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3.2.3 Consistent Physical Parameters

The high-field mobility formulas (3.29) and (3.31) need some further investigation to fully understand their impact on device modeling. In general the effective mobilities are defined by

$ \mu_{\nu}^{}$ = q$ {\frac{\tau_{m,\nu}}{m^*_{\nu}}}$ (3.35)

with $ \tau_{m,\nu}^{}$ being the momentum relaxation time and m*$\scriptstyle \nu$ being the effective carrier mass of the respective carrier. By evaluating the moments of the distribution functions [29] the momentum relaxation times $ \tau_{m,\nu}^{}$ can be approximated as functions of the average carrier energies w$\scriptstyle \nu$ = 3kBT$\scriptstyle \nu$/2 to give (3.31).

For DD T$\scriptstyle \nu$ = TL is assumed which obviously cannot be used to calculate the carrier energy and its influence on the mobilities. However, the carrier temperatures are coupled to the electric field via (3.9) and (3.10). Assuming a homogeneously doped semiconductor, divB>S$\scriptstyle \nu$ = 0, and the energy balance equations (3.9) and (3.10) degenerate to the so-called local energy balance equations which read

E2 . $ \mu_{\nu}^{}$(T$\scriptstyle \nu$) = $ {\frac{3\cdot \mathrm{k_{B}}\cdot \Delta T_{\nu}}{2 \cdot \mathrm{q}\cdot \tau_{\epsilon,\nu}}}$ (3.36)

Under the assumption of field- and carrier temperature-independent $ \tau_{\epsilon,\nu}^{}$ equation (3.36) in combination with (3.31) gives a direct, local dependence of the carrier temperatures and the electric field

$ \Delta$T$\scriptstyle \nu$(E) = $ {\frac{1}{2 \cdot \alpha_{\nu}}}$ . $ \left(\vphantom{\sqrt{1+\left(\frac{2 \cdot \mu^{\mathrm{LIS}}\cdot E}{v^{\mathrm{sat}}_{\nu}}\right)^{2}} - 1}\right.$$ \sqrt{1+\left(\frac{2 \cdot \mu^{\mathrm{LIS}}\cdot E}{v^{\mathrm{sat}}_{\nu}}\right)^{2}}$ - 1$ \left.\vphantom{\sqrt{1+\left(\frac{2 \cdot \mu^{\mathrm{LIS}}\cdot E}{v^{\mathrm{sat}}_{\nu}}\right)^{2}} - 1}\right)$ (3.37)

which can be substituted into (3.31) to find
$\displaystyle \mu^{\mathrm{LISF}}_{\nu}$(E) = $\displaystyle {\frac{2\cdot\mu^{\mathrm{LIS}}_{\nu}}{1+\sqrt{1+ \displaystyle{\...
...{2\cdot\mu^{\mathrm{LIS}}_{\nu}\cdot E}
{v^{\mathrm{sat}}_{\nu}}\right)}^{2}}}}$ (3.38)

which is similar to (3.29) with $ \beta_{\nu }^{}$ = 2. However, in (3.29) the driving force
F$\scriptstyle \nu$ = $\displaystyle \left\vert\vphantom{~{\rm grad}~\psi + \frac{s_{\nu}}{\nu}\cdot\f...
...athrm{k_{B}}}{\mathrm{q}}\cdot
{\rm grad}~{\left(T_{L}\cdot \nu\right)}}\right.$ grad $\displaystyle \psi$ + $\displaystyle {\frac{s_{\nu}}{\nu}}$ . $\displaystyle {\frac{\mathrm{k_{B}}}{\mathrm{q}}}$ . grad $\displaystyle \left(\vphantom{T_{L}\cdot \nu}\right.$TL . $\displaystyle \nu$ $\displaystyle \left.\vphantom{T_{L}\cdot \nu}\right)$$\displaystyle \left.\vphantom{~{\rm grad}~\psi + \frac{s_{\nu}}{\nu}\cdot\frac{...
...m{k_{B}}}{\mathrm{q}}\cdot
{\rm grad}~{\left(T_{L}\cdot \nu\right)}}\right\vert$ (3.39)

is used instead of the electrical field E. This is done for the following reasons. (3.38) has been derived under the assumption of a homogeneous semiconductor. In this case (3.39) reduces to F$\scriptstyle \nu$ = $ \left\vert\vphantom{~{\rm grad}~\psi}\right.$ grad $ \psi$ $ \left.\vphantom{~{\rm grad}~\psi}\right\vert$ = E which is consistent with the assumptions given above. On the other hand, (3.38) would give no velocity saturation for large carrier gradients and low electric fields. This is obviously unphysical as the carrier velocity for the resulting diffusion current is limited by the thermal velocity. Equation (3.38) is very similar to expressions found on a purely empirical basis. The most common ones are

$\displaystyle \mu^{\mathrm{LISF}}_{\nu}$(F$\scriptstyle \nu$) = $\displaystyle {\frac{\mu^{\mathrm{LIS}}_{\nu}}{\left(1+ \displaystyle{\left(\fr...
..._{\nu}}
{v^{\mathrm{sat}}_{\nu}}\right)}^{\beta_{\nu}}\right)^{1/\beta_{\nu}}}}$ (3.40)

as given by Caughey and Thomas [5] and
$\displaystyle \mu^{\mathrm{LISF}}_{\nu}$(F$\scriptstyle \nu$) = $\displaystyle {\frac{2\cdot\mu^{\mathrm{LIS}}_{\nu}}{1+\left(1+ \displaystyle{\...
..._{\nu}}
{v^{\mathrm{sat}}_{\nu}}\right)}^{\beta_{\nu}}\right)^{1/\beta_{\nu}}}}$ (3.41)

as given by Reiser [48]. Both (3.40) and (3.41) are claimed to perfectly fit measured data with an appropriate $ \beta_{\nu }^{}$, normally about $ \beta_{n}^{}$ = 2 for electrons and $ \beta_{p}^{}$ = 1 for holes. For $ \beta_{\nu }^{}$ = 1 both models are equivalent.

One problem becomes obvious when comparing (3.40) and (3.41) with (3.31). Under homogeneous conditions different mobilities and hence currents will be obtained for DD and HD simulations except when using (3.41) with $ \beta_{\nu }^{}$ = 2. This is of fundamental importance when comparing DD with HD simulations but unfortunately this problem is generally overlooked in available device simulators.

In general the problem can be stated as follows. In the homogeneous situation the electric field and the carrier temperatures are related by the local energy balance equation. The two parameters, $ \mu$ and $ \tau_{\epsilon,}^{}$ have to be modeled properly to guarantee consistency between the DD and the HD model. In principle one could start with either $ \mu^{\mathrm{LIST}}_{\nu}$(T$\scriptstyle \nu$) or $ \mu^{\mathrm{LISF}}_{\nu}$(E) to derive the other appropriate model. To derive $ \mu^{\mathrm{LIST}}_{\nu}$(T$\scriptstyle \nu$) from the scattering term of the Boltzmann equation several assumptions on the distribution function are necessary to get a closed form solution like (3.31). From a practical point of view it might seem simpler to start with $ \mu^{\mathrm{LISF}}_{\nu}$(E) as measured data are more easily available. Nevertheless, as long as closed form solutions exist, the order is of course irrelevant.

The following discussion will assume $ \mu^{\mathrm{LISF}}_{\nu}$(E) of the form

$\displaystyle \mu^{\mathrm{LISF}}_{\nu}$(E) = $\displaystyle {\frac{\mu^{\mathrm{LIS}}_{\nu}}{\xi+\left(\zeta^{\beta_{\nu}}+ \...
...cdot E}
{v^{\mathrm{sat}}_{\nu}}\right)}^{\beta_{\nu}}\right)^{1/\beta_{\nu}}}}$   . (3.42)

Figure 3.1: Mobility vs. electric field in dependence of the basic parameters $ \xi$ and $ \beta_{\nu }^{}$.
\begin{figure}
\begin{center}
\resizebox{11.4cm}{!}{
\psfrag{mobGeneral1.0_0.0_1...
...egraphics[width=11.4cm,angle=0]{figures/mobGeneral.eps}}\end{center}\end{figure}

With $ \xi$ = 0 and $ \zeta$ = 1 one obtains (3.40), and with $ \xi$ = 1/2 and $ \zeta$ = 1/2 (3.41). As these values give the most common expressions, only $ \xi$ will be used to distinguish between them and an explicit dependence

$ \zeta$ = 1 - $ \xi$ (3.43)

will be assumed. It is to note that (3.43) is only used to simplify reference to the more familiar formulas (3.40) with $ \xi$ = 0 and (3.41) with $ \xi$ = 1/2.

For these values and $ \beta_{\nu }^{}$ = 1 and $ \beta_{\nu }^{}$ = 2 the mobility vs. field dependence is shown in Fig. 3.1. In the approach given above the only assumption made about $ \tau_{\epsilon,\nu}^{}$ was to be independent of the carrier energy. This approximation is valid for large carrier energies [16].

In the following the approach outlined above will be generalized starting from (3.42). Substituting (3.42) in (3.36) results in

$ {\frac{E^{2} \cdot \mu^{\mathrm{LIS}}_{\nu}}{\xi+\left(\zeta^{\beta_{\nu}} + \...
...\cdot E}{v^{\mathrm{sat}}_{\nu}}\right)}^{\beta_{\nu}}\right)^{1/\beta_{\nu}}}}$ = $ {\frac{3\cdot \mathrm{k_{B}}\cdot \Delta T_{\nu}}{2 \cdot \mathrm{q}\cdot \tau_{\epsilon,\nu}}}$ (3.44)

which must be solved for E(T$\scriptstyle \nu$), which is then inserted into (3.42). The dependence of T$\scriptstyle \nu$ vs. E is shown in Fig. 3.2. Unfortunately (3.44) cannot be explicitly solved in general. However, it can be solved for the most important cases $ \xi$ = 0 with arbitrary $ \beta_{\nu }^{}$ and for $ \xi$ = 1/2 with $ \beta_{\nu }^{}$ = 1 or $ \beta_{\nu }^{}$ = 2. After some algebra one obtains
$\displaystyle \xi$ = 0 :  *0.0cmE(T$\scriptstyle \nu$) = $\displaystyle {\frac{v^{\mathrm{sat}}_{\nu}}{2^{1/\beta_{\nu}} \cdot \mu^{\mathrm{LIS}}_{\nu}}}$ . $\displaystyle \left(\vphantom{a_{\nu}^{\beta_{\nu}}
+ \sqrt{a_{\nu}^{\beta_{\nu}} \cdot \left(4 + a_{\nu}^{\beta_{\nu}} \right)}}\right.$a$\scriptstyle \nu$$\scriptstyle \beta_{\nu}$ + $\displaystyle \sqrt{a_{\nu}^{\beta_{\nu}} \cdot \left(4 + a_{\nu}^{\beta_{\nu}} \right)}$ $\displaystyle \left.\vphantom{a_{\nu}^{\beta_{\nu}}
+ \sqrt{a_{\nu}^{\beta_{\nu}} \cdot \left(4 + a_{\nu}^{\beta_{\nu}} \right)}}\right)^{1/\beta_{\nu}}_{}$ (3.45)
$\displaystyle \xi$ = $\displaystyle {\textstyle\frac{1}{2}}$$\displaystyle \beta_{\nu}^{}$ = 1 : E(T$\scriptstyle \nu$) = $\displaystyle {\frac{v^{\mathrm{sat}}_{\nu}}{2 \cdot \mu^{\mathrm{LIS}}_{\nu}}}$ . $\displaystyle \left(\vphantom{a_{\nu}
+ \sqrt{a_{\nu} \cdot \left(4 + a_{\nu} \right)}}\right.$a$\scriptstyle \nu$ + $\displaystyle \sqrt{a_{\nu} \cdot \left(4 + a_{\nu} \right)}$ $\displaystyle \left.\vphantom{a_{\nu}
+ \sqrt{a_{\nu} \cdot \left(4 + a_{\nu} \right)}}\right)$ (3.46)
$\displaystyle \xi$ = $\displaystyle {\textstyle\frac{1}{2}}$$\displaystyle \beta_{\nu}^{}$ = 2 : E(T$\scriptstyle \nu$) = $\displaystyle {\frac{v^{\mathrm{sat}}_{\nu}}{\mu^{\mathrm{LIS}}_{\nu}}}$$\displaystyle \sqrt{a_{\nu} \cdot \left(1 + a_{\nu} \right)}$        with    a$\scriptstyle \nu$ = $\displaystyle \alpha_{\nu}^{}$ . $\displaystyle \Delta$T$\scriptstyle \nu$ (3.47)

Of course (3.45) with $ \beta_{\nu }^{}$ = 1 is identical to (3.46) as are the expressions resulting from (3.42) with the same parameters. Inserting (3.45), (3.46), and (3.47) into (3.42) yields
$\displaystyle \xi$ = 0 : $\displaystyle \mu^{\mathrm{LIST}}_{\nu}$(T$\scriptstyle \nu$) = $\displaystyle {\frac{2^{1/\beta_{\nu}} \cdot \mu^{\mathrm{LIS}}_{\nu}}{\left(2 ...
..._{\nu}} \cdot \left(4 + a_{\nu}^{\beta_{\nu}} \right)}\right)^{1/\beta_{\nu}}}}$ (3.48)
$\displaystyle \xi$ = $\displaystyle {\textstyle\frac{1}{2}}$$\displaystyle \beta_{\nu}^{}$ = 1 : $\displaystyle \mu^{\mathrm{LIST}}_{\nu}$(T$\scriptstyle \nu$) = $\displaystyle {\frac{2 \cdot \mu^{\mathrm{LIS}}_{\nu}}{2 + a_{\nu}
+ \sqrt{a_{\nu} \cdot \left(4 + a_{\nu} \right)}}}$ (3.49)
$\displaystyle \xi$ = $\displaystyle {\textstyle\frac{1}{2}}$$\displaystyle \beta_{\nu}^{}$ = 2 : $\displaystyle \mu^{\mathrm{LIST}}_{\nu}$(T$\scriptstyle \nu$) = $\displaystyle {\frac{\mu^{\mathrm{LIS}}_{\nu}}{1 + a_{\nu}}}$ (3.50)

A comparison of these mobilities is given in Fig. 3.4. The diffusivity is defined by the generalized Einstein relation

D$\scriptstyle \nu$(T$\scriptstyle \nu$) = $ {\frac{\mathrm{k_{B}}\cdot T_{\nu}}{\mathrm{q}}}$ . $ \mu^{\mathrm{LIST}}_{\nu}$(T$\scriptstyle \nu$) (3.51)

and is shown in Fig. 3.5. In Fig. 3.5 $ \alpha$ = 1/TL has been assumed which gives a 1/T$\scriptstyle \nu$ dependence of the mobility (3.50) and hence a constant diffusivity. This choice will be justified later.

Figure 3.2: Carrier temperature as a function of electric field for the approach of Hänsch as given by (3.44).
\begin{figure}
\begin{center}
\resizebox{11.4cm}{!}{
\psfrag{THaensch1.0_0.0_1.0...
...udegraphics[width=11.4cm,angle=0]{figures/THaensch.eps}}\end{center}\end{figure}

Figure 3.3: Carrier temperature as a function of electric field for the approach of Baccarani as given by (3.54).
\resizebox{11.4cm}{!}{
\psfrag{TBacc1.0_0.0_1.0.tbl:y} {$\textstyle \xi = 0,\ \ ...
...(E)\ \ \mathrm{[K]}$}
\includegraphics[width=11.4cm,angle=0]{figures/TBacc.eps}}

A different approach has been proposed by Baccarani and Wordeman [2]. From the generalized Einstein relation it follows that the carrier temperature can be directly related to the applied field as

T$\scriptstyle \nu$ = $ {\frac{\mathrm{q}}{\mathrm{k_{B}}}}$ . $ {\frac{D_{\nu}(E)}{\mu^{\mathrm{LISF}}_{\nu}(E)}}$   . (3.52)

They assumed a constant diffusivity

D$\scriptstyle \nu$(E) = D$\scriptstyle \nu$0 = $ {\frac{\mathrm{k_{B}}\cdot T_{L}}{\mathrm{q}}}$ . $ \mu^{\mathrm{LIS}}_{}$ (3.53)

which they justified with experimental data and MC simulation results. However, (3.53) overestimates D$\scriptstyle \nu$(E) for large fields [2]. Using (3.52) and (3.53) the carrier temperature may be expressed as

T$\scriptstyle \nu$ = TL . $ {\frac{\mu^{\mathrm{LIS}}}{\mu^{\mathrm{LISF}}_{\nu}(E)}}$ = TL . $ {\frac{\mu^{\mathrm{LIS}}}{\mu^{\mathrm{LIST}}_{\nu}(T_{\nu})}}$   . (3.54)

This dependence is shown in Fig. 3.3. With (3.42) and (3.54) E(T$\scriptstyle \nu$) reads
E(T$\scriptstyle \nu$) = $\displaystyle {\frac{v^{\mathrm{sat}}_{\nu}}{\mu^{\mathrm{LIS}}_{\nu}}}$ . $\displaystyle \left(\vphantom{
\left(\frac{T_{\nu}}{T_{L}} - \xi\right)^{\beta_{\nu}}
- \zeta^{\beta_{\nu}}
}\right.$$\displaystyle \left(\vphantom{\frac{T_{\nu}}{T_{L}} - \xi}\right.$$\displaystyle {\frac{T_{\nu}}{T_{L}}}$ - $\displaystyle \xi$ $\displaystyle \left.\vphantom{\frac{T_{\nu}}{T_{L}} - \xi}\right)^{\beta_{\nu}}_{}$ - $\displaystyle \zeta^{\beta_{\nu}}_{}$ $\displaystyle \left.\vphantom{
\left(\frac{T_{\nu}}{T_{L}} - \xi\right)^{\beta_{\nu}}
- \zeta^{\beta_{\nu}}
}\right)^{1/{\beta_{\nu}}}_{}$     (3.55)

Figure 3.4: Carrier mobility as a function of carrier temperature for the approach of Hänsch as given by (3.48)-(3.50).
\begin{figure}
\begin{center}
\resizebox{11.4cm}{!}{
\psfrag{uHaensch2.0_0.0_1.0...
...udegraphics[width=11.4cm,angle=0]{figures/uHaensch.eps}}\end{center}\end{figure}

Figure 3.5: Carrier diffusion coefficient as a function of carrier temperature for the approach of Hänsch.
\resizebox{11.4cm}{!}{
\psfrag{D1_1.0.crv:y} {$\textstyle \xi = 0,\ \ \beta_{\nu...
...\ \ \mathrm{[K]}$}
\includegraphics[width=11.4cm,angle=0]{figures/DHaensch.eps}}

Figure 3.6: Carrier energy relaxation time as a function of carrier temperature for the approach of Baccarani.
\begin{figure}
\begin{center}
\resizebox{11.4cm}{!}{
\psfrag{tauBacc1.0_0.0_1.0....
...ludegraphics[width=11.4cm,angle=0]{figures/tauBacc.eps}}\end{center}\end{figure}

From the local energy balance equation the energy relaxation times $ \tau_{\epsilon,\nu}^{}$ may be expressed as

$ \tau_{\epsilon,\nu}^{}$(T$\scriptstyle \nu$) = $ {\frac{3\cdot \mathrm{k_{B}}\cdot \Delta T_{\nu}}{2 \cdot \mathrm{q}\cdot E^{2} \cdot \mu^{\mathrm{LIST}}_{\nu}(T_{\nu})}}$ = $ {\frac{3\cdot \mathrm{k_{B}}\cdot \mu^{\mathrm{LIS}}}{2 \cdot \mathrm{q}\cdot {v^{\mathrm{sat}}_{\nu}}^{2}}}$ . $ {\frac{T_{\nu} \cdot \Delta T_{\nu}}{T_{L}}}$ . $ \left(\vphantom{ \left(\frac{T_{\nu}}{T_{L}} - \xi\right)^{\beta_{\nu}} - \zeta^{\beta_{\nu}}}\right.$$ \left(\vphantom{\frac{T_{\nu}}{T_{L}} - \xi}\right.$$ {\frac{T_{\nu}}{T_{L}}}$ - $ \xi$ $ \left.\vphantom{\frac{T_{\nu}}{T_{L}} - \xi}\right)^{\beta_{\nu}}_{}$ - $ \zeta^{\beta_{\nu}}_{}$ $ \left.\vphantom{ \left(\frac{T_{\nu}}{T_{L}} - \xi\right)^{\beta_{\nu}} - \zeta^{\beta_{\nu}}}\right)^{-2/{\beta_{\nu}}}_{}$ (3.56)

An interesting special case is $ \xi$ = 1/2 and $ \beta_{\nu }^{}$ = 2 in which (3.42) simplifies to (3.38). As (3.38) has been derived under the assumption of an energy independent $ \tau_{\epsilon,\nu}^{}$ one would also expect (3.56) to result in an energy independent $ \tau_{\epsilon,\nu}^{}$. This is indeed the case and one obtains

$ \tau_{\epsilon,\nu}^{}$ = $ {\frac{3\cdot \mathrm{k_{B}}\cdot \mu^{\mathrm{LIS}} \cdot T_{L}}{2\cdot \mathrm{q}\cdot {v^{\mathrm{sat}}_{\nu}}^{2}}}$ = $ \tau_{\epsilon,\nu}^{0}$   . (3.57)

Using (3.57) in (3.50) gives $ \alpha_{\nu}^{}$ = 1/TL and thus

$ \mu^{\mathrm{LIST}}_{\nu}$(T$\scriptstyle \nu$) = $ {\frac{\mu^{\mathrm{LIS}}_{\nu}}{1 + \frac{\displaystyle \Delta T_{\nu}}{\displaystyle T_{L}}}}$ = $ \mu^{\mathrm{LIS}}_{\nu}$ . $ {\frac{T_{L}}{T_{\nu}}}$ (3.58)

which is in fact (3.54). It is to note that (3.57) gives a model for $ \tau_{\epsilon,\nu}^{}$ for which both assumptions produce the same results. However, (3.57) depends on the low-field mobility which can vary dramatically in the device . Most unfortunately no measured data could be found as yet in the literature to confirm this result.

Another interesting feature of (3.56) is

$\displaystyle \lim_{T_{\nu} \rightarrow \infty}^{}$$\displaystyle \tau_{\epsilon,\nu}^{}$(T$\scriptstyle \nu$) = $\displaystyle \tau_{\epsilon,\nu}^{0}$ . $\displaystyle \lim_{T_{\nu} \rightarrow \infty}^{}$$\displaystyle {\frac{T_{\nu} \cdot \Delta T_{\nu}}{T_{L}^{2}}}$ . $\displaystyle \left(\vphantom{
\left(\frac{T_{\nu}}{T_{L}} - \xi\right)^{\beta_{\nu}}
- \zeta^{\beta_{\nu}}
}\right.$$\displaystyle \left(\vphantom{\frac{T_{\nu}}{T_{L}} - \xi}\right.$$\displaystyle {\frac{T_{\nu}}{T_{L}}}$ - $\displaystyle \xi$ $\displaystyle \left.\vphantom{\frac{T_{\nu}}{T_{L}} - \xi}\right)^{\beta_{\nu}}_{}$ - $\displaystyle \zeta^{\beta_{\nu}}_{}$ $\displaystyle \left.\vphantom{
\left(\frac{T_{\nu}}{T_{L}} - \xi\right)^{\beta_{\nu}}
- \zeta^{\beta_{\nu}}
}\right)^{-2/{\beta_{\nu}}}_{}$ (3.59)
  = $\displaystyle \tau_{\epsilon,\nu}^{0}$ . $\displaystyle \lim_{T_{\nu} \rightarrow \infty}^{}$$\displaystyle {\frac{t^{2}-t}{\left(
\left(t - \xi\right)^{\beta_{\nu}}
- \zeta^{\beta_{\nu}}
\right)^{2/{\beta_{\nu}}}}}$        with        t = $\displaystyle {\frac{T_{\nu}}{T_{L}}}$ (3.60)
  = $\displaystyle \tau_{\epsilon,\nu}^{0}$ . $\displaystyle \lim_{T_{\nu} \rightarrow \infty}^{}$$\displaystyle {\frac{1 - \frac{\displaystyle 1}{\displaystyle t}}{\left(
\left(...
...\zeta^{\beta_{\nu}}}{\displaystyle t^{\beta_{\nu}}}
\right)^{2/{\beta_{\nu}}}}}$ (3.61)
  = $\displaystyle \tau_{\epsilon,\nu}^{0}$ (3.62)

independent of $ \xi$ and $ \beta_{\nu }^{}$ > 0 as can also be seen in Fig. 3.6. With $ \mu^{\mathrm{LIS}}_{\nu}$ = 1400 cm2/Vs and vsat$\scriptstyle \nu$ = 107 cm/s equation (3.57) gives $ \tau_{\epsilon,\nu}^{0}$ = 0.54 ps. However, as it is not unrealistic for the low-field mobility to be reduced to 20% of its maximum value, $ \tau_{\epsilon,\nu}^{0}$ is reduced in the very same way.
next up previous contents
Next: 3.2.4 Velocity Overshoot Up: 3.2 Physical Parameters Previous: 3.2.2 Energy Relaxation Time
Tibor Grasser
1999-05-31