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

The duty cycle $D=t_{\mathrm{active}}/(t_{\mathrm{active}}+t_{\mathrm{sleep}})$ is the fraction of time spent active. At a constant supply voltage and with switching energy neglected,For $D=0.01$, the average current is $0.01(25\mathrm{mA})+0.99(6.1\mu \mathrm{A})=256.039\mu \mathrm{A}$. Thus $\bar{P}=512.078\mu \mathrm{W}\simeq 0.512\mathrm{mW}$. Multiplying only the active power by the duty cycle would omit the sleep contribution.

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.