Forward bias behavior

WHAT is forward bias?

WHY does the barrier drop? At equilibrium the junction has a built-in field pointing from N→P (from the exposed + donor ions to the − acceptor ions). This field is what stops diffusion. When you apply an external voltage with + on P, you create an opposing field that partially cancels the built-in field. Weaker field ⟹ narrower depletion region ⟹ lower barrier ⟹ diffusion resumes strongly.

HOW the current turns on

The number of carriers with enough thermal energy to cross a barrier of height $qV_B$ follows Boltzmann statistics: $n \propto e^{-qV_B/kT}$

Set barrier $V_B = V_{bi} - V$: $I \propto e^{-q(V_{bi}-V)/kT} = \underbrace{e^{-qV_{bi}/kT}}_{\text{const}} \cdot e^{qV/kT}$

To fix the constant, note that at $V=0$ the net current is zero (diffusion = drift). Drift current is fixed at $-I_S$ (reverse saturation). So the diffusion current at bias $V$ is $I_S\,e^{qV/kT}$, and net:

WHY the "−1"? At $V=0$: $I_S(1-1)=0$. ✔ It guarantees zero current at zero bias. In strong reverse bias $e^{V/V_T}\to 0$, giving $I\to -I_S$. ✔ The single equation covers both regions.

Deriving $V_T = kT/q$ from scratch

At $T=300$ K: $V_T = \frac{kT}{q} = \frac{(1.38\times10^{-23})(300)}{1.6\times10^{-19}} \approx 0.0259\text{ V} = 25.9\text{ mV}$

WHY care? $V_T$ sets how "sharp" the exponential is. Every $\approx 60$ mV of forward voltage multiplies the current by ~10 (since $\ln 10 \cdot V_T \approx 2.303 \times 25.9 \approx 60$ mV).

Worked Examples

Common Mistakes

Active Recall

Connections

• PN Junction at Equilibrium — where $V_{bi}$ and the depletion region come from.
• Reverse bias behavior — the opposite polarity; barrier grows, only $I_S$ flows.
• Built-in potential — the barrier $V_{bi}$ that forward bias fights against.
• Boltzmann distribution — origin of the exponential $e^{-qV_B/kT}$.
• Diode I-V characteristics — the full curve this note produces.
• Thermal voltage $V_T$ — the $kT/q$ scale used everywhere.

Concept Map

Hinglish (regional understanding)