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Electronic Devices question

2021 · 20 Jul · Shift 2 · Q68
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Electronic Devices question

2021 · 20 Jul · Shift 2 · Q68

JEE MainPhysicsElectronic DevicesNumerical+4 / −1
For the forward biased diode characteristics shown in the figure, the dynamic resistance at ID = 3 mA will be ‾Ω\underline{\hspace{2cm}}\Omega​Ω. JEE Main 2021 (Online) 20th July Evening Shift Physics - Semiconductor Question 116 English
Numerical answer
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Correct answer: 25

  1. For a diode, the dynamic resistance at a given operating point is
rd=dVdI≈ΔVΔI r_d = \frac{dV}{dI} \approx \frac{\Delta V}{\Delta I}rd​=dIdV​≈ΔIΔV​

using the slope of the forward-bias VVV–III characteristic near that point.

  1. From the graph near ID=3 mAI_D = 3\,\text{mA}ID​=3mA, we read two nearby points symmetrically around it:
  • at ID=2 mAI_D = 2\,\text{mA}ID​=2mA, VD≈0.70 VV_D \approx 0.70\,\text{V}VD​≈0.70V
  • at ID=4 mAI_D = 4\,\text{mA}ID​=4mA, VD≈0.75 VV_D \approx 0.75\,\text{V}VD​≈0.75V
  1. Hence,
ΔV=0.75−0.70=0.05 V\Delta V = 0.75 - 0.70 = 0.05\,\text{V}ΔV=0.75−0.70=0.05V

and

ΔI=4 mA−2 mA=2 mA=2×10−3 A\Delta I = 4\,\text{mA} - 2\,\text{mA} = 2\,\text{mA} = 2\times 10^{-3}\,\text{A}ΔI=4mA−2mA=2mA=2×10−3A
  1. Therefore,
rd≈0.052×10−3=25 Ω r_d \approx \frac{0.05}{2\times 10^{-3}} = 25\,\Omegard​≈2×10−30.05​=25Ω
  1. So the dynamic resistance at ID=3 mAI_D = 3\,\text{mA}ID​=3mA is
25 Ω\boxed{25\,\Omega}25Ω​
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