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.

Approach. (a)/(b) are scattering-mechanism explanations. For (c)/(d), compare the field $\varepsilon=V/L$ against $E_{sat}$ to choose the ohmic ($v=\mu_n\varepsilon$) or saturated ($v=v_{sat}$) transport law, then $I=qN_dvA$. For (e), the bar is n-type (donor-doped) so majority carriers are electrons and $R_H=-1/(qN_d)$.

1. Part (a) — mobility falls with doping. At low doping, carrier mobility is limited mainly by lattice (phonon) scattering, roughly independent of doping. As the donor concentration rises, the density of fixed, charged donor ions rises with it, and Coulombic (ionized-impurity) scattering off these ions becomes an increasingly important scattering mechanism. Since $\mu=q\tau/m^*$ and the mechanisms combine as $1/\tau=1/\tau_{lattice}+1/\tau_{impurity}$, the added scattering shortens the mean free time $\tau$ and mobility falls as doping increases.
2. Part (b) — drift velocity saturates. At high electric fields, carriers gain enough kinetic energy between collisions to efficiently emit optical phonons. This loss channel becomes very effective at high carrier energy, so any extra energy the field supplies is dumped straight back into the lattice instead of 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 field.
3. Part (c) — 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 $E_{sat}=10^4\,\text{V/cm}$, so the low-field law applies: $v_1=\mu_n\varepsilon_1=700(100)=7\times10^4\,\text{cm/s}$. The current is $$I_1=qN_dv_1A=(1.6\times10^{-19})(10^{17})(7\times10^4)(1\times10^{-6})=\boxed{1.12\ \text{mA}}$$
4. Part (d) — L = 0.5 µm (velocity-saturated regime). The field is $$\varepsilon_2=\frac{V}{L_2}=\frac{10\,\text{V}}{0.5\times10^{-4}\,\text{cm}}=2\times10^5\,\text{V/cm}$$ — twenty times $E_{sat}$, deep in the saturated region, so $v_2=v_{sat}=1\times10^7\,\text{cm/s}$ regardless of the (very large) field. The current is $$I_2=qN_dv_{sat}A=(1.6\times10^{-19})(10^{17})(1\times10^7)(1\times10^{-6})=\boxed{0.160\ \text{A}=160\ \text{mA}}$$
5. Part (e) — Hall coefficient. This is an n-type bar (donor-doped), so $n_o\approx N_d=10^{17}\,\text{cm}^{-3}$ and $p_o\ll n_o$. Using the annex formula $R_H=1/[q(p_o-n_o)]$: $$R_H=\frac{1}{q(0-N_d)}=\frac{-1}{qN_d}=\frac{-1}{(1.6\times10^{-19})(10^{17})}=\boxed{-62.5\ \text{cm}^3/\text{C}}$$ The negative sign confirms the majority carriers are electrons (a p-type bar of the same doping magnitude would give $R_H=+62.5\,\text{cm}^3/\text{C}$).