An ideal p - n junction diode has a current-voltage characteristic described by i = i_0 \! (e^ eV/kT - 1) , in which the drift current i_0 is 10 \ muA.The temperature T is taken as 300 K.Determine the voltage V_0 such

Correct answer: A. 0.12 V Given the condition: e^ eV_0/kT = 100 Taking natural logarithm on both sides: (eV_0)/(kT) = (100) (eV_0)/(kT) = 2.303 × _ 10 (100) = 2.303 × 2 = 4.606 Rearranging for V_0 : V_0 = 4.606 × (kT)/(e) Substituting the standard values k = 1.38 × 10^ -23 J/K , T = 300 K , and e = 1.6 × 10^ -19 C : V_0 = 4.606 × 1.38 × 10^ -23 × 300 1.6 × 10^ -19 V_0 = 4.606 × 0.025875 V V_0 0.119 V 0.12 V

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

An ideal p - n junction diode has a current-voltage characteristic...

An ideal

p

n

junction diode has a current-voltage characteristic described by

i = i_0 \! \left(e^{eV/kT} - 1\right)

, in which the drift current

i_0

10 \$mu\text{A}.The temperature

is taken as

300

\text{ K}.Determine the voltage

V_0

such that

e^{eV/kT} = 100

. (The term

may be neglected for voltages exceeding this value.)

Options

A $0.12 \text{ V}$
B $0.24 \text{ V}$
C $0.06 \text{ V}$
D $0.36 \text{ V}$

Correct answer

A. $0.12 \text{ V}$

Step-by-step solution

Given the condition:

e^{eV_0/kT} = 100

Taking natural logarithm on both sides:

\dfrac{eV_0}{kT} = \ln(100)

\dfrac{eV_0}{kT} = 2.303 \times \log_{10}(100) = 2.303 \times 2 = 4.606

Rearranging for

V_0

V_0 = 4.606 \times \dfrac{kT}{e}

Substituting the standard values

k = 1.38 \times 10^{-23} \text{ J/K}

T = 300 \text{ K}

, and

e = 1.6 \times 10^{-19} \text{ C}

V_0 = 4.606 \times \dfrac{1.38 \times 10^{-23} \times 300}{1.6 \times 10^{-19}}

V_0 = 4.606 \times 0.025875 \text{ V}

V_0 \approx 0.119 \text{ V} \approx 0.12 \text{ V}

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