Question 2 of 7

Question 2 (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Approach. (a) is a conceptual trap about what a voltmeter can and cannot measure across an unbiased junction; (b)/(c)/(d) use the ideal-diode law $I=I_o(e^{qV/kT}-1)$ with $I_o$ built from the diffusion-current formula.

1. Part (a) — voltmeter reading. The built-in potential itself computes to $$V_o=\frac{kT}{q}\ln\frac{N_aN_d}{n_i^2}=0.026\ln\frac{(10^{17})(10^{15})}{(1.5\times10^{10})^2}=0.697\,\text{V}$$ but a voltmeter cannot read this: the meter's two metal probes each form their own metal–semiconductor contact potential when touched to the p- and n-sides, and in thermal equilibrium the sum of ALL contact potentials around the closed loop (probe–p, p–n junction, n–probe, and back through the meter's own leads) is identically zero — otherwise a diode sitting on a shelf would drive a perpetual current with no external power. The two probe contact potentials exactly cancel the junction's built-in potential, so $$\boxed{V_{\text{voltmeter}}=0\ \text{V}}$$
2. Part (b) — why exponential. Forward bias $V>0$ lowers the junction's potential barrier from $V_o$ to $V_o-V$. The density of majority carriers able to surmount the barrier and diffuse across follows a Boltzmann distribution in the barrier height, so the injected minority-carrier concentrations at the depletion-region edges scale as $e^{qV/kT}$ — a small increase in $V$ produces a proportionally huge increase in the number of carriers energetic enough to cross, giving the diode equation's exponential dependence.
3. Part (c)/(d) — build $I_o$. Diffusion coefficients from the Einstein relation ($D=\mu V_T$, $V_T=26\,\text{mV}$): minority holes in the n-side, $D_p=\mu_{p,n\text{-side}}V_T=450(0.026)=11.7\,\text{cm}^2/\text{s}$; minority electrons in the p-side, $D_n=\mu_{n,p\text{-side}}V_T=700(0.026)=18.2\,\text{cm}^2/\text{s}$. Diffusion lengths: $L_p=\sqrt{D_p\tau_p}=\sqrt{11.7\times10\times10^{-6}}=1.08\times10^{-2}\,\text{cm}$, $L_n=\sqrt{D_n\tau_n}=\sqrt{18.2\times0.1\times10^{-6}}=1.35\times10^{-3}\,\text{cm}$. Minority concentrations: $p_{n0}=n_i^2/N_d=2.25\times10^5\,\text{cm}^{-3}$, $n_{p0}=n_i^2/N_a=2.25\times10^3\,\text{cm}^{-3}$. Then $$I_o=qA\left(\frac{D_p}{L_p}p_{n0}+\frac{D_n}{L_n}n_{p0}\right)=\boxed{4.38\times10^{-15}\ \text{A}}$$
4. Part (c) — reverse bias. For $V=-2\,\text{V}$ and $-4\,\text{V}$, $qV/kT\ll-1$ in both cases ($-76.9$ and $-153.8$), so $e^{qV/kT}\to0$ and $I\to-I_o$ regardless of exactly how reverse-biased the diode is: $$I(-2\,\text{V})=I(-4\,\text{V})=\boxed{-4.38\times10^{-15}\ \text{A}}$$ (the reverse saturation current is essentially voltage-independent once $|V|$ exceeds a few $kT/q$ — that both voltages give the same answer is the point of the question, not a coincidence.)
5. Part (d) — forward bias. $I=I_o\left(e^{qV/kT}-1\right)$: $$I(0.38\,\text{V})=4.38\times10^{-15}\left(e^{0.38/0.026}-1\right)=\boxed{1.05\times10^{-8}\ \text{A}\ (10.5\,\text{nA})}$$ $$I(0.76\,\text{V})=4.38\times10^{-15}\left(e^{0.76/0.026}-1\right)=\boxed{2.50\times10^{-2}\ \text{A}\ (25.0\,\text{mA})}$$ Doubling $V$ squares the dominant exponential term, which is why the current rises by roughly six decades for only a 2× change in forward voltage.