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Resistance of a germanium diode

Short answer

Using Shockley equation with V_T=0.02585 V, forward current = 1 µA × (e^(0.2/0.02585)−1) ≈ 2.29 mA. Static resistance R_DC = 0.2 V / 2.29 mA ≈ 87.3 Ω; dynamic resistance r_d ≈ 0.02585 V / 2.29 mA ≈ 11.3 Ω.

Find the resistance of p-n junction Germanium diode if temperature is 27°C and I₀ = 1μA for an applied forward bias of 0.2 Volt.

The diode current is given by the Shockley diode equation

I=I0(eV/ηVT−1)I = I_0\left(e^{V/\eta V_T}-1\right)

with η=1\eta = 1 for germanium and VT=kT/eV_T = kT/e.

Given: T=27∘C=300T = 27^\circ\text{C} = 300 K, I0=1 μAI_0 = 1\ \mu\text{A}, V=0.2V = 0.2 V.

VT=kTe=1.38×10−23×3001.6×10−19=0.02585 VV_T = \frac{kT}{e} = \frac{1.38\times10^{-23}\times300}{1.6\times10^{-19}} = 0.02585\ \text{V}

Forward current

VVT=0.20.02585=7.737e7.737=2291.5I=10−6(2291.5−1)=2.29×10−3 A=2.29 mA\begin{aligned} \frac{V}{V_T} &= \frac{0.2}{0.02585} = 7.737\\ e^{7.737} &= 2291.5\\ I &= 10^{-6}(2291.5-1) = 2.29\times10^{-3}\ \text{A} = 2.29\ \text{mA} \end{aligned}

Static (DC) resistance

RDC=VI=0.22.29×10−3=87.3 ΩR_{DC} = \frac{V}{I} = \frac{0.2}{2.29\times10^{-3}} = 87.3\ \Omega

Dynamic (AC) resistance

rd=dVdI=ηVTI+I0=0.025852.29×10−3+10−6=11.3 Ω\begin{aligned} r_d &= \frac{dV}{dI} = \frac{\eta V_T}{I+I_0}\\ &= \frac{0.02585}{2.29\times10^{-3}+10^{-6}} = 11.3\ \Omega \end{aligned}

Answer: forward current ≈ 2.29 mA; static resistance ≈ 87.3 Ω; dynamic resistance ≈ 11.3 Ω.

The static value is the ratio V/I at the operating point; the dynamic value is the slope resistance seen by small signals.

✓ Answer checked by Claude Opus 5.5 (high effort)

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