FET Amplifier MCQ Quiz in मल्याळम - Objective Question with Answer for FET Amplifier - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

JFET common drain amplifier is sometimes considered as a source follower and has voltage gain less than unity.

\(A_v = \frac{V_{out}}{V_{in}}\)

⇒ \(A_v = \frac{g_m R_s}{g_m R_s +1}\)

\(r_0 = \frac 1 {g_m}\)

Putting the value of g_m, we get

\(A_v = \frac{\frac{1}{r_o} R_s}{\frac{1}{r_o} R_s +1}\)

⇒ \(A_v = \frac{ R_s}{ R_s +r_o}\)

Concept:

The lower-3dB frequency, fL in the given amplifier is given by:

\({f_L} = \frac{1}{{2\pi {R_i}{C_G}}}\)

Calculation:

R_i = 10k + 1MΩ = 1.01 MΩ

\({C_G} = \frac{1}{{2\pi {R_i}{f_L}}}\)

\({C_G} = \frac{1}{{2\pi \times 1.01\;M{\Omega} \times 100\;Hz}} \)

C_G = 1.576 nF

Concept:

Transconductance indicates the amount of control the gate has on the drain current.

Mathematically, the transconductance (gm) is defined as:

\( g_m=\frac{\partial I_{D}}{\partial V_{GS}}\)

It is given the name transconductance because it gives the relationship between the input voltage and the output current.

Given: Δ V_GS = 1 V and Δ I_D = 10 mA

\(g_m=\frac{Δ I_D}{Δ V_{GS}}\)

∴ \(g_m = \frac{(10 \times 10^{-3})~A}{1~V}\)

= 0.01 mho

• In capacitively coupled amplifiers, the coupling and bypass capacitors affect the low-frequency cut-off. These capacitors form a high-pass filter with circuit resistances.
• Coupling and bypass capacitors are also called external capacitors.
• For high-frequency response, it is determined by the device’s internal capacitance and the Miller effect.
• Device internal capacitance is shunted and also called as Junction capacitors.

The correct answer is option 1
In the given figure input signal is applied at the source terminal and the output is drawn at the drain terminal. hence it is a common gate configuration
Concept:
FET : Field Effect Transistor

The transistor that uses the electric field or voltage to control flow of current in semiconductor. 

Generally it supports 3 modes of operation :

1. Common Gate :

In this input signal is applied at the source terminal and output is drawn at the drain terminal.

Gate is kept common or generally common terminal is grounded.

2. Common Drain :

In this input signal is applied at the gate terminal and output is drawn at the source terminal.

Drain is kept common or generally common terminal is grounded.

3. Common Source :

In this input signal is applied at the gate terminal and output is drawn at the drain terminal.

Common drain Amplifies with voltage divider bias and its small-signal equivalent circuit is as shown below:

Calculation:

Given:

R₁ = 1.5 MΩ, R₂ = 1 MΩ

So,

\({R_{in}} = \frac{{{R_1} \cdot {R_2}}}{{{R_1} + {R_2}}}\)

\({R_{in}} = \frac{{1.5\;M{\rm{\Omega }} \times 1M\;{\rm{\Omega }}}}{{2.5\;M{\rm{\Omega }}}}\)

Concept:

The Miller effect refers to the increase in equivalent capacitance that occurs when a capacitor is connected from the input to the output of an amplifier with a large, negative gain.

Miller capacitance is given by:

CMi = (1 + Av) Cgs

The output miller capacitance = \({{\rm{C}}_{{\rm{Mo}}}} = {{\rm{C}}_{{\rm{gs}}}}\left( {1 + \frac{1}{{\rm{A}}}} \right)\) 

Calculation:

Given that, Av = 20

Cgs = 2 pF

CMi = (1 + 20) × 2 = 42 pF

CONCEPT: 

Amplifier:

• An amplifier is a device by which the amplitude of the input ac signal (voltage/current/power) can be increased.
•  Transistors can act as amplifiers while they are functioning in the active region or when it is correctly biased.

Transistor as an amplifier (CE configuration):

• In this configuration, the emitter is common for both input and output.
• The input signal is connected in series with the voltage applied to the base-emitter junction.
•  For proper functioning of a transistor, the emitter-base junction must be forward-biased and the collector-base junction must be reverse-biased.
• The current gain is given as,

\(\Rightarrow \beta=\frac{Δ I_C}{Δ I_B}\)

• Voltage gain is given as,

\(\Rightarrow Voltage\, Gain=\frac{V_{out}}{V_{in}}=\frac{R_L}{R_E}\)

• Power gain is given as,

\(\Rightarrow Power\, Gain=\frac{P_{out}}{P_{in}}=\beta× Voltage\,Gain\)

Where ΔIC = change in collector current, ΔIB = change in base current, Vout = output voltage, Vin = input voltage, RL = signal resistance in the Collector, RE = signal resistance in the Emitter

Calculation:

Power gain = 40 × 25 = 1000

Given that λ = 0, so the small-signal output resistance will be:

r0 = ∞

Therefore, we draw the small-signal equivalent circuit as:

From the circuit, we have

v0 = gm vgs RD      ---(1)

At input, vgs = -vi

Substituting the value of vgs in equation (1), we get:

v0 = -gm vi RD

So, the voltage gain will be:

\({V_v} = \frac{{{v_0}}}{{{v_i}}} = - {g_m}{R_D}\)

= -2.6 m × 6k = -15.6

Voice frequency range: 300 to 3.5 kHz

Audio frequency range: 20 to 20 kHz