C. MESFET Operation

To understand the operation of a MESFET, we consider the section under the gate of Fig. 4.36, as shown in the left side of Fig. C.1. The source is grounded, the gate is zero or reverse biased, and the drain is zero or forward biased; that is, $V_\mathrm{G}\leq0$ and $V_\mathrm{D}\geq0$.

Figure C.1: Cross section of the channel region of a MESFET (left), and drain voltage variation along the channel (right).

The resistance of the channel is given by (3.127)

$$R=\varrho\cdot\displaystyle\frac{L_\mathrm{g}}{A}=\displaystyle\f... ...\cdot \mu_\mathrm{n}\cdot N_\mathrm{D}\cdot W_\mathrm{g}\cdot\left(a-W\right)},$$ (C.1)

where $N_\mathrm{D}$ is the donor concentration, $A$ is the cross-section area of the current flow and equals $W_\mathrm{g}(a-W)$, and $W$ is the average width $(W_1+W_2)/2$ of the depletion region of the Schottky barrier.

When no gate voltage is applied and $V_\mathrm{D}$ is small, a small current flows in the channel. The magnitude of the current is given by $V_\mathrm{D}/R$. Therefore, the current varies linearly with the drain voltage. Of course for any given drain voltage, the voltage along the channel increases from zero at the source to $V_\mathrm{D}$ at the drain. Thus, the Schottky barrier becomes increasingly reverse biased as we proceed from the source to the drain. As $V_\mathrm{D}$ is increased, $W$ increases, and the average cross-sectional area for the current flow is reduced. The channel resistance $R$ also increases. As a result, the current increases at a slower rate.

As the drain voltage further increases, eventually the depletion region touches the semi-insulating substrate. That happens when $W_{2}=a$ at the drain. We can obtain the corresponding value of the drain voltage, called the saturation voltage $V_\mathrm{Dsat}$, from

$$V_{\mathrm{Dsat}}=\displaystyle\frac{q\cdot \mu_\mathrm{n}\cdot N_{D}\cdot a^2}{2\varepsilon_{s}}-V_\mathrm{bi}. \hspace{1cm}V_\mathrm{G}=0$$ (C.2)

At this drain voltage, the source and the drain are pinched off or completely separated by a reverse-biased depletion region. At this point, a large drain current called the saturation current $I_\mathrm{Dsat}$ can flow across the depletion region. Beyond this pinch-off point, as $V_\mathrm{D}>V_\mathrm{Dsat}$, the current remains essentially at the value $I_\mathrm{Dsat}$ and is independent of $V_\mathrm{D}$.

When a gate voltage is applied to reverse bias the gate contact, the depletion-layer width $W$ increases. For small $V_\mathrm{D}$, the channel again acts as a resistor but its resistance is higher because the cross-section area available for current flow is decreased. When $V_\mathrm{D}$ is increased to a certain value, the depletion region again touches the semi-insulating substrate. The value of this $V_\mathrm{D}$ is given by $V_\mathrm{Dsat}$

$$V_{\mathrm{Dsat}}=\displaystyle\frac{q\cdot \mu_\mathrm{n}\cdot N_\mathrm{D}\cdot a^2}{2\varepsilon_\mathrm{s}}-V_\mathrm{bi}-V_\mathrm{G}.$$ (C.3)

For an n-channel MESFET, the gate voltage is negative with respect to the source, so we use the absolute value of $V_\mathrm{G}$ in (C.3). One can see from (C.3) that the application of a gate voltage $V_\mathrm{G}$ reduces the drain voltage required for the onset of pinch-off by an amount equal to $V_\mathrm{G}$.
T. Ayalew: SiC Semiconductor Devices Technology, Modeling, and Simulation