Category: Linear Integrated Circuits Objective Questions

Linear Integrated Circuit Multiple Choice Questions on “Operational Amplifier Internal Circuit – 1”.

1. Which is not the internal circuit of operational amplifier?
A. Differential amplifier
B. Level translator
C. Output driver
D. Clamper

Answer: D
Clarification: Clamper is an external circuit connected at the output of Operational amplifier, which clamp the output to desire DC level.

2. The purpose of level shifter in Op-amp internal circuit is to
A. Adjust DC voltage
B. Increase impedance
C. Provide high gain
D. Decrease input resistance

Answer: A
Clarification: The gain stages in Op-amp are direct coupled. So, level shifter is used for adjustment of DC level.

3. How a symmetrical swing is obtained at the output of Op-amp
A. Providing amplifier with negative supply voltage
B. Providing amplifier with positive voltage
C. Providing amplifier with positive& negative voltage
D. None of the mentioned

Answer: C
Clarification: For example, consider a single voltage supply +15v. During positive half cycle the output will be +5v and -10v during negative half cycle.
Therefore, the maximum peak to peak output swing, -5v (-10v) = -15v (Asymmetrical swing).
So, to get symmetrical swing both positive and negative supply voltage with bias point fixed suitably is required.

4. What is the purpose of differential amplifier stage in internal circuit of Op-amp?
A. Low gain to differential mode signal
B. Cancel difference mode signal
C. Low gain to common mode signal
D. Cancel common mode signal

Answer: D
Clarification: Any undesired noise, common to both of the input terminal is suppressed by differential amplifier.

5. Which of the following is not preferred for input stage of Op-amp?
A. Dual Input Balanced Output
B. Differential Input Single ended Output
C. Cascaded DC amplifier
D. Single Input Differential Output

Answer: C
Clarification: Cascaded DC amplifier suffers from major problem of drift of the operating point, due to temperature dependency of the transistor.

6. What will be the emitter current in a differential amplifier, where both the transistor are biased and matched? (Assume current to be I_Q)
A. I_E = I_Q/2
B. I_E = I_Q
C. I_E = (I_Q)²/2
D. I_E = (I_Q)²

Answer: A
Clarification: Due to symmetry of differential amplifier circuit, current I_Q divides equally through both transistors.

7. From the circuit, determine the output voltage (Assume α_F=1)

A. V_O1=3.9v , V_O2=12v
B. V_O1=12v , V_O2=3.9v
C. V_O1=12v , V_O2=0v
D. V_O1=3.9v , V_O2=-3.9v

Answer: B
Clarification: The voltage at the common emitter ‘E’ will be -0.7v, which make Q₁ off and the entire current will flow through Q₂.
⇒ V_O1 = VCC V_O2= VCC-α_F×I_Q×RC,
⇒ V_O1 = 12v , V_O2=12v-1×3mA×2.7k = 3.9v.

8. At what condition differential amplifier function as a switch
A. 4V_Td T
B. -2V_T ≤ V_d ≤ 2V_T
C. 0 ≤ V_dT
D. 0 ≤ V_d ≤ 2V_T

Answer: A
Clarification: For V_d > 4V_T, the output voltage are V_O1 = VCC, V_O2= VCC-α_F I_QRC. Therefore, a transistor Q₁ will be ON and Q₂ will be OFF. Similarly for V_d> -4V_T, both transistors Q₂ & Q₁ will be ON.

9. For V_d > ±4V_T, the function of differential amplifier will be
A. Switch
B. Limiter
C. Automatic gain control
D. Linear Amplifier

Answer: B
Clarification: At this condition, input voltage of the amplifier is greater than ±100mv and thus acts as a limiter.

10. Change in value of common mode input signal in differential pair amplifier make
A. Change in voltage across collector
B. Slight change in collector voltage
C. Collector voltage decreases to zero
D. None of the mentioned

Answer: A
Clarification: In differential amplifier due to symmetry, both transistors are biased and matched. Therefore, Voltage at each collector will be same.

11. Find collector current I_C2, given input voltages are V₁=2.078v & V₂=2.06v and total current I_Q=2.4mA. (Assume α=1)

A. 0.8mA
B. 1.6mA
C. 0.08mA
D. 0.16mA

Answer: A
Clarification: Collector current, I_C2=α_F×I_Q/(1+e^V_d⁄V_T),
V_T = Volts equivalent of temperature = 25mv,
⇒ V_d = V₁-V₂ =2.078v-2.06v=0.018v (equ1)
Substituting equation 1,
⇒ V_d/V_T = 0.018v/25mv = 0.72v (equ2)
Substituting equation 2,
⇒ I_C2= 1×2.4mA/(1+e^0.72) = 2.4mA/(1+2.05) = 0.8mA.

12. A differential amplifier has a transistor with β₀= 100, is biased at I_CQ = 0.48mA. Determine the value of CMRR and A_CM, if RE =7.89kΩ and RC = 5kΩ.
A. 49.54 db
B. 49.65 d
C. 49.77 db
D. 49.60 db

Answer: B
Clarification: Differential mode gain, A_DM= -g_mRC and Common mode gain,
⇒ A_CM= -(g_mRC./(1+2g_mRE)
(for β₀≫1).
Substituting the values,
⇒ g_m= I_CQ/V_T = 0.48mA/25mv=19.2×10^-3Ω^-1
⇒ A_DM= -g_m×RC= -19.2×10^-3Ω^-1×5kΩ= -96
⇒ A_CM= -(g_mRC./(1+2g_mRE)= -(19.2×10^-3Ω^-1×5kΩ) /(1+2×-⇒ 19.2×10^-3Ω^-1×7.89kΩ) = -0.3158
CMRR = -96/-0.3158= 303.976
=20log⁡303.976
=49.65db

Linear Integrated Circuit Multiple Choice Questions on “Integrated Circuit Package Type, Pin Configuration and Temperature Range – 2”.

1. Find out the op-amp which does not have same specifications and behaviour as that of N5741?
A. MC1741
B. CA3741
C. SN52741
D. None of the mentioned

Answer: D
Clarification: All these op-amp have the same they specifications and behave the same because the last three digits in each manufacturer designation is 741. For example Fairchild’s original µA741 is also manufactured by various other manufacturers under their own designation.

2. The op-amps 741C and 741E are identical to op-amps
A. 741s and 741A
B. 741 and 741A
C. 741A and 741Sc
D. 741 and 741S

Answer: B
Clarification: The 741C and 741E are identical to 741 and 741A except that the former have their performance guaranteed over a temperature range 0^o to 70^o or 75^oc.

3. Which of the following is a military grade op-amp?
A. 741
B. 741C
C. 741S
D. 741SC

Answer: C
Clarification: The 741S is a military grade op-amp with a higher slew rate (rate of change of output voltage per unit of time) than the 741 op-amp.

4. Which of the following is in correct? “First generation op-amp ”
A. Consists of hundred transistors/chip
B. Requires an external frequency compensating network for stable operation
C. Have short circuit protection
D. Has latch-up problem

Answer: C
Clarification: First generation op-amp has no short circuit protection. The op-amp is susceptible to burnout if output accidentally shorted to ground.

5. What is the disadvantage of integrated circuit?
A. Parameter within the IC cannot be modified
B. Low power requirement
C. ICs are considered to use minimum number of external connections
D. None of the mentioned

Answer: A
Clarification: The disadvantage of IC is that, a lack of flexibility an IC. It is generally not possible to modify the parameters within which an integrated circuit will operate.

6. Whenever an IC is designed manufacturers use a
A. Specific code and manufacturer’s name
B. Specific code and specific type number
C. Specific code and specific value
D. Specific type number

Answer: B
Clarification: Each manufacturer uses a specific code and assign a specific type number to the IC’s produced. That is, each manufacturer uses their own identifying initial followed by IC type number.

7. General purpose op-amp cannot be used for the application
A. Integrator
B. Audio power amplifier
C. Differentiation
D. Summing amplifier

Answer: B
Clarification: Audio power amplifier is a special purpose op-amp and is used only for the specific application they are designed for.

8. Low volume production methods are best suited to hybrid IC technology because
A. It require stipulated temperature to fabricate a circuit
B. It require several steps to fabricate a circuit
C. It require large components to fabricate a circuit
D. It require various designers to fabricate a circuit

Answer: B
Clarification: In hybrid IC, first the individual components are made, wired or metallic interconnection is done and finally they are diffused to form a single circuit.

9. Which is the different version of IC 741C?
A. 741A
B. 741E
C. 741S
D. 741SC

Answer: D
Clarification: 741SC is the different version of 741C IC, which is a commercial grade op-amp with higher slew rate and operating in the same temperature range.

10. An example of second generation IC used in greatest percentage of application
A. µA748
B. MC1558
C. µA741
D. LM101

Answer: C
Clarification: All the IC belongs to second generation op-amp. In that general purpose op-amp’741’ is used widely in greatest percentage of application.

11. What is the advantage of Hybrid Integrated Circuit?
A. Miniaturized circuits are made of individual components
B. Insulate components by protection
C. Circuit designer can choose the component value
D. All of the mentioned

Answer: D
Clarification: In hybrid integrated circuit miniaturization can be achieved and allow circuit designer a complete freedom in choosing the resistor values, whereas monolithic IC cannot use some important components on construction.

12. Find the pin configuration of µA741operational amplifier?

Answer: A
Clarification: The metal can configuration of µA741 op-amp has eight pins with pin number 8 identified by a tab. The other pins are numbered counter clockwise from pin 8, beginning with pin1.

Linear Integrated Circuit Multiple Choice Questions on “Thermal Drift”.

1. Which factor affect the input offset voltage, bias current and input offset current in an op-amp
A. Change in temperature
B. Change in supply voltage
C. Change in time
D. All of the mentioned

Answer: D
Clarification: Any change in the mentioned parameters affect the values of input offset voltage, bias current and input offset current from remaining constant.

2. Thermal voltage drift is defined as
A. △V_io/△T
B. △V_F/△T
C. △I_io/△T
D. △I_B/△T

Answer: A
Clarification: The average rate of change of input offset voltage per unit change in temperature is called thermal voltage drift, i.e. △V_io/△T.

3. A completely compensated inverting amplifier is nulled at room temperature 25^oC, determine the temperature at which the total output offset voltage will be zero?
A. 50^oC
B. 25^oC
C. 75^oC
D. 125^oC

Answer: B
Clarification: When amplifier is nulled at room temperature, the effect of input offset voltage and current is reduced to zero. Change in the total output offset voltage occurs only, if there is any change in the value of V_io and I_io. Therefore, the total output offset voltage will be zero at room temperature.

4. How the effect of voltage and current drift on the performance of an amplifier is determined?
A. △V_ooT/△T = {[1-R_F/R₁)]×(△V_io/△T)} + R_F×(△I_io/△t)
B. △V_ooT/△T = {(-R_F/R₁)×(△V_io/△T)} + R_F×(△I_io/△t)
C. △V_ooT/△T = {[1+(R_F/R₁)]×(△V_io/△T)} + R_F×(△I_io/△t)
D. None of the mentioned

Answer: C
Clarification: As the amplifier is used in inverting configuration, the effect of voltage and current drift is given as, the average change in total output offset voltage per unit change in temperature.
△V_ooT/△T = {[1+(R_F/R₁)]×(△V_io/△T)} + R_F×(△I_io/△t).

5. The error voltage in a compensating inverting amplifier is obtained by
A. Multiplying △T to total output offset voltage
B. Multiplying △T to input offset voltage
C. Multiplying △T to input offset current
D. All of the mentioned

Answer: A
Clarification: The maximum possible change in the total output offset voltage △V_ooT results from a change in temperature △t. Therefore, error voltage is obtained by multiplying △T in the average total output offset voltage.
E_v =( △V_ooT/△T)×△T = [1+(R_F/R₁)]×(△V_io/△T)×△T + R_F×(△I_io/△T)×△T.

6. A 7.5kΩ internal resistor and a 12kΩ feedback resistor are connected to an inverting amplifier. Find the error voltage, if the output voltage is 3.99mv for an input of 1.33mv.
A. ±0.6v
B. ±0.6mv
C. ± 60mv
D. ±6mv

Answer: D
Clarification: The output voltage of inverting amplifier is V_o= -(R_F/R₁)×V_in±E_v
=> E_v= ± V_o+(R_F/R₁)×V_in = 3.99mv+(12kΩ/7.5kΩ)×1.33mv = ±6.118 ≅ ±6mv.

7. Consider the amplifier is nulled at 27^oC. Calculate the output voltage , if the input voltage is 6.21mv dc at 50^oC. Assume LM307 op-amp with specification: △V_io/△T=30µV/^oC ; △I_io/△T = 300pA/^oC; V_S =±15v.

A. +0.53v or -0.68v
B. +0.52v or -0.78v
C. +0.54v or -0.90v
D. +0.51v or -0.86v

Answer: D
Clarification: Change in temperature △T = 50^oC-27^oC = 23^oC.
=> Error voltage, E_v =[1+(R_F/R₁)]×(△V_io/△T)×△T + R_F×(△I_io/△T)×△T = [1+(100kΩ/1kΩ)]×(30µv/1^oC.× 23^oC + 100kΩ×(300pA/1^oC.× 23^oC = 0.06969+ 6.9×10^-9
=> E_v= 0.0704 = 70.4mv.
For an input voltage of 6.21mv dc, the output voltage,
V_o=-(R_F/R₁)×V_in±E_v = -(100kΩ/1kΩ)×6.21mv±70.4mv = +0.69v or -0.55v.

8. The error voltage for the above circuit is 0.93v. Compute the output voltage?
A. +15v to +17v
B. +17v or -15v
C. -17v or +15v
D. None of the mentioned

Answer: B
Clarification: The output voltage for the non-inverting amplifier is
V_o=[1+(R_F/R₁+R₂)]×V_in±E_v
= [1+(50kΩ/3kΩ+10kΩ)]×3.3±0.93v = 15.99±0.93
=> V_o = +16.92v or -15.06v ≅ +17v or -15v.

Linear Integrated Circuit Multiple Choice Questions on “Instrumentation Amplifier – 1”.

1. Strain gage is an example of which device?
A. Transducer
B. Voltage follower
C. Integrator
D. Differentiator
Answer: A
Clarification: Strain gage is a device when subjected to pressure or force undergoes change in its resistance.

2. An instrumentation system does not include
A. Transducer
B. Instrumentation amplifier
C. Automatic process controller
D. Tester
Answer: D
Clarification: Except tester the remaining blocks form the input, intermediate and output stage of instrumentation system.

3. Transmission line are used for
A. Output signal
B. Input signal
C. Signal transfer
D. All of the mentioned
Answer: C
Clarification: Transmission lines are the connecting line between the blocks and permits signal transfer from unit to unit.

4. The length of the transmission lines are
A. Longer than 10 meters
B. Shorter than 10 meters
C. Equals to 10 meters
D. None of the mentioned
Answer: D
Clarification: The length of the transmission lines depends primarily on the physical quantities being monitored and on system requirement.

5. Why output of transducer is not directly connected to indicator or display?
A. Low level output is produced
B. High level output is produced
C. No output is produced
D. Input is fed directly
Answer: A
Clarification: Many transducers do not produce output with sufficient strength to permit there use directly. Therefore, the low level output signal of transducer need to be amplified.

6. What are the features of instrumentation amplifier?
A. Low noise
B. High gain accuracy
C. Low thermal and time drift
D. All of the mentioned
Answer: D
Clarification: Instrumentation amplifiers are intended for precise low level signal amplification because of the features mentioned.

7. What is the disadvantage of using LH0036 instrumentation op-amp?
A. Extremely stable
B. Relatively expensive
C. Accurate
D. All of the mentioned
Answer:b
Clarification: LH0036 is a very precise special purpose circuit in which most electrical parameters are minimized and performance is optimized. So, it is relatively expensive.

8. What instrument is used to amplify output signal of transducer
A. Peaking amplifier
B. Instrumentation amplifier
C. Differential amplifier
D. Bridge amplifier
Answer: B
Clarification: The major function of instrumentation amplifier is to amplify the low level output signal of the transducer, so that it can drive the output stages.

9. General purpose op-amps are used in applications as
A. Instrumentation amplifier
B. Differential instrumentation amplifier
C. Inverting instrumentation amplifier
D. Non-inverting instrumentation amplifier
Answer: B
Clarification: When the requirement for the application are not too strict. The general purpose op-amp can be employed in the differential mode. Such amplifiers are called as Differential instrumentation amplifier.

10. In an instrumentation amplifier using transducer bridge, which device measure the change in physical energy
A. Resistive transducer
B. Indicating meter
C. Capacitive transducer
D. Inductor circuit
Answer: A
Clarification: A resistive transducer is used to measure the change in same physical energy, which is connected to one arm of the bridge.

.

Linear Integrated Circuit Multiple Choice Questions on “First Order Low Pass Butterworth Filter”.

1. Find the voltage across the capacitor in the given circuit
A. V_O= V_in/(1+0.0314jf)
B. V_O= V_in×(1+0.0314jf)
C. V_O= V_in+0.0314jf/(1+jf)
D. None of the mentioned

Answer: A
Clarification: The voltage across the capacitor, V_O= V_in/(1+j2πfR_C)
=> V_O= V_in/(1+j2π×5k×1µF×f)
=> V_O= V_in/(1+0.0314jf).

2. Find the complex equation for the gain of the first order low pass butterworth filter as a function of frequency.
A. A_F/[1+j(f/f_H)].
B. A_F/√ [1+j(f/f_H)²].
C. A_F×[1+j(f/f_H)].
D. None of the mentioned

Answer: A
Clarification: Gain of the filter, as a function of frequency is given as V_O/ V_in=A _F/(1+j(f/f_H)).

3. Compute the pass band gain and high cut-off frequency for the first order high pass filter.
A. A_F=11, f_H=796.18Hz
B. A_F=10, f_H=796.18Hz
C. A_F=2, f_H=796.18Hz
D. A_F=3, f_H=796.18Hz

Answer: C
Clarification: The pass band gain of the filter, A_F =1+(R_F/R₁)
=>A_F=1+(10kΩ/10kΩ)=2. The high cut-off frequency of the filter, f_H=1/2πRC =1/(2π×20kΩ×0.01µF) =1/1.256×10^-3 =796.18Hz.

4. Match the gain of the filter with the frequencies in the low pass filter

Frequency	Gain of the filter
1. f H	i. VO/Vin ≅ AF/√2
2. f=fH	ii. VO/Vin ≤ AF
3. f>fH	iii. VO/Vin ≅ AF

A.1-i,2-ii,3-iii
B.1-ii,2-iii,3-i
C.1-iii,2-ii,3-i
D.1-iii,2-i,3-ii

Answer: D
Clarification: The mentioned answer can be obtained, if the value of frequencies are substituted in the gain magnitude equation |(V_o/V_in)|=A_F/√(1+(f/f_H)²).

5. Determine the gain of the first order low pass filter if the phase angle is 59.77^o and the pass band gain is 7.
A. 3.5
B. 7
C. 12
D. 1.71

Answer: A
Clarification: Given the phase angle, φ =-tan^-1(f/f_H)
=> f/f_H=- φtan(φ) = -tan(59.77^o)
=> f/f_H= -1.716.
Substituting the above value in gain of the filter, |(V_O/V_in)| = A_F/√ (1+(f/f_H)²) =7/√[1+(-1.716)²)] =7/1.986
=>|(V_O/V_in)|=3.5.

6. In a low pass butterworth filter, the condition at which f=f_H is called
A. Cut-off frequency
B. Break frequency
C. Corner frequency
D. All of the mentioned

Answer: D
Clarification: The frequency, f=f_H is called cut-off frequency, because the gain of the filter at this frequency is down by 3dB from 0Hz. Cut-off frequency is also called as break frequency, corner frequency or 3dB frequency.

7. Find the High cut-off frequency if the pass band gain of a filter is 10.
A. 70.7Hz
B. 7.07kHz
C. 7.07Hz
D. 707Hz

Answer: C
Clarification: High cut-off frequency of a filter, f_H=0.707×A_F =0.707×10
=>f_H=7.07Hz.

8. To change the high cutoff frequency of a filter. It is multiplied by R or C by a ratio of original cut-off frequency known as
A. Gain scaling
B. Frequency scaling
C. Magnitude scaling
D. Phase scaling

Answer: B
Clarification: Once a filter is designed, it may sometimes be a need to change it’s cut-off frequency. The procedure used to convert an original cut-off frequency f_H to a new cut-off frequency is called frequency scaling.

9. Using the frequency scaling technique, convert 10kHz cut-off frequency of the low pass filter to a cutoff frequency of 16kHz.(Take C=0.01µF and R=15.9kΩ)
A. 6.25kΩ
B. 9.94kΩ
C. 16kΩ
d )1.59kΩ

Answer: B
Clarification: To change a cut-off frequency from 10kHz to 16kHz,multiply 15.9kΩ resistor.
[Original cut-off frequency/New cut-off frequency] =10kHz/16kHz =0.625.
∴ R =0.625×15.9kΩ =9.94kΩ. However 9.94kΩ is not a standard value. So, a potentiometer of 10kΩ is taken and adjusted to 9.94kΩ.

10. Find the difference in gain magnitude for a filter ,if it is the response obtained for frequencies f₁=200Hz and f₂=3kHz. Specification: A_F=2 and f_H=1kHz.
A. 4.28 dB
B. 5.85 dB
C. 1.56 dB
D. None of the mentioned

Answer: C
Clarification: When f₁=200Hz, V_O(1)/V_in =A_F/√ [1+(f/f_H)²] =2/√ [1+(200/1kHz) ²] =2/1.0198.
=> V_O(1)/V_in =1.96
=>20log|(V_O/V_in)|=5.85dB.
When f=700Hz, V_O(2)/V_in= 2/√ [1+(700/1kHz) ²] =2/1.22=1.638.
=> V_O(2)/V_in =20log|(V_O/V_in|=20log(1.638) = 4.28.
Therefore, the difference in the gain magnitude is given as V_O(1)/V_in-V_O(2)/V_in =5.85-4.28 =1.56 dB.

11. Design a low pass filter at a cut-off frequency 1.6Hz with a pass band gain of 2.

Answer: A
Clarification: From the answer, it is clear that all the C values are the same . Therefore, c= 0.01µF
Given, f_H = 1kHz,
=> R= 1/(2πCf_m) = 1/2π×0.01µF×1kHz
R= 9.9kΩ ≅ 10kΩ. Since the pass band gain is 2.
=> 2=1+ (R_F/R₁). Therefore, R_F and R₁ must be equal.

Linear Integrated Circuit Multiple Choice Questions on ” A to D Converter -1″.

1. How many control lines are present in analog to digital converter in addition to reference voltage?
A. Three
B. Two
C. One
D. None of the mentioned
Answer: B
Clarification: ADC usually has two additional control lines
1. Start input-tell ADC when to start conversion.
2. EOC- end of conversion.

2. Find out the integrating type analog to digital converter?
A. Flash type converter
B. Tracking converter
C. Counter type converter
D. Dual slope ADC
Answer: D
Clarification: Other than dual slope ADC the rest belongs to direct type ADCs.

3. Which type of ADC follow the conversion technique of changing the analog input signal to a linear function of frequency?
A. Direct type ADC
B. Integrating type ADC
C. Both integrating and direct type ADC
D. None of the mentioned
Answer: B
Clarification: Integrating type ADC performs conversion in an indirect manner by first changing the analog input signal to a linear function of time or frequency and then to a digital code.

4. Which A/D converter is considered to be simplest, fastest and most expensive?
A. Servo converter
B. Counter type ADC
C. Flash type ADC
D. All of the mentioned
Answer: C
Clarification: The simplest possible A/D converter is flash type converter and is expensive for high degree of accuracy.

5. The flash type A/D converters are called as
A. Parallel non-inverting A/D converter
B. Parallel counter A/D converter
C. Parallel inverting A/D converter
D. Parallel comparator A/D converter
Answer: D
Clarification: The flash type A/D converter are also called as parallel comparator A/D converter because the purpose of the circuit is to compare the analog input voltage with each node voltage.

6. What is the advantage of using flash type A/D converter?
A. High speed conversion
B. Low speed conversion
C. Nominal speed conversion
D. None of the mentioned
Answer: A
Clarification: Flash type ADC has the advantage of high speed as the conversion takes place simultaneously rather than sequentially. Typical conversion time is 100nanosecond or less.

7. The number of comparator required for flash type A/D converter
A. Triples for each added bit
B. Reduce by half for each added bit
C. Double for each added bit
D. Doubles exponentially for each added bit
Answer: C
Clarification: The number of comparator required almost doubles for each added bit. For example – 2 -bit ADC requires three comparators, 3 -bit ADC needs seven comparators and a 4 -bit ADC requires fifteen comparators.

8. Drawback of counter type A/D converter
A. Counter clears automatically
B. More complex
C. High conversion time
D. Low speed
Answer: D
Clarification: In counter type ADC counter frequency is kept low enough to give sufficient time for DAC to settle and for the comparator for respond. So, low speed is the most serious drawback.

9. Calculate the conversion time of a 12-bit counter type ADC with 1MHz clock frequent to convert a full scale input?
A. 4.095 µs
B. 4.095ms
C. 4.095s
D. None of the mentioned
Answer: B
Clarification: conversion time = 2ⁿ -1 clock periods = (12ⁿ-1) = 4.095ms.

10. In a servo tracking A/D converter, the input voltage is greater than the DAC output signal at this condition
A. The counter count up
B. The counter count down
C. The counter back and forth
D. None of the mentioned
Answer: A
Clarification: In servo converter, the circuit consist of an up/down counter with comparator controlling direction of the count. So, if the input voltage is greater than DAC output signal, the output of comparator goes high and counter is caused to count up.

11. At what condition error occurs in the servo tracking A/D Converter?
A. Slow change input
B. Rapid change in input
C. No change in input
D. All of the mentioned
Answer: B
Clarification: As long as the analog input changes slowly, the tracking A/D converter will be within one LSB of the corrected value. When the input changes rapidly, the tracking A/D converter cannot keep up with change and error occurs.

12. How many clock pulses do a successive approximation converter requires for obtaining a digital output.
A. Twelve
B. Six
C. Eight
D. None of the mentioned
Answer: D
Clarification: The successive approximation technique uses a very efficient code search strategy to compute n-bit conversion in just n-clock period.

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