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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Semiconductors and Semiconductor Devices

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Semiconductors and Semiconductor Devices

A p - n junction diode has the characteristic i = i_0\! (e^ eV/kT -...

AAp−-njunctiondiodehasthecharacteristicjunction diode has the characteristici = i_0\! \left(e^{eV/kT} - 1\right)wherewherei_0 = 20 \muA.Thediodeisoperatedatmu\text{A}.The diode is operated atT = 300  K.Determinethecurrentthroughthediodewhenaforwardbiasvoltageof\text{ K}.Determine the current through the diode when a forward bias voltage of300 \text{ mV}$is applied across it.

Options

  1. A0.2 A0.2 \text{ A}
  2. B2 A2 \text{ A}
  3. C4 A4 \text{ A}
  4. D20 mA20 \text{ mA}

Correct answer

B. 2 A2 \text{ A}

Step-by-step solution

Given:
Reverse saturation current, i0=20μA=20×10−6 Ai_0 = 20 \mu\text{A} = 20 \times 10^{-6} \text{ A}
Forward bias voltage, V=300 mV=0.3 VV = 300 \text{ mV} = 0.3 \text{ V}
Temperature, T=300 KT = 300 \text{ K}
Boltzmann constant, k=1.38×10−23 J/Kk = 1.38 \times 10^{-23} \text{ J/K}
Elementary charge, e=1.6×10−19 Ce = 1.6 \times 10^{-19} \text{ C}

First, calculate the exponent eVkT\dfrac{eV}{kT}:
eVkT=1.6×10−19×0.31.38×10−23×300\dfrac{eV}{kT} = \dfrac{1.6 \times 10^{-19} \times 0.3}{1.38 \times 10^{-23} \times 300}

eVkT=0.48×10−194.14×10−21=484.14≈11.6\dfrac{eV}{kT} = \dfrac{0.48 \times 10^{-19}}{4.14 \times 10^{-21}} = \dfrac{48}{4.14} \approx 11.6

The current through the diode is given by:
i=i0\lleft(eeVkTi = i_0 \lleft(e^{\dfrac{eV}{kT}} - 1 )<br><br>Since<br><br>Sincee^{11.6} \gg 1,wecanneglectthe, we can neglect the-1term:<br>term:<br>i \approxi_0 e^{\dfrac{eV}{kT}}

Using e11.6≈1.09×105e^{11.6} \approx1.09 \times10^5:
i≈20×10−6×1.09×105i \approx20 \times10^{-6} \times1.09 \times10^5

i≈2.18i \approx2.18\text{ A}

The closest value among the given options is 2 A.2 \text{ A}.
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