p-n Junction Diode — Forward and reverse bias, I-V curve
Semiconductor
11
JEE Qs
8%
Hard
75
min
Master the qualitative mechanisms of current flow and depletion region changes under both biases, and accurately interpret the I-V curve including knee voltage, reverse saturation, and breakdown.
🧮 Key Formulas
✅ Key Points for JEE
- 1Forward bias reduces the depletion width and barrier potential, allowing majority carriers to diffuse across the junction, resulting in an exponential increase in current above the knee voltage.
- 2Reverse bias increases the depletion width and barrier potential, virtually stopping majority carrier flow, leading to a small, constant reverse saturation current due to minority carrier drift.
- 3The I-V characteristic curve clearly shows the non-linear behavior, with a sharp exponential rise in forward current after the knee (cut-in) voltage and a nearly constant, very small current in reverse bias until breakdown.
- 4Knee voltage (typically 0.7V for Si, 0.3V for Ge) is the minimum forward voltage required to significantly overcome the barrier potential and allow substantial current flow.
- 5Breakdown voltage in reverse bias is the point where the electric field in the depletion region becomes strong enough to cause avalanche or Zener breakdown, leading to a sharp increase in reverse current.
⚠️ Common Mistakes
- ✕Confusing the roles of majority and minority carriers in forward versus reverse bias, or assuming reverse current is zero.
- ✕Not understanding that reverse saturation current is primarily due to minority carriers and is largely independent of applied reverse voltage until breakdown.
- ✕Overlooking the significance of the knee (cut-in) voltage and assuming a linear relationship between voltage and current in forward bias.
📝 Practice Questions
See allQ27.Consider the following statements: A. The junction area of solar cell is made very narrow compared to a photo diode. B. Solar cells are not connected with any external bias. C. LED is made of lightly doped p-n junction. D. Increase of forward current results in continuous increase of LED light intensity. E. LEDs have to be connected in forward bias for emission of light. Choose the correct answer from the options given below: (1) B, E Only (2) B, D, E Only (3) A, C Only (4) A, C, E Only
Q43.Which of the following circuits has the same outpur as that of the given circuit? (1) (2) (3) (4)
Q33.Which of the following circuits represents a forward biased diode? Choose the correct answer from the options given below : (1) (A) and (D) only (2) (B), (D) and (E) only (3) (C) and (E) only (4) (B), (C) and (E) only
Q26. A B Y 0 0 1 0 1 1 1 0 0 1 1 1 To obtain the given truth table, following logic gate should be placed at G: 2025 (22 Jan Shift 2) JEE Main Previous Year Paper (1) OR Gate (2) AND Gate (3) NOR Gate (4) NAND Gate
Q28.The output of the circuit is low (zero) for : (A) X = 0, Y = 0 (B) X = 0, Y = 1 (C) X = 1, Y = 0 (D) X = 1, Y = 1 Choose the correct answer from the options given below : (1) (B), (C) and (D) only (2) (A), (B) and (C) only (3) (A), (C) and (D) only (4) (A), (B) and (D) only
Q45. In the circuit shown here, assuming threshold voltage of diode is negligibly small, then voltage VAB is correctly represented by : (1) VAB would be zero at all times (2) (3) (4)
NCERT Chapters
- Class 12 Physics Ch 14: Semiconductor Electronics: Materials, Devices and Simple Circuits