Show that for a transistor with α close to unity, if α changes by a small per-unit amount (Δα / α), the corresponding per-unit change in β is given approximately by
$$\frac{\Delta\beta}{\beta}\approx\beta\left(\frac{\Delta\alpha}{\alpha}\right)$$
I know the equations:
$$\alpha=\frac{\beta}{1+\beta} \textrm{ and } \beta=\frac{\alpha}{1-\alpha}$$
But am unsure what Δα / α really means. I tried assuming that it was a linear approximation, but after lengthy scribbling, that led me nowhere:
$$\alpha'=\alpha\left(1+\frac{\Delta\alpha}{\alpha}\right)$$
Best Answer
Alternatively to @Chu's answer, use sensitivity analysis to determine \$\beta\$'s sensitivity to changes in \$\alpha\$.
$$ S_{\alpha}^{\beta} =\lim_{\Delta \alpha \to 0}\left \{ \frac{\Delta \beta / \beta}{\Delta \alpha / \alpha} \right \}=\frac{\alpha}{\beta} \frac{\partial \beta}{\partial \alpha} = \frac {1}{1-a} $$
Now, as @Chu has already mentioned, since \$\alpha \approx 1\$ (via the problem statement),
$$ \frac {1}{1-\alpha} \biggm \rvert_{\alpha \approx 1} \approx \frac {\alpha}{1-\alpha} = \beta $$
So,
$$ \frac {\Delta \beta}{\beta} = \frac {1}{1-\alpha} \: \frac {\Delta \alpha}{\alpha} \approx \frac {\alpha}{1-\alpha} \: \frac {\Delta \alpha}{\alpha} = \beta \: \frac {\Delta \alpha}{\alpha} $$
for \$\alpha \approx 1\$ as \$\Delta \alpha \to 0\$.