Electronic Devices and Circuits MCQ Quiz - Objective Question with Answer for Electronic Devices and Circuits - Download Free PDF

Explanation:

The voltage across the capacitor in a buck-boost converter, denoted as \( V_c \), is an important parameter in the analysis of the converter's behavior. Understanding the differential equation governing this voltage is crucial for the design and analysis of such converters.

Buck-Boost Converter Overview:

A buck-boost converter is a type of DC-DC converter that can step up (boost) or step down (buck) an input voltage. The basic components of a buck-boost converter include an inductor, a capacitor, a switch (typically a transistor), and a diode. The operation of the buck-boost converter can be divided into two modes: when the switch is on and when the switch is off. The inductor stores energy when the switch is on and releases it to the capacitor and the load when the switch is off.

Derivation of the Differential Equation:

To derive the differential equation for the voltage across the capacitor \( V_c \), we need to analyze the behavior of the circuit in both modes of operation. Let's consider the following assumptions for simplicity:

• The converter operates in continuous conduction mode (CCM), meaning the inductor current never falls to zero.
• The capacitor voltage \( V_c \) is relatively stable and changes slowly compared to the switching frequency.

When the switch is on, the inductor is connected to the input voltage \( V_s \), and the inductor current increases. The voltage across the inductor \( L \) is \( V_s \), and the current through the inductor \( I_L \) increases linearly. The capacitor \( C \) is disconnected from the input, and the load is supplied by the capacitor.

When the switch is off, the inductor is connected to the capacitor and the load. The inductor current decreases, and the energy stored in the inductor is transferred to the capacitor and the load. The voltage across the inductor is \( V_o - V_c \), where \( V_o \) is the output voltage.

The differential equation for the capacitor voltage \( V_c \) can be derived by applying Kirchhoff's current law (KCL) at the capacitor node. The current through the capacitor \( I_C \) is given by:

\[ I_C = C \frac{dV_c}{dt} \]

The current through the load \( I_L \) is given by:

\[ I_L = \frac{V_o}{R_L} \]

Applying KCL at the capacitor node, we get:

\[ I_C = I_L \]

Substituting the expressions for \( I_C \) and \( I_L \), we get:

\[ C \frac{dV_c}{dt} = \frac{V_o}{R_L} \]

Dividing both sides by \( C \), we get the differential equation for the capacitor voltage \( V_c \):

\[ \frac{dV_c}{dt} = \frac{V_o}{C R_L} \]

Correct Option Analysis:

The correct option is:

Option 1: \(\rm \frac{dV_c}{dt}=\frac{V_o}{CR_L}\)

This option correctly represents the differential equation for the voltage across the capacitor \( V_c \) in a buck-boost converter, derived from the principles explained above.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 2: \(\rm \frac{dV_c}{dt}=\frac{V_s}{CR_L}\)

This option incorrectly uses the input voltage \( V_s \) instead of the output voltage \( V_o \). The differential equation for the capacitor voltage depends on the output voltage \( V_o \), not the input voltage \( V_s \).

Option 3: \(\rm \frac{dV_c}{dt}=\frac{V_S}{C}\)

This option is incorrect because it omits the load resistance \( R_L \), which is a crucial factor in the differential equation. The load resistance \( R_L \) affects the current through the capacitor and, consequently, the rate of change of the capacitor voltage.

Option 4: \(\rm \frac{dV_c}{dt}=\frac{V_o}{R_L}\)

This option is incorrect because it does not consider the capacitance \( C \). The capacitance \( C \) is essential in determining the rate of change of the capacitor voltage. The correct differential equation must include \( C \) in the denominator.

Conclusion:

Understanding the derivation of the differential equation for the voltage across the capacitor in a buck-boost converter is crucial for the analysis and design of such converters. The correct differential equation is \(\rm \frac{dV_c}{dt}=\frac{V_o}{CR_L}\), which accounts for the output voltage \( V_o \), the capacitance \( C \), and the load resistance \( R_L \). This equation helps in predicting the behavior of the capacitor voltage and designing the converter for desired performance.

Calculation:
The diode D1 is in reverse bias, hence will show infinite resistance. The circuit will be then

The total resistance will be:
R = 25 Ω  + 175 Ω + 50 Ω = 250 Ω 

The current in circuit is 

I = 10 / 250 A = 0.04 A

Answer : 2

Solution :

The correct answer for the description of electronic bands in insulators is:

Option 2 : conduction band is empty and valence band is completely filled with electrons.

In insulators, there is a large energy gap between the valence band and the conduction band. The valence band is the band of electron orbitals that electrons occupy at absolute zero temperature, and it's usually filled with electrons in an insulator. The conduction band is the band above the valence band where an electron may be excited to in order to conduct electricity. In insulators, this conduction band is empty because the energy gap (band gap) between the valence band and the conduction band is so large that electrons in the valence band do not normally have enough energy to jump into the conduction band under room temperatures or normal conditions. Hence, there is no electrical conductivity in insulators under these conditions.

Let's break down the incorrect options:

Option 1 : In insulators, the valence band is not empty. It is filled with electrons, which is in direct contradiction to what this option states.

Option 3 : In insulators, the valence band is completely filled, not partially filled. It is the complete filling of the valence band which contributes to their insulating properties, combined with the large energy gap to the conduction band.

Option 4 : The conduction band in insulators is not partially filled; it is empty due to the energy required to move electrons from the valence band to the conduction band being prohibitively large for thermal excitation.

The correct answer is 300.

Peak inverse voltage (PIV)

The maximum voltage that appears across the diode under the reverse bias condition is known as the PIV.

The PIV for half wave rectifier is V_m. 

Here, V_m is the peak value of the supply voltage.

Calculation

\(V_2=V_1({N_2\over N_1})\)

\(V_2=200({1\over 4})\)

V₂ = 50 V

This is the RMS value. 

PIV is measured in maximum value.

The maximum value is given by:

\(V_{2(m)}=\sqrt{2}\times 50\)

V_2(m) = 70.7 V

In the motor generator set, synchronous motors are used to convert voltage, phase and frequency of power. Motor generator set also find application in isolation of electrical loads from the supply line.

The correct answer is option 1): (FET)

Concept:

• Negative resistance is an uncommon property that occurs in a few nonlinear electronic components.
• In a nonlinear device, two types of resistance can be defined: 'static' or 'absolute resistance', the ratio of voltage to current v/i, and differential resistance, the ratio of a change in voltage to the resulting change in current Δv/Δi.
• The term negative resistance means negative differential resistance (NDR), Δv/Δi. Specifically, IMPATT diodes, Gunn diodes, ISIS diodes, and tunnel (Esaki) diodes have regions of negative resistance.
• Field-Effect Transistor: It is a three-terminal device that uses the input applied voltage to control the current flow.
• FET does  not have negative resistance characteristics

The Correct option is 2

Concept:

SCR

• A Silicon Controlled Rectifier is a 3-terminal and 4-layer semiconductor current-controlling device. It is mainly used in devices for the control of high power.
• The silicon-controlled rectifier is also sometimes referred to as an SCR diode, 4-layer diode, 4-layer device, or Thyristor.
• 3 terminals- Anode, Cathode & Gate
• Three junctions- J1, J2, J3
• 4 Layers- P, N, P, N

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Characteristic of SCR:

The modes of operation of an NPN transistor are as follow:

From the above table, we can conclude that for normal operation (active mode) of NPN transistor, emitter-base Junction must be forward biased and base-collector junction must be reverse biased.

The forward voltage at which the current through PN junction starts increasing rapidly is known as knee voltage. The Knee voltage of a crystal diode is approximately equal to barrier potential.

Knee voltage of “germanium” diode is 0.3 volts

Knee voltage of “silicon" diode is 0.7 volts

In the above question except ‘Gold’, all given options are the examples of insulators.

A rectifier is a circuit that converts the AC signal at its input to pulsating DC at its output.

For half-wave rectifier, the output is present is only for one half of the input signal and clipped for the other half.

Positive half wave rectifier clips the negative half of the input signal and only positive part of the input signal is present.

Negative half wave rectifier clips the positive half of the input signal and only negative part of the input signal is present

Oscillators:

These are positive feedback amplifiers, and feedback is given to the input from the output with the help of the feedback circuit.

\(Gain = \frac{{{V_0}}}{{V_1^0}} = \frac{A}{{1 - A\beta }}\)

Where A = amplifier gain

β = feedback factor

|Aβ| = 1, then the amplitude of sinusoidal output remains constant

|Aβ| < 1, amplitude decrease with time & will disappear.

|Aβ| > 1, amplitude increases with time then becomes constant (in practical)

Note:

• In oscillators, oscillations occur without any input. A small noise signal is enough to trigger oscillations.
• When |Aβ| = 1, perfect oscillations occur. (or ∠Aβ = 0° or 360°)
• This is known as the Barkhausen criteria, i.e. |Aβ| = 1 & ∠Aβ = 0° or 360°

The correct answer is option 3): (Zero) 

Concept:

• An Online UPS is a type of uninterruptible power supply that utilises either a double or delta conversion technology.
• With double conversion, network equipment does not receive electricity directly from the AC outlet. Instead, AC power travels to a rectifier, where it becomes DC power.
• In Online UPS, the output power supply always stays ON i.e. the UPS keeps charging the battery and draws current from the battery to supply the load.
• Hence, there is no switching and, therefore, no time delay in switching between its source
• In an online UPS system, the changing time from mains to battery is Zero