To calculate the threshold voltage ($V_{th}$) for the given MOSFET, let us break the problem into parts:

Given data:

1. Boron doping concentration ($N_A$): $10^{17} \, \text{cm}^{-3}$
2. Oxide thickness ($t_{ox}$): $4 \, \text{nm} = 4 \times 10^{-7} \, \text{cm}$
3. Trapped charge density ($Q_{ox}$): $10^{12} \, \text{cm}^{-2}$
4. Dielectric constant of HfO$_2$ ($k_{ox}$): $25$
5. Dielectric constant of Si ($k_{Si}$): $11.7$
6. Permittivity of vacuum ($\varepsilon_0$): $8.85 \times 10^{-14} \, \text{F/cm}$
7. Intrinsic carrier concentration ($n_i$): $1.5 \times 10^{10} \, \text{cm}^{-3}$
8. Temperature ($T$): $300 \, \text{K}$

Step 1: Oxide capacitance per unit area ($C_{ox}$)

The oxide capacitance is calculated using: $C_{ox} = \frac{\varepsilon_{ox}}{t_{ox}}$ where $\varepsilon_{ox} = k_{ox} \varepsilon_0$.

$\varepsilon_{ox} = 25 \times 8.85 \times 10^{-14} = 2.2125 \times 10^{-12} \, \text{F/cm}$

$C_{ox} = \frac{2.2125 \times 10^{-12}}{4 \times 10^{-7}} = 5.53125 \times 10^{-6} \, \text{F/cm}^2$

Step 2: Fermi potential ($\phi_F$)

The Fermi potential is given by: $\phi_F = \frac{kT}{q} \ln{\left( \frac{N_A}{n_i} \right)}$ where $\frac{kT}{q} = 0.0259 \, \text{V}$ at $300 \, \text{K}$.

$\phi_F = 0.0259 \ln{\left( \frac{10^{17}}{1.5 \times 10^{10}} \right)} = 0.0259 \ln{\left( 6.67 \times 10^6 \right)}$

$\phi_F = 0.0259 \times 15.71 = 0.407 \, \text{V}$

Step 3: Threshold voltage ($V_{th}$)

The threshold voltage is calculated as: $V_{th} = V_{FB} + 2\phi_F + \frac{\sqrt{4 \varepsilon_{Si} q N_A (2\phi_F)}}{C_{ox}}$ where:

• $V_{FB} = -\frac{Q_{ox}}{C_{ox}}$ is the flat-band voltage.
• $\varepsilon_{Si} = k_{Si} \varepsilon_0$.

Flat-band voltage ($V_{FB}$):

$V_{FB} = -\frac{Q_{ox}}{C_{ox}} = -\frac{10^{12} \times 1.6 \times 10^{-19}}{5.53125 \times 10^{-6}}$

$V_{FB} = -0.029 \, \text{V}$

Bulk charge term:

$\varepsilon_{Si} = 11.7 \times 8.85 \times 10^{-14} = 1.03545 \times 10^{-12} \, \text{F/cm}$

$\sqrt{4 \varepsilon_{Si} q N_A (2\phi_F)} = \sqrt{4 \times 1.03545 \times 10^{-12} \times 1.6 \times 10^{-19} \times 10^{17} \times 0.814}$

$= \sqrt{5.386 \times 10^{-12}} = 7.34 \times 10^{-6} \, \text{C/cm}^2$

$\frac{\sqrt{4 \varepsilon_{Si} q N_A (2\phi_F)}}{C_{ox}} = \frac{7.34 \times 10^{-6}}{5.53125 \times 10^{-6}} = 1.327 \, \text{V}$

Total $V_{th}$:

$V_{th} = -0.029 + 2(0.407) + 1.327$

$V_{th} = 2.112 \, \text{V}$

Final Answer:

The threshold voltage ($V_{th}$) is approximately 2.11 V.

Do you want details on any specific step or have further questions?

Here are five follow-up questions:

1. How does the oxide thickness affect the threshold voltage?
2. Why is the Fermi potential ($\phi_F$) dependent on doping concentration?
3. How is the flat-band voltage ($V_{FB}$) influenced by trapped charges?
4. What is the role of the dielectric constant of the oxide in determining $C_{ox}$?
5. How would the threshold voltage change if the doping concentration were reduced?

Tip: Always ensure units are consistent when calculating $V_{th}$ components.