PHT.301 Physics of Semiconductor Devices  

MESFET Gradual Channel ApproximationThe description of a MESFET in the gradual channel approximation is almost the same as for a JFET. The difference is how the builtin voltage $V_{bi}$ is calculated. Consider an $n$channel MESFET. A MESFET consists of a semiconducting channel contacted by two ohmic contacts. The metal gate forms a Schottky contact above the channel. The current in the channel flows between the depletion layer of the Schottky diode and a semiinsulating substrate. When the Schottky contact is reverse biased, the depletion width expands and the channel becomes narrower. The thickness of the conducting channel is $hx_n(y)$ where $h$ is the thickness of the $n$doped channel and $x_n(y)$ is the depletion width that depends on the position $y$ along the channel. In the figure below, the gray region of the channel is depleted. A constant current $I_D$ flows from the drain to the source. The voltage drop across a short section of the channel is $dV = I_DR$ where $R$ is the resistance of this short section. Ohm's law states, \[\begin{equation} R = \frac{\rho dy}{(hx_n(y))Z}. \end{equation}\]For an $n$doped extrinsic semiconductor, the resistivity is $\rho=1/(e\mu_nN_D)$ where $e$ is the charge of an electron, $\mu_n$ is the electron mobility, and $N_D$ is the donor concentration. This can be combined to yield, \[\begin{equation} I_Ddy = e\mu_nN_DZ(hx_n(y))dV \end{equation}\]The formula for the depletion width of a Schottky diode is the same as for a onesided $pn$junction, \[\begin{equation} x_n = \sqrt{\frac{2\epsilon_r\epsilon_0(V_{bi}V)}{eN_D}}, \end{equation}\]where $V_{bi}$ is the builtin voltage and $V$ is the forward bias voltage across the Schottky diode. In the case of a nMESFET, a positive voltage on the gate forward biases the diode and a positive voltage on the drain reverse biases the diode so the depletion with is, \[\begin{equation} x_n(y) = \sqrt{\frac{2\epsilon_r\epsilon_0(V_{bi}+V(y)V_G)}{eN_D}}, \end{equation}\]where $V(y)$ is the potential along the channel. Differentiating this expression we find, \[\begin{equation} \frac{dx_n(y)}{dV} = \frac{1}{2\sqrt{\frac{2\epsilon_r\epsilon_0(V_{bi}+V(y)V_G)}{eN_D}}}\frac{2\epsilon_r\epsilon_0}{eN_D}=\frac{\epsilon_r\epsilon_0}{eN_Dx_n(y)}. \end{equation}\]Thus, \[\begin{equation} dV =\frac{eN_Dx_n}{\epsilon_r\epsilon_0}dx_n. \end{equation}\] Substituting this into Eq. 2 yield a first order differential equation that can be solved for the depletion width as a function of position along the channel. \[\begin{equation} I_Ddy = \frac{e^2\mu_nN_D^2Z}{\epsilon_r\epsilon_0}(hx_n)x_ndx_n. \end{equation}\]Integrating from source to drain, \[\begin{equation} I_D\int\limits_0^L dy = \frac{e^2\mu_nN_D^2Z}{\epsilon_r\epsilon_0}\int\limits_{x_L}^{x_R}(hx_n)x_ndx_n. \end{equation}\]The depletion width at the source is $x_L$, the depletion width at the drain is $x_R$. \[\begin{equation} \frac{I_D\epsilon_r\epsilon_0L}{e^2\mu_nN_D^2Z}= \frac{hx_R^2}{2} \frac{x_R^3}{3}\frac{hx_L^2}{2} +\frac{x_L^3}{3}. \end{equation}\]Using the expresions for the depletion widths at the source and the drain and the definition of the pinchoff voltage $V_p$, \[\begin{equation} x_L = \sqrt{\frac{2\epsilon_r\epsilon_0(V_{bi}V_G)}{eN_D}},\qquad x_L = \sqrt{\frac{2\epsilon_r\epsilon_0(V_{bi}+V_DV_G)}{eN_D}},\qquad h = \sqrt{\frac{2\epsilon_r\epsilon_0V_p}{eN_D}}, \end{equation}\]the drain current in the linear regime can be expressed in terms of the gate voltage and drain voltage, \[ \begin{equation} \boxed{I_D=I_p\left[\frac{V_D}{V_p}\frac{2}{3}\left(\frac{V_{bi}+V_DV_G}{V_p}\right)^{3/2}+\frac{2}{3}\left(\frac{V_{bi}V_G}{V_p}\right)^{3/2}\right].} \end{equation} \]Here, \begin{align} I_p =\frac{\mu_n {N_D}^2 Z e^2 h^3}{2L\epsilon_r\epsilon_0}, \qquad V_p =\frac{e {N_D} h^2}{2 \epsilon_r\epsilon_0}, \qquad eV_{bi} =\phi_b  k_B T \ln \left(\frac{N_c}{N_D}\left(\frac{T}{300}\right)^{3/2}\right). \end{align}Here $\phi_b$ is the Schottky barrier height. Eq. 11 for the drain is valid in the linear regime before the channel is pinchedoff. After pinchoff the MESFET is in the saturation regime and the current is, \[ \begin{equation} I_{D,sat}=I_p\left[\frac{1}{3}\frac{V_{bi}V_G}{V_p}+\frac{2}{3}\left(\frac{V_{bi}V_G}{V_p}\right)^{3/2}\right]. \end{equation} \]This derivation assumed that the Schottky diode is reverse biased. For a nchannel MESFET, $V_G$ < 0 and $V_D$ > 0 in this regime. The gray area of the plot below shows the part of the channel that is depleted.
$V_{bi}=$ V; $I_p=$ mA; $V_p=$ V $I_d=$ mA $x_L=$ μm $x_R=$ μm. 