Linear Integrated MCQ Number 01143

Linear Integrated MCQ Set 1

1. Voltage shunt feedback amplifier forms
a) A negative feedback
b) A positive feedback
c) Both positive and negative
d) None of the mentioned

Answer

Answer: a [Reason:] A voltage shunt feedback amplifier forms a negative feedback because, any increase in the output signal results in a feedback signal into the inverting input causing a decrease in the output signal.

2. The value of feedback resistor and resistor connected in series with the input signal source are equal to 10kΩ and 3.3kΩ. Calculate the closed loop voltage gain?
a) -6.7
b) -33
c) -13.3
d) -3.33

Answer

Answer: d [Reason:] Closed loop voltage gain, A_F = -R_F/R₁ = -10kΩ/3.3kΩ = -3.33.

3. Write the formula for closed loop voltage gain of inverting amplifier with feedback using open loop voltage gain and gain of feedback circuit.
a) A_F= A/(1+AB)
b) A_F= -A/(1+AB)
c) A_F= -B/(1+AB)
d) None of the mentioned

Answer

Answer: c [Reason:] The closed loop voltage gain of the amplifier is A_F= -Ak/(1+AB), where k is a voltage attenuation factor. In addition to phase inversion, A_F is k times the closed loop gain of the non-inverting amplifier where k<1.

4. Voltage shunt feedback amplifiers are also called as
a) Non-inverting amplifier with feedback
b) Non-inverting amplifier without feedback
c) Inverting amplifier with feedback
d) Inverting amplifier without feedback

Answer

Answer: c [Reason:] The input and output signal in voltage series feedback amplifier are 180^o out of phase (or of opposite polarities). Due to this phase inversion, the configuration is also called as inverting amplifier with feedback.

5. Find the block diagram representation for inverting amplifier with feedback

Answer

Answer: b [Reason:] The block diagram of non-inverting amplifier is identical to that of inverting amplifier except for the voltage attenuation factor or block. However, the major difference is that a voltage summing junction is being used as a model for what is actually a current summing.

6. The inverting input inverting of the voltage shunt feedback resistor is a commonly named as
a) Terminal ground
b) Virtual ground
c) Virtual input
d) Resistive input

Answer

Answer: b [Reason:] Ideally, the difference between input voltages is zero. Therefore, the voltage at the inverting terminal is approximately equal to that of non-inverting terminal. In other words, the inverting terminal voltage is approximately at ground potential and it is said to be virtual ground.

7. Compute R_IF for an inverting amplifier with feedback, where the value of input resistance of op-amp is 4.7kΩ.
a) 4.7kΩ
b) 9.4kΩ
c) 2.35kΩ
d) Data given is insufficient

Answer

Answer: a [Reason:] In voltage shunt feedback amplifier, the input resistance with feedback is given as R_IF = R₁ (ideally).

8. Specification of op-amp 741c is given below:
A=200000; R_i=2MΩ; R_o=75Ω; Supply voltages= ±15v; output voltage swing =±13v; f_o=5hz.
Compute the value of output resistance, bandwidth and closed loop voltage gain for the circuit shown.
a) R_OF=8.6mΩ , f_F= 53005hz and A_F=-9.36
b) R_OF=4.12mΩ , f_F= 53005hz and A_F=-11.78
c) R_OF=7.1mΩ , f_F = 53005hz and A_F=-16.95
d) R_OF=1.9mΩ , f_F= 53005hz and A_F=-10

Answer

Answer: c [Reason:] Output resistance of the amplifier, R_OF= R_o/(1+AB) =, where B= R₁+R_F = 330Ω/330Ω+6.2kΩ = 0.053.
=> R_OF= ±75/(1+200000*0.053) = 53005hz.
Closed loop voltage gain, A_F= -(A*K)/(1+AB), Where k= R_F/(R₁+_F) = 6.2kΩ/(330Ω+6.2kΩ) = 0.949
=> A_F = -(200000*0.949)/[1+(200000*0.0535)] = -16.95.

9. What is the break frequency of the op-amp?
a) f_o = Unity Gain Bandwidth /closed loop voltage gain
b) f_o = Unity Gain Bandwidth / open loop voltage gain
c) f_o = Unity Gain Bandwidth /Gain of feedback circuit
d) All of the mentioned

Answer

Answer: c [Reason:] The mentioned formula is the general break frequency of any operational amplifier.

10. The total voltage offset voltage with feedback (V_ooT) equation for inverting amplifier is
a) Same as that of non-inverting amplifier
b) k times the non-inverting amplifier, k-> voltage attenuation factor
c) Twice the equation of non-inverting amplifier
d) All of the mentioned

Answer

Answer: a [Reason:] V_oot equation for inverting amplifier is the same as that of the non-inverting amplifier because, when the input signal is reduced to zero, both inverting and non-inverting amplifier results in the same circuit.

11. Which among the following is not a special case of voltage shut feedback amplifier?
a) Voltage follower
b) Current to voltage connector
c) Inverter
d) None of the mentioned

Answer

Answer: a [Reason:] A voltage follower is a special case of non-inverting amplifier ( or voltage series feedback amplifier) and it has a gain of unity.

12. Compute the output voltage for the given circuit

a) -2.6v
b) -27.8v
c) -26.7v
d) -0.267v

Answer

Answer: c [Reason:] The given circuit is a current to voltage converter. Since V₁ =0v and V₁= V₂.
=> i_in = V_in/R₁ = 4/ 1.5kΩ =2.67mA.
The output voltage V_o = -i_in*R_F = -2.67mA*10kΩ = -26.7v.

13. At what condition an inverting amplifier works as an inverter

a) R₁ = R_F+ R_L
b) R_F =( R₁*V_in)/R_L
c) R₁ = R_F
d) R₁ = V_o/(V_in*R_L)

Answer

Answer: c [Reason:] If R₁ = R_F, the inverting amplifier will work as an inverter.

14. Determine the output waveform for the given input signal

Answer

Answer: a [Reason:] Given, V_in= 3Vpp sinewave at 1khz (Therefore F=1/T=1/ms =1khz)
=> i_in= V_in/R₁ = 3/ 470=6.4mA
=> V_o=- i_in*R_F=6.4mA*4.7kΩ = 30 Vpp sinewave at 1khz.

Linear Integrated MCQ Set 2

1. Determine the output from the following circuit

a) 180^o in phase with input signal
b) 180^o out of phase with input signal
c) Same as that of input signal
d) Output signal cannot be determined

Answer

Answer: b [Reason:] The input signal is given to the inverting input terminal. Therefore, the output V_o is 180^o out of phase with input signal V₂.

2. Which of the following electrical characteristics is not exhibited by an ideal op-amp?
a) Infinite voltage gain
b) Infinite bandwidth
c) Infinite output resistance
d) Infinite slew rate

Answer

Answer: c [Reason:] An ideal op-amp exhibits zero output resistance so that output can drive an infinite number of other devices.

3. An ideal op-amp requires infinite bandwidth because
a) Signals can be amplified without attenuation
b) Output common-mode noise voltage is zero
c) Output voltage occurs simultaneously with input voltage changes
d) Output can drive infinite number of device

Answer

Answer: a [Reason:] An ideal op-amp has infinite bandwidth. Therefore, any frequency signal from 0 to ∞ Hz can be amplified without attenuation.

4. Ideal op-amp has infinite voltage gain because
a) To control the output voltage
b) To obtain finite output voltage
c) To receive zero noise output voltage
d) None of the mentioned

Answer

Answer: b [Reason:] As the voltage gain is infinite, the voltage between the inverting and non-inverting terminal (i.e. differential input voltage) is essentially zero for finite output voltage.

5. Determine the output voltage from the following circuit diagram?

a)
b)
c)
d) None of the mentioned

Answer

Answer: c [Reason:] In an ideal op-amp when the inverting terminal is zero. The output will be in-phase with the input signal.

6. Find the output voltage of an ideal op-amp. If V₁ and V₂ are the two input voltages
a) V_O= V₁-V₂
b) V_O= A×(V₁-V₂)
c) V_O= A×(V₁+V₂)
d) V_O= V₁×V₂

Answer

Answer: b [Reason:] The output voltage of an ideal op-amp is the product of gain and algebraic difference between the two input voltages.

7. How will be the output voltage obtained for an ideal op-amp?
a) Amplifies the difference between the two input voltages
b) Amplifies individual voltages input voltages
c) Amplifies products of two input voltage
d) None of the mentioned

Answer

Answer: a [Reason:] Op-amp amplifies the difference between two input voltages and the polarity of the output voltage depends on the polarity of the difference voltage.

8. The signal to an inverting terminal of an ideal op-amp is zero. Find the output voltage, if the other input voltage is

a)
b)
c)
d) Data provided is insufficient

Answer

Answer: a [Reason:] Although the output is 180^o out of phase with input signal, the gain of the amplifier is not given.

9. Which is not the ideal characteristic of an op-amp?
a) Input Resistance –> 0
b) Output impedance –> 0
c) Bandwidth –> ∞
d) Open loop voltage gain –> ∞

Answer

Answer: a [Reason:] Input resistance is infinite so almost any signal source can drive it and there is no loading of the preceding stage.

10. Find the ideal voltage transfer curve of a normal op-amp.
a)
b)
c)
d)

Answer

Answer: c [Reason:] The ideal voltage transfer curve would be almost vertical because of the very large value of gain.

11. Find the input voltage of an ideal op-amp. It’s one of the inputs and output voltages are 2v and 12v. (Gain=3)
a) 8v
b) 4v
c) -4v
d) -2v

Answer

Answer: d [Reason:] The output voltage, V_O = (Vin₁– Vin₂)
=> 12v=3×(2- Vin₂)
=> Vin₂= -2v.

12. Which factor determine the output voltage of an op-amp?
a) Positive saturation
b) Negative saturation
c) Both positive and negative saturation voltage
d) Supply voltage

Answer

Answer: c [Reason:] Output voltage is proportional to input voltage only until it reaches the saturation voltage. The output cannot exceed the positive and negative saturation voltage. These saturation voltages are specified by an output voltage swing rating of the op-amp for given values of supply voltage.

Linear Integrated MCQ Set 3

1. To obtain a faster slew rate the op-amp should have
a) High current and large compensating capacitor
b) Small compensating capacitor
c) High current or small compensating capacitor
d) Low current or large compensating capacitor

Answer

Answer: c [Reason:] The slew rate is given as, SR =dV_c/dt|_max = I/C
Therefore the higher current should be given a small compensating capacitor is used internally or outside an op-amp.

2. Find the expression for full power response.
a) f_max(Hz) =(slew rate×10⁶)/(6.28×V_m)
b) f_max(Hz) =slew rate /(628×10⁶×V_m)
c) f_max(Hz) =(slew rate×V_m×10⁶)/6.28
d) f_max(Hz) =(6.28×V_m× 10⁶)/ slew rate

Answer

Answer: a [Reason:] The maximum input frequency at which an undistorted output voltage can be obtained for a peak value of V_m is given as f_max(Hz) =(slew rate×10⁶)/(6.28×V_m)
where f_max is called as full power response.

3. Calculate the time taken by the output to swing from +14v to -14v for a 741C op-amp having a slew rate of 0.5V/µs?
a) 22µs
b) 42µs
c) 56µs
d) 70µs

Answer

Answer: c [Reason:] Slew rate = dv/dt
=> Time taken = 14-(-14)/ 0.5V/µs = 28v/0.5V/µs = 56µs.

4. Consider a square wave having a peak to peak amplitude of 275mv and it is amplified to a peak to peak amplitude of 4v, with rise time of 5.2µs. Calculate the slew rate?
a) 0.615 v/µs
b) 0.712 v/µs
c) 0.325 v/µs
d) None of the mentioned

Answer

Answer: a [Reason:] From the definition of rise time, the change in the output voltage is 5.2µs
△v= (90%-10%)×4v= (0.9-0.1)×4v =3.2v.
Therefore, slew rate = 3.2v/5.2µs =0.615v/µs.

5. Determine the maximum input signal to be applied to an op-amp to get distortion free output. If the op-amp used is an inverting amplifier with a gain of 50 and maximum output amplitude obtained is 4.2V sine wave?
a) 159mv
b) 0.168mv
c) 207mv
d) 111mv

Answer

Answer: b [Reason:] Given, V_m= 4.2V_peak
∴ the output voltage = 4.2+4.2 =8.4 V peak to peak.
Hence for the output to be undistorted sine wave, the maximum input signal should be less than => 8.4/50= 0.168 = 168mVpeak to peak.

6. What happens if the frequency or amplitude of the input signal is increased to exceed slew rate of the op-amp?
a) All of the mentioned
b) High frequency output
c) Distorted output
d) Large amplitude output

Answer

Answer: c [Reason:] Slew rate determines the maximum frequency of operation for a desired output swing. If the slew rate is greater than 2πfV_m /10⁶ then the output is distorted, whereas an increase in the frequency /amplitude of input signal distort the output.

7. Compute the peak output amplitude, when the voltage gain verses frequency curve of 741C is flat upto 25Hz.
a) 4V_peak
b) 9V_peak
c) 20V_peak
d) None of the mentioned

Answer

Answer: d [Reason:] The slew rate of 741C op-amp = 0.5V/µs. So, the maximum output voltage at 25kHz is SR= (2πfV_m)/ 10⁶ V/µs
=> V_m = (SR×10⁶)/(2πf ) = (0.5×10⁶)/(2π×25kHz)
V_m = 3.18V_peak.

8. Calculate the maximum input frequency at which the output will be distorted from the given specifications
V_o = 30 V_pp ; Slew rate = 0.6v/µs.
a) 1000Hz
b) 10kHz
c) 1kHz
d) 10kHz

Answer

Answer: d [Reason:] The minimum time between the two zero crossing is given as
=> 30v/(0.6v/µs) =50µs. Hence the maximize input frequency f_max at which the output get distorted is f_max = 1/(2×50µs) =10000 =10kHz.

9. Match AC parameter of the op-amp in column 1 with the column 2.

Column 1	Column-2
1. Bandwidth	i . a large signal phenomenon
2. Transient response	ii. Rise time is related to bandwidth and overshoots measure stability
3. Slew rate	iii. Depends on compensating components and closed loop gain

a) 1-i 2-iii 3-ii
b) 1-ii 2-iii 3-i
c) 1-iii 2-ii 3-i
d) 1-iii 2-i 3-ii

Answer

Answer: c [Reason:] Analysis of difference between three AC parameter; bandwidth, transient response and slew rate.

Linear Integrated MCQ Set 4

1. What is the frequency of oscillation of wein bridge oscillator?
a) f_o = 1/(2πRC)
b) f_o = 2π/RC
c) f_o = RC/2π
d) f_o = 2πRC

Answer

Answer: a [Reason:] The frequency of oscillation of wein bridge oscillator is f_o=2πRC.

2. Sustained oscillation in wein bridge oscillator is possible when the value of ß is
a) 3
b) 1/3
c) 1
d) None of the mentioned

Answer

Answer: b [Reason:] The gain |A|≥3, for oscillation to keep growing ( Since, |Aß|≥1 for sustained oscillation).

3. Determine the value of f_o, ß and R_f from the following circuit diagram.

a) f_o = 80Hz, ß = 0.162 and R_f = 7kΩ
b) f_o = 100Hz, ß = 1.62 and R_f = 7kΩ
c) f_o = 60Hz, ß = 0.0162 and R_f = 7kΩ
d) f_o = 120Hz, ß = 16.2 and R_f = 7kΩ

Answer

Answer: c [Reason:] The frequency of oscillation for the circuit is given as,
f_o = 1/(2π×√(R₁R₂C₁C₂))
= 1/(2π×√(2.7kΩ×5kΩ×0.1µF×6µF) = 1/(2π×2.85×10^-3)
=> f_o = 55.8 = 60Hz.
The value of ß = (R₂C₁) / (R₁C₁ + R₂C₂ + R₂C₁)
= (5kΩ×0.1µF)/(2.7kΩ×0.1µF+5kΩ×6µF+5kΩ×0.1µF)= 0.0162.
R_f =2 R₃
=> R_f = 2×3.5kΩ=7kΩ.

4. What is the problem faced by the wein bridge oscillator?
a) Output sinewave get clipped
b) Output sinewave remain constant without growing
c) Output sinewave keep on decreasing and die out
d) All of the mentioned

Answer

Answer: a [Reason:] The gain of wein bridge oscillator is greater than 3, sometimes this may keep the oscillations growing and it may clip the output sinewave.

5. Find the type of oscillator shown in the diagram

a) Quadrature oscillator
b) Biphasic oscillator
c) RC phase shit oscillator
d) None of the mentioned

Answer

Answer: d [Reason:] The circuit shown is the practical wein bridge oscillator with adaptive negative feedback.

6. Calculate the value of capacitance in wein bridge oscillator, such that f_o =1755Hz and R=3.3kΩ.
a) 2.7µF
b) 0.91µF
c) 0.03µF
d) 0.05µF

Answer

Answer: c [Reason:] The frequency of oscillation is given as f_o = 0.159/RC
=> C = 0.159/R×f_o = 0.159/3.3kΩ×1755Hz
=> C = 0.027µF = 0.03µF.

7. Quadrature oscillators have signals with
a) Different frequency
b) Same frequency
c) Opposite frequency
d) Parallel frequency

Answer

Answer: b [Reason:] In Quadrature oscillators, signals have same frequency but have phase shift with respective to each other.

8. Which of the following component is not used for audio frequency?
a) RC oscillator
b) Wein bridge oscillator
c) LC oscillator
d) None of the mentioned

Answer

Answer: c [Reason:] RC and wein bridge oscillator are suitable for audio frequency range because of size of R and C components becomes very large for generating low frequencies.

9. Find the signal waveform for Quadrature oscillators?
a)
b)
c)
d) All of the mentioned

Answer

Answer: a [Reason:] Quadrature signals are the signals that are of same frequency but have a phase shift of 90^o with respect to each other. The mentioned waveform have a phase shift of π/2- π=90^o phase shift between sine waveform and cosine waveform.

10. If the resistor and capacitor values are same in Quadrature oscillator. Find its frequency of oscillation for R=50kΩ and C=0.01µF.
a) 112Hz
b) 275Hz
c) 159Hz
d) 318Hz

Answer

Answer: d [Reason:] Frequency of Quadrature oscillator, f_o = 1/(2πRC)
=> f_o= 1/(2π×50kΩ×0.01µF)= 318Hz.

11. What is the possible method used in Quadrature oscillator to remove distortion from the output waveform?
a) Replace the resistor at the input of non-inverting type amplifier with a potentiometer
b) Replace the resistor at the output of non-inverting type amplifier with a potentiometer
c) Replace the resistor at the input of inverting type amplifier with a potentiometer
d) None of the mentioned

Answer

Answer: a [Reason:] The resistor at the input of non-inverting type amplifier is replaced with a potentiometer in order to eliminate any possible distortion in the output waveform.

Linear Integrated MCQ Set 5

1. The process involved in photolithography is
a) Making of a photographic mask only
b) Photo etching
c) Both photo etching and making of photographic mask
d) None of the mentioned

Answer

Answer: c [Reason:] Photolithographic involves both processes in sequence. First photographic mask is used for artwork and reduction. Then Photo etching for removal of SiO₂ from designed region.

2. How will be the initial artwork done for a normal IC?
a) Smaller than the final dimension of chip
b) Same as that of final dimension of chip
c) Larger than the final dimension of chip
d) None of the mentioned

Answer

Answer: c [Reason:] The initial artwork of an IC is done at a scale several hundred times longer than the final dimensions. This is because for a tiny chip, larger the artwork, more accurate is the final mask.

3. Find the area of artwork done for a monolithic chip of area 30mil × 30mil.
a) 16 cm × 16 cm
b) 60 cm × 60 cm
c) 12 cm × 12 cm
d) 36 cm × 36 cm

Answer

Answer: d [Reason:] Drawings are magnified by a factor 500.
=> 1mil = 25µm
Therefore, 500mil = 1.2cm.
In an area of 30mil × 30mil, the area for artwork required = 30mil × 1.2cm = 36cm × 36cm.

4. Mylar coated with a sheet of red photographic Mylar is used for artwork (layout) because,
a) It is used to get a colourful layout
b) It can be easily peeled off from layout
c) It is recommended colour for layouts
d) It is used for highlighting layout

Answer

Answer: b [Reason:] For photographic purpose, artwork should not contain any line drawing but must be of alternate clear and opaque region. The red layer can be easily peeled off thus exposing clear areas with a knife edge from regions where impurities have to be diffused.

5. Find the coating material used for photo etching process along with its thickness range.
a) Kodak photoresist (5000-10000Å)
b) Kodak photoresist (1000-5000Å)
c) Kodak photo etchant (1000-5000 Å)
d) Kodak photo etchant (500-1000 Å)

Answer

Answer: a [Reason:] The coating material is Kodak photoresist. It is a photosensitive emulsion film coated on wafer to remove SiO₂ from desired region.

6. Which type of etching process is preferred to make the photoresist immune to etchants?
a) None of the mentioned
b) Wet etching
c) Plasma etching
d) Chemical etching

Answer

Answer: c [Reason:] Plasma etching is also called as dry etching. The major advantage of dry etching process is that, it is possible to achieve smaller line openings (<1µm) compared to other process.

7. Which of the following statement is not true?
a) X-ray and Electron beam lithography technique, produce device dimensions down to submicron range.
b) Ultraviolet lithography has limitation due to diffraction effects of wavelength.
c) The cost of X-ray or Electron beam is less compared to Ultraviolet photolithography.
d) The exposure time is less in Ultraviolet compared to X-ray or Electron beam lithography.

Answer

Answer: c [Reason:] The cost of X-ray or Electron beam is very high and thus, it is an expensive process. Therefore, it is used only when very small device dimension (<1 µm) are needed.

8. For photographic purpose usually coordinatograph is preferred for artwork because,
a) It is a precision drafting machine
b) Cutting head can be positioned accurately
c) It can be moved along two perpendicular axes
d) All of the mentioned

Answer

Answer: d [Reason:] The coordinatograph is a drafting machine that outlines the pattern cutting through the red Mylar without damaging the clear layer underneath.

9. Which of the following is added as an impurity to p-type material in diffusion process?
a) Phosphorous pentaoxide (P₂O₅)
b) Phosphorous oxychloride (POcl₃)
c) Boron oxide (B₂O₃)
d) None of the mentioned

Answer

Answer: c [Reason:] Boron is a p-type material, whereas Phosphorous is an n-type material.

10. In the fabrication of monolithic ICs, Boron chloride is added as an impurity in the diffusion process. Find the diffusion time, if the furnace is heated up to 1200^oc.
a) 1 hour
b) 2 hours
c) 45 minutes
d) 30 minutes

Answer

Answer: b [Reason:] In diffusion process, the depth of diffusion of impurities depends upon the time taken for diffusion, which normally extends for more than 2 hours.

11. Which component is not used as an impurity in diffusion process?
a) Phosphorous
b) Boron chloride
c) Phosphorous pentaoxide
d) Boron oxide

Answer

Answer: a [Reason:] Elemental form of Phosphorous is not added directly as an impurity in diffusion process.

12. In ion implantation method, penetrating the ions into the silicon wafer depends upon
a) Accelerating voltage
b) Accelerating speed
c) Accelerating current
d) All of the mentioned

Answer

Answer: a [Reason:]The depth of penetration of any particular type of ion increases with increasing accelerating voltage.

13. What is the advantage of using Ion implantation process?
a) Lateral spreading is more
b) Performed at high temperature
c) Beam current controlled from outside
d) Performed at low temperature

Answer

Answer: c [Reason:] In diffusion process, temperature has to be controlled over a large area inside the oven, whereas in ion implantation technique, accelerating potential and the beam current are electrically controlled from outside.