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JEE Main 2014 · P-N Junction Diode

P-N Junction Diode question from JEE Main 2014

Q103JEE Main 2014Difficult

P-N Junction Diode

The current voltage relation of diode is given by I=(e1000V/T−1) mA,I = \left(e^{1000V/T} - 1\right)\text{ mA}, where the applied voltage VV is in volts and the temperature TT is in degree kelvin. If a student makes an error measuring ±0.01 V\pm 0.01\,V while measuring the current of 5 mA5\,mA at 300 K,300\,K, what will be the error in the value of current in mAmA?

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Answer: (a)

  1. Differentiate: dIdV=1000Te1000V/T=1000T(I+1)\frac{dI}{dV} = \frac{1000}{T} e^{1000V/T} = \frac{1000}{T}(I + 1) mA/V.
  2. At I=5I = 5 mA and T=300T = 300 K: dIdV=1000300×6=20\frac{dI}{dV} = \frac{1000}{300} \times 6 = 20 mA/V.
  3. ΔI=20×0.01=0.2\Delta I = 20 \times 0.01 = 0.2 mA.
  4. So the answer is (a) 0.20.2 mA.
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