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JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. The typical ionisation energy of a donor in silicon is
(a) 10.0 eV
(b) 1.0 eV

(c) 0.1 eV
(d) 0.001 eV

Correct Answer is option (c)
When donor impurity (+5 valence) added to a pure silicon (+4 valence), the +5 donor atom sits in the place of +4 valence silicon atom. so it has a net additional +1 electronic charge. The four valence electron form covalent bond and get fixed in the lattice. The fifth electron (with net –1 electronic charge) can be approximated to revolve around +1 additional charge. The situation is like the hydrogen atom for which energy is given by JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced. For the case of hydrogen, the permittivity was taken as ε0. However, if the medium has a permittivity εr , relative to ε0 , then JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced.
For Si, εr = 12 and for n ¾ 1, E = 0.1 eV

Q.2. In the circuit given below, V(t) is the sinusoidal voltage source, voltage drop VAB(t) across the resistance R is 
(a) Is half wave rectified 
(b) Is full wave rectified
(c) Has the same peak value in the positive and negative half cycles
(d) Has different peak values during positive and negative half cycle

Correct Answer is option (d)
In positive half cycle one diode is in forward biasing and other is in reverse biasing while in negative half cycle their polarity reverses, and direction of current is opposite through R for positive and negative half cycles so out put is not rectified.  Since R1 and R2 are different hence the peaks during positive half and negative half of the input signal will be different.

Q.3. In the following circuits PN-junction diodes D1, D2 and D3 are ideal for the following potential of A and B, the correct increasing order of resistance between A and B will be
JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced(i) – 10 V, –5 V
(ii) –5 V, – 10 V
(iii) – 4 V, –12 V
(a) (i) < (ii) < (iii)
(b) (iii) < (ii) < (i)

(c) (ii) ¾ (iiii) < (i)
(d) (i) ¾ (iii) < (ii)

Correct Answer is option (c)
(i) VA =  -10 V and VB = -5 V  Diodes D1 and D3 are reveres biased and D2 is forward biased.
JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced
JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced(ii) When VA = -5V and VB = -10 V  Diodes D2 is reverse biased D1 and D3 are forward biased
JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced
(iii) In this case equivalent resistance between A and B is also R.
JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & AdvancedHence (ii) ¾ (iii) < (i).  

Q.4. A piece of copper and the other of germanium are cooled from the room temperature to 80 K, then which of the following would be a correct statement  
(a) Resistance of each increase  
(b) Resistance of each decrease  
(c) Resistance of copper increases while that of germanium decrease  
(d) Resistance of copper decrease while that of germanium increase

Correct Answer is option (d)
Resistance of conductors (Cu) decreases with decrease in temperature while that of semiconductors (Ge) increases with decrease in temperature. 

Q.5. Given below are symbols for some logic gates  The XOR gate and NOR gate respectively are 
JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced(a) 1 and 2
(b) 2 and 3
(c) 3 and 4
(d) 1 and 4 

Correct Answer is option (b)

Q.6. Which of the following gates will have an output of 1 
(a) JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced
(b) JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced
(c) JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced
(d) JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced

Correct Answer is option (c)
For 'NAND' gate (option c), output = JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced

Q.7. This symbol represents
JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced(a) NOT gate
(b) OR gate
(c) AND gate
(d) NOR gate

Correct Answer is option (a)
JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced - NOT Gate

Q.8. Which logic gate is represented by the following combination of logic gates?
JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced(a) OR
(b) NAND

(c) AND
(d) NOR

Correct Answer is Option (c)
JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced
JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced
According to De Morgan's theorem
JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced
This is output equation of 'AND' gate.

Q.9. For a given plate-voltage, the plate current in a triode is maximum when the potential of
(a) The grid is positive and plate is negative
(b) The grid is positive and plate is positive
(c) The grid is zero and plate is positive
(d) The grid is negative and plate is positive

Correct Answer is option (b)
When grid is given positive potential more electrons will cross the grid to reach the positive plate P. Hence current increases.
JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced

Q.10. A silicon specimen is made into a P-type semiconductor by doping, on an average, one Indium atom per 5 x 107 silicon atoms. If the number density of atoms in the silicon specimen is 5 x 1028 atoms / m3 then the number of acceptor atoms in silicon per cubic centimetre will be
(a) 2.5 x 1030 atoms / cm3
(b) 1.0 x 1013 atoms / cm3
(c) 1.0  x 1015 atoms / cm3
(d) 2.5 x 1036 atoms / cm3 

Correct Answer is option (c)
Number density of atoms in silicon specimen = 5 x 1028 atom/m3  = 5  x 1022 atom/cm3  Since one atom of indium is doped in =5 x 107 Si atom. So number of indium atoms doped per cm3 of silicon JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advancedatom/cm3

Q.11. The probability of electrons to be found in the conduction band of an intrinsic semiconductor at a finite temperature
(a) Decreases exponentially with increasing band gap
(b) Increases exponentially with increasing band gap
(c) Decreases with increasing temperature
(d) Is independent of the temperature and the band gap

Correct Answer is option (a)
The probability of electrons to be found in the conduction band of an intrinsic semiconductor JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced where k = Boltzman's constant.
Hence, at a finite temperature, the probability decreases exponentially with increasing band gap.

Q.12. A transistor is used as an amplifier in CB mode with a load resistance of 5 kW the current gain of amplifier is 0.98 and the input resistance is 70 W, the voltage gain and power gain respectively are
(a) 70, 68.6
(b) 80, 75.6  

(c) 60, 66.6
(d) 90, 96.6 

Correct Answer is option (a)
In common base mode α = 0.98 , R = 5 kΩ , Rin ¾ 70Ω
∴ voltage gain  
JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced
Power gain ¾ Current gain x  Voltage gain=0.98 x 70 = 68.6

Q.13. The circuit shown in following figure contains two diode D1 and D2 each with a forward resistance of 50 ohms and with infinite backward resistance. If the battery voltage is 6 V, the current through the 100 ohm resistance (in amperes) is
JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced(a) Zero
(b) 0.02 
(c) 0.03
(d) 0.036

Correct Answer is option (b)
According to the given polarity, diode D1 is forward biased while D2 is reverse biased. Hence current will pass through D1 only.
So, current
JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced

Q.14. In the circuit, if the forward voltage drop for the diode is 0.5 V, the current will be
JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced(a) 3.4 mA
(b) 2 mA
(c) 2.5 mA
(d) 3 mA

Correct Answer is option (a)
The voltage drop across resistance ¾ 8 – 0.5 ¾ 7.5 V
∴ Current JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced  = 3.4 mA

Q.15. The diode used in the circuit shown in the figure has a constant voltage drop of 0.5 V at all currents and a maximum power rating of 100 milliwatts. What should be the value of the resistor R, connected in series with the diode for obtaining maximum current?
JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced(a) 1.5 W
(b) 5 W
(c) 6.67 W
(d) 200 W

Correct Answer is option (b)
The current through circuit JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced  = 0.2 A
∴ voltage drop across resistance = 1.5 - 0.5 = 1 V
JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced

Q.16. The junction diode in the following circuit requires a minimum current of 1 mA to be above the knee point (0.7 V) of its I-V characteristics curve. Te voltage across the diode is independent of current above the knee point. If VB ¾ 5 V, then the maximum value of R so that the voltage is above the knee point, will be
JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced(a) 4.3 kΩ
(b) 860 kΩ
(c) 4.3 Ω
(d) 860 Ω

Correct Answer is option (a)
At knee point voltage across the diode is 0.7 V.
Hence voltage across resistance R is 5 - 0.7 = 4.3 V
⇒ using V = iR ⇒ 4.3 = 1 x 10-3  x  R ⇒ R = 4.3 kΩ

Q.17. Find VAB
JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced(a) 10 V
(b) 20 V
(c) 30 V
(d) None of these 

Correct Answer is option (a)
Diode is in forwards biasing hence the circuit can be redrawn as follows
JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & AdvancedJEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced

Q.18. The combination of the gates shown in the figure below produces
JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced(a) NOR gate
(b) OR gate
(c) AND gate
(d) XOR gate 

Correct Answer is option (b)
JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & AdvancedJEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced
This output equation is equivalent to OR gate.

Q.19. The output of OR gate is 1
(a) If both inputs are zero
(b) If either or both inputs are 1
(c) Only if both input are 1
(d) If either input is zero 

Correct Answer is Option (b)
The output of OR gate is Y = A + B

Q.20. In a transistor configuration b-parameter is
(a) 1b / 1c
(b) 1c / 1b
(c) 1c / 1a
(d) 1a / 1c

Correct Answer is option (b)
JEE Advanced (Single Correct Type): Semiconductor Electronics & Communication Systems | Chapter-wise Tests for JEE Main & Advanced

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