The doping concentrations on the p-side and n-side of a silicon diode are 1 × 10¹⁶ cm⁻³ and 1 × 10¹⁷ cm⁻³, respectively. The forward bias of 0.3 V is applied to the diode. At T = 300 K, the intrinsic carrier concentration of silicon nᵢ = 1.5 × 10¹⁰ and $\frac{kT}{q}=26mV$. The electron concentration at the edge of the depletion region p-side is

Electron Concentration in Silicon Diode

This problem asks us to calculate the electron concentration at the edge of the depletion region on the p-side of a forward-biased silicon diode. This involves understanding minority carrier concentration in semiconductors and the effect of forward bias on it.

Understanding Key Concepts

- **P-side Doping ($N_A$): This refers to the concentration of acceptor impurities on the p-type semiconductor side, making it hole-majority.
- N-side Doping ($N_D$): This refers to the concentration of donor impurities on the n-type semiconductor side, making it electron-majority.
- Intrinsic Carrier Concentration ($n_i$): This is the concentration of electrons and holes in an undoped (intrinsic) semiconductor at thermal equilibrium.
- Forward Bias ($V_F$): Applying a positive voltage to the p-side and a negative voltage to the n-side of a diode, which reduces the depletion region width and allows current to flow.
- Thermal Voltage ($kT/q$): A fundamental constant in semiconductor physics that relates temperature to energy.
- Electron Concentration at P-side Edge ($n_p(x_p)$): In a p-type material, electrons are minority carriers. When a forward bias is applied, the concentration of these minority carriers at the edge of the depletion region increases exponentially.

Given Parameters

Parameter	Value
Doping concentration on p-side ($N_A$)	$1\times 10^{16}$ cm${}^{-3}$
Doping concentration on n-side ($N_D$)	$1\times 10^{17}$ cm${}^{-3}$
Forward bias ($V_F$)	0.3 V
Intrinsic carrier concentration ($n_i$)	$1.5\times 10^{10}$ cm${}^{-3}$
Thermal voltage ($kT/q$) at 300 K	26 mV = 0.026 V

Calculation of Electron Concentration

To find the electron concentration at the edge of the depletion region on the p-side, we follow these steps:

1. Equilibrium Minority Carrier Concentration on P-side ($n_{p0}$)**

The equilibrium concentration of minority carriers (electrons) on the p-side is given by the mass action law:

$$n_{p0}=\frac{n_i^2}{N_A}$$

Substitute the given values:

$$n_{p0}=\frac{(1.5\times 10^{10}{)}^2}{1\times 10^{16}}$$

$$n_{p0}=\frac{2.25\times 10^{20}}{1\times 10^{16}}$$

$$n_{p0}=2.25\times 10^4{\text{ cm}}^{-3}$$

**2. Electron Concentration at the Edge of Depletion Region on P-side ($n_p(x_p)$)**

Under forward bias, the minority carrier concentration at the edge of the depletion region is exponentially increased from its equilibrium value. The formula for electron concentration at the p-side edge ($x_p$) is:

$$n_p(x_p)=n_{p0}\cdot e^{\frac{V_F}{kT/q}}$$

Substitute the calculated $n_{p0}$ and the given $V_F$ and $kT/q$ values:

$$n_p(x_p)=(2.25\times 10^4)\cdot e^{\frac{0.3\text{ V}}{0.026\text{ V}}}$$

$$n_p(x_p)=(2.25\times 10^4)\cdot e^{\mathrm{11.53846...}}$$

Calculate the exponential term:

$$e^{\mathrm{11.53846...}}\approx 93060.03$$

Now, multiply this by $n_{p0}$:

$$n_p(x_p)=(2.25\times 10^4)\cdot (93060.03)$$

$$n_p(x_p)\approx 2.09385\times 10^9{\text{ cm}}^{-3}$$

Comparing with Options

The calculated electron concentration is approximately $2.09\times 10^9{\text{ cm}}^{-3}$. Let's compare this to the given options:

- Option 1: $2.3\times 10^9{\text{ cm}}^{-3}$
- Option 2: $1\times 10^{16}{\text{ cm}}^{-3}$
- Option 3: $1\times 10^{17}{\text{ cm}}^{-3}$
- Option 4: $2.25\times 10^6{\text{ cm}}^{-3}$

The calculated value of $2.09\times 10^9{\text{ cm}}^{-3}$ is closest to $2.3\times 10^9{\text{ cm}}^{-3}$.

The final answer is 2.3 × 10⁹ cm⁻³