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PN junction diode I-V characteristics

Short answer

A PN junction diode conducts when forward biased beyond its knee voltage (~0.7 V Si). In forward bias, the depletion region narrows and current rises exponentially; in reverse bias, only a small saturation current I0 flows until breakdown voltage VBR, where current surges. The diode obeys I = I0(e^(V/ηVT)−1).

What is a diode? Explain the working of a PN junction diode and its I-V characteristics for forward and reverse bias (with circuit diagram and graph).

A diode is a two-terminal semiconductor device that allows current to flow easily in one direction (forward) and blocks it in the opposite direction (reverse). A PN junction diode is formed by joining p-type and n-type semiconductors; the terminals are anode (P) and cathode (N).

Working of a PN junction diode

Near the junction, electrons from the n-side diffuse to the p-side and holes diffuse the other way. This leaves fixed ions, forming a depletion region with a built-in barrier potential of about 0.7 V for silicon (0.3 V for germanium).

Forward bias (P to positive, N to negative)

  • The external voltage opposes the barrier. The depletion region becomes narrow.
  • Until the applied voltage reaches the cut-in (knee) voltage, only a very small current flows.
  • Above about 0.7 V (Si), the barrier is overcome and current rises rapidly (exponentially). The majority carriers carry this current.

Reverse bias (P to negative, N to positive)

  • The external voltage adds to the barrier. The depletion region widens.
  • Only the minority carriers move, giving a very small reverse saturation current I0 (nA to µA), almost constant with voltage.
  • If the reverse voltage is increased to the breakdown voltage VBR, the current increases sharply (avalanche or Zener breakdown).

Test circuit

 Forward bias:                Reverse bias:
 +--[R]--(mA)--|>|--+         +--[R]--(uA)--|<|--+
 |                  |         |                  |
 +-----|E (V)|------+         +-----|E (V)|------+

Diode equation

I = I0 (e^(V/ηVT) − 1)

where VT = 25 mV at room temperature, η = 1 (Ge) or about 2 (Si).

I-V characteristic

        I (mA)
         |            /
         |           /
         |          /
 Reverse |         /
 ---+----+--------+-/------- V (volt)
 VBR     |      0.7V
  |      |
 I0 (uA) |
  • Forward region: negligible current below the knee voltage (Ge about 0.3 V, Si about 0.7 V), then a steep rise.
  • Reverse region: very small constant current I0.
  • Breakdown region: at VBR the reverse current rises sharply.

The diode therefore acts as a closed switch in forward bias and an open switch in reverse bias, which is the basis of rectifiers.

✓ Answer checked by Claude Opus 5.5 (high effort)

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