Verified Solution • 2025–26 Session moderate

QPages 419-421: Q22: HC Verma: Concepts of Physics - Volume 2 Semiconductors and Semiconductor Devices Pages 419-421: Q22 Solution: Consider a p-n junction diode having the characteristic where . …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently calculate the voltage required for the current to be double the saturation current ($i_0$) instead of double the *operating* current. To find the voltage increase needed to double the current, you must solve for the potential difference $\Delta V = \frac{kT}{e} \ln 2$ and add it to the initial voltage, rather than solving $e^{eV/kT} = 2$ from scratch.

Consider a p-n junction diode having the characteristic i=i0eeV/kT−1)i = i_0 e^{eV/kT} - 1) where i0=20 μAi_0 = 20 \ \mu\text{A}. The diode is operated at T = 300 K. a) Find the current through the diode when a voltage of 300 mV is applied across it in forward bias. b) At what voltage does the current double?

Step-by-Step Verified Solution

**Answer:** (a) i = 2 A (b) V = 318 mV. **Step-by-Step Explanation:** Given: i_ = 20 _A, T = 300 K. (a) At V = 300 mV: i = i_(e^(eV/kT) - 1). eV/kT = 0.3/0.02585 = 11.6. e^11.6 = 109000. i = 20 _ 10__ _ 109000 = 2.18 A � 2 A. (b) For current to double: i_/i_ = 2. e^(eV_/kT)/e^(eV_/kT) = 2 _ e^(e(V_-V_)/kT) = 2. _V = (kT/e)ln2 = 0.02585 _ 0.693 = 0.0179 V. V_ = 300 + 17.9 = 318 mV .

All Questions in this Chapter

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