Switching energy

The contribution to the power dissipation due to the charge and discharge of internal nodes for each MOSFET can be defined as the integral of the voltage across the MOSFET times the current flowing through.

Theorem 4.2   The switching energy in generic n-networks and p-networks can be written as:

$$E_{sw_n} = \frac{1}{2} \sum_{i=1}^N C^i \left(V_i^{\prime\;2} - V_i^{\prime\prime\;2}\right)$$ (4.12)

$$E_{sw_p} = \frac{1}{2}\sum_{j=1}^P C_j \left[ \left(V_{DD} - V_j^{\prime}\right)^2 - \left(V_{DD}- V_j^{\prime\prime} \right)^2 \right]$$ (4.13)

where $C^i$ is the generic total capacitance of node $i$-th and $V_i^{\prime}$, $V_i^{\prime\prime}$ are, respectively, the initial and final value of the voltage swing at the same node.

Corollary 4.2.1   If the voltage swing of each node of the network is the full swing $\Delta V = V_{DD} - 0$, then equations (3.11), (3.12) can be written as:

$$E_{sw_n} = \frac{1}{2} \sum_{i=1}^N C^i \Delta V^2$$ (4.14)

$$E_{sw_p} = \frac{1}{2} \sum_{i=1}^P C^i \Delta V^2$$ (4.15)

Proof. [Proof of theorem 3.2]

Since the internal voltages and currents are known from the delay analysis, the energy for the n-MOS network can be written by summing all the contributions of internal nodes (see figure 3.3)

$$E_{sw_n} = \sum_{i=1}^N \int\left[ V_{D_n}^{i+1}(t)-V_{D_n}^i(t) \right] I_{D_n}^i(t)dt$$

where the notation of figure 3.3 is adopted.

This equation can be written in this way:

$$E_{sw_n} = \int\biggl\{ V_{D_n}^N(t) I_{D_n}^N(t) + \sum_{i=1}^{N-1} V_{D_n}^i(t)\left[ I_{D_n}^i(t)-I_{D_n}^{i+1}(t)\right] \biggr\} dt$$ (4.16)

It is possible to rewrite the previous equations by noting that in general:

$$I_{D_n}^{i+1}-I_{D_n}^i = C^i\frac{d V_{D_n}^i}{d t}$$

and, in particular, if we neglect the current of the p-MOS chain above the node N,

$$-I_{D_n}^N= C^N\frac{d V_{D_n}^N}{d t}\; .$$

Thus, for the n network it is possible to define the $E_{sw_n}$ energy in the following way:

$$\begin{split}E_{sw_n}& = - \sum_{i=1}^N C^i \int_{t_{0}^{\pri... ...left(V_i^{\prime\;2} - V_i^{\prime\prime\;2}\right) \end{split}$$

If we integrate the equation (3.11) (page ) only when the argument of the integrals are non zero, then the first integral in this equation goes from $t_{0}^{\prime}=t_{0_n}^i$ to $t_{0}^{\prime\prime}=\tau_{o_n}^i$, so that the second integral goes from $V_{i}^\prime=V_{D_n}^i(t_{0_n}^i)$ to $V_i^{\prime\prime}=V_{D_n}^i(\tau_{o_n}^i)$. Since $V_{D_n}^i(\tau_{o_n}^i)=0$, we have $E_{sw_n} = \frac{1}{2} \sum_{i=1}^N C^i V_i^{\prime\;2}$, where $V_i^\prime$ is the actual voltage swing at the node $i$.

The energy dissipated in the p network ($E_{sw_p}$) can be calculated with similar considerations leading to

$$\begin{split}E_{sw_p} &= \sum_{j=1}^P C^j \int_{t_{0}^{\prime... ... \left(V_{DD}- V_j^{\prime\prime} \right)^2 \right] \end{split}$$

Again, $V_{j}^\prime=V_{D_p}^j(t_{0_n}^i)$ and $V_j^{\prime\prime}=V_{D_p}^j(\tau_{o_p}^j)$, and in the same way $V_j^{\prime\prime} = V_{DD}$, so that $E_{sw_p } = \frac{1}{2} \sum_{j=1}^P C^j (V_{DD}-V_j^{\prime\;2})$, where $(V_{DD}-V_j^{\prime\;2})$ is the voltage swing at the node $j$.
$\qedsymbol$

In the equations (3.11) and (3.12) (page ) the voltage variation of capacitance must be included, obtaining expression for $E_{sw_{n,p}}$ slightly more complicated, but still in closed form.