Gate 2004 Questions on Electronic Devices

Gate 2004 (2 marks)

Question 3:

Consider an abrupt p – junction. Let Vbi be the built – in potential of the junction and VR be the applied reverse bias. If the junction capacitance (Cj) is 1 pF for Vbi+VR= 1V ,then for Vbi+VR = 4 V , Cjwill be

a)4pF

b)2pF

c)0.25pF

d)0.5pF

Answer : d

Explanation:

Cj ∝ Vj-1/2

Cj2Cj1 =V1V2 =14 =12

Cj2 =Cj22 = 1pF2 = 0.5pF

Question 4:

Consider the following statement S1and S2

S1 : The threshold voltage (VT)of MOS capacitor decreases with increases in the gate oxide thickness.

S2 : The threshold voltage (VT)of MOS capacitor decreases with increases in the substrate doping concentration.

Which one of the following is correct ?

a)S1 is false and S2 is true.

b)both S1 and S2are true.

c)both S1 and S2 are false.

d)S1 is true and S2 is false.

Answer : d

Explanation:

S2 : VT of N-MOS is increased by adding extra p – type impurities i.e. VT of a MOS increases by increase in substrate doping concentration .(True)

S1 :Cox = ϵoxtox tox = gate oxide thickeners.

VT = ϕGC-2ϕFQRCox- QoxCox

If Cox increases QBCox and QoxCox decreases and VT increases.

Cox Decreases when tox increases.

Question 5: 

The drain of an n-channel MOSFET is shorted to the gate so that VGS=VDS. The threshold voltage (VT) of MOSFET is 1 V. If drain current (ID) is 1mA for VGS= 2V , then for VGS = 3 V,ID is

a)2mA

b)3mA

c)9mA

d)4mA

Answer : d

Explanation

ID = K (VGS-VT)2

1mA =K (2-1)2

K = 1mA/V2

Again , ID = K (VGs-VT)2

= 1mAV2 (3-1)2 *V2

ID = 4mA

Question 6:

The longest wavelength that can be absorbed by silicon , which has the band width of 1.12 eV is 1.1µm . If the longest wavelength that can be absorbed by another material is 0.87 µm , then the band gap of this material is

a)1.416 eV

b)0.886 eV

c)0.854 eV

d)0.706 eV

Answer: b
Explanation

Eg = 1.24λ(μm) eVs

= 1.240.87µm eV = 1.425 eV

Question 7

The neutral base width of a bipolar transistor , biased in an active region , is 0.5µm .The maximum electron concentration and diffusion constant in the base are 1014 /cm3 and Dn = 25 cm3/ sec respectively . Assuming negligible recombination in the base, the collector current density is ( the electron charge is 1.6 * 10-19 coulomb)

a)800 A/cm2

b)8 A/cm2

c)200 A/cm2

d)2 A/cm2

Answer : b

Explanation:

JC = qDndndx

=1.6 *10-19 *25*10140.5*10-4

JC = 8 A/cm2