Skip to main content

Small signal model, AC resistance of diode

Short answer

The small-signal model replaces the diode at its Q-point by a dynamic resistance rd. From the exponential diode equation, the slope gives rd = dV/dI = ηVT/ID. At room temperature with η≈1 and VT≈25 mV, rd ≈ 25 mV/ID (ID in mA gives Ω).

Explain the small signal model of a PN junction diode and derive the expression for AC or dynamic resistance (AC resistance in the forward biased region).

Small signal model

When a small ac signal vd (a few mV) is superimposed on a dc bias VD, the diode operates only on a very short part of its I-V curve around the bias point (the Q-point). That part is almost a straight line, so the diode behaves like a linear resistance called the dynamic (ac or incremental) resistance rd. In the small signal model the diode is replaced by rd (with the dc bias handled separately; the junction capacitance may be added in parallel at high frequency).

 Diode I-V curve       Small-signal model
 I|      /             o---[ rd ]---o
  |     /  slope=1/rd
  |  Q /
  +---+------ V

Derivation of rd

Diode equation: ID = IS (e^(VD/ηVT) − 1), where VT = kT/q ≈ 25 mV at 300 K and η = 1 to 2.

In forward bias with VD ≫ ηVT, ID ≈ IS e^(VD/ηVT).

The conductance at the Q-point is the slope of the curve:

1/rd = dID/dVD = IS e^(VD/ηVT) / (ηVT) = ID/(ηVT)

rd = ηVT/ID

For η = 1 and VT = 25 mV: rd = 25 mV/ID (ID in mA gives rd in ohms).

Example: ID = 5 mA gives rd = 25 mV/5 mA = 5 Ω. At ID = 1 mA, rd = 25 Ω.

Result

  • rd falls as the forward current rises, so it is not fixed but depends on the bias.
  • The ac current is id = vd/rd, and the total current is ID + id.
  • In reverse bias rd is very large (open circuit).

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Basic Electronics Engineering chapter 2 questions →