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

NEET REVISION 14. SEMICONDUCTOR ELECTRONICS- MATERIALS- DEVICES AND SIMPLE CIRCUITS(H) NEET REVISION Date: March 18, 2025 Dura on: 1:00:00 Total Marks: 180 INSTRUCTIONS INSTRUCTIONS PHYSICS 1. () : Explana on Using De - Morgan's law Thus, it is equivalent to gate The truth table for the given circuit will be 2. () : Explana on If is the number of charge carriers at tempera‐ ture and at , then, and . The frac onal increase as the temperature is raised from to is, Now, Given that, 3. () : Explana on The diode is in reverse biasing so current through it is zero. 4. () : Explana on or ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ A + B = A ̄ ⋅ B ̄ ∴ Y = ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ a ̄ + ̄b = ̄ ̄ ̄a ̄ ⋅ ̄ ̄ ̄b = a ⋅ b AND n1 T1 n2 T2 n1 = n0e (−Eg/2kT1) n2 = n0e (−Eg/2kT2) T1 T2 Δn = = ( − 1) n2 − n1 n1 n2 n1 ∴ Δn = ⎡ ⎢⎢ ⎣ e ⎛ ⎝ − ⎞ ⎠ − 1 ⎤ ⎥⎥ ⎦ Eg 2k 1 T1 1 T2 (1) ( − ) Eg 2k 1 T1 1 T2 = ( − ) 0.64×1.6×10 −19 2×1.4×10 −23 1 300 1 384 = ( ) 32×8×10 2 7 84 384×300 = = 32×12×8 384×3 8 3 ∴ e ⎛ ⎝ − ⎞ ⎠ = e Eg 2k 1 T1 1 T2 8 3 ln(14.4) = 8 3 ∴ e = 14.4 8 3 ∴ Δn = (14.4 − 1) [From(1)] ∴ Δn = 13.4 nenh = n 2 ne = = n 2 nh 10 16×10 16 4.5×10 22 = 10 m−3 32 4.5×10 22

NEET REVISION 15. () : Explana on 16. () : Explana on Transistor as a common emi er amplifier, Voltage gain, Resistance gain Power gain Resistance gain 17. () : Explana on Given, and Current gain, we know: 18. () : Explana on 19. () : Explana on Here, change in base current, Change in collector current, The current gain is 20. () : Explana on Given, cri cal wavelength Energy band gap, So, 21. () : Explana on If is increased, will decrease. Since it will result in decrease in 22. () : Explana on When poten al is 80 the drop across Zener and Load is 50 Load current and current across Similarly when poten al is 120 23. () : Explana on Here Now, or Also, or or 24. () : Explana on Applying to the closed part of the circuit con‐ taining and emi er base junc on, we have Applying to the closed part of the circuit containing and base collector junc on, we have As both are in forward biasing condi on, transistor is in satura on state. 25. () : Explana on and = eneμe 1 ρ ρ = 1 1.6×10 −19×10 −13×1200 = 520.9 Ω cm AV = ΔVo ΔVi = βAC× = β × RL r = ΔP0 ΔPi = β 2 AC× = β 2 × RL r β = 49 ΔIB = 5μA β = ΔIC ΔIB ⇒ ΔIC = β ΔIB = 49 × 5 = 245μA IE = IB + IC ⇒ ΔIE = ΔIB + ΔIC = (245 + 5)μA = 250μA E = = (ev) = = 0.5eV hc λ 12000 24800A0 1 2 ΔIB = 300 μA − 100μA = 200 μA ΔIC = 20 mA − 10 mA = 10 mA β = = = = 50 ΔIC ΔIB 10 mA 200 μA 10×10 −3 A 200×10 −6 A λc = = 660 nm Eg = = hc λ 6.6 × 10 −34 × 3 × 10 8 660 × 10 −9 = 3 × 10 −19 J = eV 3×10−19 1.6×10−19 = = eV 30 16 15 8 ⇒ X = 15 IB = . VBB−VBE R1 R1 IB IC = βIB, IC. V V ∴ = = 5mA 50 10k 5kΩ = 6mA 30V 5kΩ iZener + iload = 6mA iZener = 6mA − iload = 6mA − 5mA = 1mA V iR = = = 14mA 120−50 5k 70 5k iZ = iR − iload = 14 − 5 = 9mA Vi = 10 V , VBE = 0, VCE = 0, VCC = 10 V . RB = 400 kΩ = 400 × 10 3Ω RC = 3 kΩ = 3 × 10 3Ω Vi − VBE = RBIB ∴ 10 − 0 = (400 × 10 3) IB IB = = 25 × 10 −6 A 10 400×10 3 VCC − VCE = ICRC 10 − 0 = IC × 3 × 10 3 IC = = 3.33 × 10 −3 A 10 3×10 3 ∴ β = = = 133 IC IB 3.33 × 10 −3 25 × 10 −6 Ib = = = 1.5 × 10 −5A IC β 1.5×10−3 100 KV L Vcc = Vbe + IbRb Vbe = Vcc − IbRb = 24 − (1.5 × 10 −5) (220 × 10 3) = 20.7 V KV L Rb , Rc Rb VBC − IbRb + IcRc = 0 VBC = IbRb − IcRc = 3.75V VBE&VBC Y = (A + B ̄) ⋅ (A ̄ ⋅ B) = A ⋅ A ̄ ⋅ B + B ̄ ⋅ A ̄ ⋅ B = [∵ A ⋅ A ̄ = 0 B ⋅ B ̄ = 0]