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৪৯তম বিসিএস ⎯ তথ্য ও যোগাযোগ প্রযুক্তি (EEE) [ ৮৯২]

পরীক্ষা৪৯তম বিসিএস ⎯ তথ্য ও যোগাযোগ প্রযুক্তি (EEE) [ ৮৯২]তারিখতারিখ অনির্ধারিতসময়35 minutes
মোট প্রশ্ন৫০
সিলেবাস
Exam 12 i) Analog Electronics: Field Effect Transistor (FET); Feedback Amplifiers ii) Operational Amplifiers (Op-Amp) iii)Oscillators [Source: Classes 9–10 and relevant books]
ঘনত্ব
উত্তর
উত্তরিতবর্তমানপুনরায় দেখুনঅসম্পূর্ণ

৪৯তম বিসিএস ⎯ তথ্য ও যোগাযোগ প্রযুক্তি (EEE) [ ৮৯২]

৪৯তম বিসিএস ⎯ তথ্য ও যোগাযোগ প্রযুক্তি (EEE) [ ৮৯২] · তারিখ অনির্ধারিত · ৫০ প্রশ্ন

.
A 741 op amp has an open-loop voltage gain of 2 × 105, input resistance of 2 ΩM , and output resistance of 50Ω . The op amp is used in the circuit of Figure. Find the closed-loop gain vo/vs

  1. -1.999
  2. -2.89998
  3. 0.9999
  4. 1.9999
ব্যাখ্যা

.
Refer to the op amp in Figure If vi 0.5 V, calculate the output voltage vo

  1. -1.55 V
  2. -1.5 V
  3. -1.25 V
  4. -2.25 V
ব্যাখ্যা

vₒ / vᵢ = - Rf / R1 = - 25 / 10 = - 2.5

vₒ = -2.5 vᵢ = -2.5 (0.5) = -1.25 V

.
Calculate vo in the op amp circuit in shown in figure

  1. 10V
  2. 8V
  3. -10 V
  4. -8 V
ব্যাখ্যা

This is a summer with two inputs. 

vₒ = - [10 / 5 * (2) + 10 / 2.5 * (1)] = - (4 + 4) = -8 V

.
An instrumentation amplifier shown in Figure is an amplifier of low level signals used in process control or measurement applications and commercially available in single-package units. Determine vo

  1. R2/R1{1+(2R3/R4 )}(v2-v1)
  2. R1/R2{1+(2R3/R4 )}(v2-v1)
  3. R2/R1{1+(2R3/R4 )}(v1-v2)
  4. R1/R2{1+(2R3/R4 )}
ব্যাখ্যা

We recognize that the amplifier A3 in Fig  is a difference amplifier.
Thus, from Eq. (5.20),

    vo = R2 / R1 (vo2 - vo1) ------ 1    

Since the op amps A1 and A2 draw no current, current i flows through the three resistors as though they were in series. Hence, 

    vo1 - vo2 = i(R3 + R4 + R2) = i(2R3 + R4) ------ 2

But

    i = (va - vb) / R4
    and va = v1, vb = v2. Therefore,

    i = (v1 - v2) / R4 ------- 3

Inserting Eqs. (2) and (3) into Eq. (1) gives

    vo = R2 / R1 ( 1 + 2R3 / R4 ) (v2 - v1)

.
find v0  in the circuit of the figure

  1. 350mV
  2. 500mV
  3. 25mV
  4. 100nV
ব্যাখ্যা

This circuit consists of two noninverting amplifiers cascaded. At the output of the first op amp, 
v_{a} = (1 + 12/3)(20) = 100mV

At the output of the second op amp, v_{o} = (1 + 10/4) * v_{a} = (1 + 2.5) * 100 = 350mV

.
In Figure , let R 10 k , v1 =2.011 V, and v2=2.017 V. If RG is adjusted to 500 ohm, determine the output voltage

  1. 39mV
  2. 41mV
  3. 246mV
  4. 259mV
ব্যাখ্যা

The voltage gain is
A_{v} = 1 + (2R)/R_{G} = 1 + (2 * 10000)/500 = 41
The output voltage is
v_{o} = A_{v}(v_{2} - v_{1}) = 41(2.017 - 2.011) = 41(6) * mV = 246mV

.
The two input terminals of an op amp are labeled as:
  1. high and low.
  2.  positive and negative.
  3. inverting and noninverting.
  4.  differential and nondifferential.
ব্যাখ্যা

Operational amplifiers (op amps) are fundamental components in electronics, widely used for amplifying electrical signals. These amplifiers have two input terminals that play a critical role in determining the behavior and function of the op amp in various applications. The correct labeling of these terminals is crucial for understanding the operation of the op amp and ensuring its proper application in circuits.

The two input terminals of an op amp are labeled as inverting and noninverting. The inverting terminal is the one where the input signal is applied in such a way that the output signal becomes inverted, or 180 degrees out of phase with the input. The signal applied at this terminal is subtracted from the op amp’s output, and the degree of inversion depends on the gain determined by external components like resistors.

On the other hand, the noninverting terminal allows the input signal to pass through without any phase inversion. When the signal is applied to this terminal, the output signal is in phase with the input, meaning it maintains the same polarity. The behavior at the noninverting terminal is essential in configurations where the goal is to amplify the input signal without altering its phase.

The labeling as "inverting" and "noninverting" provides clear distinctions in how the signal will behave based on which input terminal is used. This setup is critical in circuit design, as it affects the feedback loop configuration, signal gain, and the overall functionality of the circuit.

Other potential labels such as “high” and “low,” “positive” and “negative,” or “differential” and “nondifferential” are not accurate in describing the roles of the op amp’s input terminals. These terms do not reflect the specific operational principles of how the signals interact with the two input terminals. Understanding these labels—inverting and noninverting—is crucial for anyone designing or analyzing op amp circuits.


So, The two input terminals of an operational amplifier (op amp) are typically labeled as the inverting terminal (marked as "-") and the noninverting terminal (marked as "+").

.
For an ideal op amp, which of the following statements are not true?
  1.  The differential voltage across the input terminals is zero.
  2.  The current into the input terminals is zero.
  3. The input resistance is zero.
  4. The output resistance is not zero.
ব্যাখ্যা

The current from the output terminal is zero.: Not true. The output terminal of an ideal op amp does supply current to drive the load, depending on the output voltage and the load resistance. It is not zero. The output current is determined by the load connected to the output.

.
Which of these amplifiers is used in a digital-to-analog converter?
  1. non inverter
  2. voltage follower
  3. summer
  4. difference amplifier
ব্যাখ্যা

A summer (also known as a summing amplifier) is used in digital-to-analog converters (DACs). A DAC converts a digital signal (composed of binary values) into an analog signal, and a summer is typically used to sum the weighted contributions of each binary input in the circuit. In such applications, the input signals are usually weighted according to their binary values, and the summer adds them together to produce the corresponding analog output.

Here’s why the other options are not used for this purpose:

(ক) Noninverter: A noninverting amplifier is used to amplify a signal without inverting its polarity. While it may be used in other types of circuits, it is not typically used in DACs.
(খ) Voltage follower: A voltage follower (also known as a buffer) is used to provide high input impedance and low output impedance, typically used for impedance matching rather than conversion tasks.
(ঘ) Difference amplifier: A difference amplifier is used to amplify the difference between two input signals. It is not used in DACs for signal conversion.

১০.
An ideal op-amp requires infinite bandwidth because
  1. Signals can be amplified without attenuation
  2. Output common-mode noise voltage is zero
  3. Output voltage occurs simultaneously with input voltage changes
  4. Output can drive infinite number of device
ব্যাখ্যা

 An ideal op-amp has infinite bandwidth. Therefore, any frequency signal from 0 to ∞ Hz can be amplified without attenuation।

১১.
Determine the output voltage from the following circuit diagram?
 
  1. None of the mentions
ব্যাখ্যা

 In an ideal op-amp when the inverting terminal is zero. The output will be in-phase with the input signal.

১২.
How will be the output voltage obtained for an ideal op-amp?
  1. Amplifies the difference between the two input voltages
  2. Amplifies individual voltages input voltages
  3. Amplifies products of two input voltage
  4. None of the mentioned
ব্যাখ্যা

Op-amp amplifies the difference between two input voltages and the polarity of the output voltage depends on the polarity of the difference voltage.

১৩.
Which of the following is true about the frequency response of a low-pass filter using an op-amp?
  1. It allows low-frequency signals to pass and blocks high-frequency signals.
  2. It allows high-frequency signals to pass and blocks low-frequency signals.
  3. It passes all frequencies with equal gain.
  4. It is used for amplifying high-frequency signals.
ব্যাখ্যা

A low-pass filter allows signals with frequencies lower than the cutoff frequency to pass through while attenuating signals with frequencies higher than the cutoff frequency.

১৪.
Which of the following is a key characteristic of a band-pass filter?
  1. It only passes a narrow range of frequencies.
  2. It passes all frequencies above a certain cutoff.
  3. It passes all frequencies below a certain cutoff.
  4. It amplifies all frequencies equally.
ব্যাখ্যা

A band-pass filter allows signals within a specific frequency range to pass through, blocking signals both below and above this range. The filter has a "passband" centered around a specific frequency, and the bandwidth determines how wide this range is.

১৫.
In a second-order high-pass filter with an op-amp, what is the impact of increasing the quality factor (Q)?
  1. The bandwidth of the filter increases.
  2. The bandwidth of the filter decreases.
  3. The gain of the filter increases.
  4. The filter becomes a low-pass filter.
ব্যাখ্যা

In a second-order high-pass filter, the quality factor (Q) determines the sharpness of the filter's frequency response. A higher Q results in a narrower bandwidth, meaning the filter becomes more selective in passing frequencies around the cutoff point. As Q increases, the bandwidth of the filter decreases, and the filter's resonance becomes sharper. Conversely, a lower Q increases the bandwidth, making the filter more wideband.

So, increasing the Q value results in a decrease in the bandwidth of the filter.

১৬.
Which of the following is the primary characteristic of a second-order filter?
  1. It has a linear phase response.
  2. It can be designed to achieve a specific resonant peak.
  3. It allows all frequencies to pass through equally.
  4. It does not exhibit a cutoff frequency.
ব্যাখ্যা

A second-order filter is a type of filter characterized by its response to frequencies, which is influenced by the nature of its transfer function. These filters are widely used in various applications such as signal processing, audio systems, and communications, where precise frequency selection is required.

One of the primary characteristics of a second-order filter is that it can be designed to exhibit a specific resonant peak at its cutoff frequency. This is particularly important for band-pass filters, where the filter allows signals within a narrow frequency band to pass and attenuates signals outside this range. The resonant peak refers to the sharp increase in gain around the filter's central frequency, where the filter is most effective in passing signals. This sharp peak is a distinguishing feature of second-order filters and makes them valuable in applications like tuning, signal analysis, and audio equalization, where emphasizing a specific frequency band is needed.

The quality factor (Q) of the filter plays a crucial role in determining the magnitude of the resonant peak. A high Q factor results in a narrow, sharp peak, while a low Q factor produces a broader, less pronounced peak. By adjusting the Q factor and the cutoff frequency, second-order filters can be tailored for specific applications, providing fine control over the frequency response.

১৭.
What is the primary difference between JFET and MOSFET?
  1. JFET uses a gate that is insulated from the channel by a dielectric material, while MOSFET uses a metal gate.
  2. JFET is normally on, whereas MOSFET is normally off.
  3. JFET requires a positive voltage to turn on, while MOSFET requires a negative voltage.
  4. MOSFET operates on a depletion mode, while JFET operates on an enhancement mode.
ব্যাখ্যা

The primary distinction between a JFET and a MOSFET lies in their gate structure and how they operate. In JFET, the gate is directly connected to the semiconductor, and the device is normally on when no voltage is applied to the gate, as the gate-source junction is forward biased in most conditions. In contrast, a MOSFET (especially in the enhancement mode) is normally off when no voltage is applied to the gate. The gate in a MOSFET is isolated from the channel by a dielectric material, which makes it more sensitive to applied voltages than the JFET.

১৮.
In a JFET, what happens when the gate-source voltage (Vgs) becomes more negative?
  1. The channel conducts more current.
  2. The channel resistance decreases.
  3. The channel is pinched off, and current conduction is reduced.
  4. The channel becomes more conductive.
ব্যাখ্যা

In a JFET, when a negative voltage is applied to the gate relative to the source (i.e., making Vgs more negative), it creates an electric field that narrows the conducting channel. As the negative gate-source voltage increases, the channel becomes pinched off, which means the current conduction is reduced or eventually stopped. The current is only conducted through the narrow region near the drain, and the behavior becomes saturated once pinching off occurs, significantly reducing the flow of current.

১৯.
What is the key feature of MOSFET construction that differentiates it from JFET?
  1. MOSFET uses an insulated gate, whereas JFET uses a conductive gate.
  2. MOSFET has a longer channel than JFET.
  3. MOSFET is made of a single layer of semiconductors, while JFET is made of multiple layers.
  4. MOSFET does not have a drain, unlike JFET.
ব্যাখ্যা

The most significant construction difference between MOSFET and JFET is the gate. In MOSFET, the gate is insulated from the semiconductor channel by a dielectric material (often silicon dioxide). This allows for better control and less leakage current. In contrast, in a JFET, the gate is not insulated but is directly connected to the semiconductor channel, which limits the ability to control the current via the gate voltage.

২০.
Which of the following is true for the h-parameters of a JFET?
  1.  The h-parameters are used to model the behavior of a MOSFET and cannot be applied to JFETs.
  2. The h-parameters for JFETs involve parameters like input impedance and output conductance.
  3.  h-parameters are primarily used to model JFET in a voltage-controlled manner only.
  4. The h-parameters for JFETs are fixed and do not vary with changes in biasing.
ব্যাখ্যা

The h-parameters are a set of four parameters used to model the behavior of JFETs and other transistors. These parameters include:

h11: Input impedance (related to the gate-source voltage).
h12: Reverse voltage gain (related to the interaction between output and input).
h21: Forward current gain (showing the ability to control output current with input voltage).
h22: Output admittance (related to the current flowing through the drain terminal).
These parameters help describe the small-signal characteristics of the transistor, which vary based on the biasing conditions.

২১.
In a MOSFET, what is the role of the oxide layer between the gate and the channel?
  1. It ensures that the gate is electrically connected to the channel.
  2. It controls the threshold voltage of the MOSFET.
  3.  It allows current to flow freely between the gate and channel.
  4. It isolates the gate from the channel to prevent current flow between them.
ব্যাখ্যা

The oxide layer (usually made of silicon dioxide) in a MOSFET is an insulating layer between the gate and the semiconductor channel. This insulation ensures that no direct current can flow from the gate to the channel, which is a key feature of MOSFETs. This isolation allows the MOSFET to have a high input impedance, meaning that very little current is required to control the transistor's operation. The oxide layer enables the electric field from the gate to modulate the conductivity of the channel without a direct electrical connection, making the MOSFET highly efficient for switching applications.

২২.
For a FET when will maximum current flows?
  1. Vgs = 0V
  2. Vgs = 0v and Vds >= |Vp|
  3. VDS >= |Vp|
  4. Vp = 0
ব্যাখ্যা

Explanation: For a FET the current reaches maximum that is IDSS occurs when Vgs = 0V and VDS >= |Vp|

২৩.
The threshold voltage of an n-channel enhancement mode MOSFET is 0.5 V. When the device is biased at a gate voltage of 3 V, pinch-off would occur at a drain voltage of:
  1. 2 V
  2. 2.5 V
  3. 3 V
  4. 1.5 V
ব্যাখ্যা

Pinch-off voltage
The voltage at which the channel of the MOSFET closes is called the "pinch-off voltage".
When V DS is increased beyond the pinch-off voltage, V GS controls the channel current and V DS has little or no effect i.e. remains constant. 

From the above characteristic of the MOSFET, the pinch-off voltage is given by:
V DS =V p =V GS -V T
where, V DS = Drain to source voltage
V GS = Gate to source voltage
VT = Threshold voltage
Calculation
Given, V GS =3 V
VT = 0.5V
The pinch-off voltage is:
V DS =V p =V GS -V T
V DS =V p =3-0.5
V DS =V p =2.5 V

২৪.
For an n-channel E-MOSFET Vth = 5V, what is the condition to turn ON the device?
  1. VDS > 5V
  2. VGS < 5V
  3. VGS > 5V
  4. VDS = 5V
ব্যাখ্যা

Cutoff region
VGS < Vth
I_{D} = 0
Active/ linear/ Ohmic/ Triode region:
VGS > Vth
VDS  GS-Vth
Saturation region:
VGS > Vth
VDS > VGS-Vth
Where
VG = Gate voltage
VD drain voltage
Vs = Source voltage
VGS = Gate to source voltage.
VDS = Drain to source voltage.
Vth = Threshold voltage.
ID = Drain current.
Application:
The condition to turn ON the device is:
VGS > Vth
VGS > 5V

২৫.
The transconductance of an n-channel mosfet in the linear region is:
  1. {(µnCoxW/L} (Vgs-Vth
  2. {(µnCoxW)/L} Vgs
  3. {(µnCoxW)/L} Vds
  4. {(µnCoxW)/L} Vth
ব্যাখ্যা

২৬.
Which of the following is true for the triode region?

  1.  VDG > Vtp
  2. VSD < VOV
  3. ID ∝ VOV
  4. None of the mentioned
ব্যাখ্যা

VDG > |Vtp|: In the triode region, the drain-to-gate voltage (VDG) must be greater than the absolute value of the threshold voltage (|Vtp|) for an n-channel MOSFET to operate. This condition ensures that the MOSFET is turned on.
VSD < |VOV|: In the triode region, the source-to-drain voltage (VSD) is less than the absolute value of the overdrive voltage (|VOV|). VOV is the difference between the gate-to-source voltage (VGS) and the threshold voltage (Vth), and this condition is essential for the MOSFET to operate in the triode region, where it behaves like a resistor.

২৭.
Determine the VGSQ for the network of  the figure where ID = 4 mA

  1. -3V
  2. -4V
  3. 4V
  4. 0.4V
ব্যাখ্যা

The gate-to-source voltage is determined by
VGS =-IDRS    

Choosing  ID = 4 mA,    
we obtain
VGS = -(4 mA)(1 k ) = -4 V

২৮.
Determine the Ifor the network of the figure

  1. 0.21mA
  2. 1.41mA
  3. 1.32mA
  4. 1.21mA
ব্যাখ্যা

For the given transfer characteristics, the equation is as follows:
    
- ID = IDSS /4 = 8/4 =2mA: , we get VGS = -4 V/2 = -2 V respectively.
    
- The Shockley's equation for the resulting curve appears in Fig. 7.22.
    
The network equation is defined as:
    
V_G = (R2 * VDD) / (R1 + R2)
    
Given values:
- R2 = 270 kΩ
- VDD = 16 V
- R1 = 2.1 MΩ
- R2 = 0.27 MΩ
   
Solving for V_G:
V_G = (270 k Ω* 16 V) / (2.1 MΩ + 0.27 MΩ) = 1.82 V
    
Calculation for VGS and ID:
The equation for VGS is defined by:
V_GS = V_G - I_D * R_S
    
Substituting the known values:
V_GS = 1.82 V - I_D (1.5 kΩ)
    
When I_D = 0 mA, and V_GS = 1.82 V:
    
When V_GS =0       and I_D =1.82/1.5 =1.21mA

২৯.
Determine ID from the figure

  1. -22.95mA
  2. -21.95mA
  3. 21.95mA
  4. 22.95mA
ব্যাখ্যা

Given Equations:
    
VG = (R2 * VDD) / (R1 + R2)
Substituting the given values:
VG = (18 MΩ * 40 V) / (22 MΩ + 18 MΩ) = 18 V
    
Eq. (7.40):
    
VGS = VG - ID * RS
Substituting the given values:
VGS = 18 V - ID * (0.82 kΩ)

    
When ID = 0 mA:
VGS = 18 V (as appearing on Fig.)
   
When VGS = 0 V:
VGS = 18 V - ID * (0.82 kΩ) (as appearing on Fig.)
    
Finding ID when VGS = 18 V:
ID = 18 V / (0.82 kΩ) = 21.95 mA
    
Final equation:
0 = 18 V - ID * (0.82 kΩ)

Final Answer
ID = 21.95mA

৩০.
In CMOS circuits, what is the typical state of the output when both PMOS and NMOS transistors are in the "on" state?
  1. High voltage
  2. Low voltage
  3. No voltage
  4. Both high and low voltages
ব্যাখ্যা

To understand this, let's consider how CMOS logic gates work. CMOS technology uses complementary pairs of PMOS (P-type Metal-Oxide-Semiconductor) and NMOS (N-type Metal-Oxide-Semiconductor) transistors. When these transistors are in a CMOS configuration, they work in complementary fashion: one transistor is on while the other is off, depending on the input voltage.

PMOS Transistor: A PMOS transistor turns on when the gate voltage is lower than the source voltage, and it conducts current from the source to the drain. This transistor pulls the output to a high voltage when it is "on."
NMOS Transistor: An NMOS transistor turns on when the gate voltage is higher than the source voltage, and it conducts current from the drain to the source. This transistor pulls the output to a low voltage when it is "on."
In a CMOS circuit, if both PMOS and NMOS transistors are on simultaneously, it creates a direct path between the supply voltage (VDD) and ground. This short circuit causes the output to be pulled to a low voltage. Therefore, the correct state is low voltage, which is typically considered a logic 0. This is an undesirable condition in CMOS circuits and must be avoided to prevent excessive current flow and power dissipation.

৩১.
Consider a process technology for which L min =0.4 mu m, tox =8 nm, µn = 450cm2/Vs , and Vt = 0.7V
 For a MOSFET with W/L=8µm/0.8µm calculate the values of V GS needed to operate the transistor in the saturation region with a dc current ID = 100µA 
  1. -1.22 V
  2. 1.22 V
  3. 1.02 V
  4. 2.01 V
ব্যাখ্যা

The oxide capacitance per unit area, Cox, is calculated using the formula:
    
    Cox = (εox) / (tox)
    

    The permittivity of the oxide, ε_ox, is 3.45×10⁻¹¹ F/m and the oxide thickness, t_ox, is 8×10⁻⁹ m.
    
    Cox = (3.45×10⁻¹¹) / (8×10⁻⁹) = 4.32×10⁻³ F/m²
    Cox = 4.32 fF/µm²
    
    The oxide capacitance per unit area is 4.32×10⁻³ F/m² or 4.32 fF/µm². The transconductance parameter, k', is calculated using the formula:
    
    k' n= μn × Cox
    

    The electron mobility, μn, is 450 cm²/V·s and Cox is 4.32 fF/µm².
        k' n= 450×10⁸ m²/V·s (since 1 cm² = 10⁸ m²) *4.32(fF/µm2)
  k' n= 194 A/V² = 194 µA/V²
    

The drain current (iD) in the saturation region for a MOSFET is given by the equation:

    iD = (1/2) * k'n * (W/L) * (VGS - Vt

    Given values:
    - iD = 100 A
    - k'n = 194 A/V²
    - W/L = 8/0.8 = 10
    - Vt = 0.7 V

   
    100 = (1/2) * 194 * 10 * (V_GS - 0.7)²
    100 = 970 * (V_GS - 0.7)²
    (VGS - 0.7)² = 100 / 970 ≈ 0.103
    VGS - 0.7 = √(0.103) ≈ 0.32 V


    VGS = 0.32 + 0.7 = 1.02 V

    

৩২.
An oscillator converts .............
  1. a.c. power into d.c. power
  2. d.c. power into a.c. power
  3. mechanical power into a.c. power
  4. none of the above
ব্যাখ্যা

An oscillator is a device that converts direct current (D.C.) power into alternating current (A.C.) power. This is because oscillators generate a periodic, oscillating signal, typically a sine wave or square wave, which alternates between positive and negative values, thereby producing A.C. power.

Working Principle of Oscillators
Oscillators operate on the principle of feedback, where part of the output is fed back into the input to maintain continuous operation. The feedback loop typically involves a combination of electronic components, such as capacitors, inductors, resistors, and amplifiers, to create the oscillations. The key function of an oscillator is to produce a continuous output signal without the need for an external input signal.

The process starts with the D.C. power source that provides a stable voltage to the oscillator circuit. The circuit then converts this steady D.C. voltage into an oscillating waveform (A.C.). In an electronic oscillator, the D.C. power is supplied to components like transistors or operational amplifiers that generate periodic waveforms. These waveforms can take the form of sine waves, square waves, or triangle waves, depending on the oscillator design.

Types of Oscillators
There are several types of oscillators based on their function and output characteristics:

Sinusoidal Oscillators: These generate smooth sine waves and are typically used in applications where a pure A.C. signal is needed, such as in radio transmitters or audio generators.
Square Wave Oscillators: These generate square waves and are used in digital circuits, clock generation, and pulse-width modulation systems.
Triangle Wave Oscillators: These are used for signal generation in waveform generators and other electronic applications.

Applications of Oscillators

Oscillators have broad applications in electronics. They are used in radio-frequency (RF) communication, signal processing, timing devices like clocks and watches, and even in power supplies where A.C. is required. For example, in power inverters, oscillators are used to convert the low-frequency D.C. input from a battery or solar panel into high-frequency A.C. that can be used by electronic devices.

৩৩.
In an LC oscillator, the frequency of oscillator is L or C.
  1. proportional to square of
  2. directly proportional to
  3. independent of the values of
  4. inversely proportional to square root of
ব্যাখ্যা

In an LC oscillator, the frequency of oscillation is inversely proportional to the square root of the product of inductance (L) and capacitance (C). This relationship can be expressed by the following formula:

f=1/{2*pi*√(LC)}

f is the frequency of oscillation,
Lis the inductance,
C is the capacitance.

In an LC oscillator, the frequency depends on both the inductance (L) and capacitance (C) of the circuit. The formula shows that the frequency is determined by the square root of the product of these two components.

Inductance (L) represents the ability of a coil or inductor to store energy in a magnetic field.
Capacitance (C) represents the ability of a capacitor to store energy in an electric field.
The interaction between the magnetic field (from the inductance) and the electric field (from the capacitance) results in the oscillations. The energy stored in the capacitor is periodically transferred to the inductor, and vice versa, creating a continuous oscillation.

Why is it Inversely Proportional?
If the inductance (L) increases, the frequency of the oscillator decreases because the system takes longer to transfer energy between the capacitor and inductor, leading to slower oscillations.
Similarly, if the capacitance (C) increases, the frequency also decreases because a larger capacitor stores more energy, again slowing down the rate of oscillation.
Therefore, the frequency is inversely proportional to the square root of the product of L and C. As either L or C increases, the frequency of the oscillator decreases, and as L or C decreases, the frequency increases.( microelectronic circuit by sedra/ sm ith) 

৩৪.
An LC oscillator cannot be used to produce frequencies.
  1. high
  2. microwave
  3. very low
  4. very high
ব্যাখ্যা

In an LC oscillator, the frequency of oscillation is inversely proportional to the square root of the product of inductance (L) and capacitance (C). This relationship can be expressed by the following formula:

f=1/{2*pi*√(LC)}

f is the frequency of oscillation,
Lis the inductance,
C is the capacitance.


While LC oscillators are effective in generating a range of frequencies, they have limitations when it comes to very low frequencies (VLF). This is primarily due to the following reasons:

Physical Size of Inductors and Capacitors:

At very low frequencies, the values of inductance (L) and capacitance (C) required to produce such frequencies become impractically large. For example, to generate frequencies in the VLF range (tens to hundreds of Hz), the inductance values would need to be extremely high, and the capacitance values would need to be very large. These components would become physically large and inefficient for practical use.
Large inductors would require large coils, which are cumbersome and difficult to implement in compact, practical circuits.

Energy Losses:

At low frequencies, LC circuits are more susceptible to energy losses due to the resistance in the inductor and the leakage current in the capacitor. These losses affect the efficiency of the oscillator, making it difficult to sustain oscillations at very low frequencies.
The damping effect increases as the frequency decreases, which can suppress the oscillations.

Resonance:

LC circuits work by resonating at specific frequencies, and as the frequency decreases, the circuit's ability to resonate efficiently diminishes. Very low frequencies might not efficiently produce the sustained oscillations needed for practical applications.

৩৫.
Hartiey oscillator is commonly used in .........
  1. radio receivers
  2. radio transmitters
  3. TV receivers
  4. none of the above
ব্যাখ্যা

The Hartley oscillator is a type of electronic oscillator used to generate high-frequency signals, typically in the radio frequency (RF) range. It is commonly used in radio receivers due to its ability to provide stable frequency oscillations.

Working Principle:
The Hartley oscillator consists of an LC circuit (inductor and capacitor), where the inductor is usually split into two parts and the capacitor is placed between them. This configuration allows the oscillator to produce an output signal at a specific frequency, which can be adjusted by varying the values of the inductor and capacitor.

The oscillator works on the principle of feedback, where a portion of the output signal is fed back into the input to maintain continuous oscillation. The frequency of oscillation is determined by the values of the inductance (L) and capacitance (C), with the formula:
f=1/{2*pi*√(LC)}

Applications in Radio Receivers:
In radio receivers, the Hartley oscillator is used as a local oscillator to produce a stable frequency that is mixed with incoming radio signals. This mixing process allows the receiver to tune to different frequencies (radio stations) by changing the local oscillator's frequency.

The Hartley oscillator is preferred in such applications because it can generate frequencies over a wide range, from medium frequencies (MF) to high frequencies (HF).
It is also favored for its simplicity, stability, and the ease with which it can be adjusted to different frequencies, making it ideal for use in tuning circuits in radio receivers.
Why Not Radio Transmitters or TV Receivers?
While the Hartley oscillator could theoretically be used in radio transmitters, other types of oscillators (such as the Colpitts or crystal oscillators) are often preferred due to their superior frequency stability and power handling capabilities. Similarly, for TV receivers, oscillators with different configurations (such as phase-locked loops) are more commonly used for the required signal processing.

৩৬.
In a phase-shift oscillator, the frequency determining elements are ......
  1. L and C
  2. R, L and C
  3. R and C
  4. none of the above.
ব্যাখ্যা

In a phase-shift oscillator, the frequency of oscillation is primarily determined by resistors (R) and capacitors (C). This type of oscillator uses a feedback network made up of three RC stages that provide the necessary phase shift to produce sustained oscillations.

Working Principle of Phase-Shift Oscillator:
The phase-shift oscillator typically consists of:

An amplifying stage (usually an op-amp or transistor).
A feedback network made up of three RC (resistor-capacitor) stages.
The role of the RC network is to create a total phase shift of 180°, while the amplifying stage provides an additional 180° phase shift. For sustained oscillations, the total phase shift around the loop must be 360° (or 0°, effectively). Therefore, the phase-shift network must contribute 180° of phase shift, and the amplifier provides the other 180°, completing the loop for oscillation.

The frequency of oscillation in a phase-shift oscillator depends on the values of the resistors (R) and capacitors (C) in the feedback network. The frequency is given by the formula:

f = 1/{2*pi*√(LC)}

f is the frequency of oscillation,
Lis the inductance,
C is the capacitance.

Why Not R, L, and C?
L (inductor) is not used in a phase-shift oscillator. The primary frequency-determining components in this oscillator are the resistors (R) and capacitors (C) in the RC network.
R and C are sufficient for generating the oscillations in a phase-shift oscillator, as they create the necessary phase shift for oscillation.

৩৭.
A phase shift oscillator uses. ........................... feedback.
  1. negative
  2. positive
  3. both positive and negative
  4. none of the above
ব্যাখ্যা

A phase-shift oscillator uses negative feedback to produce sustained oscillations. Negative feedback is essential for stabilizing the amplitude of the oscillations and ensuring that the system doesn't experience uncontrolled growth or distortion.

Why Negative Feedback?
Negative feedback reduces the overall gain of the system, which helps maintain the oscillation at a steady amplitude. Without negative feedback, the amplifier would amplify the signal uncontrollably, leading to distortion or saturation.
In a phase-shift oscillator, the feedback network (composed of resistors and capacitors in three stages) introduces a phase shift of 180° (or π radians), and the amplifier stage adds another 180° of phase shift. The total phase shift becomes 360° (or 0°), satisfying the Barkhausen criterion for sustained oscillations.
The negative feedback keeps the loop gain stable and allows the oscillator to start and maintain oscillations without producing excessive amplitude that could distort the waveform.

৩৮.
A tungsten lamp is used in Wien bridge oscillator to control ............ of the oscillator.
  1. operating frequency
  2. gain and operating frequency
  3. output
  4. none of the above
ব্যাখ্যা

In a Wien Bridge oscillator, a tungsten lamp is often used in the feedback loop to control the amplitude of the output signal, not the frequency or gain directly. The lamp acts as a variable resistor, and its resistance changes with temperature.

Working of the Tungsten Lamp in the Wien Bridge Oscillator:
The Wien Bridge oscillator is a type of RC oscillator that uses a Wien Bridge network for frequency determination and oscillation.
The frequency of the oscillation in a Wien Bridge oscillator is determined by the values of the resistors and capacitors in the bridge network, which are typically fixed and not controlled by the tungsten lamp.
The tungsten lamp is used to stabilize the amplitude of the oscillations. As the oscillator starts, the amplitude of the signal may initially grow uncontrollably. The tungsten lamp, placed in the feedback network, has a positive temperature coefficient, meaning its resistance increases as it heats up. As the lamp heats up due to the power dissipated in it, its resistance increases, which reduces the overall gain of the oscillator, stabilizing the amplitude of the output signal.

৩৯.
The material with the piezoelectric effect is
  1. quartz
  2. tourmaline
  3. Rochelle salt
  4. all of above
ব্যাখ্যা

The piezoelectric effect refers to the ability of certain materials to generate an electrical charge in response to mechanical stress. This effect is exhibited by several materials, and in this case, quartz, tourmaline, and Rochelle salt are all materials that show the piezoelectric effect.

Quartz: Quartz is one of the most well-known materials with piezoelectric properties. It is commonly used in various applications, including crystals for oscillators, sensors, and frequency control devices.
It generates an electrical charge when subjected to mechanical stress, making it ideal for use in devices like piezoelectric sensors.

Tourmaline:
Tourmaline is another crystal that exhibits piezoelectric properties. It is often used in pressure sensors and electronic devices that require the conversion of mechanical energy into electrical energy.

Rochelle Salt:
Rochelle salt (sodium potassium tartrate) is a salt that also exhibits piezoelectric properties. It was historically used in early piezoelectric devices due to its strong piezoelectric response.

৪০.
The fp and fs of a crystal are 412 kHz and 411 kHz respectively. What will be the frequency of oscillations when the crystal is used in an oscillator?
  1. between 411 kHz and 412 kHz
  2. above 412 kHz
  3. below 411 kHz
  4. none of the above
ব্যাখ্যা

When a crystal is used in an oscillator, the frequency of oscillation is determined by the crystal’s series resonant frequency (fs) and parallel resonant frequency (fp). These two frequencies represent different characteristics of the crystal's behavior in the circuit:

fp (parallel resonant frequency): This is the frequency at which the crystal behaves like a parallel resonant circuit, where the impedance is minimum.
fs (series resonant frequency): This is the frequency at which the crystal behaves like a series resonant circuit, where the impedance is maximum.
When the crystal is used in an oscillator circuit, the frequency of oscillation will typically lie between the series resonant frequency (fs) and parallel resonant frequency (fp). This is because the oscillator typically operates in the region between the two resonant frequencies, where the circuit is most stable and produces a reliable oscillation.

Since the fp of the crystal is 412 kHz and the fs is 411 kHz, the frequency of oscillation will be between 411 kHz and 412 kHz.

৪১.
The kind of oscillator found in an electronic wrist watch is
  1. Hartley oscillator
  2. Colpitt's oscillator
  3. phase-shift oscillator
  4. crystal oscillator
ব্যাখ্যা

In an electronic wristwatch, the frequency of oscillation needs to be highly stable and precise to keep accurate time. This is typically achieved using a crystal oscillator. Here's why:

Crystal Oscillator: Crystal oscillators use piezoelectric crystals (usually quartz crystals) to generate precise oscillations at a specific frequency.The crystal's resonant frequency is determined by its physical dimensions and material properties, making crystal oscillators highly stable and accurate. In wristwatches, the crystal oscillator typically operates at a frequency of 32.768 kHz, which is ideal for dividing down to a 1-second time interval.

Other Oscillators:

Hartley and Colpitt's oscillators are used in other applications like radio-frequency (RF) circuits but are not as commonly found in low-power devices like wristwatches.
Phase-shift oscillators are used for generating low-frequency signals, but crystal oscillators are more suitable for precise and stable frequency generation needed in timekeeping applications.

৪২.
When negative voltage feedback is applied to an amplifier, its voltage gain......
  1. is increased
  2. is reduced
  3. remains the same
  4. none of the above
ব্যাখ্যা

When negative voltage feedback is applied to an amplifier, the voltage gain of the amplifier decreases. This happens because negative feedback involves feeding a portion of the output signal back to the input in such a way that it opposes the original input signal.

The feedback signal subtracts from the input signal, which reduces the overall gain of the amplifier. While this reduces the voltage gain, it has several beneficial effects, such as increasing stability, improving bandwidth, reducing distortion, and lowering the sensitivity of the amplifier to component variations.

Negative feedback forces the amplifier to adjust its output to make the input and feedback signals more balanced. This reduces the overall amplification provided by the device.
While the gain is reduced, the overall performance (in terms of linearity and stability) improves, and the amplifier becomes less sensitive to fluctuations in its components.

৪৩.
If the output of an amplifier is 10 V and 100 mV from the output is fed back to the input, then feedback fraction is .......
  1. 10
  2. 0.1
  3. 0.01
  4. 0.15
ব্যাখ্যা

The feedback fraction is the ratio of the amount of voltage fed back to the input of the amplifier to the total output voltage.

Given:

Output voltage = 10 V
Feedback voltage = 100 mV = 0.1 V
The feedback fraction (β) is calculated as:

β=Feedback voltage/Output voltage
Substituting the values:

β=0.1 V/10 V=0.01

৪৪.
The gain of an amplifier without feedback is 100 dB. If a negative feedback of 3 dB is applied, the gain of the amplifier will become.
  1. 101.5 dB
  2. 300 dB
  3. 103 dB
  4. 97 dB
ব্যাখ্যা

When negative feedback is applied to an amplifier, the gain of the amplifier is reduced. The gain reduction due to negative feedback can be calculated using the following formula:

New Gain (in dB)=Original Gain (in dB)−Feedback (in dB)
Given:

The original gain of the amplifier is 100 dB.
The negative feedback applied is 3 dB.
Now, applying the formula:

New Gain=100 dB−3 dB=97 dB

৪৫.
The gain of an amplifier with feedback is known as ................... gain.
  1. resonant
  2. open loop
  3. closed loop
  4. none of the above
ব্যাখ্যা

When feedback is applied to an amplifier, the gain is referred to as the closed-loop gain. This is because the feedback loop creates a feedback path from the output to the input, which affects the overall gain of the amplifier.

Closed-loop gain is the gain of the amplifier when the feedback network is active and is typically lower than the open-loop gain.Open-loop gain, on the other hand, is the gain of the amplifier without any feedback applied. It is usually much higher than the closed-loop gain.

৪৬.
If the voltage gain of an amplifier without feedback is 20 and with negative voltage feedback it is 12, then feedback fraction is........
  1. 5/3
  2. 3/5
  3. 1/5
  4. 0.033
ব্যাখ্যা

Avf = 12
Av = 20

mv = 1/Avf = 1/Av = 1/12 + 1/20 = 0.033

৪৭.
An amplifier has an open loop voltage gain of 1,00,000. With negative volage feedback, the voltage gain is reduced to 100. What is the sacrifice factor?
  1. 1000
  2. 100
  3. 5000
  4. none of the above
ব্যাখ্যা

The sacrifice factor is the ratio of the open-loop gain to the closed-loop gain. It shows how much the amplifier's gain is reduced due to the application of negative feedback.

Formula for Sacrifice Factor:
Sacrifice Factor=Open-loop Gain-Closed-loop Gain

Given:
Open-loop voltage gain (Av​) = 100,000,
Closed-loop voltage gain (Av′​) = 100.
Now, substitute the values into the formula:

Sacrifice Factor = 100,000/100 = 1000.

৪৮.
The negative current feedback in an amplifier .................. of the amplifier.
  1. increases voltage gain
  2. reduces voltage gain
  3. increases power gain
  4. none of above
ব্যাখ্যা

Negative current feedback works similarly to negative voltage feedback in that it reduces the overall voltage gain of the amplifier. However, it improves the stability and linearity of the amplifier and reduces distortion. In the case of current feedback, the feedback signal is related to the output current rather than the output voltage, and this feedback is applied to the input to stabilize the gain. While negative feedback of both types (voltage and current) lowers the gain, the key advantage is that it enhances other parameters such as bandwidth and linear performance. Therefore, negative current feedback results in a reduced voltage gain.

৪৯.
With negative voltage feedback to an amplifier, the lower cut-off frequency of the amplify
  1. is decreased
  2. is increased
  3. becomes zero
  4. is not changed
ব্যাখ্যা

Negative voltage feedback in an amplifier typically improves the frequency response and lowers the lower cut-off frequency. Here's how:

Negative feedback reduces the overall gain of the amplifier, but it also improves its stability and increases bandwidth.
By applying negative feedback, the gain-bandwidth product is conserved, which means that while the gain is reduced, the amplifier can maintain a larger frequency range for operation, leading to a lowered lower cut-off frequency.
As a result, the amplifier can handle lower frequencies without distortion, improving its performance at the lower end of the frequency spectrum.

৫০.
When voltage feedback (negative) is applied to an amplifier, its input impedance .......
  1. is decreased
  2. is increased
  3. remains the same
  4. none of the above
ব্যাখ্যা

When negative voltage feedback is applied to an amplifier, it typically increases the input impedance of the amplifier. Here's why:

Without feedback, the input impedance of an amplifier depends primarily on the characteristics of the active devices (like transistors or op-amps) used within the amplifier.

With negative feedback, the overall behavior of the amplifier is modified, and the feedback reduces the dependence of the output on the input. This leads to increased input impedance. This increase in input impedance is beneficial for minimizing the loading effect when the amplifier is connected to a signal source.

In essence, the negative feedback forces the amplifier to behave in a more linear and predictable way, leading to higher input impedance and lower output impedance.