PHT.301 Physics of Semiconductor Devices

## n-channel JFET

The expression for the drain current of a n-channel JFET in the linear regime is,

$$$I_D=I_p\left[\frac{V_D}{V_p}-\frac{2}{3}\left(\frac{V_{bi}+V_D-V_G}{V_p}\right)^{3/2}+\frac{2}{3}\left(\frac{V_{bi}-V_G}{V_p}\right)^{3/2}\right],$$$

where,

\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} =k_B T \ln \left(\frac{N_A N_D}{{n_i}^2}\right) \qquad n_i=\sqrt{N_cN_v\left(\frac{T}{300}\right)^{3}}\exp\left(\frac{-E_g}{2k_BT}\right). \end{align}

In the saturation regime the current is,

$$$I_D=I_p\left[\frac{1}{3}-\frac{V_{bi}-V_G}{V_p}+\frac{2}{3}\left(\frac{V_{bi}-V_G}{V_p}\right)^{3/2}\right].$$$

These expressions are valid assuming that the pn junction is reverse biased. For a n-channel JFET, $V_G$ < 0 and $V_D$ > 0 in this regime.

 ID [mA] VD [V]
 $N_c=$ cm-3 @ 300 K $N_v=$ cm-3 @ 300 K $E_g=$ eV $N_D=$ cm-3 $N_A=$ cm-3 $\mu_n=$ cm2/Vs $h=$ μm $L=$ μm $Z=$ μm $\epsilon_r=$ $T=$ K $V_{D} (\text{max})$ = V $V_g$ [1] = V $V_g$ [2] = V $V_g$ [3] = V $V_g$ [4] = V $V_g$ [5] = V $V_g$ [6] = V

$E_g=$ eV;  $n_i=$ cm-3;  $V_{bi}=$ V;  $I_p=$ A;  $V_p=$ V.