= Solution
The zero-bias gate period in the first plot is $\Delta V_g=20\,\mathrm{mV}$. In the <constant interaction model>, changing the gate-induced island charge by one electron gives
$$
C_g=\frac e{\Delta V_g}=8.0\,\mathrm{aF}.
$$
The tip is at $|V_{sd}|=5.0\,\mathrm{mV}$ above or below zero; use this half-height, not the full $10\,\mathrm{mV}$ vertical extent. The addition energy is $E_{\mathrm{add}}=e(5.0\,\mathrm{mV})=e^2/C_\Sigma$, giving
$$
C_\Sigma=32\,\mathrm{aF},\qquad C_g/C_\Sigma=0.25.
$$
To distinguish the junction capacitances, explicitly adopt the circuit convention $V_s=V_{sd}$ and $V_d=0$. The diamond with zero-bias endpoints $20$ and $40\,\mathrm{mV}$ has its positive tip near $(V_g,V_{sd})=(35,5)\,\mathrm{mV}$. Its two edge slopes are therefore $+1/3$ and $-1$.
Writing the island electrochemical potential as a charge-dependent constant minus $e(C_sV_{sd}+C_gV_g)/C_\Sigma$, alignment with the drain gives slope $-C_g/C_s$, while alignment with the source gives $C_g/(C_\Sigma-C_s)$. Thus
$$
C_s=C_g=8\,\mathrm{aF},\qquad
C_d=C_\Sigma-C_s-C_g=16\,\mathrm{aF}.
$$
The source/drain names are exchanged if the opposite lead is biased under the voltage convention. The figure alone establishes the pair $\{8,16\}\,\mathrm{aF}$ unless that wiring convention is specified. Symmetric-looking diamond heights do not imply equal junction capacitances.
Back to article page
