Last modified on 11 May 2010, at 17:06

# Semiconductors/MESFET Transistors

## MESFET OperationEdit

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 Β 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 ModelEdit

$I_{ds}=\frac{3}{2}\beta W_{00}^2\bigg[\frac{(V_{gs}-v_{to})^2}{W_{00}^2}-\frac{(V_{gd}-v_{to})^2}{W_{00}^2}\bigg]$
$g_m=3\beta W_{00}(V_{gs}-V_{to})$
$G_{ds}=3\beta W_{00}(V_{gd}-V_{to})$

### General power law:Edit

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.

$I_{ds}=\left\{ \begin{matrix} 0&V_{gs}V_{gd} \end{matrix}\right.$

## ReferencesEdit

[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.