Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2019 / Paper 1 / Section B / 12 / v / b / Solution

For $D=0.001$, the same duty cycle model gives $\bar{I}=25\mu \mathrm{A}+0.999(6.1\mu \mathrm{A})=31.0939\mu \mathrm{A}$. Thus $\bar{P}=62.1878\mu \mathrm{W}$. The tenfold reduction in active fraction reduces average power by only about $8.23$ because the sleep current remains. The ideal limit as $D\to 0$ is $V{I}_s=12.2\mu \mathrm{W}$, before leakage elsewhere or wake-up energy. Active and sleep contributions become comparable at a duty cycle of roughly $I_s/(I_a+I_s)=0.0244\%$; below that scale lowering sleep current is particularly valuable.