Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
98-Phys-A5 — Semiconductor Devices & Circuits — National Exams, May 2017
3 hours duration. Closed book exam (useful constants, equations and device models are annexed to the exam paper). Any FIVE (5) of the SEVEN (7) questions constitute a complete exam paper; all seven are answered here as a complete study resource.
Reference texts: A. S. Sedra & K. C. Smith, Microelectronic Circuits, 8th ed. (semiconductor/diode physics Ch. 3–4, op-amp and active-filter circuits Ch. 2 & 12, MOSFET small-signal amplifiers Ch. 7, data converters Ch. 17, precision rectifiers Ch. 4); M. M. Mano & M. D. Ciletti, Digital Design, 6th ed. (CMOS logic-family gates Ch. 10); C. Kittel, Introduction to Solid State Physics, 8th ed. (semiconductor carrier transport and statistics Ch. 8).
Given. An abrupt Si P-N junction, $A=10^{-4}\,\text{cm}^2$, $T=300^\circ\text{K}$, $n_i=p_i=2\times10^{10}\,\text{cm}^{-3}$, band diagram showing $\phi_p=E_i-E_F=0.323\,\text{eV}$ (p-side) and $\phi_n=E_F-E_i=0.461\,\text{eV}$ (n-side), $V_T=kT/q=26\,\text{mV}$, plus the mobility/lifetime table above.
[Figure not reproduced: Figure P2 (exam figure). See the official exam paper or the cited reference text.]
Figure P2 (exam figure) — equilibrium band diagram (left) and diode I-V characteristic (right).
Approach. (a) $V_o$ is exactly the total $E_i$ band-bending, $\phi_p+\phi_n$, read straight off the diagram. (b) invert the mass-action relations $\phi_p=V_T\ln(N_a/n_i)$, $\phi_n=V_T\ln(N_d/n_i)$ to recover the doping (=majority concentration) each side implies. (c) apply $np=n_i^2$ on each side using the majority concentrations from (b). (d) plug the stated adjusted minority concentrations directly into the diode saturation-current formula (22), using diffusion lengths from the given mobility/lifetime data.
Part (a) — contact potential. At equilibrium $E_F$ is flat across the whole structure; the diagram shows $E_i$ sitting $\phi_p=0.323\,\text{eV}$ above $E_F$ in the bulk p-region and $\phi_n=0.461\,\text{eV}$ below $E_F$ in the bulk n-region. The total drop in $E_i$ from the n-side bulk to the p-side bulk is exactly the built-in potential:
$$V_o=\phi_p+\phi_n=0.323+0.461=\boxed{0.784\ \text{V}}$$
Part (b) — majority concentrations. Inverting $\phi_p=V_T\ln(N_a/n_i)$ and $\phi_n=V_T\ln(N_d/n_i)$ for the doping (which equals the majority concentration, $p_p\approx N_a$, $n_n\approx N_d$):
$$p_p\approx N_a=n_ie^{\phi_p/V_T}=(2\times10^{10})e^{0.323/0.026}=\boxed{4.97\times10^{15}\ \text{cm}^{-3}}$$
$$n_n\approx N_d=n_ie^{\phi_n/V_T}=(2\times10^{10})e^{0.461/0.026}=\boxed{1.00\times10^{18}\ \text{cm}^{-3}}$$
Part (c) — minority concentrations. Using the mass-action law $n_p p_p=p_n n_n=n_i^2$ on each side:
$$n_p=\frac{n_i^2}{p_p}=\frac{(2\times10^{10})^2}{4.97\times10^{15}}=\boxed{8.05\times10^4\ \text{cm}^{-3}}$$
$$p_n=\frac{n_i^2}{n_n}=\frac{(2\times10^{10})^2}{1.00\times10^{18}}=\boxed{399\ \text{cm}^{-3}}$$
(This is a P-N$^+$ junction: $N_d\gg N_a$, so the p-side is far more lightly doped and its minority-electron population $n_p$ is correspondingly larger than the n-side's minority-hole population $p_n$.)
Part (d) — $I_S$ for the adjusted minority concentrations. The diffusion constants and lengths depend only on mobility and lifetime (unchanged by the doping adjustment):
$$D_n=\mu_{n,p\text{-side}}V_T=700(0.026)=18.2\,\text{cm}^2/\text{s},\quad L_n=\sqrt{D_n\tau_n}=\sqrt{18.2\times0.1\times10^{-6}}=1.349\times10^{-3}\,\text{cm}$$
$$D_p=\mu_{p,n\text{-side}}V_T=450(0.026)=11.7\,\text{cm}^2/\text{s},\quad L_p=\sqrt{D_p\tau_p}=\sqrt{11.7\times10\times10^{-6}}=1.082\times10^{-2}\,\text{cm}$$
Using formula (22) with the stated adjusted $n_p=1.35\times10^5\,\text{cm}^{-3}$, $p_n=324\,\text{cm}^{-3}$:
$$I_S=qA\left(\frac{D_nn_p}{L_n}+\frac{D_pp_n}{L_p}\right)=(1.6\times10^{-19})(10^{-4})\left(\frac{18.2(1.35\times10^5)}{1.349\times10^{-3}}+\frac{11.7(324)}{1.082\times10^{-2}}\right)$$
$$I_S=(1.6\times10^{-19})(10^{-4})(1.821\times10^9+3.50\times10^5)=\boxed{2.91\times10^{-14}\ \text{A}\ (29.1\ \text{fA})}$$
Electron injection into the p-side again dominates by roughly 5000:1, as expected for the heavily-asymmetric P-N$^+$ doping.
