## 2 Feedback Field-Effect Transistors (FBFETs)
### Positive Feedback Mechanism
\begin{array}{c}\Delta I_{e}=I_{0}\left(e^{[\Delta Y/(kT)]}-1\right)\\ I_{e}=I_{ei}\left(e^{[\phi\phi/(kT)]}-1\right)\\ \Delta I_{e}=I_{ei}\left(e^{[\phi\phi/(kT)]}-1\right)\left(e^{[\Delta Y/(kT)]} -1\right)\end{array}  \tag{3}\]  
Using the Taylor's expansion on the \(e[\Delta Y/(kT)]\) for the first two terms yields the following:  
\[ \begin{array}{c}e^{[\Delta Y/(kT)]}=1+[\Delta Y/(kT)]\\ \Delta I_{e}=I_{ei}\left(e^{[\phi\phi/(kT)]}-1\right)\left[\Delta Y/(kT)\right] \end{array}  \tag{4}\]  
Assuming that no recombination occurs in the channel region, the electrons flowing from the N+ region will accumulate in the potential wall next to the P+ region. Some carriers will be stored in the wall, while others will cross the potential wall barrier and have sufficient kinetic energy to be injected into the drain and be recombined. Letting the charge stored in the potential wall be \(\Delta Q-\), the electron carrier lifetime be \(\tau_{-}\), the barrier potential seen by the electrons in the potential wall be \(\phi\)Bh, and A be a parameter, the following can be obtained:  
\[\Delta I_{e}=\Delta Q-\tau_{-}^{-1}+\Delta I_{e}\;\mathrm{A}e^{[-\phi\mathrm{ Bh}/(kT)]} \tag{5}\]  
The charge, \(\Delta Q-\), from the electrons stored in the potential wall can be given by:  
\[\Delta Q-=\Delta I_{e}\;\tau_{-}\left(1-\mathrm{A}e^{[-\phi\mathrm{Bh}/(kT)]}\right) \tag{6}\]  
Step 2. \(V_{\mathrm{Bh}}\) will be decreased and \(\phi_{\mathrm{h}}\) will be increased by the \(\Delta Q-\) affected by the stored electrons. More holes will be injected into the channel from the collapsing potential barrier. The changes in the hole barrier (\(\Delta Z\)) and hole current (\(\Delta\mathrm{I_{h}}\)) are given as follows:  
\[\Delta Z=\Delta Q-C_{p}{}^{-1} \tag{7}\]  
\[\Delta I_{h}=I_{hi}\left(e^{[\phi\mathrm{h}/(kT)]}-1\right)\left(e^{[\Delta Z /(kT)]}-1\right) \tag{8}\]  
Figure 3: Positive feedback mechanism: the operational principles are divided into two steps [65]. The capacitance under the P+ spacer is C\({}_{\text{p}}\). Again, using the Taylor's expansion on e[\(\Delta\)Z/(kT)] and only including the first two terms yields the following:  
\[\Delta I_{\text{h}}=I_{\text{hi}}\ (\text{e}^{[\phi\text{h/(kT)]}}-1)\ [\Delta\text{Z/(kT)}] \tag{9}\]  
Through the same processes and conditions as perturbation of energy band next to the drain, the hole current flows to the source side and is trapped in a potential wall near the N+ source, affecting the valance band (Ev) potential. Thus, V\({}_{\text{Be}}\) decreases by \(\delta\)Y.  
\[\Delta Q+=\Delta I_{\text{h}}\ \tau_{+}\ (1-B\ \text{e}^{[-\phi\text{Be/(kT)]}}) \tag{10}\]  
\[\delta\text{Y}=\Delta Q+\text{C}_{\text{n}}{}^{-1} \tag{11}\]  
Let the charge stored in the potential wall be \(\Delta Q+\) near the N+ region, C\({}_{\text{n}}\) be the capacitance under the potential wall conditioning with trapped electrons region, the hole carrier lifetime be \(\tau_{+}\), the barrier potential seen by the hole in the potential wall be \(\phi_{\text{Be}}\), and B be a parameter. The equation for gain can be described with these parameters:  
\[\Delta\text{Y}=(kT\ \Delta I_{e})(\text{e}^{[\phi\text{e/(kT)]}}-1)^{-1} \tag{12}\]  
\[\text{Gain}=\delta\text{Y}/\Delta\text{Y}=\Delta Q+(I_{\text{ei}}\ (\text{e}^{[\phi\text{e/(kT)]}}-1)\ (kT\ \Delta I_{e}\ \text{C}_{\text{n}})^{-1}\]  
\[\Delta Q-=\Delta I_{e}\;\tau_{-}\left(1-\mathrm{A}e^{[-\phi\mathrm{Bh}/(kT)]}\right) \tag{6}\]0  
where,  
\[\Delta Q-=\Delta I_{e}\;\tau_{-}\left(1-\mathrm{A}e^{[-\phi\mathrm{Bh}/(kT)]}\right) \tag{6}\]1  
When the gain is \(\geq\)1, FBFET will be operated with steep-switching by the positive feedback. As the positive feedback mechanism uses both electrons and holes, the type of transistor is determined by which of the two potential barriers blocks the flow of holes/electrons and applies a negative/positive voltage to initiate a feedback operation (see Figure 4) [63; 64; 65]. \[\Delta Q-=\Delta I_{e}\;\tau_{-}\left(1-\mathrm{A}e^{[-\phi\mathrm{Bh}/(kT)]}\right) \tag{6}\]2 Illustration for positive feedback mechanism.