Field Effect Transistors MCQ Quiz - Objective Question with Answer for Field Effect Transistors - Download Free PDF

Metal Oxide Silicon Field Effect Transistors (MOSFET)

• It is a voltage-controlled three-terminal device that is used for switching and amplification purposes.
• The terminals are the drain, gate, and source.

Transfer characteristics of power MOSFET

The transfer characteristics of power MOSFET, shows the variation of drain current (I_D) as a function of gate-source voltage (V_GS).

Concept

The forward current gain (α) of a power transistor is defined as:

\(α ={I_C\over I_E}\)

where, I_C = Collector current

I_E = Emitter current

For power transistors, the forward current gain (α) is typically high, usually in the range of 0.95 to 0.99. This indicates that most of the emitter current flows to the collector, with very little loss to the base current.

\(β ={α\over 1-α}\)

where β is the common-emitter current gain, having α close to 1 ensures that power transistors operate efficiently with high current gain.

Thus, the correct approximate value of forward current gain for power transistors is 0.95 to 0.99.

Concept

The drain current in a JFET is given by:

\(I_D=I_{DSS}({1-{V_{GS}\over V_{P}}})^2\) 

where, I_D = Drain current

I_DSS = Saturation drain current

V_GS = Gate -source voltage

V_P = Pinch off voltage

Calculation

Given, V_p = - 8 V

IDSS = 30 mA

VGS = - 4 V

\(I_D=30({1-{-4\over -8}})^2\)

I_D = 7.5 mA

Explanation:

Correct Option Analysis:

The correct option is:

Option 4: 1.8 mA

We are given an N-channel E-MOSFET with the following parameters:

• ID(ON) = 5 mA at VGS = 10 V
• VGS(th) = 5 V

We need to calculate its drain current (ID) for VGS = 8 V.

The drain current for an E-MOSFET in the saturation region can be calculated using the following equation:

ID = ID(ON) * ((VGS - VGS(th)) / (VGS(ON) - VGS(th)))^2

Given:

• ID(ON) = 5 mA
• VGS(ON) = 10 V
• VGS(th) = 5 V
• VGS = 8 V

Substitute the given values into the equation:

ID = 5 mA * ((8 V - 5 V) / (10 V - 5 V))^2

Calculate the terms inside the parentheses first:

(8 V - 5 V) = 3 V

(10 V - 5 V) = 5 V

Now, substitute these values back into the equation:

ID = 5 mA * (3 V / 5 V)^2

Simplify the fraction:

ID = 5 mA * (0.6)^2

Calculate the square of 0.6:

(0.6)^2 = 0.36

Now, multiply this by 5 mA:

ID = 5 mA * 0.36

ID = 1.8 mA

Therefore, the drain current for VGS = 8 V is 1.8 mA, making option 4 the correct answer.

Additional Information

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

Option 1: 3.2 mA

Using the same method, if we substitute the values into the equation:

ID = 5 mA * ((8 V - 5 V) / (10 V - 5 V))^2

ID = 5 mA * (3 V / 5 V)^2

ID = 5 mA * (0.6)^2

ID = 5 mA * 0.36

ID = 1.8 mA

Clearly, the calculation shows that the correct drain current is 1.8 mA and not 3.2 mA, so option 1 is incorrect.

Option 2: 4 mA

Again, using the same method:

ID = 5 mA * ((8 V - 5 V) / (10 V - 5 V))^2

ID = 5 mA * (3 V / 5 V)^2

ID = 5 mA * (0.6)^2

ID = 5 mA * 0.36

ID = 1.8 mA

The calculation confirms that 1.8 mA is the correct value, so option 2 is incorrect as well.

Option 3: 2.6 mA

Using the same method:

ID = 5 mA * ((8 V - 5 V) / (10 V - 5 V))^2

ID = 5 mA * (3 V / 5 V)^2

ID = 5 mA * (0.6)^2

ID = 5 mA * 0.36

ID = 1.8 mA

The calculation clearly shows that 2.6 mA is not the correct drain current, thus option 3 is also incorrect.

Conclusion:

Understanding the operation of an N-channel E-MOSFET and the application of the drain current equation are crucial for correctly identifying the drain current for a given gate-source voltage. By carefully substituting the given values and performing the calculations, we have determined that the correct drain current for VGS = 8 V is 1.8 mA. This confirms that option 4 is the correct answer.

Explanation:

To determine the input resistance of a JFET given the gate-source leakage current (IGss) and the gate-source voltage (VGS), we can use Ohm's Law. The input resistance (Rin) can be found using the formula:

Rin = VGS / IGss

Given:

• IGss = -2 nA (nanoamperes) = -2 × 10^-9 A
• VGS = -20 V (volts)

Note that the negative signs indicate the direction of current and voltage but do not affect the magnitude of the resistance calculation. We can ignore the negative signs for the purpose of calculating the resistance.

Substituting the given values into the formula:

Rin = VGS / IGss

Rin = 20 V / 2 × 10^-9 A

Rin = 20 / 2 × 10^9

Rin = 10 × 10^9 ohms

Rin = 10 GΩ (gigaohms)

Since 1 GΩ = 1000 MΩ (megaohms), we can convert the result to megaohms:

Rin = 10 × 1000 MΩ

Rin = 10000 MΩ

Therefore, the correct input resistance is 10000 MΩ.

Correct Option Analysis:

The correct option is:

Option 3: 10000 MΩ

This option correctly represents the input resistance of the JFET based on the given values of IGss and VGS.

Additional Information

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

Option 1: 1000 MΩ

This option is incorrect because it underestimates the input resistance. The calculation clearly shows that the input resistance is much higher than 1000 MΩ.

Option 2: 10000 kΩ

This option is also incorrect. Although 10000 kΩ is equal to 10 MΩ, it still underestimates the input resistance by a factor of 1000.

Option 4: 1000 kΩ

This option is incorrect as well. 1000 kΩ is equal to 1 MΩ, which is significantly lower than the calculated input resistance of 10000 MΩ.

Conclusion:

In summary, the input resistance of the JFET is determined by the ratio of the gate-source voltage to the gate-source leakage current. The calculated input resistance of 10000 MΩ matches option 3, making it the correct choice. Understanding the principles of Ohm's Law and the conversion between different units of resistance is essential for accurately solving such problems.

MOSFET (Metal Oxide Semiconductor Field Effect Transistor)

MOSFET transistor is a semiconductor device which is used for amplifying and switching electronic signals in electronic devices.

MOSFET is of two types:

1. Enhancement MOSFET:

Concept:

Small signal current gain is defined in common drain amplifier as

\(\rm A_F = \frac{I_s}{I_g} = \frac{Source \ current}{Gate \ current}\)

For FET, I_g = 0

∴ \(\rm A_i = \frac{I_s}{0}\) = Infinite

The correct answer is option 1):(N type; P type)

Concept:

• The schematic of an n-channel JFET along with its circuit symbol is shown in Figure
• The n-channel JFET has its major portion made of n-type semiconductors.
• The mutually-opposite two faces of this bulk material are from the source and the drain terminals.
• There are two relatively-small p-regions embedded into this substrate which are internally joined together to form the gate terminal
• Thu the source and the drain terminals are of n-type while the gate is of p-type.
• For n-JFET, the channel is a N-type channel and gates are P type
• Due to this, two pn junctions will be formed within the device, whose analysis reveals the mode in which the JFET works

Concept:

The drain current in an N-channel JFET is given by:

\(I_D = I_{DSS}({1-{V_{GS}\over V_p}})^2\)

where, I_D = Drain current

I_DSS = Drain current when the gate to the source is equal to zero

V_GS = Gate to source voltage

V_P = Pinch-off voltage

Calculation:

Given, IDSS = 12 mA

VP = −7V, VGS = −3.5V

\(I_D = 12({1-{-3.5\over -7}})^2\)

\(I_D = 12({1-{1\over 2}})^2\)

I_D = 3 mA

Three regions of MOSFET operation (nMOS) are:

Cut off region:

VGS < Vth

ID = 0

VGS = Gate to source voltage

Vth = Threshold voltage

Linear/ Ohmic/ Triode region:

VGS > Vth

VDS < VGS – Vth

The current equation for a MOSFET in the linear region is given by:

\({{I}_{D}}={{\mu }_{n}}{{C}_{ox}}\frac{W}{L}\left\{ \left( {{V}_{GS}}-{{V}_{th}} \right){{V}_{DS}}-\frac{V_{DS}^{2}}{2} \right\}\)

Saturation region:

VGS > Vth

VDS > VGS – Vth

The current equation for a MOSFET in the saturation region is given by:

\({{I}_{D}}=\frac{1}{2}{{\mu }_{n}}{{C}_{OX}}\frac{W}{L}{{\left( {{V}_{GS}}-{{V}_{th}} \right)}^{2}}\)

Analysis:

At the edge of saturation:

\({V_{DS}} = {V_{GS}} - {V_T}\)

V_DS = 2 – (6)

V_DS = - 4 Volts

This is explained with the help of the V-I characteristics as shown:

Concept:

1) The thermal runway is not possible in FET because as the temperature of the FET increases, the mobility decreases, i.e. if the Temperature (T) ↑, the carries Mobility (μn or μ­p) ↓, and Ips↓

2) Since the current is decreasing with an increase in temperature, the power dissipation at the output terminal of a FET decreases or we can say that it’s minimum.

So, there will be no Question of thermal Runway at the output of the FET.

• The thermal runaway takes place in a BJT.
• Thermal Runway in BJT is a process of self-damage of BJT because of overheating at the collector junction due to an increase in Ic with Ico
• If T↑, then Ico (Reverse separation current) ↑, which results in an increase in the collector current, i.e. Ic ↑.
• Power dissipation at the collector junction increases in the form of heat which again raises the temperature and the cycle continues.
• If the above cycle becomes repetitive then the collector junction gets overheated and thereby thermal runway takes place.

As shown in the above figure in a power MOSFET, pinch-off occurs when V_DS = V_GS - V_T

Where, V_DS = Drain to source voltage

​V_GS = Gate to source voltage

V_T = Threshold voltage

• In power MOSFET when V_DS < V_GS - V_T, then power MOSFET works in the triode region.
• In power MOSFET when V_DS > V_GS - V_T, then power MOSFET works in the saturation region.

• FET is a voltage-driven/controlled device, i.e. the output current is controlled by the electric field applied.
• The current through the two terminals is controlled by a voltage at the third terminal (gate).
• It is a unipolar device (current conduction is only due to one type of majority carrier either electron or hole)
• It has a high input impedance.

• For a FET, either

  \({I_D} = {I_{{D_{SS}}}}{\left( {1 - \frac{{{V_{GS}}}}{{{V_P}}}} \right)^2}\)in case of JFET

  or

  \({I_D} = K{\left( {{V_{GS}} - {V_T}} \right)^2}\)in case of MOSFET

  So, a FET is a voltage-controlled current source.

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