Electronic Devices 5

10 Questions MCQ Test RRB JE for Electronics & Communication Engineering | Electronic Devices 5

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The figure shows the VTC characteristics of an NMOS inverter with three varying resistive loads (R).

The correct statement is


As Vi is increased voltage drop across RL increases and V0 decreases.

The sharpness of the transition region increases with the increasing load resistance.

Hence R3 > R2 > R1.

*Answer can only contain numeric values

Assume that the zero for electrostatic potential is in the semiconductor bulk at large x and that there is no metal semiconductor work function difference. The relative dielectric constant for the oxide is ϵr = 11.8. If the intrinsic concentration is 1010/cm3. The doping density ND is _____ × 1017/cm3 (KT = 0.026 V)


The doping density is given by

= 1010 e(0.437/0.026)

= 1.99 × 1017


The figure shows MOS capacitor variation with applied gate voltage for n-type body/substrate

The flat band voltage is



Flat band voltage is the voltage where there is no charge present in oxide or oxide-semiconductor interface.

i.e. No thickness of channel

Cox is constant and decreased subsequently

Note: Here, VG is negative (for PMOS)

Hence, voltage is decreased from positive (VFB) to negative.

*Answer can only contain numeric values

The insulator capacitance Ci of an ideal MOS capacitor with 10-nm gate oxide (εr = 3.9) on p-type Si with Na = 1016 cm-3 is ________ × 10-7 F/cm2


The insulator capacitance is the capacitance under strong accumulation

= 3.453 × 10-7 F/cm2



Which of the following curves represents the correct C – V characteristics of an NMOS transistor having an oxide layer thickness of 10 nm and a maximum depletion thickness of 100 nm. [Assume ϵs = permittivity of semiconductor, ϵox = permittivity of oxide and 



tox = 10 nm

d = 100 nm

Cmin = series combination of Cox and Cdep

= 0.09 ϵA


The mobility of hole is 0.4 times the mobility of electron. What must be the ratio of width of n-channel to p-channel MOSFET if they are to have equal drain currents when operated in saturation mode with same magnitude of overdrive voltage:


ID-n = ID-p

μn Wn = μp Wp


The circuit shown uses an NMOS transistor to implement a current source. For the transistor VTN = 1V and =12.5 uA/V2. The required value of VGS to get IDC = 25 μA and corresponding compliance voltage is


2 = (VGS - 1)2‑

VGS = √2 + 1 = 2.414 V

To function as a current source, the transistor must be in saturation or

VDS > VDS (sat).

VDS (sat) = VGS - VTN

= 2.414 – 1

= 1.414.

Hence compliance voltage = 1.414 V.

Compliance voltage is the minimum voltage required to achieve the desired performance.

*Answer can only contain numeric values

The output resistance R0 of the NMOS circuit if ID = 0.5 mA, λ = 0.02 V-1  _____ kilo ohms.


The small signal model with a test voltage Vx is shown.


The output resistance is given by 

From the circuit

Vgs = Vx

Applying KCL


For a MOSFET with gate plate area 0.5 × 10-2 cm2 and oxide layer thickness 80 nm, the value of MOS capacitance and its break down voltage are, (assume relative di-electric constant of sio2, ϵr = 4 and ϵ0 = 8.854 × 10-14 F/cm and dielectric strength of sio2 film is 5 × 106 V/cm) 



Given gate plate area, which is also MOS capacitor area A = 0.5 × 10-2cm2

Oxide layer thickness tox = 80 nm

The value of MOS capacitance 


= 0.22 μF

Dielectric breakdown happens of field  greater than dielectric strength (5 × 106 V/cm)

V = 5 × 108 × 80 × 10-9 V = 40V

*Answer can only contain numeric values

In the circuit shown in Figure, Transistors are characterized by  and λ = 0

The output voltage V0 is _______V


Gate is connected to drain Both Transistor are in Saturation

For M1 Transistor

⇒ (V - 8)2 = 4

V - 8 = ± 2

Taking +ve

V = 8 + 2 = 10 V (Not Possible)

Taking -ve

V = 8 - 2 = 6 V

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