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High-speed heterostructure devices: from device concepts to circuit modeling

Patrick Roblin, Hans Rohdin

Chapter 2

Semiclassical theory of heterostructures - all with Video Answers

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Chapter Questions

01:08

Problem 1

p-GaAs-n-AlGaAs heterojunction: Consider a p-GaAs-n- $\mathrm{Al}_{0.3} \mathrm{Ga}_{0.7} \mathrm{As}$ heterojunction at $300 \mathrm{~K}$. The acceptor doping on the $\mathrm{p}$ side is $N_A=10^{16} \mathrm{~cm}^{-3}$. The donor doping on the $\mathrm{n}$ side is $N_D=10^{16} \mathrm{~cm}^{-3}$. Assume that all donor and acceptor impurities are ionized and that the doping is non-degenerate. The temperature is $300 \mathrm{~K}$.
(a) Calculate the conduction and valence-band discontinuities. Use the $68 / 32 \%$ rule and the $\mathrm{AlGaAs}$ material parameters given in Table 2.3.
(b) Sketch the band diagram.
(c) Calculate the built-in potential $V_0$. Rank by order of importance the various terms contributing to $V_0$.

Chai Santi
Chai Santi
Numerade Educator
06:02

Problem 2

An efficient one-dimensional Poisson solver: We wish to solve the Poisson equation in a heterostructure. The heterostructure is divided into series of atomic layers $i$ of width $a_i$ and centered at the position $x_i$. The dielectric constant $\epsilon_i$ is uniform in each atomic layer $i$.
(a) We first assume that the charge distribution $\rho(x)$ and dielectric constant $\epsilon_i$ are uniform in each atomic layer $i$ :
$$
\rho(x)=\sum_i \rho_i\left[\mathrm{u}\left(x-x_i+\frac{a_i}{2}\right)-\mathrm{u}\left(x-x_i-\frac{a_i}{2}\right)\right],
$$
Fig. 2.9 can't copy. Charge distribution $\rho(x)$ for the step approximation.
where $\rho_i$ is the value of $\rho(x)$ in the interval $i$ (see Figure 2.9). We assume initially that the charge distribution $\rho_i$ is known.
Verify that the following relations are obtained if we integrate the Poisson equation exactly from the site 1 to the site $n$ :
$$
\begin{aligned}
& \epsilon_n F\left(X_n^{+}\right)=\epsilon_1 F\left(X_1^{+}\right)+\sum_{j=1}^{n-1} \rho_j a_j, \\
& V\left(X_n\right)=V\left(X_1\right)-\sum_{j=1}^{n-1} F\left(X_j^{+}\right) a_j-\frac{1}{2} \sum_{j=1}^{n-1} \frac{\rho_j a_j^2}{\epsilon_j}
\end{aligned}
$$
with $X_i=x_i-a_i / 2$.
(b) Assume now that the charge distribution in each atomic layer is given by an impulse function (see Figure 2.10):
$$
\rho(x)=\sum_i \rho_i a_i \delta\left(x-x_i\right)
$$

Compare the field and potential distribution. Do the expressions given in (a) still hold? Assume that $\rho_i$ is known and is the same as in (a).
(c) The charge distribution $\rho(x, V)$ is now assumed to depend on the voltage $V(x)$ as well as position $x$. The charge distribution is assumed to be $\rho_i=\rho\left(x_i, V\left(x_i\right)\right)$ for both the uniform charge distribution (a) and the impulse distribution (b). Verify that a close-form solution can easily be obtained with method (b) but not with method (a).
Note: This method of integrating a non-linear Poisson equation using the initial conditions $F\left(X_i^{+}\right)$and $V\left(X_i\right)$ gives results quite similar to the Runge-Kutta algorithm. This new algorithm, however, is much simpler and therefore faster than the Runge-Kutta algorithm. Note that the solution obtained is exact. However, the uniform charge distribution (a) is easier to justify than the impulse distribution (b).
Fig. 2.10 can't copy. Charge distribution $\rho(x)$ for the impulse approximation.

Chai Santi
Chai Santi
Numerade Educator
01:16

Problem 3

Ballistic electrons in the HBT (after Pelouard et al. [8]): Consider a n-p-n HBT of base width $W$. We wish to study the impact of the quasi-ballistic electrons launched at the emitter-base heterojunction upon the performance of the HBT.
The total emitter current which flows in the base includes two components: a diffusion current and a quasi-ballistic current:
$$
\begin{aligned}
& J_E=J_R(x)+J_{Q B}(x)=J_s\left[\exp \left(\frac{q V_{B E}}{k_B T}\right)-1\right], \\
& J_R(x)=q D_n \frac{d n_R}{d x}, \\
& J_{Q B}(x)=J_s \exp \left(\frac{q V_{B E}}{k_B T}\right) \exp \left(\frac{-x}{\lambda}\right),
\end{aligned}
$$
where $D_n$ is the diffusion constant for the relaxed electron population and $\lambda$ is the mean free path of the quasi-ballistic electrons. Note that the ballistic current $J_{Q B}$ in the base is maximum at $x=0$, which is the location of the emitter-base heterojunction. The $x$ axis is assumed to be oriented from the emitter to the collector. Therefore $J_s$ must be negative for this $\mathrm{n}-\mathrm{p}-\mathrm{n}$ transistor.
(a) Express the thermalized electron concentration $n_R(x)$ at position $x$ in the base in terms of the electron concentration $n_R(W)$.
(b) Calculate the equilibrium $\left(V_{B E}=V_{B C}=0\right)$ electron concentration $n_{R 0}(x)$ at the position $x$ in the base in terms of $n_{R 0}(W)$.
(c) Calculate the base current $J_B$ resulting from the recombination of electrons with holes in the base
$$
J_B=q \int_0^W \frac{n_R(x)-n_{R 0}(x)}{\tau_n} d x
$$
with $\tau_n$ the lifetime of electrons in the base. Note that the excess of electrons at the edge of the collector-base depletion region is controlled by the collector-base diode
$$
\Delta n_P=n_R(W)-n_{R 0}(W)=n_P\left[\exp \left(\frac{q V_{B C}}{k_B T}\right)-1\right] \text {, }
$$
assuming low injection.
(d) Verify that the common base current gain is
$$
\alpha=1-\frac{W^2}{2 L_n^2}\left\{1+\frac{2 \lambda}{W} \exp \left(\frac{-W}{\lambda}\right)+2\left(\frac{\lambda}{W}\right)^2\left[\exp \left(\frac{-W}{\lambda}\right)-1\right]\right\} \text {, }
$$
where $L_n^2=D_n \tau_n$ is the diffusion length of the electrons in the base. Note that for $\lambda=0$ the current gain $\alpha$ reduces to the classical bipolar expression for the base transport $B=$ $\operatorname{sech}\left(W / L_n\right) \simeq 1-W^2 / 2 L_n^2$. The emitter efficiency $\gamma$ is assumed to be $1(\alpha=B)$ which is quite reasonable for an HBT.
(e) Plot the ratio of the common emitter current gains $\beta(\lambda) / \beta(\lambda=0)$ for a base width $W$ varying from 0.6 to $0.1 \mu \mathrm{m}$ and $\lambda=0.19 \mu \mathrm{m}$. Assume that the gate length is much smaller than the diffusion length so that we can use the approximation $\beta=\alpha /(1-\alpha) \simeq 1 /(1-\alpha)$.

Chai Santi
Chai Santi
Numerade Educator