# 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 $V_{gc}(x)$.
    $W_{n}(x)={\sqrt {\frac {2\varepsilon _{0}\varepsilon _{r}(\Psi -V_{gc}(x))}{qN_{d}}}}$
    $W_{n}(x)^{2}={\frac {2\varepsilon _{0}\varepsilon _{r}(\Psi -V_{gc}(x))}{qN_{d}}}$
    ${\frac {W_{n}(x)^{2}qN_{d}}{2\varepsilon _{0}\varepsilon _{r}}}=\Psi -V_{gc}(x)$
    $V_{gc}(x)=\Psi -{\frac {W_{n}(x)^{2}qN_{d}}{2\varepsilon _{0}\varepsilon _{r}}}$
    ${\frac {dV_{gc}(x)}{dW_{n}(x)}}=-{\frac {2W_{n}(x)qN_{d}}{2\varepsilon _{0}\varepsilon _{r}}}$ (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,
    $I_{n}(x)=-\sigma aW{\bigg1-{\frac {W_{n}(x)}{a}}{\bigg )}{\frac {dV_{gc}(x)}{dWn(x)}}{\frac {dWn(x)}{dx}}$
    $\int _{0}^{L}I_{n}(x)\,dx=\int _{0}^{L}-\sigma aW{\bigg1-{\frac {W_{n}(x)}{a}}{\bigg )}{\frac {dV_{gc}(x)}{dW_{n}(x)}}{\frac {dW_{n}(x)}{dx}}\,dx$
    $I_{n}\cdot L=-\sigma aW\int _{Wn(0)}^{W_{n}(L)}{\bigg1-{\frac {W_{n}(x)}{a}}{\bigg )}{\frac {dV_{gc}(x)}{dW_{n}(x)}}\,dW_{n}(x)$
Substituting from equation 1:
    $I_{n}={\frac {-\sigma aW}{L}}\int _{W_{n}(0)}^{W_{n}(L)}{\bigg1-{\frac {W_{n}(x)}{a}}{\bigg )}{\bigg-{\frac {2W_{n}(x)qN_{d}}{2\varepsilon _{0}\varepsilon _{r}}}{\bigg )}\,dWn(x)$
    $I_{n}={\frac {\sigma aW2qN_{d}}{2\varepsilon _{0}\varepsilon _{r}L}}\int _{W_{n}(0)}^{W_{n}(L)}{\biggW_{n}(x)-{\frac {W_{n}(x)^{2}}{a}}{\bigg )}\,dWn(x)$
    $I_{n}={\frac {2\sigma aWqN_{d}}{2\varepsilon _{0}\varepsilon _{r}L}}{\bigg{\frac {W_{n}^{2}(x)}{2}}-{\frac {W_{n}^{3}(x)}{3a}}{\bigg ]}_{W_{n}(0)}^{W_{n}(L)}$
    $I_{n}={\frac {2\sigma aWqN_{d}}{2\varepsilon _{0}\varepsilon _{r}L}}{\bigg{\frac {W_{n}^{2}(L)-W_{n}^{2}(0)}{2}}-{\frac {W_{n}^{3}(L)-W_{n}^{3}(0)}{3a}}{\bigg ]}$
    $I_{n}={\frac {2\sigma aWqN_{d}a^{2}}{6L\cdot 2\varepsilon _{0}\varepsilon _{r}}}{\bigg{\frac {3(W_{n}^{2}(L)-W_{n}^{2}(0))}{a^{2}}}-{\frac {2(W_{n}^{3}(L)-W_{n}^{3}(0))}{a^{3}}}{\bigg ]}$
One defines constant Β as the channel conductance with no depletion. And the work function to deplete the channel W$_{00}$ [1]:
    $W_{00}=\Psi -V_{to}={\frac {qN_{d}a^{2}}{2\varepsilon _{0}\varepsilon _{r}}}$
    $\beta ={\frac {\sigma a}{3LW_{00}}}$
We now define V$_{to}$, 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:
    $I_{n}=W\cdot {\frac {\sigma a\cdot W_{00}}{3L}}{\big3(d^{2}-s^{2})-2(d^{3}-s^{3}){\big ]}$
    $I_{n}=W\cdot \beta W_{00}^{2}{\big3(d^{2}-s^{2})-2(d^{3}-s^{3}){\big ]}$ (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. Β is also empirically chosen and is proportion to the previous Β
    $\beta {\mbox{ proportial to }}{\frac {\sigma aW}{3LW_{00}}}$
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.
    $I_{ds}=\left\{{\begin{matrix}0&V_{gs}<V_{to}\\\beta {\big(V_{gs}-V_{to})^{Q}-(V_{gd}-V_{to})^{Q}{\big ]}&V_{gs}\leq V_{gd}\\\beta (V_{gs}-V_{to})^{Q}&V_{gs}>V_{gd}\end{matrix}}\right.$

## References

[1] A. E. Parker. Design System for Locally Fabricated Gallium Arsenide Digital Integrated Circuits. PhD thesis, Sydney University, 1990.
[2] W. Shockley. A unipolar field-effect transistor. IEEE Trans/ Electron Devices, 20(11):1365–1376, November 1952.
[3] I. Richer and R.D. Middlebrook. Power-law nature of field-effect transistor experimental characteristics. Proc. IEEE, 51(8):1145–1146, August 1963.