Semiconductors/MESFET Transistors

MESFET Operation
Assume an N channel MESFET with uniform doping and sharp depletion region shown in figure 1.

The depletion region $$W_n$$ is given by the depletion width for a diode. Where the voltage is the voltage from the gate to the channel, where the channel voltage is given for a position x along the channel as $$V_{gc}(x)$$.


 * $$W_n(x)=\sqrt{\frac{2\varepsilon_0\varepsilon_r(\Psi-V_{gc}(x))}{qN_d}}$$


 * $$W_n(x)^2=\frac{2\varepsilon_0\varepsilon_r(\Psi-V_{gc}(x))}{qN_d}$$


 * $$\frac{W_n(x)^2qN_d}{2\varepsilon_0\varepsilon_r}=\Psi-V_{gc}(x)$$


 * $$V_{gc}(x)=\Psi-\frac{W_n(x)^2qN_d}{2\varepsilon_0\varepsilon_r}$$


 * $$\frac{dV_{gc}(x)}{dW_n(x)}=-\frac{2W_n(x)qN_d}{2\varepsilon_0\varepsilon_r}$$            (1)

The current density in the channel is given by:


 * $$J_n=\sigma\xi$$


 * $$I_n(x)=\sigma \xi\cdot W\cdot b(x)$$


 * $$I_n(x)=-\sigma \frac{dV_{gc}(x)}{dx}W(a-W_n(x))$$

where:


 * $$\xi=-\frac{dV_{gc}(x)}{dx}$$

Therefore,


 * $$I_n(x)=-\sigma aW\bigg(1-\frac{W_n(x)}{a}\bigg)\frac{dV_{gc}(x)}{dWn(x)}\frac{dWn(x)}{dx}

$$


 * $$\int_0^L I_n(x)\, dx=\int_0^L-\sigma aW\bigg(1-\frac{W_n(x)}{a}\bigg)\frac{dV_{gc}(x)}{dW_n(x)}\frac{dW_n(x)}{dx}\, dx$$


 * $$I_n\cdot L= -\sigma aW\int_{Wn(0)}^{W_n(L)}

\bigg(1-\frac{W_n(x)}{a}\bigg)\frac{dV_{gc}(x)}{dW_n(x)}\, dW_n(x)$$

Substituting from equation 1:


 * $$I_n= \frac{-\sigma aW}{L}\int_{W_n(0)}^{W_n(L)}

\bigg(1-\frac{W_n(x)}{a}\bigg)\bigg(-\frac{2W_n(x)qN_d}{2\varepsilon_0\varepsilon_r}\bigg)\, dWn(x)$$


 * $$I_n= \frac{\sigma aW2qN_d}{2\varepsilon_0\varepsilon_rL}\int_{W_n(0)}^{W_n(L)}

\bigg(W_n(x)-\frac{W_n(x)^2}{a}\bigg)\, dWn(x)$$


 * $$I_n= \frac{2\sigma aWqN_d}{2\varepsilon_0\varepsilon_rL}

\bigg[\frac{W_n^2(x)}{2}-\frac{W_n^3(x)}{3a}\bigg]_{W_n(0)}^{W_n(L)} $$


 * $$I_n= \frac{2\sigma aWqN_d}{2\varepsilon_0\varepsilon_rL}

\bigg[\frac{W_n^2(L)-W_n^2(0)}{2}-\frac{W_n^3(L)-W_n^3(0)}{3a}\bigg] $$


 * $$I_n= \frac{2\sigma aWqN_da^2}{6L\cdot

2\varepsilon_0\varepsilon_r} \bigg[\frac{3(W_n^2(L)-W_n^2(0))}{a^2}-\frac{2(W_n^3(L)-W_n^3(0))}{a^3}\bigg] $$

One defines constant &Beta; as the channel conductance with no depletion. And the work function to deplete the channel W00 [1]:


 * $$W_{00}=\Psi-V_{to}=\frac{qN_da^2}{2\varepsilon_0\varepsilon_r}$$


 * $$\beta = \frac{\sigma a}{3LW_{00}}$$

We now define Vto, the voltage such that the channel is pinched off. d is the ratio of channel depletion to maximum depletion for the drain. s the ratio of channel depletion to maximum depletion for the source.


 * $$d=\frac{W_n(L)}{a}=\frac{\sqrt{\frac{2\varepsilon_0\varepsilon_r(\Psi-V_{gd})}{qN_d}}}{\sqrt{\frac{2\varepsilon_0\varepsilon_r(\Psi-V_{to})}{qN_d}}}=\sqrt{\frac{\Psi-V_{gd}}{W_{00}}}

$$


 * $$s=\frac{W_n(0)}{a}=\frac{\sqrt{\frac{2\varepsilon_0\varepsilon_r(\Psi-V_{gs})}{qN_d}}}{\sqrt{\frac{2\varepsilon_0\varepsilon_r(\Psi-V_{to})}{qN_d}}}=\sqrt{\frac{\Psi-V_{gs}}{W_{00}}}

$$

Substituting:


 * $$I_n= W\cdot \frac{\sigma a\cdot W_{00}}{3L}

\big[3(d^2-s^2)-2(d^3-s^3)\big]$$


 * $$I_n= W \cdot\beta W_{00}^2 \big[3(d^2-s^2)-2(d^3-s^3)\big]$$ (2)

Equation 2 is Shockley's expression [2] for drain current in the linear region. When the device enters saturation, one end is pinched off(normally the drain). Thus $d=1$ and one may derive the equation for the saturation region:


 * $$I_{sat}=\beta W_{00}^2(1-3s^2+2s^3)$$


 * $$g_m=3\beta W_{00}(s-1)$$


 * $$G_{DS}=3\beta W_{00}(1-d)$$

Simpler Model

 * $$I_{ds}=\frac{3}{2}\beta

W_{00}^2\bigg[\frac{(V_{gs}-v_{to})^2}{W_{00}^2}-\frac{(V_{gd}-v_{to})^2}{W_{00}^2}\bigg] $$


 * $$g_m=3\beta W_{00}(V_{gs}-V_{to})$$


 * $$G_{ds}=3\beta W_{00}(V_{gd}-V_{to})$$

General power law:
It was found that a general power law provided a better fit for real devices [3].


 * $$I_{ds}=\beta\big[(V_{gs}-V_{to})^Q-(V_{gd}-V_{to})^Q\big]$$

Where Q is dependent on the doping profile and a good fit is usually obtained for Q between 1.5 and 3. A general power law is approximately equal to Shockley's equation for Q = 2.4. &Beta; is also empirically chosen and is proportion to the previous &Beta;


 * $$\beta \mbox{ proportial to } \frac{\sigma aW}{3LW_{00}}$$

Modelling the various regions is done though model binning. This however infers that a sharp transition exists from one region to another, which may not be accurate.


 * $$I_{ds}=\left\{

\begin{matrix} 0&V_{gs}V_{gd} \end{matrix}\right. $$