Category: Linear Integrated Circuits Objective Questions

Linear Integrated Circuit Multiple Choice Questions on “Fabrication of FET”.

1. JFET is similar to that of fabrication of
A. Diode fabrication
B. BJT fabrication
C. FET fabrication
D. None of the mentioned

Answer: B
Clarification: The basic processes used are as same as BJT fabrication. Epitaxial layer (collector of BJT) is used as the n-channel of JFET. The p⁺ is formed in n-channel by process of diffusion and n⁺ region formed under drain and source provide good ohmic contact.

2. What are the types of MOSFET devices available?
A. P-type enhancement type MOSFET
B. N-type enhancement type MOSFET
C. Depletion type MOSFET
D. All of the mentioned

Answer: D
Clarification: MOSFET are available as Enhancement type and depletion type MOSFET. These are further classified into n-type and p-type device.

3. Which insulating layer used in Fabrication of MOSFET?
A. Aluminium oxide
B. Silicon Nitride
C. Silicon dioxide
D. None of the mentioned

Answer: C
Clarification: Silicon dioxide is used as insulating layer in MOSFET Fabrication. It gives an extremely high input resistance in the order of 10¹⁰ to 10¹⁵ Ω for MOSFET.

4. Which of the following plays an important role in improving device performance of MOSFET?
A. Dielectric constant
B. Threshold voltage
C. Power supply voltage
D. Gate to drain voltage

Answer: B
Clarification: In MOSFET, the threshold voltage is typically 3 to 6v. This large voltage is not compatible with 5v supply used in digital IC. So, to improve device performance, magnitude of threshold voltage should be reduced.

5. A technique used to reduce the magnitude of threshold voltage of MOSFET?
A. Use of complementary MOSFET
B. Use of Silicon nitride
C. Using thin film technology
D. None of the mentioned

Answer: B
Clarification: Silicon nitride is sandwiched between two SiO₂ layer and provide necessary barrier .The dielectric constant of Si₃N₄ is 7.5, whereas that of SiO₂ is 4. This increase in overall dielectric constant reduces threshold voltage.

6. Find the sequence of steps involved in fabrication of poly silicon gate MOSFET?
Step 1: Entire wafer surface of a Si₃N₄is coated and it is etched away with the help of mask to include source, gate and drain.
Step 2: The contact areas are defined using photolithographic process
Step3: Selective etching of Si₃N₄ and thin oxide growth
Step 4: Deposition of poly silicon gate
Step 5: thick oxide growth called field oxide and P ^{implantation
Step 6: Metallization and interconnection between substrate and source
A. 1->5->3->4->2->6
B. 1->3->4->2->5->6
C. 1->5->4->3->2->6
D. 1->4->2->5->3->6
View Answer}

Answer: A
Clarification: The mentioned steps are the sequence of steps involved in the fabrication of poly silicon gate MOSFET.

7. What is used to higher the speed of operation in MOSFET fabrication?
A. Ceramic gate
B. Silicon dioxide
C. Silicon nitride
D. Poly silicon gate

Answer: D
Clarification: In conventional metal gate, small overlap capacitance is present, which lowers the speed of operation. Due to self aligning property of poly silicon gate, it eliminates this capacitance.

8. Why MOSFET is preferred over BJT in IC components?
A. MOSFET has low packing density
B. MOSFET has medium packing density
C. MOSFET has high packing density
D. MOSFET has no packing density

Answer: c
Clarification: No isolation island is required in MOSFET structure because, the drain of an n-mos device is held positive with respect to source. This cutoff the drain to substrate diode and the source to substrate diode formed due to p⁺ region. In BJT, the isolation diffusion occupies extremely large percentage of chip area.

9. Which of the following statement is true?
A. Fabrication of p-mos transistor require few additional steps compared to n-mos transistor
B. Fabrication of n-mos transistor require few additional steps compared to p-mos transistor
C. Fabrication on n-mos is same as that of p-mos transistor
D. Fabrication on n-mos is different from that of p-mos transistor

Answer: a
Clarification: There are two additional steps required in the formation of p-mos transistor compared to n-mos transistor. Such as, the formation of n-region and ion implantation of p-type source and drain regions.

10. Find complementary MOSFET from the given circuit diagram?

Answer: B
Clarification: Complementary MOSFET is combination of n-mos and p-mos enhancement device such that the source of p-mos is connected to Vdd and source of n-mos is connected to ground.

Linear Integrated Circuit Multiple Choice Questions on “Input Bias Current”.

1. Input bias current is defined as
A. Average of two input bias current
B. Summing of two input bias current
C. Difference of two input bias current
D. Product of two input bias current

Answer: A
Clarification: Input bias current is the average of two input bias current flowing into the non-inverting and inverting input of an op-amp.

2. Although the value of input bias current is very small, it causes
A. Output voltage
B. Input offset voltage
C. Output offset voltage
D. All of the mentioned

Answer: C
Clarification: Even a very small value of input bias current can cause a significant output offset voltage in circuits using relatively large feedback resistors.

3. The formula for output offset voltage of an op-amp due to input bias current
A. V_OIB= R_F*I_B
B. V_OIB= (R_F+R₁)/I_B
C. V_OIB= (1+R_F)*I_B
D. V_OIB= [1+(R_F/R₁)]*I_B

Answer: A
Clarification: The output offset voltage due to input bias current is V_OIB = R_F*I_B.

4. Find the input bias current for the circuit given below

A. 10mA
B. 2mA
C. 5mA
D. None of the mentioned

Answer: C
Clarification: Input bias current, I_B=(I_B1+ I_B2)/2
=> I_B =(4mA+6mA./2 = 5mA.

5. Mention a step to reduce the output offset voltage caused due to input bias current?
A. Use small feedback resistor and resistance at the input terminal
B. Use small feedback resistors
C. Reduce the value of load resistors
D. None of the mentioned

Answer: B
Clarification: Since the output offset voltage is proportional to feedback resistor and input bias current. The amount of V_OIB can be reduced by reducing the value of feedback resistor.

6. Given below is a differential amplifier in which V₁=V₂. What happens to V_OIB at this condition?

A. V_OIB= 0
B. V_OIB= V_OIB×10^-10
C. V_OIB= V_OIB/2
D. V_OIB= -1

Answer: A
Clarification: The voltage V₁ and V₂ are caused by the current I_B1 and I_B2. Although this bias current are very small, if they are made equal, then there will be no output voltage V_OIB.

7. Name the resistor that is connected in the non-inverting terminal of op-amp which is in parallel combination of resistor connected in inverting terminal and feedback resistor.
A. Random minimizing resistor
B. Offset minimizing resistor
C. Offset reducing resistors
D. Output minimizing resistors

Answer: B
Clarification: The voltage is product of resistors and input bias current. Therefore, the value of the resistors are adjusted such that the resistors are connected at the inverting input terminal is made equal to resistor connected in non-inverting input terminal. The use of this resistors minimize the amount of output offset voltage and therefore, they are referred to as offset minimizing resistors.

8. Calculate R_OM, if the value of I_B1 = I_B2 in the given circuit.

A. 1173.11Ω
B. 171.31Ω
C. 1171.43Ω
D. 1071.43Ω

Answer: D
Clarification: Offset minimizing resistor, R_OM =(R₁* R_F)/( R₁+R_F).
=> R_OM = (1.2kΩ*10kΩ)/(1.2kΩ+10Ω) = 1071.43Ω.

9. Calculate the output voltage for the given circuit using the specification: R₁ = 820Ω; R_OM=811.882Ω; V_in=10mVpp; V_OIB≅0.

A. 1.025Vpp
B. 1.8Vpp
C. 1Vpp
D. 2Vpp

Answer: C
Clarification: Offset minimizing resistor, R_OM = (R₁*R_F)/(R₁+ R_F)
=> R_F = (R_OM* R₁)/( R₁– R_OM) = (812Ω*811.882Ω)/(820Ω-811.882Ω) = 82kΩ.
∴ V_o = -(R_F/ R₁)* V_in = -(82kΩ/820Ω)*10mVpp = 1Vpp.

10. Analyse the given circuit and determine the correct option

A. V_oo ≥ V_IOB
B. V_oo = V_IOB
C. V_oo >> V_IOB
D. V_ooIOB

Answer: C
Clarification: 741op-amp has V_io = 6mvdc and I_B =500nA.
The output offset voltage due to input offset voltage is given as V_oo =[1+(R_F/R₁)]*V_io = [1+(4.7kΩ/47Ω)]*6mv = 0.606v.
The output offset voltage due to input bias current is given as V_IOB = R_F*I_B =4.7kΩ*500nA = 2.35mv.
=>∴ V_oo >> V_IOB.

11. The specification for LM101A op-amp is given as I_B =75nA. Determine the value of V_IOB– V₁.

A. 0.112v
B. 0.750v
C. 0.374v
D. 0.634v

Answer: A
Clarification: The voltage at non-inverting terminal is given as V₁ = R_OM*I_B1 = 148Ω*7.5nA = 1.11µv.
=> ∵ R_OM = (R₁*R_F)/(R₁+ R_F) = (15kΩ*150Ω)/(15kΩ+150Ω) =148Ω
The output offset voltage is given as V_IOB = R_F*I_B
=> V_IOB = 15kΩ*7.5nA = 112.5mv
=> ∴ V_IOB– V₁ = 0.112v.

Linear Integrated Circuit Multiple Choice Questions on “Summing, Scaling & Averaging Amplifier – 1”.

1. In which amplifier the output voltage is equal to the negative sum of all the inputs?
A. Averaging amplifier
B. Summing amplifier
C. Scaling amplifier
D. All of the mentioned
Answer: B
Clarification: In summing amplifier the output voltage is equal to the sum of all input. Since the total input is a sum of negative input, the amplifier is an inverting summing amplifier.

2. Determine the expression of output voltage for inverting summing amplifier consisting of four internal resistors? (Assume the value of internal resistors to be equal)
A. V_o = -(R_f/R )×(V_a +V_b+V_c+V_d)
B. V_o = (R_F/R)×(V_a +V_b+V_c+V_d)
C. V_o = (R/ R_F)×(V_a +V_b+V_c+V_d)
D. None of the mentioned
Answer: A
Clarification: If the internal resistors of the circuit is same i.e R_a=R_b=R_c=R_d=R (since there are four internal resistor)
Then, the output voltage for inverting amplifier is given as V_o= -(R_f/R)×(V_a +V_b+V_c+V_d).

3. An inverting amplifier with gain 1 have different input voltage: 1.2v,3.2v and 4.2v. Find the output voltage?
A. 4.2v
B. 8.6v
C. -4.2v
D. -8.6v
Answer: D
Clarification: When the gain of the inverting summing amplifier gain is 1 then, the internal resistors and feedback resistors have the same value. So, the output is equal to the negative sum of all input voltages.
V_O= -(V_a+V_b+V_c) =-(1.2+3.2+4.2)= -8.6v.

4. In which type of amplifier, the input voltage is amplified by a scaling factor
A. Summing amplifier
B. Averaging amplifier
C. Weighted amplifier
D. Differential amplifier
Answer: C
Clarification: The weighted amplifier is also called as scaling amplifier. Here each input voltage is amplified by a different factor i.e. R_a,R_b and R_c are different in values ( which are the input resistors at each input voltage).

5. An inverting scaling amplifier has three input voltages V_a, V_b and V_c. Find it output voltage?
A. V_O= – {[(R_F/R_a)×V_a] +[(R_F/R_b)×V_b]+[(R_F/R_c)×V_c]}
B. V_O= – [(R_F/R_a)+(R_F/R_b)+(R_F/R_c)]×[( V_a +V_b+V_c)].
C. V_O = – {[(R_a/R_F)×V_a] +[(R_b/R_F)×V_b]+[(R_c/R_F)×V_c]}
D. None of the mentioned
Answer: A
Clarification: Since three input voltages are given assume the input resistors to be R_a,R_b and R_c. In a scaling amplifier, the input voltages are amplified by a different factor
=> ∴ R_F/R_a ≠ R_F/R_b ≠ R_F/R_c
Therefore, output voltage V_o = -{[(R_F/R_a) V_a] +[(R_F/R_b) V_b]+[(R_F/R_c) V_c]}.

6. An amplifier in which the output voltage is equal to average of input voltage?
A. Summing amplifier
B. Weighting amplifier
C. Scaling amplifier
D. Averaging amplifier
Answer: D
Clarification: An averaging amplifier can be used as an averaging circuit, in which the output voltage is equal to the average of all the input voltages.

7. Find out the gain value by which each input of the averaging amplifier is amplified ?( Assume there are four inputs)
A. 0.5
B. 0.25
C. 1
D. 2
Answer: B
Clarification: In an averaging amplifier, the gain by which each input is amplified must be equal to lower number of input.
=> R_F /R =1/n , where n=number of inputs
∴ R_F /R=1/4 = 0.25 (Four inputs)
So, each input in the averaging amplifier must be amplified by 0.25.

8. 3v, 5v and 7v are the three input voltage applied to the inverting input terminal of averaging amplifier. Determine the output voltage?
A. -5v
B. -10v
C. -15v
D. -20v
Answer: A
Clarification: The output voltage, V_o = -[(V_a+V_b+V_c)/3] = -[(3+5+7)/3] =-5v.

9. The following circuit represents an inverting scaling amplifier. Compute the value of R_oM and V_O?

A. V_O = -0.985v ; R_oM = 111.11Ω
B. V_O = -2.567v ; R_oM = 447.89Ω
C. V_O = -1.569v ; R_oM = 212.33Ω
D. V_O = -1.036v ; R_oM = 320.56Ω
Answer: D
Clarification: V_O = – {[(R_F/R_a)×V_a]+[(R_F/R_b)×V_b] + [(R_F/R_c)×V_c]}
= – {[(10kΩ/1kΩ)×3.3mv] + [(10kΩ/1.25kΩ)×5mv] + [(10kΩ/820Ω)×7.9mv]} = -1.036v.
R_oM = [R_a||R_b||R_c||R_F]
= [(R_a×R_b)/(R_a+ R_b)] || [(R_c×R_F)/( R_c+ R_F)] = [(1kΩ×1.25kΩ)/(1kΩ+1.25kΩ)] || [(820Ω×10kΩ)/(820Ω+10kΩ)] = 555.55||757.85 =[(555.55 ×757.85)/(555.55+757.85)] = 320.56Ω.

.

target=”_blank” rel=”noopener noreferrer”>

Linear Integrated Circuit Multiple Choice Questions on “Active Filters – 1”.

1. An electrical filter is a
A. Phase-selective circuit
B. Frequency-selective circuit
C. Filter-selective circuit
D. None of the mentioned

Answer: B
Clarification: An electric filter is often a frequency selective circuit that passes a specified band of frequencies and blocks or alternates signal of frequencies outside this band.

2. Filters are classified as
A. Analog or digital
B. Passive or active
C. Audio or radio frequency
D. All of the mentioned

Answer: D
Clarification: Filters are classified based on the design technique (analog or digital), elements used for construction (active or passive) and operating range (audio or radio frequency).

3. Why inductors are not preferred for audio frequency?
A. Large and heavy
B. High power dissipation
C. High input impedance
D. None of the mentioned

Answer: A
Clarification: At audio frequencies, inductor becomes problematic, as the inductors become large, heavy and expensive.

4. The problem of passive filters is overcome by using
A. Analog filter
B. Active filter
C. LC filter
D. A combination of analog and digital filters

Answer: B
Clarification: The active filters enclose as a capacitor in the feedback loop and avoid using inductors, this way inductorless active filter are obtained.

5. What happens if inductors are used in low frequency applications?
A. Enhance inductor usage
B. No losses occurs
C. Degrades inductor performance
D. Low power dissipation

Answer: C
Clarification: For low frequency applications more number of turns of wire must be used, which in turn adds to the series resistance degrading inductor’s performance.

6. Find out the incorrect statement about active and passive filters.
A. Gain is not attenuated in active filter
B. Passive filters are less expensive
C. Active filter does not cause loading of source
D. Passive filters are difficult to tune or adjust

Answer: B
Clarification: Typically active filters are more economical than passive filters. This is because of the variety of cheaper op-amp and the absence of inductor’s.

7. What are the most commonly used active filters?
A. All of the mentioned
B. Low pass and High pass filters
C. Band pass and Band reject filters
D. All-pass filters

Answer: A
Clarification: All the mentioned filters use op-amp as active element and capacitors & resistors as passive elements.

8. Choose the op-amp that improves the filter performance.
A. µA741
B. LM318
C. LM101A
D. MC34001

Answer: B
Clarification: LM318 is a high speed op-amp that improves the filter’s performance through increased slew rate and higher unity gain-bandwidth.

9. Ideal response of filter takes place in
A. Pass band and stop band frequency
B. Stop band frequency
C. Pass band frequency
D. None of the mentioned

Answer: C
Clarification: The ideal response indicates the practical filter response and it lies within the pass band frequencies.

10. Find out the low pass filter from the given frequency response characteristics.

Answer: A
Clarification: A low pass filter has a constant gain from 0Hz to high cut-off frequency f_H.

Linear Integrated Circuit Multiple Choice Questions on “Basic DAC Techniques – 1”.

1. Express the output voltage of digital to analog converter?
A. V_o =KV_FS(d₁2^-1+d₂2^-2+….d_n2^-n)
B. V_o =V_FS/k(d₁2^-1+d₂2^-2+….d_n2^-n)
C. V_o =V_FS(d₁2^-1+d₂2^-2+….d_n2^-n)
D. V_o =K(d₁2^-1+d₂2^-2+….d_n2^-n)

Answer: A
Clarification: The input is an n-bit binary word D and is combined with the reference voltage V_R to give on analog output signal. Mathematically it is described as
V_o =KV_FS(d₁2^-1+d₂2^-2+….d_n2^-n) where, K -scaling factor, V_FS-full scale output voltage.

2. Why the switches used in weighted resistor DAC are of single pole double throw (SPDT) type?
A. To connect the resistance to reference voltage
B. To connect the resistance to ground
C. To connect the resistance to either reference voltage or ground
D. To connect the resistance to output

Answer: c
Clarification: SPDT are electronic switches controlled by a binary word. If the binary input to a switch is 1, it connects the resistance to the reference voltage and if the input is 0, the switch connects the resistor to ground.

3. Determine the output current for an n-bit weighted resistor DAC?

A. (V_R/R )× (d_o/2 +d₁/2² + ……d_n/2ⁿ)
B. (V_R/R )× (d₁/2¹ +d₂/2² + ……d_n/2ⁿ)
C. (V_R/R )× (d₀²/2 +d₁²/2² + ……d_n²/2ⁿ)
D. None of the mentioned

Answer: B
Clarification: The output current, I_o= I₁+I₂+….I_n
I_o= (V_R/2R )×(d₁) +(V_R/2²R)× (d₂) ….+(V_R/2ⁿR )×(d_n)
I_o =(V_R/R)× (d₁/2¹ +d₂/2² + ……d_n/2ⁿ).

4. In a D-A converter with binary weighted resistor, a desired step size can be obtained by
A. Selecting proper value of V_FS
B. Selecting proper value of R
C. Selecting proper value of R_F
D. All of the mentioned

Answer: C
Clarification: The size of the steps depends on the value of R_F, provided that the maximum output voltage does not exceed the saturation level of an op-amp.

5. Determine the Full scale output in a 8-bit DAC for 0-15v range?
A. Full scale output=15.1v
B. Full scale output=15.2v
C. Full scale output=14.5v
D. Full scale output=14.94v

Answer: D
Clarification: Full scale output = (Full scale voltage -LSB.
= [15v-(15v/2⁸)] = (15v-0.0586) = 14.94v.

6. Pick out the incorrect statement “In a 3 bit weighted resistor DAC”
A. Although the op-amp is connected in inverting mode, it can also be connected in non-inverting mode
B. The op-amp simply work as a current to voltage converter
C. The polarity of the reference voltage is chosen in accordance with the input voltage
D. None of the mentioned

Answer: C
Clarification: The polarity of the reference voltage is accordance with the type of the switch used. For example, in TTL switches, the reference voltage should be +5v and the output will be negative.

7. What is the disadvantage of binary weighted type DAC?
A. Require wide range of resistors
B. High operating frequency
C. High power consumption
D. Slow switching

Answer: A
Clarification: For better resolution of output, the input binary word length has to be increased. As the number of bit increases, the range of resistance value increases.

8. The smallest resistor in a 12 bit weighted resistor DAC is 2.5kΩ, what will be the largest resistor value?
A. 40.96MΩ
B. 10.24MΩ
C. 61.44 MΩ
D. 18.43MΩ

Answer: B
Clarification: The largest resistor value for 12-bit DAC= 2ⁿ×R = 2¹²×2.5kΩ = 4096×2.5kΩ =10.24MΩ.

9. CMOS inverter is used as SPDT switch in resistor DAC and is connected to the op-amp line. Find the output of CMOS, if the input applied is 1
A. Resistance is connected to ground
B. Resistance is connected to input line
C. Resistance is connected to bit line
D. None of the mentioned

10. How to overcome the limitation of binary weighted resistor type DAC?
A. Using R-2R ladder type DAC
B. Multiplying DACs
C. Using monolithic DAC
D. Using hybrid DAC

Answer: A
Clarification: Usage wide range of resistors is the limitation of binary weighted resistor type DAC, this can be avoided by using R-2R ladder type DAC Where only two value of resistor are required.

11. Find output voltage equation for 3 bit DAC converter with R and 2R resistor?
A. V_o= -R_F [(b₂/8R) +(b₁/4R) +(b₀/2R)].
B. V_o= -R_F [(b₂/R) +(b₁/2R) +(b₀/4R)].
C. V_o= -R_F [(b₂/2R)+(b₁/4R) +(b₀/8R)].
D. V_o= -R_F [(b₀/4R)+(b₁/2R) +(b₂/R)].

Answer: C
Clarification: The output voltage corresponding to all possible combination of binary input in a 3-bit R-2R DAC is given as
V_o=-R_F [(b₂/2R) +(b₁/4R) +(b₀/8R)].

Linear Integrated Circuit Multiple Choice Questions on “Power Amplifiers”.

1. Output current in general purpose op-amp can be increased using
A. Power comparator
B. Power amplifier
C. Power resistor
D. Power booster
Answer: D
Clarification: A simple method of increasing output current of a general purpose op-amp is to connect a power booster in series with the op-amp.

2. Which type of power transistor is chosen for a discrete power booster?
A. Collector follower stage
B. Emitter follower stage
C. Base follower stage
D. None of the mentioned
Answer: B
Clarification: A discrete power booster is formed by an emitter follower stage using a power transistor. Because of unity gain, the emitter follower helps to retain the voltage gain characteristics of the general purpose op-amp.

3. What is the power dissipation of power transistor?
A. ≅ 0.5W
B. ≤ 0.5W
C. > 0.5W
D. ≠ 0.5W
Answer: C
Clarification: Transistors with power dissipation larger than 1/2 (0.5)W are called power transistors.

4. Burr-Brown 3553 power amplifier is suited for
A. Line driving applications
B. Power supply requirement
C. Bandwidth adjustments
D. High frequency application
Answer: A
Clarification: Burr-Brown 3553 power amplifier is suited for line driving applications in which fast pulses or wideband signals are involved.

5. Find the IC that can be used for short circuit proof protection?
A. All of the mentioned
B. IC LM384
C. ICL8063
D. ICLM380
Answer: C
Clarification: The intersil ICL8063 monolithic power transistor driver and amplifier has a built-in safe area protection and short-circuit proof protection.

6. Which among the following is ideal for consumer applications?
A. NE5018
B. LM380
C. MC1408
D. SE5018
Answer: B
Clarification: LM380 is a power audio amplifier designed to deliver a minimum of 2.5w to an 8Ω load and hence is ideal for consumer applications.

7. Bridge power audio amplifier can deliver power upto
A. Twice as much as output of single LM380 amplifier
B. Thrice as much as output of single LM380 amplifier
C. Four times as much as output of single LM380 amplifier
D. Half of the output of single LM380 amplifier
Answer: C
Clarification: Two LM380s are used in the bridge configuration.Therefore, it produce a maximum output voltage swing which is equal to twice that of a single LM380 amplifier.

8.

What precautionary measure should be taken if the circuit is used in a RF-sensitive environment?
A. Connect a parallel combination of RC at the output
B. Connect a series combination of Resistors at the output
C. Connect parallel capacitor at the output
D. Connect a series combination of RC at the output
Answer: D
Clarification: When LM380 audio power amplifier is used in an RF-sensitive environment, an RC combination should be added at the output terminal (Pin 8) to eliminate 5 to 10-MHz oscillation.

9. Which amplifier provides twice output swing as that of LM380 amplifier?
A. Hybrid power amplifier
B. Bridge power audio amplifier
C. Monolithic power audio amplifier
D. Dual power amplifier
Answer: B
Clarification: In bridge power audio amplifier, two LM380s are used in bride configuration. Therefore, the maximum output voltage swing will be twice that of a single LM380 amplifier.

10. A LM380 power amplifier is used in a intercom system with amplifier gain = 50 and the transformer turns ratio is given as 35.Find the overall gain of the circuit.
A. 1880
B. 1750
C. 1370
D. 1580
Answer: B
Clarification: Given, N₁/N₂=35; Gain of amplifier = 50.Therefore, the overall gain of the circuit = 50×35 =1750.

11. Determine the work done by the intercom system depending on the position of the switch

A. Remote speaker act as microphone
B. Master speaker act as microphone
C. Remote and master speaker act as microphone
D. None of the mentioned
Answer: A
Clarification: When the switch of the intercom system is in the listen mode, the remote speaker acts as the microphone.

.