## 2 Field-Effect-Transistor Switch  
A metal-oxide-semiconductor-field-effect-transistor (MOSFET) is the archetypal (electronic) binary switch that encodes the bits 0 and 1 in its two conductance states, high (ON) and low (OFF). In the ON state, charges flood into the channel providing a conduction path between the source and the drain to turn the transistor on. In the OFF state, these charges are driven out of the channel in order to disrupt the conduction path and turn the transistor off. Therefore, the two states are ultimately encoded in two _different amounts of charge_ (\(Q_{1}\) and \(Q_{2}\)) in the channel. The switching action changes the amount of charge from \(Q_{1}\) to \(Q_{2}\), or vice versa, resulting in the (time-averaged) flow of a current:  
\[I=|Q_{1}-Q_{2}|/\Delta t=\Delta Q/\Delta t \tag{1}\]  
where \(\Delta t\) is the amount of time it takes for the channel charge to change from \(Q_{1}\) to \(Q_{2}\) (or vice versa). This current will cause energy dissipation of the amount:  
\[E_{d}=I^{2}R\Delta t=(\Delta Q/\Delta t)IR\Delta t=\Delta QIR=\Delta Q\Delta V \tag{2}\]  
where \(R\) is the resistance in the path of the current and \(\Delta V=IR\). We can think of \(\Delta V\) as the amount of voltage needed to be imposed at the transistor's gate in order to change the charge in the channel by the amount \(\Delta Q\). Note that the energy dissipation given in Equation (2) is _not_ independent of the switching time, because \(\Delta V\) depends on the switching time for a fixed \(\Delta Q\) and \(R\) (\(\Delta V=\Delta QR/\Delta t\)). We can rewrite the energy dissipation in Equation (2) as \(E_{d}=(\Delta Q)^{2}R/\Delta t\), which clearly shows that for a fixed \(\Delta Q\) and \(R\), we will dissipate _more_ energy if we switch _faster_ (smaller \(\Delta t\)). Therefore, a more meaningful quantity to benchmark energy-efficiency is the _energy-delay product_ which is \(E_{d}\Delta t=(\Delta Q)^{2}R\). For a fixed \(R\), we can reduce this quantity by reducing \(\Delta Q\), but that increasingly blurs the distinction between \(Q_{1}\) and \(Q_{2}\) and thereby impairs our ability to distinguish between bits 0 and 1. If \(\Delta Q\) is too small, then thermal generation and recombination can randomly change the amount of charge in the channel by an amount comparable to \(\Delta Q\), causing random switching. Therefore, a larger \(\Delta Q\) translates to both stronger error-resilience and better reliability. This makes it obvious that there is a direct relation between reliability and energy-delay product; if we reduce the energy dissipation or energy-delay product by reducing \(\Delta Q\), then we will invariably make the switch less reliable. We can make this argument a little more precise by noting that \(\Delta Q=C_{g}\Delta V\), where \(C_{g}\) is the gate capacitance. The thermal voltage fluctuation at the gate terminal is given by \(\sqrt{kT/C_{g}}\), where \(kT\) is the thermal energy [3]. Hence, the thermal charge fluctuation in the channel is:  
\[\Delta Q|_{\text{fluctuation}}=\sqrt{C_{g}kT} \tag{3}\]  
This quantity must be much smaller than the \(\Delta Q\) one needs to switch the conductance state of the transistor, and hence \(\Delta Q|_{\text{fluctuation}}<<\Delta Q\). Let us define a quantity \(\eta\) such that \(\eta=\Delta Q/(\left.\Delta Q\right|_{\text{fluctuation}})\). Clearly, \(\eta\) is a measure of the'switching reliability'; the larger its value, the more reliable is the switch. From Equation (2), we can now obtain:  
\[ \begin{array}{c}E_{d}=\frac{(\Delta Q)^{2}}{C_{g}}=\frac{\eta^{2}(\Delta Q|_ {\text{fluctuation}})^{2}}{C_{g}}=\eta^{2}kT\\ E_{d}\Delta t=(\Delta Q)^{2}R=\eta^{2}kTRC_{g}\end{array}  \tag{4}\]  
which immediately shows that we have to tolerate more energy dissipation, \(E_{d}\), and larger energy-delay product, \(E_{d}\Delta t\), if we desire more reliability (i.e., a larger \(\eta\)) [4].