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Exercise Problems · Q1

Q.A p-n junction diode has a reverse saturation current rating of 100 nA at 50∘^\circC. What should be the value of the forward current for a forward voltage drop of 0.6 V?

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[!TLDR]

With VT=kT/q≈27.88V_T = kT/q \approx 27.88 mV at 323 K, the Shockley equation gives I≈223.72I \approx 223.72 A.

This is a common II PUC Electronics question-bank problem. The forward current follows the Shockley diode equation I=IS[eV/VT−1]I = I_S\left[e^{V/V_T} - 1\right], with the thermal voltage VT=kT/qV_T = kT/q evaluated at the diode's temperature.

Given IS=100×10−9I_S = 100 \times 10^{-9} A, V=0.6V = 0.6 V and T=273+50=323T = 273 + 50 = 323 K:

VT=1.381×10−23×3231.6×10−19≈27.88 mVV_T = \frac{1.381 \times 10^{-23} \times 323}{1.6 \times 10^{-19}} \approx 27.88\ \text{mV}

I=100×10−9[e0.6/0.02788−1]≈223.72 AI = 100 \times 10^{-9}\left[e^{0.6/0.02788} - 1\right] \approx 223.72\ \text{A}

[!ANSWER]

I≈223.72I \approx 223.72 A

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