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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 VV opposes the built-in potential V0V_0, lowers the barrier to e(V0−V)e(V_0 - V), and allows a large diffusion current of majority carriers.

Energy band diagram

In equilibrium EFE_F is flat and the bands bend by eV0eV_0. Under forward bias the n-side bands are raised by eVeV relative to the p side (the n side is at lower potential), so the Fermi levels separate by eVeV and the bending is reduced to e(V0−V)e(V_0 - V).

 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

W=2ε(Na+Nd)(V0−V)eNaNdW = \sqrt{\frac{2\varepsilon(N_a+N_d)(V_0-V)}{eN_aN_d}}

2. Minority carrier injection (law of the junction). Using the Boltzmann relation across the lowered barrier:

pn(0)=pn0exp⁡(eVkT),np(0)=np0exp⁡(eVkT)p_n(0) = p_{n0}\exp\left(\frac{eV}{kT}\right),\qquad n_p(0) = n_{p0}\exp\left(\frac{eV}{kT}\right)

with pn0=ni2/Ndp_{n0} = n_i^2/N_d and np0=ni2/Nan_{p0} = n_i^2/N_a.

3. Excess carriers decay by recombination in the neutral n region:

Δpn(x)=Δpn(0) e−x/Lh,Lh=Dhτh\Delta p_n(x) = \Delta p_n(0)\,e^{-x/L_h},\qquad L_h = \sqrt{D_h\tau_h}

4. Diffusion current of holes at the edge x=0x = 0:

Jh=−eDhdΔpndx∣0=eDhLh ni2Nd(eeV/kT−1)J_h = -eD_h\frac{d\Delta p_n}{dx}\Big|_{0} = \frac{eD_h}{L_h}\,\frac{n_i^2}{N_d}\left(e^{eV/kT}-1\right)

Similarly for electrons in the p region:

Je=eDeLe ni2Na(eeV/kT−1)J_e = \frac{eD_e}{L_e}\,\frac{n_i^2}{N_a}\left(e^{eV/kT}-1\right)

5. Total current — Shockley diode equation

I=I0(eeV/ηkT−1),I0=Aeni2(DhLhNd+DeLeNa)I = I_0\left(e^{eV/\eta kT}-1\right),\qquad I_0 = Aen_i^2\left(\frac{D_h}{L_hN_d}+\frac{D_e}{L_eN_a}\right)

where η=1\eta = 1 for ideal diffusion current (Ge) and η≈2\eta \approx 2 when recombination in the depletion layer dominates (Si at low current).

Key results

  • The current rises exponentially with V; for V≫kT/eV \gg kT/e, I≈I0eeV/kTI \approx I_0e^{eV/kT}.
  • 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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