PHT.301 Physics of Semiconductor Devices

## MOS Capacitor - Depletion approximation

The charge density, electric field, and electrostatic potential can be calculated for a MOS capacitor in the depletion approximation. This approximation is valid in depletion but not in accumulation or inversion. It is useful for estimating the threshold voltage and for estimating the concentration of charge carriers in the channel in the subthreshold regime.

Consider a MOS capacitor with a p-substrate. We will ground the substrate and apply a voltage $V_g$ to the metal gate. Initially the gate voltage is set to the flat-band voltage $V_{fb}$ so that the hole concentration is constant in the semiconductor all the way to the oxide. The flat-band voltage is determined by the work functions of the materials as well as trapped charge in the oxide or at the semiconductor-oxide interface. To bring the MOS capacitor into depletion, the voltage is made more positive than the flat-band voltage to push the holes back exposing the negatively charged acceptors. The basic assumption of the depletion approximation is to assume that the charge density in the depletion region has the form,

$$\rho(x) = \begin{cases} 0 & \mbox{for } -t_{\text{ox}} \lt x\lt 0, \\ -eN_A & \mbox{for } 0\lt x\lt x_p, \\ 0 & \mbox{for } x \gt x_p.\end{cases}$$

The electric field is the integral of the charge density,

$$E(x) = \begin{cases} \frac{eN_Ax_p}{\epsilon_{\text{ox}}} & \mbox{for } -t_{\text{ox}} \lt x\lt 0, \\ \frac{eN_A}{\epsilon_s}(x_p-x) & \mbox{for } 0\lt x\lt x_p, \\ 0 & \mbox{for } x \gt x_p.\end{cases}$$

The electric field is constant in the oxide and there is an abrupt jump in the field at the interface between the semiconductor and the oxide. The condition at this interface is that $\epsilon_{\text{ox}}E(0-) = \epsilon_sE(0+)$ where $E(0-)$ is the electric field next to the interface on the oxide side and and $\epsilon_{\text{ox}}$ is the dielectric constant in the oxide while $E(0+)$ is the electric field next to the interface on the semiconductor side and $\epsilon_s$ is the dielectric constant in the semiconductor.

The electric field can be integrated to obtain the electrostatic potential,

$$\phi(x) = \begin{cases} -\frac{eN_Ax_px}{\epsilon_{\text{ox}}} + \frac{eN_A}{2\epsilon_s}x_p^2 & \mbox{for } -t_{\text{ox}} \lt x\lt 0, \\ \frac{eN_A}{2\epsilon_s}(x_p - x)^2 & \mbox{for } 0\lt x\lt x_p, \\ 0 & \mbox{for } x \gt x_p.\end{cases}$$

The voltage at the gate is the potential at $x = -t_{\text{ox}}$,

$$V_g - V_{fb} = \frac{eN_Ax_pt_{\text{ox}}}{\epsilon_{\text{ox}}} + \frac{eN_A}{2\epsilon_s}x_p^2.$$

This can be solved for the depletion width,

$$x_p = -\frac{\epsilon_s}{\epsilon_{\text{ox}}}t_{\text{ox}} + \sqrt{\left( \frac{\epsilon_s}{\epsilon_{\text{ox}}}t_{\text{ox}} \right)^2 + \frac{2\epsilon_s}{eN_A}(V_g - V_{fb})}$$

The depletion width goes to zero at $V_g = V_{fb}$.

The electron concentration and the hole concentration are,

$$n(x) = N_c\exp\left(\frac{E_F - E_c(x)}{k_BT}\right), \qquad p(x) = N_v\exp\left(\frac{E_v(x) - E_F}{k_BT}\right),$$

where

$$E_F - E_c(x) = \begin{cases} k_BT\ln\left(\frac{n_i^2}{N_cN_A}\right) + \frac{e^2N_A}{2\epsilon_s}(x_p - x)^2 & \mbox{for } 0 \lt x\lt x_p, \\ k_BT\ln\left(\frac{n_i^2}{N_cN_A}\right) & \mbox{for } x\gt x_p,\end{cases},$$

and

$$E_v(x) - E_F = \begin{cases} k_BT\ln\left(\frac{N_A}{N_v}\right) - \frac{e^2N_A}{2\epsilon_s}(x_p - x)^2 & \mbox{for } 0 \lt x\lt x_p, \\ k_BT\ln\left(\frac{N_A}{N_v}\right) & \mbox{for } x\gt x_p,\end{cases}$$

therefore,

$$n(x) = \begin{cases}\frac{n_i^2}{N_A}\exp\left(\frac{e^2N_A(x_p - x)^2}{2\epsilon_sk_BT}\right) & \mbox{for } 0 \lt x\lt x_p, \\ \frac{n_i^2}{N_A} & \mbox{for } x\gt x_p,\end{cases}$$

and

$$p(x) = \begin{cases}N_A\exp\left(-\frac{e^2N_A(x_p - x)^2}{2\epsilon_sk_BT}\right) & \mbox{for } 0 \lt x\lt x_p, \\ N_A & \mbox{for } x\gt x_p.\end{cases}$$

The form below will plot the bands, the charge density, and the electric field in the depletion approximation.

 $t_{ox}$ = nm $\epsilon_{ox}$ = $N_c(300)$ = 1/cm³ $T$ = K $E_g$ = eV $\epsilon_s$ = $N_v(300)$ = 1/cm³ $N_A$ = 1/cm³
 $V_g$ =  V $V_{fb}$ =

Band diagram

 Energy[eV] x [μm]
 $E_g=$  eV $n_i=$  1/cm³ $x_p=$  μm $V_T=$  V $V_{\text{ox}}= V_g - \phi(x=0) =$  V At the oxide/semiconductor interface $n(x=0)=$  1/cm³ $\phi(x=0)=$  V $E(0-)=$  V/m $E(0+)=$  V/m $n_{2d}=$  1/cm²

Charge density

 ρ×10-15[e/cm³] x [μm] The is an obvious inconsistency here. We assumed $\rho = -eN_A$ in the depletion region but clearly $p$ also contributes to the charge density. This is resolved by solving the Poisson equation.

Electric field

 $E$[MV/m] x [μm]

### Threshold voltage

At the interface between the semiconductor and the oxide, the electron density is,

$$n(0) = \frac{n_i^2}{N_A}\exp\left(\frac{e^2N_Ax_p^2}{2\epsilon_sk_BT}\right).$$

The condition for the threshold voltage is $n(0) = N_A$. At this point the channel begins to form and the depletion region does not increase in width anymore. In inversion, the depletion width has a constant value $x_{p\text{(max)}}$ and any extra positive charges that appear on the gate are compensated by mobile electrons in the channel.

$$N_A = \frac{n_i^2}{N_A}\exp\left(\frac{e^2N_Ax_{p\text{(max)}}^2}{2\epsilon_sk_BT}\right).$$

Solving for $x_{p\text{(max)}}$,

$$x_{p\text{(max)}} = 2\sqrt{\frac{\epsilon_s}{e^2N_A}k_BT\ln\left(\frac{N_A}{n_i}\right)}.$$

This can now be substituted into the expression for the gate voltage,

$$V_T - V_{fb} = \frac{eN_At_{\text{ox}}}{\epsilon_{\text{ox}}}x_{p\text{(max)}} + \frac{eN_A}{2\epsilon_s}x_{p\text{(max)}}^2.$$

The threshold voltage is $$V_T = \frac{2t_{\text{ox}}}{\epsilon_{\text{ox}}}\sqrt{\epsilon_sN_Ak_BT\ln\left(\frac{N_A}{n_i}\right)} + \frac{2k_BT}{e}\ln\left(\frac{N_A}{n_i}\right) + V_{fb}.$$

### Electron concentration at subthreshold

Between weak inversion where $n=p$ and strong inversion where $n= N_A$, the electron concentration is maximum at the interface to the oxide and falls to its equilibrium value like,

$$n(x) = \frac{n_i^2}{N_A}\exp\left(\frac{e^2N_A(x_p - x)^2}{2\epsilon_sk_BT}\right).$$

Most of the electrons are near the oxide and there the $x^2$ term can be neglected and the electron density is approximately,

$$n(x) \approx \frac{n_i^2}{N_A}\exp\left(\frac{e^2N_A(x_p^2 - 2x_px)}{2\epsilon_sk_BT}\right).$$

This expression can be integrated over $x$ to determine the 2-D electron density at the interface,

$$n_{2d} = \frac{n_i^2}{N_A}\exp\left(\frac{e^2N_Ax_p^2}{2\epsilon_sk_BT}\right)\frac{\epsilon_sk_BT}{e^2N_Ax_p}$$

The 2-D electron density as a function of the gate voltage can be calculated by using the expression for $x_p$ from above. The electron density falls approximately exponentially as a function of the gate voltage below the threshold voltage.

Two-dimensional electron density

 $\log (n_{2d})$cm-2 $V_g$ [V]

This plot shows the logarithm of the electron density in the gate voltage range between weak inversion and strong inversion.

### Subthreshold current

The dominant current mechanism in the subthreshold regime is diffusion where the diffusion current density is given by $\vec{j}=eD_n\nabla n$. If the transistor is biased so that the channel is pinched off at the drain, then the electron density is $n_{2d}$ near the source and nearly zero near the drain and the gradient of the electron density can be approximated as,

$$\nabla n \approx -\frac{n_{2d}}{L}$$

where $L$ is the gate length. The subthreshold current is,

$$I_D \approx \frac{k_BT\mu Wn_{2d}}{L} = \frac{n_i^2\mu \epsilon_sk_B^2T^2W}{e^2N_A^2x_pL}\exp\left(\frac{e^2N_Ax_p^2}{2\epsilon_sk_BT}\right).$$

Here $W$ is the width of the transistor and $\mu$ is the electron mobility.

$W$ =  m,  $L$ =  m,   $\mu$ =  cm²/Vs

 $\log (I_D)$ A $V_g$ [V]

The drain current decreases exponentially with the gate voltage. The subthreshold slope is the amount of gate voltage needed to reduce drain current by a factor of 10. For this calculation, the subthreshold slope is mV/decade.