- 2069 Asar · 8 marks
Forward biased pn junction band diagram
Short answer
Forward bias raises the n‑side bands, lowering the barrier to e(V₀−V). The junction law p_n(0)=p_n0 exp(eV/kT) describes minority injection. The total current is I = I₀(e^(eV/ηkT)−1) with η=1 for ideal diffusion; depletion width shrinks as √(V₀−V).
Explain with energy band diagram, the forward biased P-N junction and derive mathematical expressions for the same.
Forward bias means connecting the p side to the positive terminal and the n side to the negative terminal. The applied voltage opposes the built-in potential , lowers the barrier to , and allows a large diffusion current of majority carriers.
Energy band diagram
In equilibrium is flat and the bands bend by . Under forward bias the n-side bands are raised by relative to the p side (the n side is at lower potential), so the Fermi levels separate by and the bending is reduced to .
Equilibrium Forward bias V
p n p n
Ec_ Ec__
\___ Ec \__ Ec
EF ------- EF EFp --- -.- EFn
Ev_ barrier: eV gap
\___ Ev eV0 Ev__
\__ Ev
barrier e(V0-V)
Mathematical expressions
1. Barrier and width
2. Minority carrier injection (law of the junction). Using the Boltzmann relation across the lowered barrier:
with and .
3. Excess carriers decay by recombination in the neutral n region:
4. Diffusion current of holes at the edge :
Similarly for electrons in the p region:
5. Total current — Shockley diode equation
where for ideal diffusion current (Ge) and when recombination in the depletion layer dominates (Si at low current).
Key results
- The current rises exponentially with V; for , .
- Each 60 mV increase in V raises the current about ten times at 300 K.
- Noticeable conduction starts at the cut-in voltage (~0.7 V Si, ~0.3 V Ge).
✓ Answer checked by Claude Opus 5.5 (high effort)
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