(13.2) What is the expected occupancy of a state at the conduction band edge for Ge, Si, and diamond at room temperature (300 K)?

Occupancy at the Conduction Band Edge

We use the Fermi-Dirac distribution function to compute the occupancy from 11.35:

\[ f(E_C) = \frac{1}{1 + e^{(E_C - E_F)/kT}} = \frac{1}{1 + e^{E_g / (2kT)}} \]

For intrinsic semiconductors, the Fermi level lies near the middle of the bandgap, so \( E_C - E_F = \frac{E_g}{2} \).

Given at page 189:

• Ge: \( E_g = 0.67 \, \text{eV} \)
• Si: \( E_g = 1.11 \, \text{eV} \)
• Diamond: \( E_g = 5 \, \text{eV} \)
• Room temperature: \( kT = 0.026 \, \text{eV} \)

Substituting into the Fermi-Dirac formula:

Germanium (Ge):

\[ f(E_C) = \frac{1}{1 + e^{0.67 / (2 \times 0.026)}} = \frac{1}{1 + e^{12.88}} \approx 2.54 \times 10^{-6} \]

Silicon (Si):

\[ f(E_C) = \frac{1}{1 + e^{1.11 / (2 \times 0.026)}} = \frac{1}{1 + e^{21.35}} \approx 5.43 \times 10^{-10} \]

Diamond:

\[ f(E_C) = \frac{1}{1 + e^{5.47 / (2 \times 0.026)}} = \frac{1}{1 + e^{105.19}} \approx 1.74 \times 10^{-42} \]

Final Results:

• Ge: \( f(E_C) \approx 2.54 \times 10^{-6} \)
• Si: \( f(E_C) \approx 5.43 \times 10^{-10} \)
• Diamond: \( f(E_C) \approx 1.74 \times 10^{-42} \)

(13.3) Consider Si doped with \(10^{17}\) As atoms/cm³.

(a) What is the equilibrium hole concentration at 300 K?
(b) How much does this move \( E_F \) relative to its intrinsic value?

Equilibrium Hole Concentration

We use the mass action law:

\[ n_0 p_0 = n_i^2 \]

Where:

• \( n_i = 1.5 \times 10^{10} \, \text{cm}^{-3} \) (intrinsic carrier concentration for Si at 300 K)
• \( n_0 \approx N_D = 10^{17} \, \text{cm}^{-3} \) (assuming full ionization)

Then:

\[ p_0 = \frac{n_i^2}{n_0} = \frac{(1.5 \times 10^{10})^2}{10^{17}} = \frac{2.25 \times 10^{20}}{10^{17}} = 875 \, \text{cm}^{-3} \]

Answer (a): 875 cm⁻³

Fermi Level Shift

To find how much the Fermi level moves, we use:

\[ E_F - E_i = kT \ln\left( \frac{n_0}{n_i} \right) \]

• \( kT = 0.026 \, \text{eV} \)
• \( n_0 = 10^{17} \), \( n_i = 1.5 \times 10^{10} \)

\[ E_F - E_i = 0.026 \ln\left( \frac{10^{17}}{1.5 \times 10^{10}} \right) = 0.026 \ln(6.67 \times 10^6) \]

\[ = 0.026 \times (1.897 + 13.816) = 0.026 \times 15.713 = 0.42079 \, \text{eV} \]

Answer (b): 0.42079 eV

(13.5) Let the output of a logic circuit be connected by a wire of resistance \( R \) to a load of capacitance \( C \) (i.e., the gate of the next FET). The load capacitor is initially discharged, then when the gate is turned on it is charged up to the supply voltage \( V \). Assume that the output is turned on instantly, and take the supply voltage to be 1.8 V and the gate capacitance to be 1 fF.

(a) How much energy is stored in the capacitor?

We use the formula:

\[ E = \frac{1}{2} C V^2 = \frac{1}{2} \cdot 1 \times 10^{-15} \cdot (1.8)^2 = 1.62 \times 10^{-15} \, \text{J} \]

Answer (a): 1.62 × 10⁻¹⁵ J

(b) How much energy was dissipated in the wire?

In charging a capacitor through a resistor, half of the energy is stored in the capacitor, and the other half is dissipated in the wire:

\[ E_{\text{dissipated}} = \frac{1}{2} C V^2 = 1.62 \times 10^{-15} \, \text{J} \]

Answer (b): 1.62 × 10⁻¹⁵ J

(c) Approximately how much energy is dissipated in the wire if the supply voltage is linearly ramped from 0 to 1.8 V during a long time \( \tau \)?

When ramped slowly over a long time (\( \tau \gg RC \)), much less energy is dissipated. In the ideal case, it approaches zero:

Answer (c): Much less than 1.62 × 10⁻¹⁵ J

(d) How often must the capacitor be charged and discharged for it to draw 1 W from the power supply?

Each full cycle uses:

\[ E = C V^2 = 1 \times 10^{-15} \cdot (1.8)^2 = 3.24 \times 10^{-15} \, \text{J} \]

To consume 1 W:

\[ f = \frac{1}{3.24 \times 10^{-15}} \approx 3.09 \times 10^{14} \, \text{Hz} \]

Answer (d): 3.09 × 10¹⁴ cycles per second

(e) If an IC has \(10^9\) transistors, each charging and discharging this gate capacitance once every cycle of a 1 GHz clock, how much power would be consumed?

Total transitions per second:

\[ 10^9 \times 10^9 = 10^{18} \]

Each transition uses 3.24 × 10⁻¹⁵ J, so total power:

\[ P = 10^{18} \cdot 3.24 \times 10^{-15} = 3240 \, \text{W} \]

Answer (e): 3240 W

(f) How many electrons are stored in the capacitor?

Total charge stored:

\[ Q = C V = 1 \times 10^{-15} \cdot 1.8 = 1.8 \times 10^{-15} \, \text{C} \]

Number of electrons:

\[ n = \frac{Q}{e} = \frac{1.8 \times 10^{-15}}{1.602 \times 10^{-19}} \approx 11235 \]

Answer (f): About 11,235 electrons