C12-Ch14-JEE Semiconductor and Electronics HOTS PDF

Semiconductor and Electronics – JEE HOTS 1. A silicon speciman is made into a P-type semi-conductor by dopping, on an average, one Indium atom per 5  10 7 silicon atoms. If the number density of atoms in the silicon specimen is 5  10 28 atoms / m 3 then the number of acceptor atoms in...

Semiconductor and Electronics – JEE HOTS 1. A silicon speciman is made into a P-type semi-conductor by dopping, on an average, one Indium atom per 5  10 7 silicon atoms. If the number density of atoms in the silicon specimen is 5  10 28 atoms / m 3 then the number of acceptor atoms in silicon per cubic centimetre will be [MP PMT 1993, 2003] (a) 2.5  10 atoms / cm 30 3 (b) 1.0  10 atoms / cm 13 3 (c) 1.0  10 atoms / cm 15 (d) 2.5  10 36 atoms / cm 3 3 2. The probability of electrons to be found in the conduction band of an intrinsic semiconductor at a finite temperature [IIT-JEE 1995; DPMT 2004] (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 3. The typical ionisation energy of a donor in silicon is [IIT-JEE 1992] (a) 10.0 eV (b) 1.0 eV (c) 0.1 eV (d) 0.001 eV 4. In PN-junction diode the reverse saturation current is 10 −5 amp at 27 C. The forward current for a voltage of 0.2volt is [MP PMT 1993] −3 −3 (a) 2037.6  10 amp (b) 203.76  10 amp (c) 20.376  10 −3 amp (d) 2.0376  10 3 amp [exp(7.62) = 2038.6, K = 1.4  10 −23 J / K] 5. When a potential difference is applied across, the current passing through [IIT- JEE 1999] (a) An insulator at 0 K is zero (b) A semiconductor at 0 K is zero (c) A metal at 0 K is finite (d) A P-N diode at 300 K is finite, if it is reverse biased 6. A 2V battery is connected across the points A and B as shown in the figure given below. Assuming that the resistance of each diode is zero in forward bias and infinity in reverse bias, the current supplied by the battery when its positive terminal is connected to A is [UPSEAT 2002] 10  (a) 0.2 A 10  (b) 0.4 A (c) Zero (d) 0.1 A A B 7. In the circuit, if the forward voltage drop for the diode is 0.5V, the current will be [UPSEAT 2003] 0.5V (a) 3.4 mA (b) 2 mA 8V 2.2K 1|Page  (c) 2.5 mA (d) 3 mA 8. A P-type semiconductor has acceptor levels 57 meV above the valence band. The maximum wavelength of light required to create a hole is (Planck’s constant h = 6.6  10 −34 J-s) [MP PET 1995] (a) 57 Å (b) 57  10 −3 Å (c) 217100 Å (d) 11.61  10 −33 Å 9. Current in the circuit will be [CBSE PMT 2001] 5 (a) A 40 20  5 (b) A 50 30  5 (c) A i 10 5 20  5V (d) A 20 10. 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 [CBSE PMT 1997] R 0.5 V (a) 1.5  (b) 5  (c) 6.67  (d) 200  1.5 V 11. For a transistor amplifier in common emitter configuration for load impedance of 1 k (hfe = 50 and hoe = 25 A/V) the current gain is [AIEEE 2004] (a) – 5.2 (b) – 15.7 (c) – 24.8 (d) – 48.78 12. In the following common emitter configuration an NPN transistor with current gain  = 100 is used. The output voltage of the amplifier will be [AIIMS 2003] (a) 10 mV (b) 0.1 V 10K 1K 1mV Vout (c) 1.0 V (d) 10 V 13. In semiconductor the concentrations of electrons and holes are 8  1018/m3 and 5  1018/m respectively. If the mobilities of electrons and hole are 2.3 m2/volt-sec and 0.01 m2/volt-sec respectively, then semiconductor is (a) N-type and its resistivity is 0.34 ohm-metre (b) P-type and its resistivity is 0.034 ohm-metre (c) N-type and its resistivity is 0.034 ohm-metre (d) P-type and its resistivity is 3.40 ohm-metre 2|Page 14. 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 E0= 200 Volt R (a) 200 (b) 100 200 (c) (d) 280 2 15. 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 characteristic curve. The voltage across the diode is independent of current above the knee point. If VB = 5 V, then the maximum value of R so that the voltage is above the knee point, will be R 0.7 V VB (a) 4.3 k (b) 860 k (c) 4.3  (d) 860  16. In the circuit given below, V(t) is the sinusoidal voltage source, voltage drop VAB(t) across the resistance R is [IIT 1993] D1 D2 R VAB R1=100  R2=150  V(t) (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 17. The peak voltage in the output of a half-wave diode rectifier fed with a sinusoidal signal without filter is 10 V. The dc component of the output voltage is [CBSE PMT 2004] (a) 10 / 2 V (b) 10/ V (c) 10 V (d) 20/ V 18. A transistor is used as an amplifier in CB mode with a load resistance of 5 k  the current gain of amplifier is 0.98 and the input resistance is 70 , the voltage gain and power gain respectively are [Pb. PET 2003] 3|Page  (a) 70, 68.6 (b) 80, 75.6 (c) 60, 66.6 (d) 90, 96.6 19. The Bohr radius of the fifth electron of phosphorus (atomic number = 15) acting as dopant in silicon (relative dielectric constant = 12) is (a) 10.6 Å (b) 0.53 Å (c) 21.2 Å (d) None of these 20. 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 D1 R D2 R D3 R A B (i) – 10 V, – 5V (ii) – 5V, – 10 V (iii) – 4V, – 12V (a) (i) < (ii) < (iii) (b) (iii) < (ii) < (i) (c) (ii) = (iii) < (i) (d) (i) = (iii) < (ii) 21. 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 [IIT-JEE 1997] 150 D1 50 D2 100 6V (a) Zero (b) 0.02 (c) 0.03 (d) 0.036 22. Find VAB [RPMT 2000] (a) 10 V 10 (b) 20 V 30V (c) 30 V VAB 10 10 (d) None of these 23. A diode is connected to 220 V (rms) ac in series with a capacitor as shown in figure. The voltage across the capacitor is (a) 220 V 4|Page 220 V C ac  (b) 110 V (c) 311.1 V 220 (d) V 2 24. A potential difference of 2V 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  1019/m3 and mobilities m2 m2 of electrons and holes are 0.36 and 0.14 respectively, then the current flowing volt − sec volt − sec through the plate will be (a) 0.25 A (b) 0.45 A (c) 0.56 A (d) 0.64 A 25. The contribution in the total current flowing through a semiconductor due to electrons and holes 3 1 5 are and respectively. If the drift velocity of electrons is times that of holes at this 4 4 2 temperature, then the ratio of concentration of electrons and holes is (a) 6 : 5 (b) 5 : 6 (c) 3 : 2 (d) 2 : 3 26. 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 [Based on Roorkee 2000] Ge V0 12 V Si 5 k (a) 0.2 V (b) 0.4 V (c) 0.6 V (d) 0.8 V 27. In the circuit shown in figure the maximum output voltage V0 is + Vi D1 D2 10 V 2k T 0 + – T/2 t V0 2k 2k – (a) 0 V (b) 5 V 5 (c) 10 V (d) V 2 28. In the following circuit find I1 and I2 2k (a) 0, 0 i2 (b) 5 mA, 5 mA 10 V 14k 12k 5|Page i1  (c) 5 mA, 0 (d) 0, 5 mA 29. For the transistor circuit shown below, if  = 100, voltage drop between emitter and base is 0.7 V then value of VCE will be 100  (a) 10 V C 8.6 k 18 V (b) 5 V VCE B (c) 13 V E (d) 0 V 5V 30. In NPN transistor, 10 electrons enters in emitter region in 10–6 sec. If 2% electrons are lost in 10 base region then collector current and current amplification factor () respectively are (a) 1.57 mA, 49 (b) 1.92 mA, 70 (c) 2 mA, 25 (d) 2.25 mA, 100 31. The following configuration of gate is equivalent to [AMU 1999] (a) NAND OR A B AND (b) XOR G1 Y NAND (c) OR G3 (d) None of these G2 32. Figure gives a system of logic gates. From the study of truth table it can be found that to produce a high output (1) at R, we must have (a) X = 0, Y = 1 X P (b) X = 1, Y = 1 Y R (c) X = 1, Y = 0 (d) X = 0, Y = 0 O 33. The combination of gates shown below produces (a) AND gate A G1 (b) XOR gate G3 G4 Y (c) NOR gate B G2 (d) NAND gate 34. The shows two NAND gates followed by a NOR gate. The system is equivalent to the following logic gate A X (a) OR (b) AND Z B (c) NAND (d) None of these C Y 6|Page 35. The diagram of a logic circuit is given below. The output F of the circuit is represented by (a) W.( X + Y ) W X (b) W  (X  Y ) F (c) W + (X  Y ) W (d) W + (X + Y ) Y 1 c 2 a 3 c 4 c 5 abd 6 a 7 a 8 c 9 b 10 b 11 d 12 c 13 a 14 b 15 a 16 d 17 b 18 a 19 a 20 c 21 b 22 a 23 d 24 d 25 a 26 b 27 b 28 d 29 c 30 a 31 b 32 c 33 d 34 b 35 c 7|Page