2.7.1 Ballistic Transport

For ballistic transport one can consider an ideal case where electrons are not scattered on the wire of length $L$ connected to two electrodes, $1$ and $2$ (see Fig. 2.13). Since two electrodes have a large electron capacity, the FERMI energy for electrodes $1$ and $2$ are constants denoted by $E_{\mathrm{F}_1}$ and $E_{\mathrm{F}_2}$, ( $E_{\mathrm{F}_1}>E_{\mathrm{F}_2}$), respectively. If there are no reflections of electrons at the electrodes, $k>0$ states are occupied primarily by electrons coming from the left contact while the $k<0$ states are occupied primarily by electrons coming from the right contact. Consequently the occupation factors for the $k>0$ and $k<0$ states are given by the FERMI functions for the left and right contacts, respectively.

Figure 2.13: A ballistic conductor with length $L$ is connected to two electrodes $1$ and $2$ with FERMI energies $E_{\mathrm{F}_1}$ and $E_{\mathrm{F}_2}$, respectively. $M$ is the number of channels for electrons to propagate from the electrode $1$ to $2$.

Because of the confinement of electronic states in the direction perpendicular to the current flow, there are several energy subbands $E^{\nu}(k)$. Thus the total current is given by the sum of the microscopic currents of all the subbands $E^{\nu}(k)$. The subbands are also called channels. The number of channels is a function of energy, which is denoted by $M(E)$. An electron which has a velocity of $\upsilon=\hbar^{-1}(\partial E/\partial k)$ in an unoccupied state contributes to the microscopic current $I=\ensuremath {\mathrm{q}}/t_\mathrm{t}$, in which $t_\mathrm{t}$ is the carrier transit time $t_\mathrm{t}=L/\upsilon$. Then the total current is given by [60]

$$\begin{array}{ll} I \ & \displaystyle = \ \frac{\ensuremath {... ...\mathrm{F}_2}\right]}{\ensuremath {\mathrm{q}}} \ , \end{array}$$ (2.21)

where the sum over $k$ is converted to the integral. In (2.21) a spin degeneracy of $2$ and the inverse of the level spacing $L/2\pi$ is introduced and $M(E)$ is assumed to be constant over the integration range. One can easily show that for $T=0~\mathrm{K}$ only states with $E_{\mathrm{F}_2}<E<E_{\mathrm{F}_1}$ contribute to the total current. If the width of a wire is very small (less than $1~\mathrm{nm}$), $M=1$ even for $E_{\mathrm{F}_1}-E_{\mathrm{F}_2}=1~\mathrm{eV}$. On the other hand, if the width of a wire is on the order of $1~\mathrm{\mu m}$ and $E_{\mathrm{F}_1}-E_{\mathrm{F}_2}=1~\mathrm{eV}$, the number of channels $M$ becomes very large ($10^6$).

Since $V=(E_{\mathrm{F}_1}-E_{\mathrm{F}_2})/\ensuremath {\mathrm{q}}$ is the voltage between the electrodes, the resistance of the ballistic conductor is given by

$$\begin{array}{l}\displaystyle R_\mathrm{c} \ = \ \frac{\left(... ...m{h}}}{2\ensuremath {\mathrm{q}}^2} \frac{1}{M} \ , \end{array}$$ (2.22)

where $R_\mathrm{c}$ is called the contact resistance and $\ensuremath{\mathrm{h}}/2q^2$ is the quantized resistance

$$\begin{array}{l}\displaystyle R_\mathrm{0} \ = \ \frac{\ensur... ...suremath {\mathrm{q}}^2} \ \approx 12.9 k\Omega \ . \end{array}$$ (2.23)

This contact resistance arises from the mismatch of the numbers of conduction channels in the mesoscopic conductor and the macroscopic metal lead [60]. In addition to this quantum-mechanical contact resistance, there are other sources of contact resistance, such as that produced by poor coupling between the mesoscopic conductor and the leads. The inverse of (2.22) gives the contact conductance $G_\mathrm{c}=G_{0}M$ where $G_\mathrm{0}=2\ensuremath {\mathrm{q}}^2/\ensuremath{\mathrm{h}}$ denotes the quantized conductance. Thus in a wire without scattering the conductance is proportional to $M$. The quantized resistance and conductance can be observed in clean semiconductors at very low temperature on samples which have a small number of channels $M$ [61]. If one considers the range of $M(E)$ for a SW-CNT with a diameter of $1-2~\mathrm{nm}$, one finds that $M=1$ under low bias conditions, $E_{\mathrm{F}_1}-E_{\mathrm{F}_2}<1~\mathrm{eV}$. In a zigzag SW-CNT close to the FERMI energy the bands are doubly degenerate, and thus the total conductance is $G=2G_0$.

In case of coherent transport, the wave-function is determined by the SCHRÖDINGER equation. The phase and amplitude of the wave-function at electrode $2$ can be obtained from those at electrode $1$. The resistance and the conductance are thus given by

$$\begin{array}{l}\displaystyle R \ = \ \frac{\ensuremath{\math... ...}^2}{\ensuremath{\mathrm{h}}} \ M \ \mathcal{T} \ , \end{array}$$ (2.24)

where $\mathcal{T}$ is the transmission probability for a channel extending from electrode $1$ to electrode $2$. Here it is assumed again that $\mathcal{T}$ is constant near the FERMI energy. Equation (2.24) is known as the LANDAUER formula. It can be applied only if the wave-function spreads over the whole sample.

The resistance $R_\mathrm{w}$ for a single channel of a mesoscopic wire is given in terms of the transmission probability $\mathcal{T}$ as

$$\begin{array}{l}\displaystyle R_\mathrm{w} \ = \ R \ - \ R_\m... ...equiv \ R_{0} \ \frac{\mathcal{R}}{\mathcal{T}} \ . \end{array}$$ (2.25)

The reflected wave-function, which is proportional to $\mathcal{R}=1-\mathcal{T}$, causes a voltage drop in the wire.

M. Pourfath: Numerical Study of Quantum Transport in Carbon Nanotube-Based Transistors