A 2D analytical cylindrical gate tunnel FET (CG-TFET) model: impact of shortest tunneling distance

The TFET is a p-i-n diode operating in a reverse bias condition. The structural schematic diagram of the p-channel CG-TFET is shown in figure 1. The source and drain regions of the CG-TFET model are heavily doped by pentavalent impurity and trivalent impurity, respectively. However, the channel region of the device is made up of intrinsic material. For the present model, the channel is assumed to be very lightly doped of the order of 10¹⁵. The potential distribution along the channel can be estimated by solving two-dimensional Poisson's equation under parabolic approximation [10].

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The 2D Poisson's equation in the cylindrical coordinate is expressed as

$$\begin{eqnarray}&&\frac{1}{r}\frac{\partial }{\partial r}\left( r\frac{\partial }{\partial r}\varphi (r,z) \right)+\frac{{{\partial }^{2}}\varphi (r,z)}{\partial {{z}^{2}}}=\frac{q{{N}_{c}}}{{{\epsilon }_{si}}},\end{eqnarray} \tag{ 1 }$$

where ${{N}_{c}}$ is the doping concentration of the thin silicon cylindrical body, $\varphi (r,z)$ is the electrostatic potential distribution in the intrinsic channel, ${{\epsilon }_{si\,}}$ is the dielectric permittivity of silicon and $q$ is the elementary electron charge. As the device body is very lightly doped (order of ∼10¹⁵), the effect of ${{N}_{c}}$ upon the solution of equation (1) is negligible. So equation (1) takes the form of a 2D Laplace equation as

$$\begin{eqnarray}&&\frac{1}{r}\frac{\partial }{\partial r}\left( r\frac{\partial \varphi (r,z)}{\partial r} \right)+\frac{{{\partial }^{2}}\varphi (r,z)}{\partial {{z}^{2}}}=0.\end{eqnarray} \tag{ 2 }$$

The potential distribution along the channel is approximately defined as a parabolic equation with the boundary conditions of GAA-MOSFET [12]

$$\begin{eqnarray}&&\varphi \left( r,z \right)={{a}_{0}}\left( z \right)+r{{a}_{1}}\left( z \right)+{{r}^{2}}{{a}_{2}}\left( z \right).\end{eqnarray} \tag{ 3 }$$

The boundary conditions are essential to predict the coefficients a₀, a₁, a₂ of the 2nd order polynomial and are finally able to find the solution of the 2D Laplace equation given by equation (2). The required boundary conditions in the intrinsic channel region are approximated as [10].

• (1)  
  The potential at the center of the cylinder structure varies w.r.t. 'z' only

  $$\begin{eqnarray}&&{{\left. \varphi \left( r,z \right) \right]}_{r=0}}={{\varphi }_{c}}\left( z \right)={{a}_{0}}\left( z \right).\end{eqnarray} \tag{ 4 }$$
• (2)  
  The electric field at the center of the thin silicon body is minimal and has a negligible effect on the potential distribution in the channel

  $$\begin{eqnarray}&&{{\left. E\left( r,z \right) \right]}_{r=0}}=0={{a}_{1}}\left( z \right).\end{eqnarray} \tag{ 5 }$$
• (3)  
  The electric field at the gate oxide–channel interface is maximal and is represented as

$$\begin{eqnarray}\begin{array}{lllllllllllllll} {{\left. E\left( r,z \right) \right]}_{r=\frac{{{t}_{si}}}{2}}} & = & -{{\left. \frac{{\rm d}\varphi (r,z)}{{\rm d}r} \right]}_{r=\frac{{{t}_{si}}}{2}}} \\ {} & = & \frac{{{C}_{ox}}({{\varphi }_{GS}}-{{\varphi }_{s}}(z))}{{{\epsilon }_{si}}} \\ {} & = & {{t}_{si}}{{a}_{2}}\left( z \right), \\ \end{array}\end{eqnarray} \tag{ 6 }$$

where

$$\begin{eqnarray*}\begin{array}{lllllllllllllll} {{\left. \varphi \left( r,z \right) \right]}_{r=\frac{{{t}_{si}}}{2}}} & = & {{\varphi }_{s}}\left( z \right), \\ {{\varphi }_{GS}} & = & {{V}_{GS}}-{{V}_{FB}}, \\ {{C}_{ox}} & = & \frac{2{{\epsilon }_{ox}}}{{{t}_{si}}{\rm ln} \left( 1+\frac{2{{t}_{ox}}}{{{t}_{si}}} \right)\;}. \\ \end{array}\end{eqnarray*}$$

Here ${{t}_{si}}$ is the diameter of the cylindrical channel, ${{C}_{ox}}$ is the oxide capacitance, ${{t}_{ox}}$ is the thickness of the SiO₂ layer, ${{\epsilon }_{ox}}$ is the permittivity of the oxide layer, ${{V}_{GS}}$ is the gate-to-source voltage and ${{V}_{FB}}$ is the flat-band voltage.

Using the boundary conditions in equation (3), we get

$$\begin{eqnarray}&&\varphi \left( r,z \right)={{\varphi }_{c}}(z)+{{r}^{2}}\frac{{{C}_{ox}}({{\varphi }_{GS}}-{{\varphi }_{s}}(z))}{{{\epsilon }_{si}}{{t}_{si}}}.\end{eqnarray} \tag{ 7 }$$

Equation (2) can be solved for the center of the cylindrical body (r = 0). Substituting equation (7) in (2) and setting r = 0, we obtain the following one-dimensional (1D) differential equation

$$\begin{eqnarray}&&\frac{{{\partial }^{2}}{{\varphi }_{c}}(z)}{\partial {{z}^{2}}}+\frac{4({{\varphi }_{GS}}-{{\varphi }_{s}}(z)){{C}_{ox}}}{{{\epsilon }_{si}}{{t}_{si}}}=0.\end{eqnarray} \tag{ 8 }$$

The surface potential ${{\varphi }_{s}}(z)$ is related to the center potential ${{\varphi }_{c}}(z)$ as [13]

$$\begin{eqnarray}&&{{\varphi }_{c}}\left( z \right)=\left( H+1 \right){{\varphi }_{s}}\left( z \right)-H{{\varphi }_{GS}}\;,\end{eqnarray} \tag{ 9 }$$

where

$$\begin{eqnarray*}&&H=\frac{{{\epsilon }_{ox\,{{t}_{si}}}}}{4{{\epsilon }_{si}}{{t}_{o{{x}^{'}}}}}\;,\;{{t}_{ox^{\prime} }}=\frac{{{t}_{si}}}{2}\;{\rm ln} \left( 1+\frac{2{{t}_{ox}}}{{{t}_{si}}} \right).\end{eqnarray*}$$

Differentiating the previous equation

$$\begin{eqnarray}&&\frac{{{\partial }^{2}}{{\varphi }_{c}}(z)}{\partial {{z}^{2}}}=\frac{{{\partial }^{2}}{{\varphi }_{s}}(z)}{\partial {{z}^{2}}}\;,\end{eqnarray} \tag{ 10 }$$

now equation (8) becomes

$$\begin{eqnarray}&&\frac{{{\partial }^{2}}{{\varphi }_{s}}(z)}{\partial {{z}^{2}}}+\frac{4({{\varphi }_{GS}}-{{\varphi }_{s}}(z)){{C}_{ox}}}{{{\epsilon }_{si}}{{t}_{si}}}=0,\end{eqnarray} \tag{ 11 }$$

$$\begin{eqnarray}&&\frac{{{\partial }^{2}}{{\varphi }_{s}}}{\partial {{z}^{2}}}+\frac{{{\varphi }_{GS}}-{{\varphi }_{s}}}{{{\lambda }_{c}}^{2}}=0,\end{eqnarray} \tag{ 12 }$$

where ${{\lambda }_{c}}$ is known as the characteristic length of the cylindrical gate and is mathematically defined as

$$\begin{eqnarray}&&{{\lambda }_{c}}=\sqrt{\frac{{{\epsilon }_{si}}{{t}_{si}}}{4{{C}_{ox}}}}=\sqrt{\frac{{{t}_{si}}^{2}{{\epsilon }_{si}}\;{\rm ln} \left( 1+\frac{2{{t}_{ox}}}{{{t}_{si}}} \right)}{8{{\epsilon }_{ox}}}}\;.\end{eqnarray} \tag{ 13 }$$

The differential equation (12) can be solved by using an auxiliary equation to obtain the surface potential of the cylindrical channel as

$$\begin{eqnarray}&&{{\varphi }_{s}}\left( z \right)={{c}_{1}}{{e}^{kz}}+{{c}_{2}}{{e}^{-kz}}-{{\varphi }_{d}}\;,\end{eqnarray} \tag{ 14 }$$

where ${{\varphi }_{d}}=-{{\varphi }_{GS}}$ and $k=1/{{\lambda }_{c}}.$

For the previous expression the coefficients c₁ and c₂ can be calculated by using the boundary conditions at the source and drain ends. The surface potential at the source end (z = 0) is defined as the built-in potential, which is the amount of band bending of valence band and conduction band at the source-channel interface for ${{V}_{GS}}=0.$ Similarly, the potential at the channel-drain interface (z = z_max = L) is the sum of the built-in potential (V_bi) and drain voltage (V_DS), as the channel is inverted at the drain end.

The necessary sufficient boundary conditions are [18]

$$\begin{eqnarray*}&&{{\left. \;\;{{\varphi }_{s}}\left( z \right) \right]}_{z=0}}={{V}_{bi}},\end{eqnarray*}$$

$$\begin{eqnarray}&&{{\left. {{\varphi }_{s}}\left( z \right) \right]}_{z={{z}_{max}}=L}}={{V}_{bi}}+{{V}_{DS}}.\end{eqnarray} \tag{ 15 }$$

Equating equation (14) using the previous boundary conditions

$$\begin{eqnarray}&&\begin{array}{lllllllllllllll} {{c}_{1}}=\frac{1}{2\;{\rm sinh} (kL)}\left[ {{V}_{bi}}\left( 1-{{e}^{-kL}} \right) \right. \\ \left. \quad \;\;\;+{{\varphi }_{d}}\left( 1-{{e}^{-kL}} \right)+{{V}_{DS}} \right], \\ \end{array}\end{eqnarray} \tag{ 16 }$$

and

$$\begin{eqnarray}&&\begin{array}{lllllllllllllll} {{c}_{2}}=-\frac{1}{2\;{\rm sinh} (kL)}\left[ {{V}_{bi}}\left( 1-{{e}^{kL}} \right) \right. \\ \quad \;\;\;\;\left. +{{\varphi }_{d}}\left( 1-{{e}^{kL}} \right)+{{V}_{DS}} \right]. \\ \end{array}\end{eqnarray} \tag{ 17 }$$

The electric field of the channel can be derived by differentiating electrostatic potential distribution w.r.t. r and z ($0\leqslant z\leqslant L,\;0\leqslant r\leqslant {{t}_{si}}/2$).

The radial and lateral electric fields are expressed as

$$\begin{eqnarray}\begin{array}{lllllllllllllll} {{E}_{r}}\left( r,z \right) & = & -\frac{\partial \varphi (r,z)}{\partial r} \\ {} & = & -2r{{a}_{2}}\left( z \right) \\ {} & = & -2r\frac{({{\varphi }_{GS}}-{{\varphi }_{s}}){{C}_{ox}}}{{{t}_{si}}{{\epsilon }_{si}}}\;, \\ \end{array}\end{eqnarray} \tag{ 18 }$$

$$\begin{eqnarray}\begin{array}{lllllllllllllll} {{E}_{z}}\left( r,z \right) & = & -\frac{\partial \varphi \left( r,z \right)}{\partial z} \\ {} & = & -\frac{\partial {{\varphi }_{s}}}{\partial z} \\ {} & = & -{{c}_{1}}k{{e}^{kz}}+{{c}_{2}}k{{e}^{-kz}}. \\ \end{array}\end{eqnarray} \tag{ 19 }$$

However, for a TFET model, the tunneling process is dominant in the lateral z-direction compared to the radial direction [27]. So the lateral electric field along the length of the channel is considered in the next section to calculate the tunneling current.

The tunneling of the charge carrier at the source end starts only when the conduction band of the source gets aligned with the valence band of the intrinsic channel region. The shortest tunneling distance (L_tunn) is the lateral distance between the source-channel interface (z = 0) and the point in the channel where the surface potential changes by an amount of unit bandgap energy per elementary charge (E_g/q). However, the shortest tunneling distance plays a crucial role in finding the amplitude of the tunneling current in the device.

The gate voltage has a significant effect on the tunneling phenomena as shown in figure 2. At low gate voltage, there is a small amount of band bending of the valence band and conduction band. So the current is negligible due to the absence of BTBT of the charge carrier. For a critical value of the gate voltage known as the threshold voltage (V_th), the conduction band of the source and the valence band of the channel are in line to each other. For V_GS > V_th, the tunneling probability increases, which improves the BTBT current in the channel. The BTBT drain current increases exponentially due to the improvement in tunneling volume as depicted by the area under the solid lines in figure 2(b). L_tunn is calculated for the values of z at which the potential energy changes by E_G/q. The critical threshold condition is expressed as

$$\begin{eqnarray}&&{{\left. {{\varphi }_{scond}}\left( z \right) \right]}_{z=0}}={{\left. {{\varphi }_{sval}}\left( z \right) \right]}_{z={{L}_{tunn}}}}.\end{eqnarray} \tag{ 20 }$$

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The corresponding potential values for z = 0 and z = L_tunn are obtained from the defined boundary conditions

$$\begin{eqnarray}&&{{\left. {{\varphi }_{scond}}\left( z \right) \right]}_{z=0}}={{\varphi }_{s}}\left( 0 \right)={{V}_{bi}}.\end{eqnarray} \tag{ 21 }$$

$$\begin{eqnarray}\begin{array}{lllllllllllllll} {{\left. {{\varphi }_{sval}}\left( z \right) \right]}_{z={{L}_{tunn}}}} & = & {{\varphi }_{s}}\left( {{L}_{tunn}} \right)+\frac{{{E}_{g}}}{q} \\ {} & = & {{c}_{1}}{{e}^{k{{L}_{tunn}}}}+{{c}_{2}}{{e}^{-k{{L}_{tunn}}}} \\ {} & {} & -{{\varphi }_{d}}+\frac{{{E}_{g}}}{q}\;, \\ \end{array}\end{eqnarray} \tag{ 22 }$$

where ${{\varphi }_{scond}}\left( z \right)$ and ${{\varphi }_{sval}}\left( z \right)$ are the conduction potential of the source and valence potential of the channel.

The value of the shortest tunneling distance can be obtained from the critical threshold condition as

$$\begin{eqnarray}&&{{L}_{tunn}}=\frac{1}{k}ln\left( \frac{A-\sqrt{{{A}^{2}}-4{{c}_{1}}{{c}_{2}}}}{2{{c}_{1}}} \right)\;,\end{eqnarray} \tag{ 23 }$$

where A is the device-dependent constant and expressed as

$$\begin{eqnarray*}&&A={{\varphi }_{d}}-\frac{{{E}_{g}}}{q}+{{V}_{bi}}.\end{eqnarray*}$$

However, the shortest tunneling distance can be reduced by using low bandgap material and increasing the gate potential (V_GS), which increases the tunneling volume in the channel region. The higher the tunneling volume, the greater the ON current of the device. For a constant gate voltage, the tunneling volume is not affected by drain voltage.

The drain current of the proposed model is calculated analytically by Kane's model, which uses a nonlocal BTBT approach. However, Kane's model [28] is used to find the rate of generation of carrier (${{G}_{tunn}})$ tunneling from source to drain along the channel as:

$$\begin{eqnarray}&&{{G}_{tunn}}={{A}_{kane}}{{\left( {{E}_{avg}} \right)}^{\beta }}{{e}^{-\left( \frac{{{B}_{kane}}}{{{E}_{avg}}} \right)}},\end{eqnarray} \tag{ 24 }$$

where $\;{{A}_{kane}}$ and ${{B}_{kane}}$ are tunneling-dependent parameters and the default values of ${{A}_{kane}}=4\times {{10}^{15}}{{{\rm m}}^{-1/2}}{{{\rm V}}^{-5/2}}{{{\rm s}}^{-1}}$ and ${{B}_{kane}}=1.9\times {{10}^{9}}{\rm V}{{{\rm m}}^{-1}}$ [29]. $\beta$ is the material-dependent parameter and its value depends on the type of tunneling, i.e., 2.5 for the indirect tunneling and 2 for direct tunneling [29]. However, in the present analysis the value of $\beta$ is considered as 2. ${{E}_{avg}}$ is the average electric field over the tunneling volume and is expressed as

$$\begin{eqnarray}&&{{E}_{avg}}=\frac{{{E}_{g}}}{q{{L}_{path}}\;}\;,\end{eqnarray} \tag{ 25 }$$

here ${{L}_{path}}$ is the length of the BTBT tunneling path, which varies over the tunneling volume in the z-direction. From figure 2, the ${{L}_{path}}$ is described as the path between two lateral points z₁ = L_tunn and z₂.

Now the drain current can be computed with the help of the BTBT generation rate

$$\begin{eqnarray}&&\begin{array}{lllllllllllllll} {{I}_{D}}=\mathop \int \limits^{} {{G}_{tunn}}{\rm d}v \\ \quad \;=q\int \limits_{0}^{{{t}_{si}}} \left( \int \limits_{{{z}_{1}}}^{{{z}_{2}}} {{A}_{kane}}{{E}_{z}}{{\left( {{E}_{avg}} \right)}^{\beta -1}}{{e}^{-\left( \frac{{{B}_{kane}}}{{{E}_{avg}}} \right)}}{\rm d}z \right)\;{\rm d}r. \\ \end{array}\end{eqnarray} \tag{ 26 }$$

Substituting the value of the lateral electric field and the average tunneling field from equations (19) and (25)

$$\begin{eqnarray*}&&\begin{array}{ccccccccccccccc} {{I}_{D}}=\frac{q{{t}_{si}}{{E}_{g}}^{\beta -1}{{A}_{kane}}}{{{q}^{\beta -1}}}\left[ \int \limits_{{{z}_{1}}}^{{{z}_{2}}} \frac{-{{c}_{1}}k{{e}^{\left( k-\frac{q{{B}_{kane}}}{{{E}_{g}}} \right)z}}}{{{z}^{\beta -1}}}{\rm d}z \right. \\ \end{array}\end{eqnarray*}$$

$$\begin{eqnarray}&&\begin{array}{lllllllllllllll} \left. \quad \quad +\int \limits_{{{z}_{1}}}^{{{z}_{2}}} \frac{{{c}_{2}}k{{e}^{-\left( k+\frac{q{{B}_{kane}}}{{{E}_{g}}} \right)z}}}{{{z}^{\beta -1}}}{\rm d}z \right]. \\ \end{array}\end{eqnarray} \tag{ 27 }$$

Here, the tunneling of the charge carrier takes place between two lateral points z₁ = L_tunn and z₂. Between the two boundaries along the z-direction, the effect of the polynomial term (${{z}^{\beta -1}})$ is negligible as compared to the exponential term ${{e}^{-\left( k+\frac{q{{B}_{kane}}}{{{E}_{g}}} \right)z}}.$ So the drain current is obtained by integrating equation (27) over the tunneling interval and neglecting the polynomial term; thus, we get

$$\begin{eqnarray}&&\begin{array}{lllllllllllllll} {{I}_{D}}=\frac{q{{t}_{si}}{{E}_{g}}{{A}_{kane}}}{q}\left[ \left( \frac{-{{c}_{1}}k}{k-\frac{q{{B}_{kane}}}{{{E}_{g}}}} \right)\left( {{P}_{{{z}_{2}}}}-{{P}_{{{z}_{1}}}} \right) \right. \\ \left. \quad \quad -\left( \frac{{{c}_{2}}k}{k+\frac{q{{B}_{kane}}}{{{E}_{g}}}} \right)\left( {{Q}_{{{z}_{2}}}}-{{Q}_{{{z}_{1}}}} \right) \right], \\ \end{array}\end{eqnarray} \tag{ 28 }$$

where ${{P}_{z}}$ and ${{Q}_{z}}$ are expressed as

$$\begin{eqnarray}&&{{P}_{z}}=\frac{{{e}^{\left( k-\frac{q{{B}_{kane}}}{{{E}_{g}}} \right)z}}}{z},\;{{Q}_{z}}=\frac{{{e}^{-\left( k+\frac{q{{B}_{kane}}}{{{E}_{g}}} \right)z}}}{z}\;.\end{eqnarray} \tag{ 29 }$$

As ${{z}_{2}}\gt {{z}_{1,}}$ it is noted that the exponential coefficient ${{P}_{{{z}_{2}}}}\ll {{P}_{{{z}_{1}}}}$ and ${{Q}_{{{z}_{2}}}}\ll {{Q}_{{{z}_{1}}}}.$ In the present model, the current analysis is carried out by neglecting the minimal effects of ${{P}_{{{z}_{2}}}}$ and ${{Q}_{{{z}_{2}}}}$ and instead using the shortest tunneling distance (${{z}_{1}}={{L}_{tunn}}$).

$$\begin{eqnarray}&&{{I}_{D}}={{I}_{0}}\left[ {{P}_{{{z}_{1}}}}\left( \frac{{{c}_{1}}k}{k-\frac{q{{B}_{kane}}}{{{E}_{g}}}} \right)+{{Q}_{{{z}_{1}}}}\left( \frac{{{c}_{2}}k}{k+\frac{q{{B}_{kane}}}{{{E}_{g}}}} \right) \right]\;,\end{eqnarray} \tag{ 30 }$$

where I₀ is the dc current that only depends on constant parameters and is defined as

$$\begin{eqnarray}&&{{I}_{0}}=\frac{q{{t}_{si}}{{E}_{g}}{{A}_{kane}}}{q}.\end{eqnarray} \tag{ 31 }$$

SS is a feature of a MOSFET's current–voltage characteristic in the sub-threshold region (V_GS < V_th) and is defined as the slope of logarithmic drain current w.r.t. gate voltage at a constant drain voltage. The inverse of this slope is usually referred to as SS in mV/decade

$$\begin{eqnarray}&&SS=\frac{\partial {{V}_{GS}}}{\partial (log{{I}_{D}})}={{\left( \frac{\partial (log{{I}_{D}})}{\partial {{V}_{GS}}} \right)}^{-1}}.\end{eqnarray} \tag{ 32 }$$

The optimum value of the SS of a conventional MOSFET is 60 mV/decade at room temperature (300 K). It is always desirable to have a small SS, as it improves the $\frac{{{I}_{ON}}}{{{I}_{OFF}}}$ ratio by reducing the leakage of the OFF state current. Also the low SS or steep subthreshold slope of a device indicates the faster switching behavior, i.e., faster transition between OFF state and ON state. SS can be computed by substituting the value of I_D in equation (32)

$$\begin{eqnarray}&&\begin{array}{lllllllllllllll} SS={{\left( \frac{\partial (log{{I}_{D}})}{\partial {{V}_{GS}}} \right)}^{-1}}=\left( \frac{\partial }{\partial {{V}_{GS}}}\left[ \left( \frac{{{c}_{1}}k{{I}_{0}}{{P}_{{{z}_{1}}}}}{k-\frac{q{{B}_{kane}}}{{{E}_{g}}}} \right) \right. \right. \\ {{\left. \left. \quad \quad +\left( \frac{{{c}_{2}}k{{I}_{0}}{{Q}_{{{z}_{1}}}}}{k+\frac{q{{B}_{kane}}}{{{E}_{g}}}} \right) \right] \right)}^{-1}}, \\ \end{array}\end{eqnarray} \tag{ 33 }$$

$$\begin{eqnarray}&&SS=\frac{2\;{\rm sinh} (kL)\left( {{Y}_{1}}{{c}_{1}}+{{Y}_{2}}{{c}_{2}} \right)}{-{{Y}_{1}}\left( 1-{{e}^{-kL}} \right)+{{Y}_{2}}\left( 1-{{e}^{kL}} \right)}\;,\end{eqnarray} \tag{ 34 }$$

where

$$\begin{eqnarray}&&{{Y}_{1}}=\frac{{{I}_{0}}k{{P}_{{{z}_{1}}}}}{k-\frac{q{{B}_{kane}}}{{{E}_{g}}}},\;{{Y}_{2}}=\frac{{{I}_{0}}k{{Q}_{{{z}_{1}}}}}{k+\frac{q{{B}_{kane}}\;\;}{{{E}_{g}}}}.\end{eqnarray} \tag{ 35 }$$

It is noted that the model offers an SS of 26 ∼ 40 mV/decade compared to the GAA-MOS limit of 60 mV/decade over a wide range of gate voltages, as shown in table 1. This leads to a reasonable amount of improvement in $\frac{{{I}_{ON}}}{{{I}_{OFF}}}.$

Transconductance (${{g}_{m}})$ is the transfer characteristics of the device and is calculated by differentiating equation (30) w.r.t. V_GS at constant V_DS.

$$\begin{eqnarray}&&\begin{array}{lllllllllllllll} {{g}_{m}}={{\left. \frac{\partial {{I}_{D}}}{\partial {{V}_{GS}}} \right]}_{{{V}_{DS}}=cont.}} \\ \quad \;\;=\frac{-{{Y}_{1}}\left( 1-{{e}^{-kL}} \right)+{{Y}_{2}}\left( 1-{{e}^{kL}} \right)}{2\;{\rm sinh} (kL)}\;. \\ \end{array}\end{eqnarray} \tag{ 36 }$$

As gate oxide thickness reduces, the effective gate control over the channel improves and conversely increases transconductance of CG-TFET as depicted in figures 12 and 13.