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Forward and reverse biased pn junction

Short answer

Forward bias (p positive) reduces the barrier to V₀−V, injects majority carriers and gives a large diffusion current I = I₀(e^(eV/kT)−1). Reverse bias raises the barrier to V₀+Vr, only minority carriers are swept across, resulting in a tiny saturation current I₀.

With the help of P-N junction, explain the phenomena of forward and reverse biased.

A pn junction in equilibrium has a depletion region of width W0W_0 and a built-in barrier V0V_0 that stops majority carriers from diffusing across. Applying an external voltage changes this barrier: lowering it is forward bias, raising it is reverse bias.

Forward bias (p to +, n to −)

   + |  p  |-|+|  n  | -
     | --> holes      |
     |      electrons<--
          W small, barrier V0 - V
  • The applied voltage VV opposes the built-in field. Barrier becomes V0−VV_0 - V; depletion width shrinks:
W=2ε(Na+Nd)(V0−V)eNaNdW = \sqrt{\frac{2\varepsilon(N_a+N_d)(V_0-V)}{eN_aN_d}}
  • Many more majority carriers can now cross. Holes are injected into the n side and electrons into the p side, where they become excess minority carriers:
pn(0)=pn0eeV/kT,np(0)=np0eeV/kTp_n(0) = p_{n0}e^{eV/kT},\qquad n_p(0) = n_{p0}e^{eV/kT}
  • These carriers diffuse away from the junction and recombine, giving a diffusion current that rises exponentially with V.
  • Large current (mA) flows once V exceeds the cut-in voltage (~0.6–0.7 V Si, ~0.2–0.3 V Ge).

Reverse bias (p to −, n to +)

   - |  p  |--|++|  n  | +
     |  majority pulled away  |
     | minority swept across  |
          W large, barrier V0 + Vr
  • The applied voltage adds to the built-in field; barrier becomes V0+VrV_0 + V_r and the depletion region widens.
  • Majority carriers cannot cross. Only minority carriers generated thermally near the junction are swept across by the field.
  • A tiny, almost constant reverse saturation current I0I_0 (nA in Si, µA in Ge) flows. It increases with temperature.
  • At a large reverse voltage, breakdown (Zener or avalanche) occurs and the current rises sharply.

Diode equation and characteristic

I=I0(eeV/ηkT−1)I = I_0\left(e^{eV/\eta kT} - 1\right)
        I (mA)
         |      /
         |     /  forward
         |    /
 Vbr ____|___/______ V
     |   |-I0 (uA/nA)
     |  reverse
PointForward biasReverse bias
Connectionp to +, n to −p to −, n to +
BarrierV0−VV_0 - V (lower)V0+VrV_0 + V_r (higher)
Depletion widthDecreasesIncreases
Current carriersMajority (injected)Minority
CurrentLarge, exponentialSmall, saturates at −I0-I_0
ResistanceLowVery high

✓ Answer checked by Claude Opus 5.5 (high effort)

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