JFET MCQ Quiz in తెలుగు - Objective Question with Answer for JFET - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Last updated on Mar 12, 2025

పొందండి JFET సమాధానాలు మరియు వివరణాత్మక పరిష్కారాలతో బహుళ ఎంపిక ప్రశ్నలు (MCQ క్విజ్). వీటిని ఉచితంగా డౌన్‌లోడ్ చేసుకోండి JFET MCQ క్విజ్ Pdf మరియు బ్యాంకింగ్, SSC, రైల్వే, UPSC, స్టేట్ PSC వంటి మీ రాబోయే పరీక్షల కోసం సిద్ధం చేయండి.

Latest JFET MCQ Objective Questions

Top JFET MCQ Objective Questions

JFET Question 1:

Which of the following is involved in the operation of a JFET?

  1. Flow of minority carriers alone
  2. Flow of majority carriers alone
  3. Flow of both minority and majority carriers
  4. Use of a magnetic field

Answer (Detailed Solution Below)

Option 2 : Flow of majority carriers alone

JFET Question 1 Detailed Solution

Junction Field Effect Transistor (JFET):

F1 S.B Deepak 02.03.2020 D1

  • It is a transistor that relies on the electric field to control the shape of the channel and therefore on the conductivity of the channel. Hence referred to as field-effect transistor.
  • FET is a unipolar transistor that involves only one type of charge carrier in its operation.
  • In the case of an N-channel JFET, the operation involves a flow of electrons.
  • In the case of a p-channel JFET, the operation involves a flow of holes.
  • JFET operation involves the flow of majority carriers.
  • In JFET, the current flowing through channel b/w Drain in the source is controlled by the voltage applied at the Gate terminal, which is reverse biased.
  • As it is reverse biased, the Gate current is practically zero as it has high input impedance.

JFET Question 2:

A JFET has three terminals namely ______.

  1. Cathode, anode, grid
  2. Emitter, base, collector
  3. Source, gate, drain
  4. Emitter, gate, collector

Answer (Detailed Solution Below)

Option 3 : Source, gate, drain

JFET Question 2 Detailed Solution

The correct answer is 'option 3'

Solution:

  • The JFET is a long channel of semiconductor material, doped to contain an abundance of positive charge carriers or holes (p-type), or of negative carriers or electrons (n-type).
  • Ohmic contacts at each end form the source (S) and the drain (D).
  • A p-n junction is formed on one or both sides of the channel, or surrounding it using a region with doping opposite to that of the channel, and biased using an ohmic gate contact (G).

F1 Savita Engineering 25-4-22 D9

JFET Question 3:

A JFET has high input impedance because-

  1. it is made of semiconductor material
  2. input in reverse biased
  3. of impurity atoms
  4. none of these

Answer (Detailed Solution Below)

Option 2 : input in reverse biased

JFET Question 3 Detailed Solution

The correct answer is 'option 2'

Solution:

  • The high impedance of JFET is due to the reason that the Gate junction is reverse biased and because there is no minority carrier contribution to the flow through the device.
  • The control element for the JFET comes from depletion of charge carriers from the n-channel. When the Gate is made more negative, it depletes the majority carriers from a larger depletion zone around the gate.
  • This reduces the current flow for a given value of Source-to-Drain voltage.
  • Modulating the Gate voltage modulates the current flow through the device.

JFET Question 4:

What is the value of Rs required to self bias an N channel JFET with Vp = -10 V, IDSS = 40 mA and VGSQ = -5 V?

  1. 250 Ω
  2. 500 Ω
  3. 750 Ω
  4. 1500 Ω
  5. None of these

Answer (Detailed Solution Below)

Option 2 : 500 Ω

JFET Question 4 Detailed Solution

\(\rm I_{DS}=I_{DSS}\left(1-\frac{V_{GSQ}}{V_P}\right)^2\)

\(\rm I_{DS}=40\left(1-\frac{(-5)}{(-10)}\right)^2\)

\(\rm I_{DS}=40\left(1-\frac{1}{2}\right)^2\)

IDS = 10 mA

\(\rm R_s=\frac{-V_{GSQ}}{I_{DS}}\)

\(\rm R_s=\frac{(-5)}{10\times 10^{-3}}\)

Rs = 500 Ω

JFET Question 5:

Junction field-effect transistor (JFET) has _______ diodes.

  1. 1
  2. 2
  3. 3
  4. 4

Answer (Detailed Solution Below)

Option 2 : 2

JFET Question 5 Detailed Solution

The junction FET, or JFET uses a reverse biased diode junction to provide the gate connection.

JFET (Junction Field Effect Transistor) is a uni-polar voltage controlled device that consists of three terminals called drain, source and gate

JFET has two diodes.

The structure consists of a semiconductor channel which can be either N-type or P-type.

A semiconductor diode is then fabricated onto the channel in such a way that the voltage on the diode affects the FET channel.

In operation this is reverse biased and this means that it is effectively isolated from the channel - only the diode reverse current can flow between the two.

The JFET is the most basic type of FET, and the one that was first developed. However it still provides excellent service in many areas of electronics.

JFET Question 6:

For transconductance amplifier, input and output resistances are, respectively:

  1. ∞ and 
  2. ∞ and 0
  3. 0 and 
  4. 0 and 0

Answer (Detailed Solution Below)

Option 1 : ∞ and 

JFET Question 6 Detailed Solution

Feedback Amplifiers:

Feedback amplifiers are the type of amplifiers in which a part of the output is given back to the input.

F3 Savita Engineering 30-5-22 D2

Feedback amplifiers are basically classified into 2 categories:

1.) Positive Feedback amplifier:

  • It is a type of amplifier in which the source signal and the feedback signal are in the same phase.
  • Thus, the feedback signal applied increases the strength of the input signal.

F3 Savita Engineering 30-5-22 D3

2.) Negative Feedback amplifier:

  • In this type of amplifier source signal and the feedback signal is out of phase with each other. 
  • Thus, the feedback signal is applied to decrease the strength of the input signal.

F3 Savita Engineering 30-5-22 D4

Classification of feedback amplifier on basis of input and output resistance:

Amplifier

Input Resistance

Output Resistance

Transconductance amplifier

∞ 

∞ 

Transresistance amplifier

0

0

Voltage amplifier

∞ 

0

Current amplifier

0

∞ 

 

From the above table, the input and output resistances of transconductance amplifier are ∞ and ∞ respectively.

JFET Question 7:

Which of the following is correct?

  1. JFET has zero offset value.
  2. Compared with MOSFETs, JFETs are easier to fabricate
  3. The drain resistance of MOSFET is higher than that of JFET
  4. The input resistance of MOSFET is higher than that of JFET

Answer (Detailed Solution Below)

Option 4 : The input resistance of MOSFET is higher than that of JFET

JFET Question 7 Detailed Solution

MOSFET: Input impedance (typically of the order of >1010 Ω

BJT: input impedance is of the order of >104 Ω

JFET: Input impedance is of the order of >108 Ω

MESFET: Input impedance is of the order of >1012 Ω

Hence, option 4 is correct.

Note: At low or medium frequency MOSFET is offering high input impedance than JFET but at very higher frequencies, JFET has more input impedance than MOSFET. (Usually, We use MOSFET and JFET at low or medium frequencies. Hence, by default the Input impedance of MOSFET > Input impedance of JFET.

26 June 1

The difference between FET and BJT is explained in the following table:

FET

BJT

Unipolar device: Uses only one type of charge carrier

Bipolar device: Uses both electron and hole

Voltage-controlled device: voltage between gate and source control the current through the device.

Current-controlled device: Base current control the amount of collector current

High input resistance

High input impedance

Slower in switching

Faster in switching

JFET Question 8:

For a given JFET, the typical values of amplification factor and transconductance are given as 40 and 100 μS respectively. The dynamic drain resistance of JFET will be:

  1. 200 Ω
  2. 400 Ω
  3. 200 kΩ
  4. 400 kΩ

Answer (Detailed Solution Below)

Option 4 : 400 kΩ

JFET Question 8 Detailed Solution

JFET Parameters:

In a JFET, the drain current ID is a function of VGS and VDS.

Dynamic drain resistance (rd): This is the ratio of the change in the drain to source voltage to the change of drain current keeping the gate to source voltage constant, i.e.

\({r_d} = \frac{{\partial {V_{DS}}}}{{\partial {I_D}}},\) at constant VGS

Transconductance (gm): It is the ratio of the change in drain current to the change in the gate to source voltage keeping drain to source voltage constant, i.e.

\({g_m} = \frac{{\partial {I_D}}}{{\partial {V_{GS}}}},\) at constant VDS

Amplification factor (μ): It is the ratio of the change in drain voltage to the change in the gate voltage at a constant ID, i.e.

\(\mu = \frac{{\partial {V_{DS}}}}{{\partial {V_{GS}}}},\) at constant ID

The above can be written as:

\(\mu = \frac{{\partial {V_{DS}}}}{{\partial {I_D}}} \times \frac{{\partial {I_D}}}{{\partial {V_{GS}}}} = {r_d} \times {g_m}\)   

Application:

Given μ = 40, gm = 100 μs

Since μ = rd × gm

40 = rd × 100 μ

rd = 4 × 105 Ω

rd = 400 kΩ 

JFET Question 9:

Which of the following is NOT a type of Field Effect Transistor?

  1. JFET
  2. Depletion MOSFET
  3. Enhancement MOSFET
  4. Thyristor

Answer (Detailed Solution Below)

Option 4 : Thyristor

JFET Question 9 Detailed Solution

A thyristor is not a type of Field Effect Transistor (FET).

Thyristor

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  • A thyristor is a four-layer semiconductor device, consisting of alternating P-type and N-type materials (PNPN).
  • A thyristor usually has three electrodes: an anode, a cathode, and a gate, also known as a control electrode. The most common type of thyristor is the silicon-controlled rectifier (SCR).

Field Effect Transistor(FET)

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  • The field-effect transistor is a type of transistor that uses an electric field to control the flow of current in a semiconductor.
  • FETs have three terminals: the source (S), the drain (D), and the gate (G). It works in two modes: depletion and enhancement type.
  • Enhancement FET does not conduct at 0 volts, as there is no channel in this type to conduct. Depletion FET conducts at 0 volts as the channel is already present in it.

JFET Question 10:

How is the amplification factor (µ) for a JFET is expressed?

  1. µ = gm ×  rd
  2. µ = \(\triangle\) Vgs × rd
  3. µ = \(\triangle\) Vds ×  \(\triangle\) Vgs
  4. µ = \(\triangle\) Vds × rd

Answer (Detailed Solution Below)

Option 1 : µ = gm ×  rd

JFET Question 10 Detailed Solution

Amplification Factor (μ) of Junction-FET:

The amplification factor is defined as the ratio of change of drain voltage (δVDS) to change of gate voltage (δVGS) at a constant drain current (ID = Constant).

\(μ = \frac{{{\delta _{DS}}}}{{\delta {V_{GS}}}}\ at\,constan t\,{I_D}\)

There is a relation between transconductance (gm) and dynamic output resistance (rd) and that can be established in the following way.

\(μ = \frac{{{\delta _{DS}}}}{{\delta {V_{GS}}}} = \frac{{\delta {V_{DS}}}}{{\delta {I_D}}} × μ = \frac{{\delta {I_D}}}{{\delta {V_{GS}}}}\)

⇒ μ = rd × gm

Additional InformationTransconductance (gmo): Transconductance is the ratio of change in drain current (δID) to change in the gate to source voltage (δVGS) at a constant drain to source voltage (VDS = Constant).

\(g_m = \frac{{{\delta {I_D}}}}{{\delta {V_{GS}}}}at\,cons\tan t\,{V_D{_S}}\)

This value is maximum at VGS = 0.

This maximum value (gmo) is specified in a JFET data sheet.

The trans-conductance at any other value of gate to source voltage (gm) can be determined as follows.

The expression of drain current (ID) is

\({I_D} = {I_{DSS}}{\left[ {1 - \frac{{{V_{GS}}}}{{{V_{GS}}_{\left( {off} \right)}}}} \right]^2} \)

By partial differentiating the expression of drain current (ID) in respect of gate to source voltage (VGS)

\({g_m} = \frac{{\delta {I_D}}}{{\delta {V_{GS}}}} = \frac{{2{I_{DSS}}}}{{{V_{GS\left( {off} \right)}}}}\left[ {1 - \frac{{{V_{GS}}}}{{{V_{GS\left( {off} \right)}}}}} \right]\)

At VGS = 0, the transconductance gets its maximum value and that is

\({g_{mo}} = \frac{{2{I_{DSS}}}}{{{V_{GS\left( {off} \right)}}}}\)

Therefore, we can write,

\({g_m} = {g_{mo}}\left[ {1 - \frac{{{V_{GS}}}}{{{V_{GS\left( {off} \right)}}}}} \right]\)

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