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Diode leakage current at 60°C

Short answer

IS(60°C)=10⁻⁹×2^(3.5)≈11.3 nA. With VT=28.71 mV, ηVT=45.93 mV, ID=11.3 nA×exp(0.8/0.04593)≈0.41 A.

The leakage current of a silicon diode is Is = 10⁻⁹ A at 25°C, and the emission coefficient is η = 1.6. The operating junction temperature is Tj = 60°C. Determine (i) the leakage current Is and (ii) the diode current ID at VD = 0.8 V.

The reverse saturation (leakage) current roughly doubles for every 10 °C rise in temperature, and the thermal voltage VT = kT/q also increases with temperature.

Given

IS(25 °C) = 10⁻⁹ A, η = 1.6, Tj = 60 °C (333 K), VD = 0.8 V.

(i) Leakage current at 60 °C

IS(T2) = IS(T1) × 2^[(T2 - T1)/10]
       = 10⁻⁹ × 2^[(60 - 25)/10] = 10⁻⁹ × 2^3.5
       = 10⁻⁹ × 11.31 = 1.13×10⁻⁸ A ≈ 11.3 nA

(ii) Diode current at VD = 0.8 V

VT = kT/q = 8.617×10⁻⁵ × 333.15 = 28.71 mV
η VT = 1.6 × 28.71 mV = 45.93 mV

ID = IS [exp(VD/(η VT)) - 1]
   = 11.31×10⁻⁹ × [exp(0.8/0.04593) - 1]
   = 11.31×10⁻⁹ × [exp(17.42) - 1]
   = 11.31×10⁻⁹ × 3.66×10⁷
   ≈ 0.41 A

Answer: (i) IS ≈ 11.3 nA; (ii) ID ≈ 0.41 A (about 414 mA).

Note: such a large current only shows that 0.8 V is a high forward voltage for this diode at 60 °C; in practice an external resistor limits the current.

✓ Answer checked by Claude Opus 5.5 (high effort)

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