Gate 2004 Questions on Electronic Devices

Gate 2004 (1 mark )

Question 1

The impurity commonly used for realizing the base region of a silicon n-p-n transistor is

a)Gallium

b)Indium

c)Boron

d)Phosphorus

Answer : c

Question 2

If silicon n-p-n transistor, the base –to –emitter voltage VBE is 0.7 and the collector –to –base voltage VCB is 0.2 V , then the transistor is operating in the

a)normal active mode

b)saturation mode

c)inverse active mode

d)cutoff mode

Answer : a

Question3

Consider the following statement S1and S2

S1: The β of a bipolar transistor reduces if the base width is increased.

S2: The β of a bipolar transistor increases if the doping concentration in the base increased.

Which one of the following is correct?

a)S1is false and S2is true

b) both S1 and S2are true

c)both S1and S2are false

d)S1is true and S2 is false

Answer : d

Explanation:

β = ICIB = α1-α

when base width increases , recombination in base region increases and α decreases & hence β decreases.

If doping in base region increases , then recombination in base increases and α decreases thereby decreasing β.

Question 4

Given figure is the voltage transfer characteristic of (diagram)

a)an NMOS inverter with enhancement mode transistor as load.

b)an NMOS inverter with depletion mode transistor as load.

c)as CMOS inverter.

d)as BJT inverter.

 Answer : c

Gate 2004 (2 marks)

Question1 

In an abrupt p-n junction ,the doping concentration of the p – side and n – side are NA = 9 * 1016/cm3 and ND= 1*1016/cm3 respectively . The p- n junction is reversed biased and the total depletion width is 3μm. The depletion width on the p- side is

a)2.7μm

b)0.3μm

c)2.25μm

d)0.75μm

Answer : b

Explanation:

ωnωp = NAND ω ≅ ωn (ωn≫ωp)

ωp = ωn*NDNA = 3µm*10169*1016 = 0.3µm

Question 2 

The resistivity of a uniform doped n – type silicon sample is 0.5Ω - cm. If the electron mobility (µn) is 1250 cm2 / V – sec and the charge of an electron is 1.6 *10-19 coulomb , the donor impurity concentration (ND) in the sample is

a) 2* 1016/cm3

b)1 * 1016/cm3

c)2.5 *1015/cm3

d) 2 * 1015/cm3

Answer : b

Explanation:
ρ = 1nqµn

n = ND

ND = 1qµnρ = 11.6* 10-19*1250*0.5 = 1016 /cm3