6. The transistor working point

In Figure 5, the conductance as a function of the normalized applied electric field is plotted. The transistor can work as a digital device, where the field varies between the binary cases:

$$T\_{1,1} = \begin{cases} \mathbf{0} & \text{for} \quad F = F\_R \\ \mathbf{1} & \text{for} \quad F = \mathbf{0} \end{cases} \tag{41}$$

where

$$\frac{1}{2}F\_{\mathbb{R}}w \equiv E\_F - \left(\frac{2\pi}{w}\right)^2 + \frac{4\pi^2\sin^4(2\pi\varepsilon/w)}{w^2\ln^2(\rho\_0/R(\varepsilon))} \cong E\_F - \left(\frac{2\pi}{w}\right)^2 + \frac{\varepsilon^4(2\pi/w)^6}{\ln^2(\rho\_0\exp\left(-\gamma/2\right)/4\varepsilon)}.\tag{42}$$

Figure 5. <sup>2</sup> Plot of the conductance as <sup>a</sup> function of the applied electric field for the particles' energy <sup>E</sup> <sup>¼</sup> <sup>4</sup>ðπ=w<sup>Þ</sup> .

But the transistor can also work in an analog mode as an amplifier, in which case the applied voltage should be modulated with respect to the bias voltage F∗w, which is the center of the linear regime, i.e.,

$$F\_\* w / 2 = E\_F - \left(2\pi / w\right)^2 + \delta^2 \left[1 \pm 2 \frac{\left|\chi\_1(\varepsilon)\right|^2}{\left|\chi\_2(\varepsilon)\right|^2 \sqrt{E\_2 - E\_1}}\right]. \tag{43}$$

Using this bias voltage, the transmission coefficient can be written

$$t\_{1,1} = 1 - \left\{ 1 + ic \left[ 1 - \frac{8}{\sqrt{\delta^2 / \left(1 - c^{-1}\right)^2 + (F - F\_\*) \omega / 2}} \right] \right\}^{-1} \tag{44}$$

where

 $\sigma \equiv \frac{\sqrt{E\_2 - E\_1}}{8} \frac{|\chi\_2(\varepsilon)|^2}{|\chi\_1(\varepsilon)|^2} \cong 4\pi\sqrt{3}/8$ , and  $6 \equiv \pi |\chi\_2(\varepsilon)|^2 / \ln\left(\rho\_0/R(\varepsilon)\right) \cong 4\pi(2\pi\varepsilon)^2 / \ln\left(\rho\_0 \exp\left(-\sqrt{2}\varepsilon\right)/4\varepsilon\right)$ .  $[w^3 \ln\left(\rho\_0 \exp\left(-\chi/2\}/4\varepsilon\right)]$ .

Therefore, the transistor gain at the working point is the ratio between the change in conductance and the applied transverse voltage Δv ¼ ðF � F∗Þw, which is

$$
\text{gain} = \frac{\Delta G}{\Delta v} = \frac{c \left(1 - c^{-1}\right)^{\overline{3}}}{4\pi \delta^2} \tag{45}
$$

when the point defect is a surface one, i.e., ε < < 1, then δ < < 1 and c>>1

$$
gamma \cong \frac{\sqrt{3}}{8^3} = \frac{\sqrt{3} w^9 \ln^3(\rho\_0 \exp\left(-\gamma/2\right)/4\epsilon)}{\left(4\pi\right)^3 \left(2\pi\epsilon\right)^6},\tag{46}
$$

which can be extremely large.
