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Linear Integrated MCQ Set 1

1. Which package type is chosen for military purposes?
a) Ceramic DIP
b) Plat pack
c) Metal can pack
d) Plastic DIP

View Answer

Answer: a [Reason:] Ceramic DIP can be used for high temperature and high performance equipment.

2. A Dual-In-Line Package is usually referred to as
a) DIPn
b) nDIP
c) DIPn
d) All of the mentioned

View Answer

Answer: a [Reason:] A Dual-In-Line Package is usually referred to as DIPn. Where, n represent the number of pin terminals in the IC.

3. Which type of DIP IC dissipates more heat?
a) Ceramic DIP
b) Plastic DIP
c) Metal DIP
d) None of the mentioned

View Answer

Answer: b [Reason:] Plastic DIP are cheaper than metal or ceramic DIP, but are not regarded as satisfactory in extremes of temperature.

4. Choose the type of package used for Airborne application?
a) DIP package
b) Metal can package
c) Flat pack
d) Transistor pack

View Answer

Answer: c [Reason:] The flat pack is more reliable and lighter than a comparable DIP package and therefore is suited for airborne application.

5. How a choice is made, if all three package types are available?
a) Based on cost
b) Based on fabrication
c) Based on Experimentation usage
d) All of the mentioned

View Answer

Answer: d [Reason:] When all three packages are available for a specific application, the choice can be made based on the relative cost, ease of fabrication and breadbording the IC.

6. How many temperature grades are available for IC?
a) Two
b) Three
c) Four
d) Five

View Answer

Answer: b [Reason:] All ICs manufactured fall into one of the three basic temperature grades. They are military, industrial and commercial temperature range.

7. ICs used for industrial application will have temperature range from
a) -55o to +85oc
b) 90o to 155oc
c) 10o to 100oc
d) -20o to +85oc

View Answer

Answer: d [Reason:] The industrial temperature range is from -20o to +85oc.

8. Find the types of temperature range used for an IC, which can be used only up to 75oc?
a) Industrial temperature range
b) Commercial temperature range
c) Military temperature range
d) All of the mentioned

View Answer

Answer: b [Reason:] Commercial grade IC can be used up to 75oc. It has the worst tolerance among the three types and is the cheapest available IC.

9. Which grade device is selected for superior quality performance?
a) Military grade IC
b) Industrial grade IC
c) Commercial grade IC
d) None of the mentioned

View Answer

Answer: a [Reason:] The military grade devices are always of superior quality, with tightly controlled parameters and consequently cost more.

10. In ordering an IC, the device type is represented as
a) Numbers
b) Symbols
c) Alphabets
d) Alphanumeric characters

View Answer

Answer: d [Reason:] The device type is a group of alphanumeric characters. For example, 741 IC is represented as µA741, LM741 and MC1741.

Linear Integrated MCQ Set 2

1. Determine the early voltage, if the output resistance is 2.5×2kΩ and input current is 2mA
a) 9.8v
b) 5.6v
c) 7.8v
d) 10v

View Answer

Answer: d [Reason:] Output resistance, Ro=VA/Iref ⇒ VA = Ro×Iref =2.5×2kΩ×2mA =10v.

2. In practical application of current mirror, early voltage is assumed to be
a) Infinite
b) Zero
c) Unity
d) None of the mentioned

View Answer

Answer: a [Reason:] Early voltage is assumed to be infinity, so that output resistance tend to infinity and the output current is constant.

3. A widlar current source is used
a) to get low value of current
b) to get high value of CMRR
c) to get low voltage of gain
d) to get high value of Output

View Answer

Answer: a [Reason:] In the widlar current source Re is added to emitter lead of transistor, which consequently results in smaller output current value.

4. What will be the value of emitter resistance in widlar current source for output current 10mA, having Iref=2.7A
a) 67/(1+1/β)Ω
b) 13/(1+1/β)Ω
c) 14/(1+1/β)Ω
d) 1.36/(1+1/β)Ω

View Answer

Answer: c [Reason:] Emitter resistor, RE=VT/(1+1/β) Iref×ln(Io/Iref) ⇒ RE= 0.025/(1+1/β)10mA × ln(12.7A/10mA) = 14/(1+1/β)Ω.

5. A current repeater having identical transistor has collector current, IC1 =0.39mA. Find IC2,IC4& IC6
a) 0.39mA, 0.39mA, 0.78mA
b) 0.78mA, 0.39mA, 0.39mA
c) 0.39mA, 0.78mA, 0.39mA
d) None of the mentioned

View Answer

Answer: d [Reason:] In current repeater, the current IC = IC1 =IC2 =⋯= IC N≅ Iref . Where, N – Number of transistors used in current repeater circuit.

6. If the reference and collector current are 0.539mA and 0.49mA respectively, how many transistors are used in current repeater circuit? (Assume β =150)
a) 11
b) 14
c) 10
d) 8

View Answer

Answer: b [Reason:] The current equation is given as IC= Iref×β/(β+1+N) ⇒ 0.49mA= 0.539mA×150/(150+1+N) ⇒ N=14.

7. For the current repeater shown in the circuit, determine IC4 value, Where β = 75.
linear-integrated-circuits-questions-answers-freshers-q7
a) 0.035mA
b) 0.028mA
c) 0.04mA
d) 0.052mA

View Answer

Answer: a [Reason:] The reference current, Iref = VCC-VBE/R1 =(15v-0.7v)/39kΩ = 0.366mA. ⇒ Iref=IC+4×IB =IC(1+1/β) ∴ IC= Iref×(1+1/β) =Iref×(1+1/β) = 0.366mA×(1+1/75) =0.347mA ⇒ IC1=IC2=IC3=0.347mA To determine IC4, RE=VT/(1+1/β)×IC4×ln(C3/IC4) ⇒ 1.62kΩ = 25mv/(1+1/75)×IC4×ln(0.347mA/IC4) ⇒ IC4=0.035mA (find using trial and error method).

8. The requirements for a good current source is the one in which, (Take Output current – IO and Output resistance – rO )
a) IO independent upon current gain and should be low
b) rO should be very high
c) IO in the circuit should be low
d) IO independent upon current gain and rO should be very high

View Answer

Answer: d [Reason:] The need for high output resistance current source can be seen because the common mode gain of the differential amplifier can only be reduced by using high resistance current sources.

9. Which current source exhibits a very high output resistance?
a) Simple current mirror
b) Wilson current mirror
c) Widlar current mirror
d) All of the mentioned

View Answer

Answer: b [Reason:] The output resistance of Wilson current mirror is substantially greater than ≅(β× Output resistance)/2 than Simple or Widlar current source.

10. What will be the overall gain in Darlington circuit, if the individual transistor gain is 200?
a) 10000
b) 40000
c) 8000
d) 1000

View Answer

Answer: b [Reason:] Overall current gain, β= β1×β2 (Multiplication of current gain of individual transistor) ⇒ β=200×200=40000

11. To increase the input resistance in differential amplifier, replace the transistor by
a) Current mirror
b) Current repeater
c) Darlington pair
d) All of the mentioned

View Answer

Answer: c [Reason:] Higher value of input resistance can be obtained by using Darlington pair in place of transistor.

12. What is the drawback in using Darlington pair in differential amplifier?
a) Large current gain
b) Output current in milli ampere
c) Gain is proportional to load resistor
d) High offset voltage

View Answer

Answer: d [Reason:] Due to cascaded stage, Darlington differential amplifier offers higher offset voltage which is two times larger than ordinary two transistor used in differential amplifier.

13. Determine the amount of shift happens in level shifter?
a) Vcc + 0.7v
b) Vcc – 0.7v
c) -0.7v
d) +0.7v

View Answer

Answer: c [Reason:] Level shifter is basically a simple type emitter follower. Hence, level shifter also act as a buffer to isolate high gain stages from the output stage. Therefore, the amount of shift obtained is VO – Vi = -VBE = -0.7v.

Linear Integrated MCQ Set 3

1. Find the epitaxial resistor from the given cross-sectional view diagram?
a) linear-integrated-circuits-questions-answers-mcqs-q1a
b) linear-integrated-circuits-questions-answers-mcqs-q1b
c) linear-integrated-circuits-questions-answers-mcqs-q1c
d) linear-integrated-circuits-questions-answers-mcqs-q1d

View Answer

Answer: a [Reason:] The mention figure is the cross sectional view of epitaxial resistor. The remaining diagrams are the cross-sectional view of pinched, thin film and diffused resistor.

2. Which integrated resistor can achieve high value of sheet resistance?
a) Pinched resistor
b) Epitaxial resistor
c) Thin film resistor
d) All of the mentioned

View Answer

Answer: a [Reason:] In a pinched resistor, the sheet resistivity can be increased by reducing its effective area. This technique is used to achieve high value of sheet resistance from ordinary diffused resistor.

3. How pinched resistor can give resistance in order of mega-ohm in a reasonably small area?
a) By increasing fabrication steps
b) By offering bulk resistance in n-region
c) By reducing conduction path
d) By limiting the thickness of are

View Answer

Answer: c [Reason:] In pinched resistor structure, one of the diode conducts in reverse direction and only a small reverse saturation current can flow through n-type material. By doing so, the effective cross-sectional area of the conduction path will be reduced and resistance between two contact lead increases.

4. Which of the following is not used as metallic film in the thin film resistor?
a) Nichrome (NiCr)
b) Tantalum (Ta)
c) Stannic oxide (SnO2)
d) Silicon dioxide (SiO2)

View Answer

Answer: d [Reason:] Silicon dioxide is the non-metallic layer on which the metallic thin films are deposited.

5. Pick out the incorrect statement
a) Sheet resistance have smaller and lesser parasitic components
b) Value of resistor can be easily adjusted after fabrication
c) Resistance in the range 100kΩ possible using nichrome resistors
d) Thin film resistors are more stable

View Answer

Answer: c [Reason:] Typically, sheet resistance value of nichrome is 40 – 400Ω/square (depending upon film thickness). So, the resistance in the range 20 to 50kΩ can only be obtained.

6. Find the equivalent circuit of junction capacitor?
a) linear-integrated-circuits-questions-answers-mcqs-q6a
b) linear-integrated-circuits-questions-answers-mcqs-q6b
c) linear-integrated-circuits-questions-answers-mcqs-q6c
d) linear-integrated-circuits-questions-answers-mcqs-q6d

View Answer

Answer: b [Reason:] The mentioned diagram is the equivalent circuit diagram of junction capacitor.

7. The capacitance of junction capacitor does not depend upon
a) Impurity concentration of p-type epitaxial layer
b) Impurity concentration of n-type epitaxial layer
c) Area of the junction
d) Voltage across the junction

View Answer

Answer: a [Reason:] There is no p-type epitaxial layer present in junction capacitor. But a p-type substrate is present and it forms one of the junctions in the junction type IC capacitor.

8. Which is used as the dielectric layer in MOS Capacitor?
a) Silicon Nitride (Si3N4)
b) Aluminium oxide (Al2O3)
c) Tantalum oxide (Ta2O5)
d) All of the mentioned

View Answer

Answer: d [Reason:] Si3N4 offers higher value of capacitance. Whereas , Al2O3 and Ta2O5 are preferred for large value of capacitance. Hence all of them are used as dielectric layer.

9. Which is considered to be a serious disadvantages of thin film capacitor, when Al2O3is used as dielectric.
a) Additional fabrication step required
b) It require over voltage protection
c) Higher dielectric constant value is required
d) All of the mentioned

View Answer

Answer: b [Reason:] One of the serious disadvantages of thin film capacitor is that it fails, when the voltage rating exceeds due to breakdown of the dielectric, which is a destructive and irreversible failure mechanism and it require over voltage protection.

10. In MOS capacitor, the preference in dielectric layer is given to Silicon Nitride (Si3N4) because
a) It makes capacitor non-polar
b) It contain a small resistance
c) It offers less processing step
d) It reduces failure mechanism

View Answer

Answer: a [Reason:] Si3N4 gives more circuit flexibility by being non-polar, that is , it does not matter which plate is positive or negative and the voltage applied.

11. Why inductor is avoided in Integrated Circuit component?
a) They provide many losses compared to other IC components
b) IC devices are essentially two dimensions
c) Device density of IC increases
d) Fabrication process of these components are complicated

View Answer

Answer: b [Reason:] Usually, IC devices are very small (~1 to 10µm). Even if IC inductor is made in form of a flat metallic thin film spirals. Very small value of the order nanohenry with low quality factor can only be obtained.

12. Which circuit is used to replace inductor in IC components?
a) RC active network
b) PN-junction diode
c) LC active network
d) None of the mentioned

View Answer

Answer: a [Reason:] Circuit designer go to great lengths to avoid the use of inductors or otherwise simulate them by using RC active networks.

13. In application such as RF and IF circuits, inductor cannot be avoided. How to manage such situation?
a) Using inductors external to IC package
b) Thin film inductor spiral are used
c) Thin film hybrid microwave can be used
d) All of the mentioned

View Answer

Answer: d [Reason:] In most cases, inductors external to IC packages are used. However, thin film hybrid Microwave IC (MIC) and thin film inductor spiral can provide inductance up to 250nH.

Linear Integrated MCQ Set 4

1. Find the input and output resistance for the circuit shown.
Specification for 741 op-amp : A=400000 ; Ri = 33MΩ; Ro = 60Ω;
RF = 11kΩ; R1 = 2kΩ; Supply voltage = ± 15v; Maximum output voltage swing = ± 13v.
linear-integrated-circuits-questions-answers-online-quiz-q1
a) RIF = 66MΩ, ROF = 30Ω
b) RIF = 30MΩ, ROF = 6kΩ
c) RIF = 15kΩ, ROF = 50MΩ
d) None of the mentioned

View Answer

Answer: a [Reason:] AF = 1+(RF/R1) = 1+(11kΩ/2kΩ) = 6.5; B= 1/ AF = 1/6.5 = 0.154; Input resistance of RIF = R1(1+AB) = 33MΩ[1+(6.5*0.154) ]= 66MΩ; Output resistance of ROF = Ro/(1+AB) = 60/[1+(6.5*0.154) ]= 29.98 ≅30Ω.

2. The output resistance of the op-amp with feedback is
a) Same as that of the output resistance without feedback
b) Greater than that of the output resistance without feedback
c) Smaller than that of the output resistance without feedback
d) None of the mentioned

View Answer

Answer: c [Reason:] In voltage series feedback amplifier, the output resistance is (1/(1+AB)) times the output resistance of the op-amp. Therefore, the output resistance of the op-amp with feedback is much smaller than the output resistance without feedback.

3. Find the output current in the voltage series feedback amplifier.
a) io ={ [Vo+(A*Vid)]/Ro}
b) io ={ [Vo-(A*Vid)]/Ro}
c) io =(Vo/Ro)*A
d) io =[A*(Vo-Vid)]/Ro

View Answer

Answer: b [Reason:] The output current in voltage series feedback amplifier is given as io ={[Vo-(A*Vid)]/Ro}.

4. Find the unity gain bandwidth for voltage series feedback amplifier?
a) UBG = Afo
b) UBG = AfF
c) UBG = Afo fF
d) UBG = AFfo

View Answer

Answer: a [Reason:] The unity gain bandwidth is given as product of open loop voltage gain and break frequency of an op-amp.

5. The bandwidth of a non-inverting amplifier with feedback is equal to
a) fo(AB)
b) fo(AB-1)
c) fo(1+AB)
d) fo(1-AB)

View Answer

Answer: c [Reason:] The bandwidth of the non-inverting amplifier with feedback is equal to its bandwidth without feedback times (1+AB). i.e. fF=fo(1+AB).

6. How are the saturation voltage specified on the manufacture’s datasheet?
a) Negative voltage
b) Output voltage swing
c) Supply voltage
d) None of the mentioned

View Answer

Answer: b [Reason:] In an open loop op-amp, the total output offset voltage (i.e. output voltage swing) is equal to either the positive or negative saturation voltage.

7. What is the formula for total output offset voltage with feedback?
a) VooT = ± Vo/(1+AB)
b) VooT = ± Vsat*(1+AB)
c) VooT = ± Vsat/(1+AB)
d) VooT = ± Vo*(1+AB)

View Answer

Answer: c [Reason:] The total output offset voltage with feedback = (Total output offset voltage witput feedback) / (1+AB). i.e. VooT = ± Vsat/(1+AB).

8. Which of the following has the same characteristic as that of non-inverting amplifier with feedback?
a) Perfect feedback amplifier
b) Voltage follower
c) Perfect voltage amplifier
d) All of the mentioned

View Answer

Answer: c [Reason:] A perfect voltage amplifier has very high input resistance, very low output resistance, stable voltage gain, large bandwidth and very little output offset voltage. From the analysis of the characteristic of non-inverting amplifier with feedback, it is clear that it exhibits the characteristics of a perfect voltage amplifier.

9. What is the gain of voltage follower?
a) Gain > ∞
b) Gain –>1
c) Gain <1
d) Gain –>∞

View Answer

Answer: b [Reason:] Voltage follower is non-inverting amplifier configured for unity gain. Such that the output voltage is equal to and in phase with the input.

10. Which is preferred to attain higher input resistance and the output amplitude equal to input?
a) Voltage follower
b) Voltage series feedback amplifier
c) Voltage shunt feedback amplifier
d) Inverter

View Answer

Answer: a [Reason:] In the voltage follower the output follow the input due to unity gain. Therefore, it is attained to get higher input resistance and output amplitude equal to input.

11. Find the input and output voltage in voltage follower circuit?
a) Vin=2v and Vout = 3v
b) Vin=10v and Vout = 11v
c) Vin=9v and Vout = 9v
d) Vin=4v and Vout = 7v

View Answer

Answer: c [Reason:] Voltage follower has input voltage equal to output voltage. The closed loop voltage gain is equal to one. For example, take the input and output voltage to be 2v, then AF = Vout/Vin = 2v/2v = 1.

12. Voltage follower is also called as
a) None of the mentioned
b) Non-inverting amplifier
c) Inverting amplifier
d) Normal buffer

View Answer

Answer: b [Reason:] The voltage follower is also called as a non-inverting buffer because, when placed between two networks, it removes the loading on the first network.

13. Find the bandwidth and total output offset voltage of a voltage follower? The following are the specifications for the op-amp 741: A=200000, fo =5hz and supply voltage =±15v.
linear-integrated-circuits-questions-answers-online-quiz-q13
a) fF = 1000hz, VooT = ± 7.5µv.
b) fF = 100khz, VooT = ± 7.5µv.
c) fF = 10khz, VooT = ±7.5µv.
d) fF = 1000khz, VooT = ± 7.5µv.

View Answer

Answer: d [Reason:] Bandwidth fF =A* fF = 200000*5= 1Mhz. Total output offset voltage, VooT= ±Vsat/A= ±15/200000 =± 7.5µv.

Linear Integrated MCQ Set 5

1. How are input offset voltage of op-amp expressed
a) Microvolts per week
b) Nanovolts per week
c) Megavolts per week
d) Millivolts per week

View Answer

Answer: a [Reason:] For long term stability the amount of change in input offset voltage with time is crucial and denoted as micro volts per week.

2. Find the expression for output offset voltage?
a) VooT = [RF/R1×Vio] + (RF×Io)
b) VooT = [1+(RF/R1)]×(Vio) + (RF×Io)
c) VooT = [1+(RF/R1)]/(Vio) + (RF×Io)
d) VooT = [RF/R1]×(Vio) + (RF/Io)

View Answer

Answer: b [Reason:] Output offset voltage is expressed as a function of input offset current and input offset voltage, when is given as VooT = [1+(RF/R1)]×(Vio) + (RF×Io).

3. Why it is necessary to calibrate all op-amp in the system periodically?
a) To maintain current drift of a system
b) To maintain output offset voltage in op-amp
c) To maintain accuracy and linearity of a system
d) None of the mentioned

View Answer

Answer: c [Reason:] In practice, the output offset voltage in all op-amp circuit will change with time. Therefore, to maintain the desired accuracy and linearity of a system, it is necessary to calibrate all op-amps in that system periodically.

4. Determine the maximum possible change in output offset voltage after 3 month if the LH0041C op-amp is initially nulled and room temperature, voltage across terminal, +Vcc & -Vee remain constant.
a) △VooT = 456.78mv
b) △VooT = 3.452mv
c) △VooT = 21.1mv
d) None of the mentioned

View Answer

Answer: d [Reason:] For LH0041C, △Vio/△t = 5µV/week and △Iio /△t = 2nA/week. Due to the time drift △t=12 weeks (3months). The maximum possible change in output offset voltage, △VooT = [[1+(RF/R1)]×(△Vio/△t)×△t]+[RF×(△Iio/△t) ×△t]= [(1+(250kΩ/1kΩ))×5µV×12] + [250kΩ×2µA×12] = 15.1mv+6mv => △VooT = 21.1mv.

5. How to minimize the drift in input signal amplifier without affecting the performance of the circuit?
a) All of the mentioned Choosing small value of external components
b) Enhancing the input signal amplifier
c) Reducing the gain of op-amp
d) All of the mentioned

View Answer

Answer: a [Reason:] The amount of drift in input signal depends on the relative values of external components. So, if relatively small values are selected for external component then drift can be minimized.

6. An amplifier has a supply voltage ±15v. Compute its peak to peak output swing
a) 36v
b) 30v
c) 26v
d) 49v

View Answer

Answer: c [Reason:] The peak to peak output swing depends on the value of the supply voltage ( which is +15v-(-15v) =30v ) and is always less than the value to maintain safe operation of op-amp.

7. Find out the incorrect statement.
a) Power consumption of op-amp decrease with increasing ambient temperature
b) Output short circuit current of op-amp decreases with increasing temperature.
c) Common mode voltage range of op-amp increases with increasing supply voltage value
d) Power consumption of op-amp decreases with increasing supply voltage

View Answer

Answer: d [Reason:] If the power consumption as a function of supply voltage curve is drawn, it can be seen that the amount of dc power required to operate the op-amp under no-load condition increases with an increase in supply voltage.

8. Choose the temperature dependent parameter
a) Common mode rejection ratio
b) Voltage gain
c) Output voltage swing
d) Absolute maximum power dissipation

View Answer

Answer: d [Reason:] Other than absolute maximum power dissipation the remaining are frequency dependent parameters.

Linear Integrated MCQ Set 6

1. The number of leads in schottky barrier diode are
a) Four
b) Three
c) Two
d) Six

View Answer

Answer: c [Reason:] Schottky barrier diode has two contact leads namely, 1. Schottky Barrier contact 2. Ohmic contact.

2. In Schottky barrier diode, which contact has similar characteristics to that of an ordinary PN diode?
a) Ohmic contact
b) Schottky barrier contact
c) Both ohmic and Schottky barrier contact
d) None of the mentioned

View Answer

Answer: b [Reason:] A metal semiconductor is formed when aluminium is deposited directly upon n-type silicon. Its characteristics is found to be same as in an ordinary PN junction diode (Although the physical mechanism is different).

3. Find the symbol for Schottky barrier diode from the given circuit diagram?
linear-integrated-circuits-questions-answers-quiz-q3

View Answer

Answer: a [Reason:] The mentioned symbol is the symbol for metal semiconductor diode or Schottky diode.

4. How the ohmic contact is formed in metal semiconductor diode? (AL-Aluminium)
a) n+ diffusion in p-region near AL lead
b) p+ diffusion in p-region near AL lead
c) n+ diffusion in n-region near AL lead
d) p+ diffusion in n-region near AL lead

View Answer

Answer: c [Reason:] Aluminium is p-type impurity in silicon. So, when it is used to make a contact with n-type silicon, its essential contact is ohmic and no PN-junction is formed. Therefore, the contact is done by making n+ diffusion in the region near the surface where aluminium is deposited.

5. The flow of current in Schottky barrier diode is due to
a) Majority and Minority carriers
b) Majority carriers
c) Minority carriers
d) None of the mentioned

View Answer

Answer: b [Reason:] When the diode is forward biased, electron flow from semiconductor to metal (where electrons are abundant). Hence, the majority carrier ‘electrons’ carry current in Schottky diode.

6. Find the application areas, where Schottky diode can be used?
a) Radio frequency
b) Power rectifier
c) Clamping diode
d) All of the mentioned

View Answer

Answer: d [Reason:] Schottky diode can be used for ideal clamping or as detector in high frequency microwave ICs. Therefore, it is used for all these applications.

7. Which of the following resistor is not used as an integrated resistor?
a) Poly gate resistor
b) Pinched resistor
c) Epitaxial resistor
d) Thin film resistor

View Answer

Answer: a [Reason:] Except poly gate resistor, other resistors are integrated resistor.

8. Which of the following is not true about diffused resistor?
a) Limitation due to small range of resistance
b) Resistance depends upon surface geometry
c) Resistance depends on diffusion characteristic of material
d) Diffused resistors are non-economical

View Answer

Answer: d [Reason:] In diffused resistor method, the resistors are very economical as no extra fabrication steps are required.

9. Determine the formula for sheet resistance (Rs).
a) R×L×W
b) R×(L×ρ)/W
c) R×(W/L)
d) R×(W×ρ)/L

View Answer

Answer: c [Reason:] The formula for sheet resistance of a material of surface dimension L and W is Rs = R×(W/L).

10. Consider a 52cm×52cm material of uniform resistivity 100Ωm and thickness 3cm. Find the area and resistance of this sheet of material.
a) 16 m2, 1.923 Ω/square
b) 8112 cm2, 1.733 Ω/square
c) 156 cm2, 33.33 Ω/square
d) 901 cm2, 3.333 Ω/square

View Answer

Answer: c [Reason:] Area = L × t= 52cm×3cm=156 cm2 => Sheet resistance RS = (ρ×L)/(L×t) = ρ/t = 100 Ω m/3m =33.33 Ω/square.

11. If a 25Ω diffused resistor is to be designed for an emitter resistor, determine the pattern in which it is fabricated?
a) 20mil long by 5mil wide
b) 25mil long by 1mil wide
c) 5mil long by 1mil wide
d) 16mil long by 4mil wide

View Answer

Answer: b [Reason:] The sheet resistance of n-type diffused resistor is 5Ω/square. => L/W=R/RS= 25Ω/5Ω = 5/1 => 5mil long by 1mil wide.

12. The number of square contained in the integrated resistor by diffused resistor method depend on ratio of
a) ρ/t
b) ρ×L/W
c) W/L×t
d) L/W

View Answer

Answer: d [Reason:] The number of square contained in the resistor depends on the surface geometry. Which is given by, the ratio L/W is called the aspect ratio of surface geometry.

13. Match the sheet resistance value for the following region in diffused resistor

1. Epitaxial Collector regioni. 200Ω/square
2. p-type base regionii. 1 to 10kΩ/square
3. n-type emitter regioniii. 5Ω/square

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

View Answer

Answer: b [Reason:] The mentioned values are sheet resistance values for respective diffused resistor.

Linear Integrated MCQ Set 7

1. The major disadvantage of PN-junction isolation technique is:
a) Formation of Parasitic Resistance
b) Formation of Parasitic Capacitance
c) Formation of Isolation island
d) None of the mentioned

View Answer

Answer: b [Reason:] The presence of Parasitic Capacitance at the isolating PN-junction, results in an inevitable capacitor coupling between the component and substrate. This also limits the performance of circuit at high frequencies.

2. Which isolation technique is used in applications like military and aeroscope?
a) Thin film isolation
b) PN-junction isolation
c) Barrier isolation
d) Dielectric isolation

View Answer

Answer: d [Reason:] In dielectric isolation a layer of solid dielectric is used for isolation purpose. This layer is thick enough such that its associated capacitance is negligible. Moreover, it is more expensive, which is justified by its superior performance.

3. Pick out the incorrect statement
Aluminium is usually used for metallization of most IC as it offers
a) Relatively a good conductor
b) High resistance
c) Good mechanical bond with silicon
d) Deposition of aluminium film using vacuum deposition

View Answer

Answer: c [Reason:] Aluminium forms low resistance (it is a good conductor of electricity). Therefore, it forms ohmic contact (Semiconductor-metal contact) with p-type silicon and heavily doped n-type silicon.

4. How the aluminium film coating is carried out in metallization process?
a) Heating and pouring aluminium in required place.
b) Aluminium is vacuum evaporated and then condensed
c) Placing the aluminium in required place and then heating it using tungsten
d) None of the mentioned

View Answer

Answer: b [Reason:] Metallization process takes place in Vacuum evaporation chamber, where the material is evaporated by focussing a high power density electron beam. Vapours then hit the substrate and get condensed to form thin film coating.

5. What type of packing is suitable for Integrated Circuits?
a) Metal can package
b) Dual-in-line package
c) Ceramic flat package
d) All of the mentioned

View Answer

Answer: d [Reason:] These packages are the three different possible packages available in Integrated Circuits. Its usage depends upon the number of leads required for application.

6. Metal can IC packages are available in
a) 42 leads
b) 16 leads
c) 12 leads
d) 24 leads

View Answer

Answer: c [Reason:] The maximum lead available in a metal can IC package is 12. The remaining lead numbers are commonly available in dual-in-line packages.

7. What process is used in semiconductor industry to fabricate Integrated Circuits?
a) Silicon wafer preparation
b) Silicon planar process
c) Epitaxial growth of silicon
d) Photolithography process

View Answer

Answer: b [Reason:] The planar process (Silicon planar technology) in semiconductor industry built individual components. It is the primary process by which Integrated Circuits are built. The other processes are the different steps involved within the planar process.

8. Which semiconductor is most widely used for fabrication of Integrated Circuit?
a) Germanium, Ge
b) Gallium Arsenide, GaAs
c) Silicon, Si
d) All of the mentioned

View Answer

Answer: c [Reason:] Silicon is abundantly available in the form of sand. It is possible to form superior stable SiO2(Which has superb insulating property). Whereas GaAs is more difficult to grow in crystal form and Ge crystal will be destroyed at high temperature.

9. What will be the next step after slicing (process) silicon wafers?
a) All of the mentioned
b) Lapping
c) Polishing
d) Chemical

View Answer

Answer: a [Reason:] When the silicon ingots are sliced for the given industrial dimension. It gives a rough surface and thus undergo lapping, polishing and chemical processing steps to get a smooth surface.

10. During ion implantation process (before the ion strike the wafer) the accelerated ions are passed through
a) Strong Electric field
b) Strong Magnetic field
c) Strong Electric and Magnetic Field
d) None of the mentioned

View Answer

Answer: b [Reason:] During arc discharge in ion implantation, the unwanted impurities gets generated. The magnetic field acts to separate unwanted impurities from dopant ions.

11. In which method shallow penetration of dopants is possible?
a) Ion implantation
b) Vertical diffusion
c) Horizontal diffusion
d) Dopants diffusion

View Answer

Answer: a [Reason:] The depth of diffusion in this method, can be easily regulated by control of the incident ion velocity and is capable of shallow penetration.

12. Which method is most suitable for silicon crystal growth in silicon wafer preparation?
a) Float zone process
b) Bridgeman-Stockbarger method
c) Czochralski crystal growth process
d) Laser heated pedestal growth

View Answer

Answer: c [Reason:] Czochralski crystal growth processes obtain single crystal of semiconductor. The most important application of this method may be growth of large cylindrical ingot of single crystal silicon.

Linear Integrated MCQ Set 8

1. The characteristics of non-inverting amplifier is identical to
a) Differential Amplifier with one op-amp
b) Differential Amplifier with two op-amp
c) Differential Amplifier with three op-amp
d) None of the mentioned

View Answer

Answer: b [Reason:] The characteristics of non-inverting amplifier are identical to differential amplifier with two op-amps as the gain of the amplifier are same. i.e. A = 1+(RF /R1).

2. How is it possible to vary the voltage from closed loop gain to the open loop gain?
a) By using differential amplifier with larger gain
b) By using differential amplifier with small gain
c) By using differential amplifier with variable gain
d) By using differential amplifier with differential gain

View Answer

Answer: c [Reason:] Variable gain in differential amplifier can be obtained by using a potentiometer. So, depending on the position of the wiper in potentiometer, voltage gain can be varied from the closed loop gain to the open loop gain.

3. A differential amplifier with two op-amps has the following specifications:
R1=R3=1.5kΩ ; RF=R2=5.7kΩ. Compute the input resistance for the following circuit.
linear-integrated-circuits-questions-bank-q3
a) RIFx = 83GΩ, RIFy = 317GΩ
b) RIFx = 90GΩ, RIFy = 400GΩ
c) RIFx = 59GΩ, RIFy = 269GΩ
d) RIFx = 36GΩ, RIFy = 156GΩ

View Answer

Answer: a [Reason:] The input resistance of the first stage is given as RIFy, => RIFy = Ri*[1+(A *R2)/( R2+ R3)] = RIFy = 2MΩ*[(1+200000*5.7kΩ)/(5.7kΩ+1.5kΩ)] = 3.166*1011 = 317GΩ. Similarly the input resistance of the second stage amplifier is given as RIFx, => RIFx = Ri*[1+(A *R1)/( R1+ RF)] = RIFx = 2MΩ*[(1+200000*1.5kΩ)/(5.7kΩ+1.5kΩ)] = 83GΩ.

4. Determine the output voltage from the diagram. Where V1=-4.3Vpp and V2 = -5.1Vpp sinewave at 1000hz.
linear-integrated-circuits-questions-bank-q4
a) 12 Vpp sinewave at 1000hz
b) 13.3 Vpp sinewave at 1000hz
c) 14 Vpp sinewave at 1000hz
d) 11 Vpp sinewave at 1000hz

View Answer

Answer: b [Reason:] In differential amplifier with two op-amp, R1=R3 and RF=R2. The closed loop voltage gain, AD = 1+( RF/ R1) = 1+(7.5kΩ/480Ω) = 16.62. Output voltage , Vo = AD(V1– V2) = 16.62(-4.3Vpp+5.1Vpp) = 13.29 =13.3 Vpp sinewave at 1000hz.

5. Determine the output resistance of differential amplifier with three op-amp. The op-amp used is 741c, with A=200000 and Ro. The output and difference of input voltages are 44 and 11.
a) 5.5mΩ
b) 3.5mΩ
c) 2.4mΩ
d) 1.5mΩ

View Answer

Answer: d [Reason:] Gain AD = Vo/Vid = 44/11 =4. The output resistance of differential amplifier is given as ROF=Ro/[1+(A/AD)]= 75/[1+(200000/4)] = 1.5mΩ.

6. Find the input resistance for the given circuit
linear-integrated-circuits-questions-bank-q6
a) RIF = Ri*[1+(A*R4+R5)/(2* R4+R5)].
b) RIF = Ri*[1+(A*R4)/(2* R4+R5)].
c) RIF = Ri*[1+(A*R4+R5)/(R4+R5)].
d) None of the mentioned

View Answer

Answer: a [Reason:] The input resistance of the op-amp with feedback is given as RIF= Ri*(1+AB). From the diagram, B=Vf/Vo = [R4+R5)/(2* R4+R5)]. Therefore, RIF = Ri*[1+(A*R4+R5)/(2* R4+R5)].

7. The bandwidth of the differential amplifier increases, if the value of
a) Open loop voltage gain decreases
b) Closed loop voltage gain decreases
c) Differential voltage gain decreases
d) All of the mentioned

View Answer

Answer: b [Reason:] The bandwidth of differential amplifier is inversely proportional to the closed loop voltage gain.

8. Calculate the bandwidth for the given circuit.(Where the 741 op-amp has gain of 200000).
linear-integrated-circuits-questions-bank-q8
a) 560hz
b) 390khz
c) 25.6khz
d) 1.5Mhz

View Answer

Answer: c [Reason:] Gain of the amplifier AD = -{[1+(2*R4/R5)]*(RF/R1)} = -{[1+(2*5kΩ/2.5kΩ)]*(3.9kΩ/500Ω)} (since, RF = R3 and R2 = R1). => AD = -5*7.8=-39. The bandwidth for the circuit is calculated as fF = (A*fo)/AD = (200000*5hz)/39 (For 741 op-amp 5hz is the break frequency) => fF = 25.6khz.

9. Calculate the output voltage. If Vx=3.9Vpp and Vy = 5.5 Vpp sinewave at 1khz for the following circuit.
linear-integrated-circuits-questions-bank-q9
a) 11 Vpp sinewave at 1Khz
b) -8Vpp Vpp sinewave at 1Khz
c) -10 Vpp sinewave at 1Khz
d) 6 Vpp sinewave at 1Khz

View Answer

Answer: b [Reason:] Voltage gain, AD=-RF/ R1 = -15kΩ/3kΩ = -5kΩ. The output voltage, Vo = – AD*(V x– V y) = 5*(3.9-5.5) = -8 Vpp sinewvae at 1Khz.