98-Phys-A5, May 2015, Q2 | Nivaar Exam Prep

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.

Given. An abrupt silicon $n^+p$ junction diode with the doping, minority-carrier lifetime and mobility data in Table T2 at 300 K.

Symbol	p-type region	n-type region
Doping	$N_a=5\times10^{15}\,\text{cm}^{-3}$	$N_d=10^{18}\,\text{cm}^{-3}$ ($n^+$)
Minority-carrier lifetime	$\tau_n=0.1\,\mu\text{s}$ (electrons)	$\tau_p=10\,\mu\text{s}$ (holes)
Majority mobility	$\mu_p=200\,\text{cm}^2/\text{V-s}$	$\mu_n=1300\,\text{cm}^2/\text{V-s}$
Minority mobility	$\mu_n=700\,\text{cm}^2/\text{V-s}$	$\mu_p=450\,\text{cm}^2/\text{V-s}$

$A=10^{-4}\,\text{cm}^2$, $T=300^\circ\text{K}$, $n_i=p_i=2\times10^{10}\,\text{cm}^{-3}$ (as stated in this question), $V_T=kT/q=26\,\text{mV}$.

Find. (a) $\phi_p=E_i-E_F$ (p-side) and $\phi_n=E_F-E_i$ (n-side); (b) contact potential $V_o$; (c) saturation current $I_S$; (d) field in the p-region far from the junction at $I=4\,\text{mA}$.

Approach. (a) mass-action law gives the Fermi-level offset from $E_i$ on each side. (b) $V_o=\phi_p+\phi_n$ is exactly the total band-bending drawn on the equilibrium diagram. (c) the ideal-diode saturation current sums minority-carrier diffusion into both sides; because $N_a\ll N_d$ (this is an $n^+p$ junction), electron injection into the lightly-doped p-side dominates. (d) far from the junction only majority-carrier drift is present, so $\varepsilon=J/\sigma$ with the majority mobility.

Part (a) — Fermi-level offsets. Using $E_i-E_F=V_T\ln(N_a/n_i)$ on the p-side and $E_F-E_i=V_T\ln(N_d/n_i)$ on the n-side: $$\phi_p=0.026\ln\!\left(\frac{5\times10^{15}}{2\times10^{10}}\right)=\boxed{0.323\ \text{eV}}\qquad \phi_n=0.026\ln\!\left(\frac{10^{18}}{2\times10^{10}}\right)=\boxed{0.461\ \text{eV}}$$ ($E_F$ sits 0.323 eV below $E_i$ in the p-region and 0.461 eV above $E_i$ in the n-region.)
Part (b) — band diagram and contact potential. At equilibrium the Fermi level $E_F$ is flat (constant) across the whole structure. Far from the junction on each side, $E_i$ sits at its own flat bulk position relative to $E_F$: 0.323 eV above $E_F$ in the p-region, 0.461 eV below $E_F$ in the n-region. Between those two flat regions, $E_i$ (and every band edge with it) bends smoothly through the depletion layer; reading the total vertical drop in $E_i$ from the n-side bulk to the p-side bulk on the diagram gives exactly the built-in potential: $$V_o=\phi_p+\phi_n=V_T\ln\!\left(\frac{N_aN_d}{n_i^2}\right)=0.323+0.461=\boxed{0.784\ \text{V}}$$
Part (c) — saturation current $I_S$. Both sides contribute minority-carrier diffusion current. On the p-side the minority carriers are electrons ($D_n=\mu_{n,p\text{-side}}V_T=700(0.026)=18.2\,\text{cm}^2/\text{s}$, $L_n=\sqrt{D_n\tau_n}=\sqrt{18.2\times0.1\times10^{-6}}=1.349\times10^{-3}\,\text{cm}$, $n_{p0}=n_i^2/N_a=8\times10^4\,\text{cm}^{-3}$). On the n-side the minority carriers are holes ($D_p=\mu_{p,n\text{-side}}V_T=450(0.026)=11.7\,\text{cm}^2/\text{s}$, $L_p=\sqrt{D_p\tau_p}=\sqrt{11.7\times10\times10^{-6}}=1.082\times10^{-2}\,\text{cm}$, $p_{n0}=n_i^2/N_d=400\,\text{cm}^{-3}$). Summing both terms: $$I_S=qA\left(\frac{D_nn_{p0}}{L_n}+\frac{D_pp_{n0}}{L_p}\right)=qA\left(1.079\times10^9+4.33\times10^5\right)\ \text{cm}^{-2}\text{s}^{-1}$$ $$I_S=(1.6\times10^{-19})(10^{-4})(1.080\times10^9)=\boxed{1.73\times10^{-14}\ \text{A}\ (17.3\ \text{fA})}$$ Electron injection into the p-side outweighs hole injection into the n-side by roughly $2500:1$ — expected, since this is an $n^+p$ diode ($N_d\gg N_a$).
Part (d) — field in the p-region far from the junction. Far from the junction the minority-carrier density has relaxed back to its tiny equilibrium value, so essentially all of the current is carried by majority-carrier (hole) drift, using the p-region's majority mobility $\mu_p=200\,\text{cm}^2/\text{V-s}$: $$\sigma_p=qN_a\mu_p=(1.6\times10^{-19})(5\times10^{15})(200)=0.160\ \text{S/cm}$$ $$\varepsilon_p=\frac{J}{\sigma_p}=\frac{I/A}{\sigma_p}=\frac{4\times10^{-3}/10^{-4}}{0.160}=\boxed{250\ \text{V/cm}}$$

Quantity	Value
$\phi_p$ (p-side)	0.323 eV
$\phi_n$ (n-side)	0.461 eV
Contact potential $V_o$	0.784 V
Saturation current $I_S$	$1.73\times10^{-14}\,\text{A}$ (17.3 fA)
Field in p-region at $I=4\,\text{mA}$	250 V/cm

Concept & Theory

Topic: pn-junction equilibrium electrostatics and minority-carrier diffusion current.

Key relations: $\phi=V_T\ln(N/n_i)$; $V_o=\phi_p+\phi_n=V_T\ln(N_aN_d/n_i^2)$; $I_S=qA\!\left(\dfrac{D_nn_{p0}}{L_n}+\dfrac{D_pp_{n0}}{L_p}\right)$; $\varepsilon=J/\sigma=J/(qN\mu)$.

Why this works: at equilibrium $E_F$ is flat everywhere, so each side's own bulk offset from $E_i$ is fixed purely by its doping; the total band bending across the junction is just the sum of those two offsets. Forward current is dominated by minority injection into whichever side is more lightly doped — here the p-side, since $N_a\ll N_d$ makes $n_{p0}=n_i^2/N_a$ far larger than $p_{n0}=n_i^2/N_d$. Far from the junction, only majority carriers exist in any real numbers, so Ohm's law with the majority mobility applies.

Common pitfall: swapping the majority/minority mobility columns in Table T2 (each side lists both, since one is needed for $I_S$ and the other for part (d)); and treating this as a symmetric $pn$ junction instead of recognizing the $n^+p$ asymmetry that makes one injection term negligible.

Source: Sedra & Smith, Microelectronic Circuits, Ch. 3 (semiconductor and diode physics); Kittel, Ch. 8.

Question 2 of 7

Question 2 (20 marks)