Tiêu đề 14. SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS.pdf ✅

14. SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS NEET PREPARATION (MEDIUM PHYSICS PAPER) Date: March 12, 2025 Dura on: 1:00:00 Total Marks: 180 INSTRUCTIONS INSTRUCTIONS PHYSICS 1. () : Explana on The typical ionisa on energy of donor in silicon is of the order of 2. () : Explana on In a reverse biased junc on, when the applied bias voltage is equal to the breakdown voltage, then voltage remains constant while current increases sharply. 3. () : Explana on In the given figure, side of junc on is at higher poten al than -side, junc on is forward biased. Taking its resistance to be zero (for ideal diode)and applying Ohm's law, we get 4. () : Explana on 5. () : Explana on The criteria for sustained stable oscilla ons is voltage amplifica on a enua on of feedback circuit. 6. () : Explana on Given that 7. () : Explana on To amplify the weak signal into a signal with a much higher voltage, the weak input signal is applied across the forward biased junc on and the output signal is taken across the reverse bi‐ ased junc on i.e., a transistor has to operate in ac ve region. So a transistor can not be used as an amplifier if the base and collector of it is in forward bias. So correct op on is (3) 8. () : Explana on Conceptual Ques on 9. () : Explana on Energy, Subs tu ng the values of , and in the above equa on As Signal of wavelength can be detected by the photodiode. 10. () : Explana on For OR gate if all inputs are zero then output is zero. If any one of the input is non zero then output is non zero. 0.1 eV . p − n p pn n ⇒ p−n I = = = 10 −2 A = 10 mA V R 5−2 300 i = = total P.D total resistence V1+V2 R+r AV β = 1 ∴ β = = = 0.02 1 AV 1 50 AV → β → RB = 1000 Ω & RL = 2000 Ω β = 100 = , Vo = 4V Ic IB = β ⇒ Vi = = 0.02 = 20 mV Vo Vi RL RB 4×1000 100×2000 E = hv = h c λ ⇒ λ = hc E h c E λ = = 5000 ∘ A 6.6×10 −34×3×10 8 2.5×1.6×10 −19 λ < Xmax = 4000 ∘ A 4000 ∘ A < 5000 ∘ A 4000 ∘ A

23. () : Explana on So, compared to , the waveform of is in‐ verted as well as amplified during the two inter‐ vals and Phase difference Asser on is true from circuit and as the tran‐ sistor is to be biased and is reverse biased. Reason is false. So correct op on is (3). 24. () : Explana on Here if and 25. () : Explana on Voltage drop across is and voltage drop across is Total current through the circuit is, Voltage across load resistance is, 26. () : Explana on In circuit ‘ ’, is connected with , which is not a series combina on of - junc on. In cir‐ cuit ‘ ’, each - junc on is forward biased. In circuit ‘ ’, each - junc on is reverse bi‐ ased, poten al will be distributed equally across two diodes in and . 27. () : Explana on Doping with a donor impurity (like ) in a semiconductor creates n-type semiconductors, where the extra free electron increases electron concentra on. So correct op on is (2). 28. () : Explana on A junc on photodiode can be operated un‐ der photovoltaic condi ons similar to that of a solar cell. The current - voltage characteris cs of a photodiode and solar cell under illumina‐ on are also similar. When light energy falls on the solar cell, then it converts solar energy into electrical energy. But the area of solar cell is larger, so that it can absorb more amount of sunlight and hence more amount of electrical energy is obtained. Hence, Asser on and Reason both are true but Reason is not the correct explana on of Asser on. 29. () : Explana on Two -junc ons placed back to back cannot work as transistor because in transistor the width and concentra on of doping of -type semiconductor is less as compared to width and doping of -type semiconductor type. Op on (4) is correct. 30. () : Explana on The symbols given in problems are (i) OR (ii) AND (iii) NOT (iv) NAND 31. () : Explana on In semiconductor, resis vity decreases with in‐ crease in temperature. 32. () : Explana on In semiconductors, the energy gap between conduc on band and valence band is small . Due to temperature rise, electron in the valence band gain thermal energy and may jumpy across the small energy gap, (to the con‐ duc on band). Thus conduc vity increases and hence resistance decreases. 33. () : Explana on The input current flowing into emi er is quite large as it is the sum of both the base current and the collector current respec vely, therefore the collector current output is less than emi er current and hence . Because, For amplifica on, the collector terminal of the transistor is reverse biased. 34. () : Explana on vi v0 (0 to ) T 2 ( to T) T 2 ⇒ = 180 ∘ = π ⇒ JCB ⇒ Y = ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ (A. B). ̄ ̄ ̄ ̄ C = ( ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ A. B) + C = ̄ ̄ ̄ ̄A + ̄ ̄ ̄ ̄B + C Y = 0 A = 1, B = 1 C = 0 D2 VB2 = 0.7 V D1 VB1 = 0.3 V . I = V − VB1 − VB2 Reff I = = = 3.5 × 10 −3 A (15−0.7−0.3) (1.5+2.5)×10 3 14 4×10 3 VL = IRL VL = 3.5 × 10 −3 × 2.5 × 10 3 ⇒ VL = 8.75 V A N N p n B p n C p n B C P, As pn pn npn p n (≈ 1eV ) α < 1 α = ΔIC ΔIE 2 1 1 1 0 0 (2) = 1 × 2 2 + 1 × 2 1 + 0 × 2 0 = 6