Category: Linear Integrated Circuits Objective Questions

Linear Integrated Circuit Questions on “Integrated Circuits, Types & Manufacturer’s Design – 2”.

1. Which package type is chosen for military purposes?
A. Ceramic DIP
B. Plat pack
C. Metal can pack
D. Plastic DIP

Answer: A
Clarification: Ceramic DIP can be used for high temperature and high performance equipment.

2. A Dual-In-Line Package is usually referred to as
A. DIP_n
B. _nDIP
C. DIPⁿ
D. All of the mentioned

Answer: A
Clarification: A Dual-In-Line Package is usually referred to as DIP_n. 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

Answer: B
Clarification: 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

Answer: C
Clarification: 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

Answer: D
Clarification: 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

Answer: B
Clarification: 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. -55^o to +85^oc
B. 90^o to 155^oc
C. 10^o to 100^oc
D. -20^o to +85^oc

Answer: D
Clarification: The industrial temperature range is from -20^o to +85^oc.

8. Find the types of temperature range used for an IC, which can be used only up to 75^oc?
A. Industrial temperature range
B. Commercial temperature range
C. Military temperature range
D. All of the mentioned

Answer: B
Clarification: Commercial grade IC can be used up to 75^oc. 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

Answer: A
Clarification: 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

Answer: D
Clarification: The device type is a group of alphanumeric characters. For example, 741 IC is represented as µA741, LM741 and MC1741.

Linear Integrated Circuit Multiple Choice Questions on “Block Diagram Representation of Feedback Configurations”.

1. A feedback amplifier is also called as
A. Open loop amplifier
B. Closed loop amplifier
C. Feedback network amplifier
D. Looped network amplifier

Answer: B
Clarification: A feedback amplifier is sometimes referred as a closed loop amplifier because the feedback forms a closed loop between input and the output.

2. How many types of configuration are available for feedback amplifier?
A. Six
B. Four
C. Two
D. Eight

Answer: B
Clarification: There are four type of configuration are available. They are voltage series feedback, voltage shunt feedback, Current series feedback and Current shunt feedback.

3. Which of the following is not a feedback configuration?
A. Current-series feedback
B. Voltage-shunt feedback
C. Current-Voltage feedback
C. Current-Shunt feedback

Answer: C
Clarification: In a feedback amplifier, either current or voltage can be fed back to the input, but both current and voltage cannot be feedback simultaneously.

4. When load current flows into the feedback circuit, the configuration is said to be
A. Current-shunt feedback
B. Voltage-shunt feedback
C. Voltage-series feedback
D. All of the mentioned

Answer: B
Clarification: In current-series and current-shunt feedback circuit, the load current flows into the feedback circuit.

5. Find the voltage-series feedback amplifier from the given diagram?

Answer: A
Clarification: The mentioned diagram is the voltage-series feedback amplifier because the voltage across load resistor is the input voltage to the feedback circuit.

6. On what criteria does the feedback amplifier are classified?
A. Signal fed back to input
B. Signal applied to input
C. Signal fed back to output
D. None of the mentioned

Answer: D
Clarification: The feedback amplifiers are classified according to whether the voltage or current is fed back to the input in series or in parallel.

7. The closed loop voltage gain is reciprocal of
A. Voltage gain of op-amp
B. Gain of the feedback circuit
C. Open loop voltage gain
D. None of the mentioned

Answer: B
Clarification: Comparing the equation of closed loop voltage gain (A_F) and the gain of the feedback circuit (B.. A_F is reciprocal of B
=> A_F = 1+( R_F/ R₁) ; B= R₁/( R₁+ R_F)
=> B = 1+( R₁/ R_F)
Therefore, A_F = 1/B.

8. Select the specifications that implies the inverting amplifier?
A. V₁ = -3v, V₂ = -4v
B. V₁ = -2v, V₂ = 3v
C. V₁ = 5v, V ₂ = 15v
D. V₁ = 0v, V₂ = 5v

Answer: D
Clarification: In inverting amplifier, the input is applied to the inverting terminal and the non-inverting terminal is grounded. So,the input applied to inverting amplifier can be V₁ = 0v, V₂ = 5v.

Linear Integrated Circuit Multiple Choice Questions on “Frequency Response of Internally Compensating and Non-Compensating Op-Amp”.

1. Open loop bandwidth of an op-amp extend its bandwidth from
A. 0Hz to f_o
B. 20dB to f_o
C. 3dB to f_o
D. 0.704dB to f_o
Answer: A
Clarification: The gain of the op-amp remains essentially constant from 0 to the break frequency f_o 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 f_o.

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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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.

.

Tough Linear Integrated Circuit Questions and Answerson “Voltage to Current Converter with Floating and Grounded Load – 2”.

1. Determine the full scale range for the input voltage if the resistance in series with meters are 1kΩ, 2kΩ, 47kΩ and full scale meter movement is 1mA in low voltage AC voltmeter?
A. 1.0 to 7.48 V_rms
B. 1.1 to 7.48 V_rms
C. 1.2 to 7.48 V_rms
D. 1.3 to 7.48 V_rms
Answer: B
Clarification: The minimum and maximum values of resistors are 1kΩ and 6.8kΩ.So, the range for the input voltages are
V_in(rms)|_min = 1.1×R₁/I_o = (1.1×1kΩ)/1mA =1.1v.
V_in(rms)|_max =1.1×R₁/ I_o =(1.1×6.8kΩ)/1mA = 7.48v.
Thus, the full scale input voltage ranges from 1.1 to 7.48 V_rms.

2. Determine the current through the diode, when the switch is in position 1, 2& 3. Assuming op-amps initially nulled.

A. I_o (LED. =4.01mA; I_o (Zener) =4.01mA; I_o (rectifier) =8.33mA
B. I_o (LED. =25mA; I_o (Zener) =4.01mA; I_o (rectifier) =4.01mA
C. I_o (LED. =16.67mA; I_o (Zener) =16.66mA; I_o (rectifier) =4.01mA
D. I_o (LED. =8.33mA; I_o (Zener) =8.33mA; I_o (rectifier) =8.33mA
Answer: D
Clarification: All the diodes are connected one after another in the feedback path. Therefore, current through the diode remains same. I_o =V_in/R₁ =1.5/180 =8.33mA.

3. A diode match finder circuit has input voltage of 2.6v and output voltage is 5.78v. Calculate the voltage drop across diode 1N4735
A. 2.22v
B. 8.38v
C. 3.18v
D. 15.02v
Answer: C
Clarification: The output voltage V_o =V_in+ V_D.
∴ the voltage drop across 1N4735, V_D =V_o – V_in = 5.78-2.6 =3.18v.

4. Find the voltage drop across the zener diode in the zener diode tester from the given specifications: I_Zk=1mA, V_Z =6.2v, input voltage= 1.2v, output voltage =3.2v and resistance in series with meter =150Ω.
A. 6.2mA
B. 8mA
C. 21.33mA
D. Cannot be determined
Answer: A
Clarification: Current through the zener I_o=V_in/R₁ =1.2v/150Ω =8mA.
Since, I_o > I_Zk the voltage across the zero will be approximately equal to 6.2v. As the current is larger than the knee current (I_Zk) of the zener, it blocks V_Z volts.

5. Which among the following is preferred to display device in digital application?
A. Matched zener diode
B. Matched LEDs
C. Matched rectifier diode
D. All of the mentioned
Answer: B
Clarification: Matched LEDs with equal brightness at a specific value of current are useful as indicators and display devices in digital applications.

6. The maximum current through the load in all application that uses voltage to current converter with floating is
A. 100mA
B. 75mA
C. 25mA
D. 50mA
Answer: C
Clarification: The maximum current through the load cannot exceed the short circuit current of the 741c op-amp which is 25mA.

7. For voltage to current converter with grounded load, establish a relation between the non-inverting input terminals and load current
A. V₁ = [V_in+V_o-(I_L×R)] /2
B. V₁ = [V_in-V_o-(I_L×R)] /2
C. V₁ = [V_in+V_o-I_L+R] /2
D. V₁ = [V_in+V_o+(I_L×R)] /2
Answer: A
Clarification: In the voltage to current converter circuit the relationship between the voltage v₁ at the non-inverting input terminal and load is given as V₁ = [V_in+V_o-(I_L×R)] /2.

8. Find the gain of the voltage to current converter with grounded load?
A. 2
B. 1
C. ∞
D. 0
Answer: A
Clarification: In voltage to current converter with grounded load all resistor must be equal in value.
∴ Gain = V_o/V_in = [1+(R_F/R₁)] = 1+R/R =1+1=2.

9. Find the output voltage and the load current for the circuit given below. Assume that the op-amp is initially nulled V₁ =2.5v

A. I_L=0.42mA, V_o =10v
B. I_L=0.42mA, V_o =3.4v
C. I_L=0.42mA, V_o =6.1v
D. I_L=0.42mA, V_o =5v
Answer: D
Clarification: The load current I_L =v_in /R =5/12kΩ =0.42mA
v_o =I_L×R, can be obtained when V_o=2×V₁ = 2×2.5 =5v.

To practice Tough questions and answers on all areas of Linear Integrated Circuit, .

Linear Integrated Circuit Multiple Choice Questions on “Basic Principles of Sine Wave Oscillator-1”.

1. What is Barkhausen criterion for oscillation?
A. Aß > 1
B. Aß < 1
C. Aß = 1
D. Aß ≠ 1

Answer: C
Clarification: The Barkhausen criterion for oscillation is Aß = 1.
Where, A-> gain of amplifier and ß-> transfer ratio.

2. At what condition the output signal can be continuously obtained from input signal?
A. When the product of input voltage and feedback voltage is equal to 1
B. When the product of amplifier gain and transfer ratio is equal to 1
C. When the product of feedback voltage and transfer ratio is equal to 1
D. When the product of amplifier gain and input voltage is equal to 1

Answer: B
Clarification: When Aß=1, the feedback signal will be equal to the input signal. At this condition, the circuit will continue to provide output, even if the external signal is disconnected. This is because the amplifier cannot distinguish between external signal and signal from the feedback circuit. Thus, output signal is continuously obtained.

3. An oscillator is a type of
A. Feedforward amplifier
B. Feedback amplifier
C. Waveform amplifier
D. RC amplifier

Answer: B
Clarification: An oscillation is a type of feedback amplifier in which a part of output is fed back to the input via a feedback circuit.

4. Find the basic structure of feedback oscillator.

Answer: C
Clarification: The above mentioned diagram is the basic structure of feedback oscillator. It consists of an amplifier, to the external input (v_i) is applied and it have a feedback network from which the feedback signal (v_f) is obtained.

5. What is the condition to achieve oscillations?
A. |Aß|=1
B. ∠Aß=0^o
C. ∠Aß=multiples of 2π
D. All the mentioned

Answer: D
Clarification: All the conditions should be simultaneously satisfied to achieve oscillations.

6. What happens if |Aß|<1
A. Oscillation will die down
B. Oscillation will keep on increasing
C. Oscillation remains constant
D. Oscillation fluctuates

Answer: A
Clarification: If |Aß| becomes less than unity, the feedback signal goes on reducing in each feedback cycle and oscillation will die down eventually.

7. How sustained oscillation can be achieved?
A. Maintaining |Aß| slightly greater than unity
B. Maintaining |Aß| equal to unity
C. Due to non-linearity of transistor
D. Due to use of feedback network

Answer: C
Clarification: When |Aß| is kept slightly greater than unity the signal, however, cannot go on increasing and get limited due to non-linearity of the device (that is transistor enters into saturation). Thus, it is the non-linearity of the transistor because of which the sustained oscillation can be achieved.

8. Why it is difficult to maintain Barkhausen condition for oscillation?
A. Due to variation in temperature
B. Due to variation in supply voltage
C. Due to variation in components life time
D. All of the mentioned

Answer: D
Clarification: The Barkhausen condition |Aß|=1 is usually difficult to maintain in the circuit as the value of A and ß vary due to temperature variations, aging of components, change of supply voltage etc.

9. Name the type of noise signal present in the oscillation?
A. Schmitt noise
B. Schottky noise
C. Saturation noise
D. None of the mentioned

Answer: B
Clarification: Schottky noise is the noise signal always present at the input of the transistor due to variation in the carrier concentration.

10. A basic feedback oscillator is satisfying the Barkhausen criterion. If the ß value is given as 0.7072, find the gain of basic amplifier?
A. 2.1216
B. 0.7072
C. 1
D. 1.414

Answer: D
Clarification: Barkhausen criterion for oscillation is given as Aß=1
=> A=1/ ß = 1/0.7072 = 1.414.

11. The feedback signal of basic sine wave oscillator is given as
A. V_f = Aß ×V_o
B. V_f = Aß ×V_i
C. V_f = Aß × (V_o/ V_i)
D. V_f = Aß × (V_i/ V_o)

Answer: B
Clarification: The feedback signal of an oscillator is given as the product of external applied signal & the loop gain of the system.
=> V_f= Aß ×V_i.

12. Express the requirement for oscillation in polar form
A. Aß =1∠360^o
B. Aß =1∠90^o
C. Aß =1∠π^o
D. Aß =1∠270^o

Answer: A
Clarification: There are two requirements for oscillation
1. The magnitude of Aß=1
2. The total phase shift of Aß=0^o or 360^o.

Linear Integrated Circuit Multiple Choice Questions on “555 timer as a Monostable Multivibrator”.

1. Determine the time period of a monostable 555 multivibrator.
A. T = 0.33RC
B. T = 1.1RC
C. T = 3RC
D. T = RC

Answer: B
Clarification: The time period of a monostable 555 timer is T = RC×ln(1/3) = 1.1.RC.

2. Find monostable vibrator circuit using 555 timer.

Answer: A
Clarification: When 555 timer is configured in monostable operation, the trigger input is applied through pin2 whereas, upper comparator threshold (pin6) & discharge (pin7) are shorted and connected at the output.

3. How to overcome mistriggering on the positive pulse edges in the monostable circuit?
A. Connect a RC network at the input
B. Connect an integrator at the input
C. Connect a differentiator at the input
D. Connect a diode at the input

4. A monostable multivibrator has R = 120kΩ and the time delay T = 1000ms, calculate the value of C?
A. 0.9µF
B. 1.32µF
C. 7.5µF
D. 2.49µF

Answer: C
Clarification: Time delay for a monostable multivibrator, T = 1.1RC
=> C = T/(1.1R) = 1000ms/(1.1×120kΩ) = 7.57µF.

5. Which among the following can be used to detect the missing heart beat?
A. Monostable multivibrator
B. Astable multivibrator
C. Schmitt trigger
D. None of the mentioned

Answer: A
Clarification: A monostable multivibrator can be used as a missing pulse detector by connecting a transistor between trigger inputs. If a pulse misses, the discharge trigger input goes high & transistor become cut-off and the output goes low. So, this type of circuit can be used to detect missing heart beat.

6. A 555 timer in monostable application mode can be used for
A. Pulse position modulation
B. Frequency shift keying
C. Speed control and measurement
D. Digital phase detector

Answer: C
Clarification: In monostable operation mode, if input trigger pulses are generated from a rotating wheel, the circuit will determine the wheel speed whenever it drops below a predetermined value. Therefore, it can be used for speed control and measurement.

7. How can a monostable multivibrator be modified into a linear ramp generator?
A. Connect a constant current source to trigger input
B. Connect a constant current source to trigger output
C. Replace resistor by constant current source
D. Replace capacitor by constant current source

Answer: C
Clarification: The resistor R of the monostable circuit is replaced by a constant current source. So, that the capacitor is charged linearly and generates ramp signal.

8. Determine time period of linear ramp generator using the specifications
R_E = 2.7kΩ, R₁ =47kΩ , R₂ 100kΩ , C= 0.1µF, V_CC =5v.

A. 8ms
B. 4ms
C. 2ms
D. 1ms

Answer: D
Clarification: The time period of the linear ramp generator, T= [(2/3)×(V_CC×R_E)×(R₁+ R₂)×C]/{(R₁×V_CC)-[V_BE×(R₁+R₂)]}
= {(2/3)×5v×[2.7kΩ×(4.7kΩ+ 100kΩ)]×(0.1µF)}/{[(47kΩ)×5v]-[(0.7)×(47kΩ+100kΩ)]}
=>T= 132.3/132.100 =1.0015×10^-3 = 1ms.

9. What will be the output, if a modulating input signal and continuous triggering signal are applied to pin5 and pin22 respectively in the following circuit?

A. Frequency modulated wave form
B. Pulse width modulated wave form
C. Both pulse and frequency modulated wave form
D. None of the mentioned

Answer: B
Clarification: On application of continuous trigger at pin22 and a modulated input signal at pin5, a series of output pulses are obtained. The duration of which depends on the modulating signal. Also in the pulse duration, only the duty cycle varies, keeping the frequency same as that of the continuous input pulse train trigger.