3.10 Body Effect Model

CAUTION: The above lateral non-uniform doping model or the body effect model are empirical and have their limits as to how much $V_{th}$ shift can be achieved without distorting the I-V curve. Over usage could lead to negative $g_m$ or negative $g_{ds}$. For example, the lateral non-uniform doping model could be used in combination with the mobility model to achieve high $V_{th}$ shift between C-V and I-V curved to avoid any distortion of higher order derivatives.

The equations showing the determination of the bulk charge (qi_acc_for_QM) are provided next. This bulk charge is critical in terms of determination of the centroid of charge in the accumulation region.

If $BULKMOD \ne 0$,

$T_9 = \dfrac{K1}{(2.0 \cdot nV_{tm})} \sqrt{V_{tm}} \qquad (3.350)$

$T_0 = \dfrac{T_9}{2} \qquad (3.351)$

$T_2 = \dfrac{V_{ge} - (\Delta \phi - E_g - V_{tm} + ln (\frac{NBODY_i}{N_c}) + DELVFBACC) }{V_{tm}} \qquad (3.352)$

where $V_{ge}$ is the gate to substrate voltage.

The following equations calculate the accumulation charge and related quantities considering QM effects.

If $(T_2 \cdot V_{tm}) > (\phi_B + T_9 \cdot \sqrt{\phi_B \cdot V_{tm}})$,

$T_1 = \sqrt{T_2 - 1.0 + T_0 \cdot T_0} - T_0 \qquad (3.353)$

$T_{10} = 1.0 + T_1 \cdot T_1 \qquad (3.354)$

Else,

$T_3 = 0.5 \cdot T_2 - 3.0 \cdot \Big( 1.0 + \dfrac{T_9}{\sqrt{2}} \Big) \qquad (3.355)$

$T_{10} = T_3 + \sqrt{T_3 \cdot T_3 + 6 \cdot T_2} \qquad (3.356)$

If $T_2 < 0$,

$T_4 = \dfrac{T_2 - T_{10}}{T_9} \qquad (3.357)$

$T_{10} = -ln (1.0 - T_{10} + T_4 \cdot T_4) \qquad (3.358)$

Else,

$T_{11} = exp(-T_{10}) \qquad (3.359)$

$T_4 = \sqrt{T_2 - 1.0 + T_{11} + T_0 \cdot T_0} - T_0 \qquad (3.360)$

$T_{10} = 1.0 - T_{11} + T_4 \cdot T_4 \qquad (3.361)$

$T_6 = exp(-T_{10}) - 1.0 \qquad (3.362)$

$T_7 = \sqrt{T_6 + T_{10}} \qquad (3.363)$

If $T_{10} > 10^{-15}$,

$e_0 = -(T_2 - T_{10}) + T_9 \cdot T_7 \qquad (3.364)$

$e_1 = 1.0 - T_9 \cdot 0.5 \cdot \dfrac{T_6}{T_7} \qquad (3.365)$

$T_8 = T_{10} - \dfrac{e_0}{e_1} \qquad (3.366)$

$T_{11} = exp(-T_8) - 1.0 \qquad (3.367)$

$T_{12} = \sqrt{T_{11} + T_8} \qquad (3.368)$

$qb\_acc\_s = -T_9 + T_{12} \cdot V_{tm} \qquad (3.369)$

Else if $T_{10} < -10^{-15}$,

$e_0 = -(T_2 - T_{10}) - T_9 \cdot T_7 \qquad (3.370)$

$e_1 = 1.0 + T_9 \cdot 0.5 \cdot \dfrac{T_6}{T_7} \qquad (3.371)$

$T_8 = T_{10} - \dfrac{e_0}{e_1} \qquad (3.372)$

$T_{12} = T_9 \cdot \sqrt{exp(-T_8) + T_8 - 1.0} \qquad (3.373)$

Else,

$T_{12} = 0.0 \qquad (3.374)$

$T_8 = 0.0 \qquad (3.375)$

Then,

$qb\_acc\_s = T_{12} \cdot V_{tm} \qquad (3.376)$

$qi\_acc\_for\_QM = T_9 \cdot exp \Big( \dfrac{-T_8}{2} \Big) \cdot V_{tm} \qquad (3.377)$

$qb\_acc\_d = qb\_acc\_s \qquad (3.378)$

$psipclamp = 0.5 \cdot (T_8 + 1.0 + \sqrt{(T_8 - 1.0) \cdot (T_8 - 1.0) + 0.25 \cdot 2.0 \cdot 2.0}) \qquad (3.379)$

$sqrtpsip = \sqrt{psiclamp} \qquad (3.380)$

$nq = 1.0 + \dfrac{T_9}{sqrtpsip} \qquad (3.381)$