Question 1.10: EXPANSION OF Si The expansion coefficient of silicon over th...
EXPANSION OF Si The expansion coefficient of silicon over the temperature range 120–1500 K is given by Okada and Tokumaru (1984) as
\lambda=3.725 \times 10^{-6}\left[1-e^{-5.88 \times 10^{-3}(T-124)}\right]+5.548 \times 10^{-10} T [1.24]
where \lambda is in K^{−1} (or °C^{−1}) and T is in kelvins. At a room temperature of 20 °C, the above gives \lambda = 2.51 × 10^{−6} K^{−1}. Calculate the fractional change ΔL / L_{0} in the length L_{o} of an Si crystal from 20 to 320 °C, by (a) assuming a constant \lambda equal to the room temperature value and (b) assuming the above temperature dependence. Calculate the mean \bar{\lambda} for this temperature range.
Assuming a constant \lambda , we have
\frac{\Delta L}{L_{o}}=\lambda\left(T-T_{0}\right)=\left(2.51 \times 10^{-6}{ }^{\circ} C ^{-1}\right)(320-20)=0.753 \times 10^{-3} \quad \text { or } \quad 0.075 \%
With a temperature-dependent \lambda (T) ,
\frac{\Delta L}{L_{o}} =\int_{T_{o}}^{T} \lambda(T) d T
=\int_{20+273}^{320+273}\left\{3.725 \times 10^{-6}\left[1-e^{-5.88 \times 10^{-3}(T-124)}\right]+5.548 \times 10^{-10} T\right\} d T
The integration can either be done numerically or analytically (both left as an exercise) with the result that
\frac{\Delta L}{L_{o}}=1.00 \times 10^{-3} \quad \text { or } \quad 0.1 \%
which is substantially more than when using a constant \lambda . The mean \bar{\lambda} over this temperature range can be found from
\frac{\Delta L}{L_{o}}=\bar{\lambda}\left(T-T_{o}\right) \quad \text { or } \quad 1.00 \times 10^{-3}=\bar{\lambda}(320-20)
which gives \bar{\lambda}=3.33 \times 10^{-6}{ }^{\circ} C ^{-1} A 0.1 percent change in length means that a 1 mm chip would expand by 1 micron.