Gate 2006 Questions on Electronic Devices

GATE – 2006 (1 mark)

Question 1.

The value of voltage (VD) across a tunnel – diode corresponding to peak and valley currents are Vpand Vv respectively . The range of tunnel – diode voltage VD for which the slope of its I – VD characteristics is negative would be

a)VD < 0

b)0 ≤ VD < Vp

c)Vp≤ VD<Vv

d)VD ≥ Vv

Answer : c

(diagram)

Question 2:

The concentration of minority carriers in an extrinsic semiconductor under equilibrium is

a) directly proportional to the doping concentration.

b)inversely proportional to the doping concentration.

c)directly proportional to the intrinsic concentration.

d)inversely proportional to the intrinsic concentration.

Answer : b

Explanation

np = ni2

ni= constant

For n- type p is minority carrier concentration

P = ni2n , P ∝ 1n

Question 3:

Under low level injection assumption , the injection minority carrier current for an extrinsic semiconductor is essentially the

a) diffusion current .

b)drift current.

c)recombination current .

d)induced current.

Answer : a

Question 4:

The phenomenon known as” Early Effect” in a bipolar transistor refers to reduction of the effective base – width caused by

a)electron - hole recombination at the base.

b)the reverse biasing of the base – collector junction.

c)the forward biasing of emitter – base junction .

d)the early removal of the stored base charge during saturation – to – cutoff switching .

Answer : b

Gate 2006 (2 mark)

Question 1

In the circuit shown below , the switch was connected to position 1 at t< 0 and at t = 0 , it is changed to position 2 . Assume that the diode has zero voltage drop and a storage time ts .For 0 < t ≤ ts, VR is given by ( all in volts )(diagram)

a) = -5

b) = +5

c)0 ≤VR < 5

d) -5 < VR < 0

Answer : a

Explanation

Diode retains resistance of forward bias condition in reverse bias (ideally zero resistance ) for the time interval of storage time.

So , VR= -5 V

Question 2

The majority carriers in n- type semiconductor have an average drift velocity V in a direction perpendicular to a uniform magnetic to field B .The electric field E induced due Hall effect acts in the direction

a) v * B

b)B * v

c)along v

d)opposite to v

Answer : b

Explanation:

Hall effect

Electric force + magnetic force = 0

qE +qv * B = 0

E = -v * B

E = B * v

Question 3

Find the correct match between group 1 and group 2

Group 1

E : Varactor diode

F : PIN diode

G : Zener diode

H : Schottky diode

Group 2

1. Voltage reference

2. High – frequency switch

3. Tuned circuits

4. Current controlled attenuator .

a) E – 4,F – 2 ,G – 1 ,H – 3.

b)E – 2,F – 4 ,G – 1 ,H – 3.

c)E – 3 ,F – 4, G – 1 ,H – 2.

d)E – 1,F – 3 ,G – 2 ,H – 4.

Answer : c

Question 4:

A heavily doped n – typed semiconductor has the following data :

Hole electron mobility ratio : 0.4

Doping concentration : 4.2 * 103 atoms/m3

Intrinsic concentration :1.5 * 104atoms/m3

The ratio of conductance of the n- type semiconductor to that of intrinsic semiconductor of same material and at the same temperature is given by

a)0.00005

b)2,000

c)10,000

d)20,000

Answer : d

Explanation

For n –type semiconductor σi = nqµn

For intrinsic semiconductor ,

σi = niq(µn+µp)

σnσi = nµnni( µn+µp)

= 4.2*108*µn1.5*104*µn ( 1+ µpµn)

= 4.2*1081.5*104*1.4 = 2 * 104