Gate 2003 Questions on Electronic Devices

Gate 2003 (2 mark)

Question 1: 

An n- type silicon bar 0.1 cm long and 100µm2 in a cross – sectional area has majority carrier concentration of 5 * 1020/m3 and carrier mobility is 0.13 m2/V-s at 300 k. If the charge of the electron is 1.6 * 1019 coulomb , then the resistance of the bar is,

a)106 ohm

b)104ohm

c)10-1ohm

d)10-4ohm

Answer : a

Explanation:

σ = nqµn

= 5 * 1020 * 1.6 * 10-19* 0.13 = 10.4

ρ = 1σ = 1 / 10.4

R = ρlA = 1*0.1*10-210.4*100*10-12 =106Ω

Question 2.

The electron concentration in a sample of uniformly doped n – type silicon at 300 k various linearly from 1017/cm3 at X = 0 to 6 * 1016/cm3 at X = 2µm . Assume a situation that electron are supplied to keep this concentration gradient constant with time. If electronic charge is 1.6 * 10-19 coulomb and the diffusion constant Dn = 35 cm2/ s, the current density in the silicon ,if no electric field is present ,is

a)zero

b)-1120A/cm2

c)-560A/cm2

d) +1120A/cm2

Answer : b

Explanation:

Jn =nqµeE +Dnqdndx

Here , E = 0

So, Jn =qDndndx

dndx = 6*1016-10172*10-4 = -2*1020

Jn =1.6*10-19*35*(-2*1020)

Jn =-1120A/cm2

Question 3:  

Match items in Group -1 with items in Group – 2, mostly suitable.

Group – 1

P.LED

Q.Avalanche photodiode

R. Tunnel diode

S. LASER

Group -2

1. Heavy doping

2. Coherent radistion

3. Spontaneous emission

4.Current gain

a)P – 1 ;Q – 2 ;R – 4 ;S – 3

b)P – 2 ;Q – 3; R – 1 ;S – 4

c)P – 3 ;Q – 4 ;R – 1 ;S – 2

d)P – 2 ;Q – 1 ;R – 4 ;S – 3

Answer : c

Question 4

A particular green LED emits light of wavelength 5490A°. The energy band gap of the semiconductor material used there is (Planck’s constant = 6.626 * 10-34 J-s)

a)2.26eV

b)1.98eV

c)1.17eV

d)0.74eV

Answer : a

Explanation:

Eg = 1.24λ(μm)

=1.245490*10-4 μm

=2.26eV

Question 5

When the gate – to –source voltage VGS of a MOSFET with threshold voltage of 400 mV, working in saturation is 900 mV, the drain current is observed to be 1 mA. Neglecting the channel width modulation effect and assuming that the MOSFET is operating at saturation , the drain current for an applied VGS of 1400 mV is

a)0.5mA

b)2.0mA

c)3.5mA

d)4.0mA

Answer : d

Explanation

ID = K(VGs - VT)2

1mA =K (900-400)2

K =1mA/V2104*25

When VGS = 1400

ID = K(VGs - VT)2

=1mA25* 104V2 (1400-400)2V2

=1mA*10025*104* 104

ID = 4mA

Question 6:

If P is Passivation ,Q is n – well implant , R is metallization and S is source/ drain diffusion , then the order in which they are carried out in standard n – well CMOS fabrication process , is

a)P –Q –R –S

b)Q –S –R –P

c)R –P –S –Q

d)S –R –Q –P

Answer : b