Integer Answer Type Questions for JEE: Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. A potential difference of 2 V is applied between the opposite faces of a Ge crystal plate of area 1 cm² and thickness 0.5 mm. If the concentration of electrons in Ge is 2 x 1019 / m³ and mobilities of electrons and holes are  and  respectively, then the current flowing through the plate will be 

Ans. 0.64

Q.2. Ge and Si diodes conduct at 0.3 V and 0.7 V respectively. In the following figure if Ge diode connection are reversed, the valve of V0 changes by

Ans. 0.4
Consider the case when Ge and Si diodes are connected as show in the given figure.  Equivalent voltage drop across the combination Ge and Si diode ¾ 0.3 V
⇒ Current =  2.34 mA
∴ Out put voltage V₀ = Ri = 5 kΩ x  2.34 mA = 11.7 V  Now consider the case when diode connection are reversed. In this case voltage drop across the diode's combination ¾ 0.7 V
⇒ Current  =  2.26 mA
∴ V₀ = iR = 2.26 mA x 5 kΩ = 11.3 V  Hence charge in the value of V₀ = 11.7 - 11.3 = 0.4 V

Q.3. In the circuit shown in figure the maximum output voltage V₀ is

Ans. 5V
For the positive half cycle of input the resulting network is shown below
⇒

Q.4. The voltage gain of the following amplifier is 

Ans. 100
Voltage gain

Q.5. In an NPN transistor the collector current is 24 mA. If 80% of electrons reach collector its base current in mA is 

Ans. 6 mA
Given

By using

Q.6. In a common base amplifier circuit, calculate the charge in base current if that in the emitter current is 2 mA and a = 0.98 

Ans. 0.04 mA

∴

Q.7. A sinusoidal voltage of peak value 200 volt is connected to a diode and resistor R in the circuit shown so that half wave rectification occurs.  If the forward resistance of the diode is negligible compared to R the rms voltage (in volt) across R is approximately

Ans. 100 V

Q.8. The antenna current of an AM transmitter is 8 A when only carrier is sent but increases to 8.96 A when the carrier is sinusoidally modulated. The percentage modulation is

Ans: 71%
We know that
Here, = I_t 8.96A and I_c = 8A
∴
or = m² / 2 = 0.254 or m² = 0.508
or m = 0.71 = 71%

Q.9. The total power content of an AM wave is 1500 W. For 100% modulation, the power transmitted by the carrier is

Ans. 1000 W

∴
∵ m = 100% = 1= 1000 W

Q.10. The bit rate for a signal, which has a sampling rate of 8 kHz and where 16 quantisation levels have been used is

Ans. 32,000 bits/sec
If n is the number of bits per sample, then number of quantisation level = 2ⁿ
Since the number of quantisation level is 16
⇒ 2ⁿ = 16 ⇒ n = 4
bit rate = sampling rate  x no. of bits per sample
= 8000  x 4 = 32,000 bits/sec.

Q.11. An audio signal of amplitude one half the carrier amplitude is used in amplitude modulation. Calculate the modulation index?

Ans. 0.5
Here, E_m = 0.5 E_c
E_max = E_c + E_m = E_c + 0-5E_c = 1 -5E_c
E_min = E_c - E_m = E_c -0.5E_c = 0.5E_c

Q.12. Show that the minimum length of antenna required to transmit a radio signal of frequency 10 MHz is 30 m.

Ans. 30m
Here, f = 10 MHz = 107Hz

= 30 m

Q.13. A T.V. tower has a height of 100 m. How much population is covered by the T.V. broadcast if the average population density around the tower is 1000 per sq. km. Radius of the earth 6.37 x 10⁶ m. By how much height of the tower be increased to double its coverage range?

Ans. 300 m
Here, h= 100 m;
R = 6.37 x 10⁶ m;
Population density, ρ = 1000 km^-2 = 1000 x 10^-6m^-2 = 10^-3m ^-2
Population covered
=
= 40 x 10⁵ = 40 lakhs
Now, d’ =
or h’ = 4h = 4 x 100 = 400 m
Increase in height of tower = h’ – h = 400 – 100 = 300 m.

Q.14. How many AM broadcast stations can be accommodated in a 100 kHz bandwidth if the highest frequency modulating a carrier is 5 kHz?

Ans. 10 stations
Given: Total BW = 100 kHz
F_{a max.} = 5 kHz
Any station being modulated by a 5 kHz signal will produce an upper – side frequency 5 kHz above its carrier and a lower – side frequency 5 kHz below its carrier, thereby requiring a bandwidth of 10 kHz. Thus,
Number of stations accommodated  = Total BW / BW per station =
Number of stations accommodated = 10 stations