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Dynamic resistance of diode, numerical

Short answer

From the Shockley equation, diode current ID = 10 pA × (e^(650/50) − 1) = 4.42 µA. Dynamic resistance rd = nVT / ID = 50 mV / 4.42 µA ≈ 11.3 kΩ.

Find the value of dynamic resistance if voltage in the diode is 650mV and IRS is 10pA = (10×10⁻¹² A) (Given n = 2 and VT = 25 mV).

The dynamic resistance is rd = n VT / ID, so first find the diode current from the Shockley equation.

Given

VD = 650 mV, IS = 10 pA = 10×10⁻¹² A, n = 2, VT = 25 mV (n VT = 50 mV).

Step 1: diode current

ID = IS [exp(VD/(n VT)) - 1]
   = 10×10⁻¹² × [exp(650/50) - 1]
   = 10×10⁻¹² × [exp(13) - 1]
   = 10×10⁻¹² × 4.424×10⁵
   = 4.42×10⁻⁶ A = 4.42 µA

Step 2: dynamic resistance

rd = n VT / ID = 0.05 V / 4.42×10⁻⁶ A ≈ 1.13×10⁴ Ω

Answer: rd ≈ 11.3 kΩ.

✓ Answer checked by Claude Opus 5.5 (high effort)

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