Question 1 of 7

Question 1 (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.

[Figure not reproduced: Figure P1a — electron/hole mobility vs. doping for Si at 300 K, redrawn from the exam figure; the read-off point used in part (c) is marked. See the official exam paper.]

[Figure not reproduced: Figure P1b — electron drift velocity vs. field for Si, redrawn; the two operating points from part (c) are marked. See the official exam paper.]

Approach. (a)/(b) are scattering-mechanism explanations. For (c), first find which transport regime each length puts the bar in by comparing the resulting field $\varepsilon=V/L$ against the knee of Figure P1b, then use $v=\mu_n\varepsilon$ (ohmic) or $v=v_{sat}$ (saturated) as appropriate, and $I=qN_dvA$.

1. Part (a) — mobility falls with doping. At low doping, carrier mobility is limited mainly by lattice (phonon) scattering, which is roughly independent of doping level. As the impurity concentration rises, the density of charged donor/acceptor ions rises with it, and Coulombic (ionized-impurity) scattering off these fixed charges becomes an increasingly important scattering mechanism. Since $\mu=q\tau/m^*$ and the two scattering mechanisms combine as $1/\tau=1/\tau_{lattice}+1/\tau_{impurity}$, the added impurity scattering shortens the mean free time $\tau$ and mobility falls — exactly the roll-off seen in Figure P1a above $N\sim10^{16}\,\text{cm}^{-3}$.
2. Part (b) — drift velocity saturates. At high electric fields, carriers gain enough kinetic energy between collisions to efficiently emit optical phonons. This energy-loss channel becomes very effective at high carrier energy, so any extra energy the field supplies is dumped straight back into the lattice rather than continuing to accelerate the carrier. The carrier reaches a maximum, scattering-limited average velocity ($v_{sat}\approx10^7\,\text{cm/s}$ for Si) that no longer rises with further increases in field — the flat plateau in Figure P1b.
3. Part (c)(i) — L = 0.1 cm (ohmic regime). The field is $$\varepsilon_1=\frac{V}{L_1}=\frac{10\,\text{V}}{0.1\,\text{cm}}=100\,\text{V/cm}$$ well below Figure P1b's knee ($\sim5\times10^4\,\text{V/cm}$), so the low-field mobility applies: $v_1=\mu_n\varepsilon_1=780(100)=7.8\times10^4\,\text{cm/s}$. The current is $$I_1=qN_dv_1A=(1.6\times10^{-19})(10^{17})(7.8\times10^4)(1\times10^{-6})=\boxed{1.25\ \text{mA}}$$
4. Part (c)(ii) — L = 0.5 µm (velocity-saturated regime). The field is $$\varepsilon_2=\frac{V}{L_2}=\frac{10\,\text{V}}{0.5\times10^{-4}\,\text{cm}}=2\times10^5\,\text{V/cm}$$ — about four times the knee field (some 0.6 decade past it), well into the flat part of Figure P1b, so $v_2=v_{sat}=1\times10^7\,\text{cm/s}$ regardless of the (very large) field. The current is $$I_2=qN_dv_{sat}A=(1.6\times10^{-19})(10^{17})(1\times10^7)(1\times10^{-6})=\boxed{0.160\ \text{A} = 160\ \text{mA}}$$