amp?
a) Formation of junction capacitor
b) Due to internally connected capacitors
c) Formation of stray capacitor
d) None of the mentioned

Answer: c [Reason:] In op-amps a number of transistors, resistors and capacitors are integrated on the same substrate. In fact substrates act as an insulator and separate these components. So, the various components are connected by conducting paths. However, when two conducting paths are separated by an insulator, it acts as a capacitor (stray capacitor).

6. Determine the high frequency model op-amp with single break frequency Answer: a [Reason:] Op-amp with only one break frequency is represented by adding a single capacitor at the output.

7. The gain of the differential amplifier is -125. Assume the voltage applied to bridge circuit Vdc=+10v and the unstrained resistance of four element of strain gage is 200Ω. When a certain weight is placed on the platform, the output voltage Vo=5v. Determine the change in resistance of each strain gage for analog weight scale. (Assuming the output is initially nulled).
a) 1Ω
b) 0.8Ω
c) 0.3Ω
d) 1.83Ω

Answer: b [Reason:] The output voltage of analog weight scale, Vo = Vdc×(△R/R)×(RF/R1) => △R=( Vo/ Vdc)×(R1/RF)×R =( 5v×200Ω)/(125×10v) =0.8Ω.

8. Assume that the increase and decrease in the resistance of the strain gage element is by the same number of △R. Determine the unbalanced voltage equation.
a) V = – Vdc×(△R/R)
b) V = – Vdc×(△R/R+△R)
c) V = Vdc×(△R/R)
d) V = – Vdc×[(△R+ R)/R].

Answer: a [Reason:] When the unstrained gage resistance are same, then the output voltage of the strain gage bridge circuit (unbalanced voltage) is given as V= -Vdc×(△R/R).

## Linear Integrated MCQ Set 5

1. The special designed comparators are compatible with
a) RTL
b) MOS Logic
c) TTL
d) All of the mentioned

Answer: d [Reason:] The special designed comparator outperforms the op-amp type. They are optimized for the most desirable promoters like speed and accuracy. So, their output is compatible with RTL, DTL, TTL and MOS Logic.

2. Response time of comparators is defined as
a) Time interval between input and output crossing upper threshold voltage
b) Time interval between input and output function
c) Time interval between input and output crossing threshold logic
d) Time interval between input and output crossing lower threshold voltage

Answer: c [Reason:] The response time of the comparator is the interval between the application of an input step function and the time output crosses the logic threshold voltage.

3. The voltage level at which the comparator saturated by a differential input equal to greater than a specified voltage is called
a) Zero output level
b) Positive output level
c) Negative output level
d) All of the mentioned

Answer: c [Reason:] The negative output level or output low voltage is the negative dc output voltage with comparator saturated by a differential input equal to or greater than specified voltage.

4. When the logic level of strobe output is at zero, then the output current is called
a) Strobe terminal current
b) Strobe current
c) Strobe threshold current
d) Strobe bias current

Answer: b [Reason:] The strobe current is the current out of the strobe terminal when it is of the zero logic current.

5. The strobe release time is defined as
a) Time required for output to rise to logic threshold
b) Time required for input to rise to logic threshold
c) Time required for output to rise to zero logic
d) Time required for output to rise to one logic

Answer: d [Reason:] The strobe release time is the time required for the output to rise to the logic threshold voltage after the strobe terminal has been driven from zero to the one logic level.

6. Which of the following comparator can operate from ±15 to down to +5v supply in op-amp?
a) µA741
b) µA311
c) µA351
d) All of the mentioned

Answer: b [Reason:] µA311 precision comparator is designed for low level signal detection and high level output drive capability. So, it can operate from ±15v op-amp supplies down to the logic +5v supply used for IC logic.

7. Find the comparator that has slow response time
a) LM1414
b) µF311
c) ICL8001
d) None of the mentioned

Answer: a [Reason:] LM1414 has slow response time typically of 30ns, where µAF311 & ICL800 has response time of 200ns and 250ns.

8. Why ICL8001 op-amp comparator is preferred for most applications?
a) Low power consumption
b) High response time
c) High input current
d) Balanced input offset voltage