# Semiconductor Junctions

This is a continuation from the previous tutorial - current density in semiconductors.

Useful semiconductor devices are made of inhomogeneous semiconductors that have either spatially nonuniform doping distribution or spatially varying bandgaps or both.

There are two categories of semiconductor junctions: homojunctions and heterojunctions. A homojunction is formed by different doping in the same semiconductor, whereas a heterojunction is formed between two different semiconductors.

In addition, a metal-semiconductor junction can be formed between a metal and a semiconductor.

The immense possibilities of such structures are witnessed by the existence of the great variety of semiconductor devices and by the constant invention of new devices.

In this tutorial, we review the basic properties of the semiconductor homojunctions and heterojunctions.

The most important homojunctions are the p-n junctions. A p-n junction is formed between a p-type region and an n-type region with different doping in the same semiconductor.

A homojunction can also be a p-i junction, which is formed between a p-type region and an undoped intrinsic region of the same semiconductor, or an i-n junction, which is formed between an undoped intrinsic region and an n-type region of the same semiconductor.

A heterojunction is normally formed between two lattice-matched semiconductors of different bandgaps.

To name a heterojunction, the conductivity type of the small-gap semiconductor is represented by a lowercase letter, n, p, or i, and the conductivity type of the large-gap semiconductor is represented by an uppercase letter, N, P, or I.

Because the two semiconductors that form a heterojunction have different bandgaps, they can be either of different conductivity types or of the same conductivity type.

Junctions formed between dissimilar semiconductors of the same conductivity type, such as p-P and n-N junctions, are isotype heterojunctions; those formed between dissimilar semiconductors of different conductivity types, such as p-N and P-n junctions, are anisotype heterojunctions.

A semiconductor junction can be either an abrupt junction, which has a sudden change of doping and/or bandgap from one region to the other region, or a graded junction, where the change of doping and/or bandgap is gradual. The basic principles of abrupt and graded junctions are the same though there are quantitative differences in the properties of these two different types.

In this tutorial, we consider only abrupt p-n, p-N, and P-n junctions. For simplicity, we assume that the p region is doped with a concentration $$N_\text{a}$$ of fully ionized acceptors, and the n region is doped with a concentration $$N_\text{d}$$ of fully ionized donors.

In the vicinity of a junction between the p and n regions, there exists a depletion layer, where majority carriers, holes on the p side and electrons on the n side, are depleted.

A junction reaches thermal equilibrium when it is not subject to an external excitation.

In thermal equilibrium, the p and n regions outside the depletion layer are homogeneous regions because the characteristics of the semiconductor in these regions approach those of homogeneous semiconductors in thermal equilibrium.

The homogeneous p region has a majority hole concentration of $$p_\text{p0}=N_\text{a}^-=N_\text{a}$$ and a minority electron concentration of $$n_\text{p0}$$, whereas the homogeneous n region has a majority electron centration of $$n_\text{n0}=N_\text{d}^+=N_\text{d}$$ and a minority hole concentration of $$p_\text{n0}$$.

The equilibrium state can be perturbed with a bias voltage. For a junction under bias, diffusion regions exist between the depletion layer and the homogeneous regions on both p and n sides.

In the diffusion regions, the characteristics of the semiconductor are dominated by diffusion of the minority carriers, which are electrons on the p side and holes on the n side.

The major differences among p-n homojunctions and p-N and P-n heterojunctions are their energy band structures. The energy bands and the built-in potential for each type of junction are considered in this tutorial to illustrate the differences among these junctions.

In addition, the difference in the electric permittivities, $$\epsilon_\text{p}$$ and $$\epsilon_\text{n}$$, respectively, of the p and n regions can be significant for a p-N or P-n heterojunction but is practically negligible for a p-n homojunction. This issue is minor because it can be easily taken care of by considering $$\epsilon_\text{p}$$ and $$\epsilon_\text{n}$$ to be generally different, even for a p-n homojunction.

Other than these differences, these junctions have similar electrical characteristics. Therefore, the discussions and the mathematical relations regarding the depletion layer, the carrier distribution, the current-voltage characteristics, and the capacitance are treated generally and are valid for both homojunctions and heterojunctions.

Energy Bands and Electrostatic Potential

As discussed in the current density tutorial, a semiconductor in thermal equilibrium is characterized by a spatially constant Fermi level. This statement is true for both homojunctions and heterojunctions.

Therefore, as shown in Figures 12-6, 12-7, and 12-8 for p-n, p-N, and P-n junctions, respectively, $$E_\text{Fp}=E_\text{Fn}=E_\text{F}$$ for a junction in thermal equilibrium, where $$E_\text{Fp}$$ and $$E_\text{Fn}$$ are the Fermi levels in the p and n regions, respectively.

Because $$E_\text{Fp}$$ lies close to the valence-band edge in the p region but $$E_\text{Fn}$$ lies close to the conduction-band edge in the n region, a constant Fermi level throughout the semiconductor in thermal equilibrium leads to bending of the energy bands across the junction, as shown in Figure 12-6 for a p-n homojunction and in Figures 12-7 and 12-8 for p-N and P-n heterojunctions, respectively.

As we shall see later, this band bending occurs primarily within the depletion layer. The energy bands remain relatively flat outside the depletion layer on both p and n sides.

For a homojunction, shown in Figure 12-6, the energy bands remain continuous and smooth across the junction because the semiconductors on the two sides of the junction have the same bandgap.

For a heterojunction, the semiconductors on the two sides of the junction have different bandgaps. At the junction where these two semiconductors are joined together, the disparity in their bandgaps results in a discontinuity of $$\Delta{E}_\text{c}$$ in the conduction-band edge and a discontinuity of $$\Delta{E}_\text{v}$$ in the valence-band edge, as shown in Figures 12-7 and 12-8.

Therefore,

$\tag{12-71}\Delta{E}_\text{c}+\Delta{E}_\text{v}=\Delta{E}_\text{g}$

where $$\Delta{E}_\text{g}$$ is the difference between the bandgaps of the two dissimilar semiconductors.

The value of $$\Delta{E}_\text{g}$$ and those of the band offsets $$\Delta{E}_\text{c}$$ and $$\Delta{E}_\text{v}$$ are intrinsic properties of a specific pair of semiconductors.

The conduction-band offset $$\Delta{E}_\text{c}$$ is determined by the difference in the electron affinities of the two semiconductors, the valence-band offset $$\Delta{E}_\text{v}$$ is then fixed by (12-71).

In practice, these parameters are measured experimentally for each given pair of semiconductors. For GaAs - AlxGa1-xAs heterojunctions, $$\Delta{E}_\text{c}\approx65\%$$ $$\Delta{E}_\text{g}$$ and $$\Delta{E}_\text{v}\approx35\%$$ $$\Delta{E}_\text{g}$$.

p-n homojunction in thermal equilibrium

Because the bandgap remains constant across a semiconductor homojunction, at any given location the conduction- and valence-band edges have the same gradient. According to (12-62) [refer to the current density in semiconductors tutorial], this spatially varying band-edge gradient creates a spatially varying built-in electric field that is seen by both electrons and holes: $$\pmb{E}_\text{e}=\pmb{E}_\text{h}$$ at any given location.

In thermal equilibrium, this built-in electric field results in a built-in electrostatic potential across the p-n junction, as shown in Figure 12-6(b). The height, $$V_0$$, of this built-in potential is called the contact potential of the junction.

Note that the energy bands plotted in Figure 12-6(a) refer to the energy of an electron. The n region has a lower energy for an electron than the p region, whereas the converse is true for a hole. Because an electron carriers a negative charge of $$q=-e$$, the built-in electrostatic potential is higher on the n side than on the p side.

As shown in Figure 12-6(a), we have, for a p-n homojunction in thermal equilibrium,

$\tag{12-72}E_\text{cp}-E_\text{cn}=E_\text{vp}-E_\text{vn}=eV_0$

where $$E_\text{cp}$$ and $$E_\text{cn}$$ are, respectively, the conduction-band edges in the homogeneous p and n regions, and $$E_\text{vp}$$ and $$E_\text{vn}$$ are, respectively, the valence-band edges in the homogeneous p and n regions.

Therefore, $$eV_0$$ is the same energy barrier for an electron on the n side to move to the p side as that for a hole on the p side to move to the n side.

In the case when both p and n regions are nondegenerate so that (12-26) and (12-27) [refer to the electron and hole concentrations tutorial] are valid, we can use (12-72) to obtain the following relation for the carrier concentrations in the homogeneous p and n regions:

$\tag{12-73}\frac{p_\text{p0}}{p_\text{n0}}=\frac{n_\text{n0}}{n_\text{p0}}=\text{e}^{eV_0/k_\text{B}T}$

By using the law of mass action, $$p_\text{p0}n_\text{p0}=p_\text{n0}n_\text{n0}=n_\text{i}^2$$, we find that the contact potential of a nondegenerate p-n homojunction in a nondegenerate semiconductor is given by

$\tag{12-74}V_0=\frac{k_\text{B}T}{e}\ln\frac{p_\text{p0}}{p_\text{n0}}=\frac{k_\text{B}T}{e}\ln\frac{n_\text{n0}}{n_\text{p0}}=\frac{k_\text{B}T}{e}\ln\frac{p_\text{p0}n_\text{n0}}{n_\text{i}^2}$

Under the condition that $$p_\text{p0}\approx{N}_\text{a}\gg{n}_\text{p0}$$ and $$n_\text{n0}\approx{N}_\text{d}\gg{p}_\text{n0}$$, the contact potential can be found as

$\tag{12-75}V_0=\frac{k_\text{B}T}{e}\ln\frac{N_\text{a}N_\text{d}}{n_\text{i}^2}=\frac{k_\text{B}T}{e}\ln\frac{N_\text{a}N_\text{d}}{N_\text{v}N_\text{c}}+\frac{E_\text{g}}{e}$

Example 12-8

An abrupt GaAs p-n homojunction is formed by creating a uniform p region on one side an a uniform n region on the other side. The p region is doped with fully ionized acceptors of a concentration $$N_\text{a}=1\times10^{23}\text{ m}^{-3}$$, and the n region is doped with fully ionized donors of a concentration $$N_\text{d}=1\times10^{22}\text{ m}^{-3}$$. Find the contact potential of this junction at $$300\text{ K}$$.

From Example 12-2 [refer to the electron and hole concentrations tutorial], we have $$n_\text{i}=2.33\times10^{12}\text{ m}^{-3}$$ for GaAs at $$300\text{ K}$$. At the given doping levels for the p and n regions, $$p_\text{p0}\approx{N}_\text{a}\gg{n}_\text{p0}$$ and $$n_\text{n0}\approx{N}_\text{d}\gg{p}_\text{n0}$$. At $$300\text{ K}$$, $$k_\text{B}T/e=25.9\text{ mV}=0.0259\text{ V}$$. Therefore, we can use the first relation in (12-75) to find the contact potential for this junction at $$300\text{ K}$$ as

$V_0=0.0259\text{ V}\times\ln\frac{1\times10^{23}\times1\times10^{22}}{(2.33\times10^{12})^2}=1.209\text{ V}$

p-N heterojunction in thermal equilibrium

For a p-N heterojunction, the semiconductor on the n side has a larger bandgap than that on the p side: $$E_\text{gn}\gt{E}_\text{gp}$$. Therefore,

$\tag{12-76}\Delta{E}_\text{c}+\Delta{E}_\text{v}=E_\text{gn}-E_\text{gp}=\Delta{E}_\text{g}$

according to (12-71).

Because of the presence of band offsets at the junction, (12-72) is not valid for a p-N junction. Instead, as can be seen in Figure 12-7(a), we have

$\tag{12-77}E_\text{cp}-E_\text{cn}=eV_0-\Delta{E}_\text{c}$

$\tag{12-78}E_\text{vp}-E_\text{vn}=eV_0+\Delta{E}_\text{v}$

where $$V_0$$ is the contact potential of the p-N junction.

In contrast to the case of a p-n homojunction, where electrons and holes have the same energy barrier of $$eV_0$$, electrons and holes have different energy barriers in the case of a p-N junction.

According to (12-77) and (12-78), the energy barrier for an electron on the n side is lowered from $$eV_0$$ by $$\Delta{E}_\text{c}$$ due to the conduction-band offset, but that for a hole on the p side is raised by $$\Delta{E}_\text{v}$$ due to the valence-band offset.

Therefore, in a p-N heterojunction the energy barrier for an electron on the n side is lower than that for a hole on the p side by the amount of the bandgap difference of $$\Delta{E}_\text{g}=E_\text{gn}-E_\text{gp}$$ between the two semiconductors.

Though the energy barriers for electrons and holes to cross a p-N junction are different, electrons and holes see the same spatially varying built-in electric field because the conduction- and valence-band edges are parallel to each other and have the same gradient at every location except at the junction where the discontinuities of the energy bands take place.

Therefore, similarly to the situation in a homojunction, we still have a common electrostatic field, $$\pmb{E}_\text{e}=\pmb{E}_\text{h}$$, at any given location for both electrons and holes.

As a result, there is a common built-in electrostatic potential, shown in Figure 12-7(b), across a p-N junction. For an abrupt p-N junction, there is a sudden change of slope in $$V(x)$$ at the junction because of the sudden change in electric permittivity from one semiconductor to the other.

For a p-N junction, (12-73) is not valid. Instead, we can use (12-77) and (12-78) to obtain the following relation in the case when both p and n regions are nondegenerate:

$\tag{12-79}\frac{p_\text{p0}}{p_\text{n0}}\frac{N_\text{vn}}{N_\text{vp}}\text{e}^{-\Delta{E}_\text{v}/k_\text{B}T}=\frac{n_\text{n0}}{n_\text{p0}}\frac{N_\text{cp}}{N_\text{cn}}\text{e}^{\Delta{E}_\text{c}/k_\text{B}T}=\text{e}^{eV_0/k_\text{B}T}$

where $$N_\text{cp}$$ and $$N_\text{vp}$$ are the effective densities of states, as defined in (12-24) [refer to the electron and hole concentrations tutorial], for the semiconductor on the p side, and $$N_\text{cn}$$ and $$N_\text{vn}$$ are the effective densities of states for the semiconductor on the n side.

From this relation, we find that the contact potential for a nondegenerate p-N junction can be expressed as

\tag{12-80}\begin{align}V_0&=\frac{k_\text{B}T}{e}\ln\left(\frac{p_\text{p0}}{p_\text{n0}}\frac{N_\text{vn}}{N_\text{vp}}\right)-\frac{\Delta{E}_\text{v}}{e}=\frac{k_\text{B}T}{e}\ln\left(\frac{n_\text{n0}}{n_\text{p0}}\frac{N_\text{cp}}{N_\text{cn}}\right)+\frac{\Delta{E}_\text{c}}{e}\\&=\frac{k_\text{B}T}{e}\ln\left(\frac{p_\text{p0}n_\text{n0}}{n_\text{in}^2}\frac{N_\text{vn}}{N_\text{vp}}\right)-\frac{\Delta{E}_\text{v}}{e}=\frac{k_\text{B}T}{e}\ln\left(\frac{p_\text{p0}n_\text{n0}}{n_\text{ip}^2}\frac{N_\text{cp}}{N_\text{cn}}\right)+\frac{\Delta{E}_\text{c}}{e}\end{align}

where $$n_\text{ip}$$ and $$n_\text{in}$$ are the intrinsic carrier concentrations for the semiconductors on the p and n sides, respectively.

Under the condition that $$p_\text{p0}\approx{N}_\text{a}\gg{n}_\text{p0}$$ and $$n_\text{n0}\approx{N}_\text{d}\gg{p}_\text{n0}$$, the contact potential can be found as

\tag{12-81}\begin{align}V_0&=\frac{k_\text{B}T}{e}\ln\left(\frac{N_\text{a}N_\text{d}}{n_\text{in}^2}\frac{N_\text{vn}}{N_\text{vp}}\right)-\frac{\Delta{E}_\text{v}}{e}=\frac{k_\text{B}T}{e}\ln\left(\frac{N_\text{a}N_\text{d}}{n_\text{ip}^2}\frac{N_\text{cp}}{N_\text{cn}}\right)+\frac{\Delta{E}_\text{c}}{e}\\&=\frac{k_\text{B}T}{e}\ln\frac{N_\text{a}N_\text{d}}{N_\text{vp}N_\text{cn}}+\frac{E_\text{gn}}{e}-\frac{\Delta{E}_\text{v}}{e}=\frac{k_\text{B}T}{e}\ln\frac{N_\text{a}N_\text{d}}{N_\text{vp}N_\text{cn}}+\frac{E_\text{gp}}{e}+\frac{\Delta{E}_\text{c}}{e}\end{align}

P-n heterojunction in thermal equilibrium

For a P-n heterojunction, the semiconductor on the p side has a larger bandgap than that on the n side: $$E_\text{gp}\gt{E}_\text{gn}$$. Therefore,

$\tag{12-82}\Delta{E}_\text{c}+\Delta{E}_\text{v}=E_\text{gp}-E_\text{gn}=\Delta{E}_\text{g}$

As shown in Figure 12-8(a), the band offsets at the junction lead to

$\tag{12-83}E_\text{cp}-E_\text{cn}=eV_0+\Delta{E}_\text{c}$

$\tag{12-84}E_\text{vp}-E_\text{vn}=eV_0-\Delta{E}_\text{v}$

where $$V_0$$ is the contact potential of the P-n junction.

Similar to the case of the p-N junction discussed above, electrons and holes do not have the same energy barrier. However, because $$E_\text{gp}\gt{E}_\text{gn}$$ in the case of a P-n heterojunction, the energy barrier for an electron on the n side is now higher than that for a hole on the p side by the amount of the bandgap difference of $$\Delta{E}_\text{g}=E_\text{gp}-E_\text{gn}$$ between the two semiconductors.

Figure 12-8(b) shows the built-in electrostatic potential across a P-n junction. For an abrupt P-n junction, there is also a sudden change of slope in $$V(x)$$ at the junction because of the sudden change in electric permittivity across the junction.

For a P-n junction, we can use (12-83) and (12-84) to obtain the following relation in the case when both p and n regions are nondegenerate:

$\tag{12-85}\frac{p_\text{p0}}{p_\text{n0}}\frac{N_\text{vn}}{N_\text{vp}}\text{e}^{\Delta{E}_\text{v}/k_\text{B}T}=\frac{n_\text{n0}}{n_\text{p0}}\frac{N_\text{cp}}{N_\text{cn}}\text{e}^{-\Delta{E}_\text{c}/k_\text{B}T}=\text{e}^{eV_0/k_\text{B}T}$

From this relation, we find the following contact potential for a nondegenerate P-n junction:

\tag{12-86}\begin{align}V_0&=\frac{k_\text{B}T}{e}\ln\left(\frac{p_\text{p0}}{p_\text{n0}}\frac{N_\text{vn}}{N_\text{vp}}\right)+\frac{\Delta{E}_\text{v}}{e}=\frac{k_\text{B}T}{e}\ln\left(\frac{n_\text{n0}}{n_\text{p0}}\frac{N_\text{cp}}{N_\text{cn}}\right)-\frac{\Delta{E}_\text{c}}{e}\\&=\frac{k_\text{B}T}{e}\ln\left(\frac{p_\text{p0}n_\text{n0}}{n_\text{in}^2}\frac{N_\text{vn}}{N_\text{vp}}\right)+\frac{\Delta{E}_\text{v}}{e}=\frac{k_\text{B}T}{e}\ln\left(\frac{p_\text{p0}n_\text{n0}}{n_\text{ip}^2}\frac{N_\text{cp}}{N_\text{cn}}\right)-\frac{\Delta{E}_\text{c}}{e}\end{align}

Under the condition that $$p_\text{p0}\approx{N}_\text{a}\gg{n}_\text{p0}$$ and $$n_\text{n0}\approx{N}_\text{d}\gg{p}_\text{n0}$$, the contact potential can be found as

\tag{12-87}\begin{align}V_0&=\frac{k_\text{B}T}{e}\ln\left(\frac{N_\text{a}N_\text{d}}{n_\text{in}^2}\frac{N_\text{vn}}{N_\text{vp}}\right)+\frac{\Delta{E}_\text{v}}{e}=\frac{k_\text{B}T}{e}\ln\left(\frac{N_\text{a}N_\text{d}}{n_\text{ip}^2}\frac{N_\text{cp}}{N_\text{cn}}\right)-\frac{\Delta{E}_\text{c}}{e}\\&=\frac{k_\text{B}T}{e}\ln\frac{N_\text{a}N_\text{d}}{N_\text{vp}N_\text{cn}}+\frac{E_\text{gn}}{e}+\frac{\Delta{E}_\text{v}}{e}=\frac{k_\text{B}T}{e}\ln\frac{N_\text{a}N_\text{d}}{N_\text{vp}N_\text{cn}}+\frac{E_\text{gp}}{e}-\frac{\Delta{E}_\text{c}}{e}\end{align}

Example 12-9

An abrupt AlGaAs/GaAs P-n heterojunction is formed with a uniform p region of Al0.3Ga0.7As and a uniform n region of GaAs. The p Al0.3Ga0.7As region is doped with fully ionized acceptors of a concentration $$N_\text{a}=1\times10^{23}\text{ m}^{-3}$$, and the n GaAs region is doped with fully ionized donors of a concentration $$N_\text{d}=1\times10^{22}\text{ m}^{-3}$$. The density of states effective masses for Al0.3Ga0.7As are $$m_\text{e}^*=0.092m_0$$ and $$m_\text{h}^*=0.62m_0$$, as compared to $$m_\text{e}^*=0.067m_0$$ and $$m_\text{h}^*=0.52m_0$$ for GaAs. Find the contact potential of this junction at $$300\text{ K}$$ by taking $$\Delta{E}_\text{c}=65\%\Delta{E}_\text{g}$$ and $$\Delta{E}_\text{v}=35\%\Delta{E}_\text{g}$$. Compare this contact potential to that of the GaAs homojunction considered in Example 12-8 above, which has the same doping profile.

We find from (12-3) [refer to the introduction to semiconductors tutorial] with $$x=0.3$$ that $$E_\text{g}=1.798\text{ eV}$$ for Al0.3Ga0.7As. Because $$E_\text{g}=1.424\text{ eV}$$ for GaAs [refer to table 12-1 in the introduction to semiconductors tutorial], we thus have $$E_\text{gp}=1.798\text{ eV}$$, $$E_\text{gn}=1.424\text{ eV}$$, and $$\Delta{E}_\text{g}=E_\text{gp}-E_\text{gn}=0.374\text{ eV}$$ for this heterostructure.

Following the procedures in Example 12-2 [refer to the electron and hole concentrations tutorial] but taking $$m_\text{e}^*=0.092m_0$$ and $$m_\text{h}^*=0.62m_0$$, we find that $$N_\text{cp}=7.00\times10^{23}\text{ m}^{-3}$$, $$N_\text{vp}=1.22\times10^{25}\text{ m}^{-3}$$, and $$n_\text{ip}=2.46\times10^9\text{ m}^{-3}$$ for the p Al0.3Ga0.7As at $$300\text{ K}$$.

Also from Example 12-2 [refer to the electron and hole concentrations tutorial], we have $$N_\text{cn}=4.35\times10^{23}\text{ m}^{-3}$$, $$N_\text{vn}=9.41\times10^{24}\text{ m}^{-3}$$, and $$n_\text{in}=2.33\times10^{12}\text{ m}^{-3}$$ for the n GaAs at $$300\text{ K}$$.

Because $$p_\text{p0}\approx{N}_\text{a}\gg{n}_\text{p0}$$ and $$n_\text{n0}\approx{N}_\text{d}\gg{p}_\text{n0}$$, we can use any one of the relations in (12-86) and (12-87) to find the contact potential. By using the first relation in (12-87), the contact potential for this junction at $$300\text{ K}$$ is found as

\begin{align}V_0&=\frac{k_\text{B}T}{e}\ln\left(\frac{N_\text{a}N_\text{d}}{n_\text{in}^2}\frac{N_\text{vn}}{N_\text{vp}}\right)+\frac{\Delta{E}_\text{v}}{e}\\&=0.0259\text{ V}\times\ln\frac{1\times10^{23}\times1\times10^{22}\times9.41\times10^{24}}{(2.33\times10^{12})^2\times1.22\times10^{25}}+0.35\times0.374\text{ V}\\&=1.333\text{ V}\end{align}

The same result is obtained by using any other relation in (12-86) or (12-87).

The contact potential of this AlGaAs/GaAs heterojunction is different from that of the GaAs homojunction of the same doping profile considered in Example 12-8 for two reasons: (a) the density-of-states effective masses are different for AlGaAs and GaAs, and (2) the band offset due to the difference in the bandgaps between AlGaAs and GaAs causes an additional adjustment for the contact potential.

Junctions under bias

A bias voltage, $$V$$, is defined as the voltage applied to the p side of a junction with respect to the n side. A bias voltage changes the electrostatic potential between the p and n regions, thus changing the difference between $$E_\text{cp}$$ and $$E_\text{cn}$$ and that between $$E_\text{vp}$$ and $$E_\text{vn}$$.

In the case of a p-n homojunction under a bias voltage $$V$$, we have

$\tag{12-88}E_\text{cp}-E_\text{cn}=E_\text{vp}-E_\text{vn}=e(V_0-V)$

For a p-N junction, we have

$\tag{12-89}E_\text{cp}-E_\text{cn}=e(V_0-V)-\Delta{E}_\text{c}$

$\tag{12-90}E_\text{vp}-E_\text{vn}=e(V_0-V)+\Delta{E}_\text{v}$

For a P-n junction, we have

$\tag{12-91}E_\text{cp}-E_\text{cn}=e(V_0-V)+\Delta{E}_\text{c}$

$\tag{12-92}E_\text{vp}-E_\text{vn}=e(V_0-V)-\Delta{E}_\text{v}$

Clearly from these relations, a bias voltage has similar effects on p-n, p-N, and P-n junctions.

A junction is under forward bias if $$V\gt0$$. A forward bias voltage raises the potential on the p side with respect to that on the n side, resulting in a lower potential barrier of $$V_0-V$$. Consequently, the energy barrier between the homogeneous p and n regions is reduced by the amount of $$eV$$.

A junction is under reverse bias if $$V\lt0$$. A reverse bias voltage lowers the potential on the p side with respect to that on the n side, thus raising the potential barrier to $$V_0-V=V_0+|V|$$. The consequence is an increase in the energy barrier between the homogeneous p and n regions by the amount of $$e|V|$$.

Figure 12-9 shows the energy bands and the electrostatic potential of a p-n homojunction (a) in thermal equilibrium, (b) under forward bias, and (c) under reverse bias.

Except for the presence of band offsets at the junction, the characteristics of p-N and P-n heterojunctions under bias are similar to those shown in Figure 12-9.

A bias voltage causes an electric current to flow in a semiconductor. The bias voltage splits the Fermi level into separate quasi-Fermi levels, $$E_\text{Fc}$$ and $$E_\text{Fv}$$, for electrons and holes, respectively, and creates spatial gradients in them to support an electric current in the semiconductor.

According to (12-69) [refer to the current density in semiconductors tutorial], when an electric current flows in a semiconductor, the gradients in $$E_\text{Fc}$$ and $$E_\text{Fv}$$ exist throughout the semiconductor but they vary from one location to another in conjunction with the variations in the local concentrations of electrons and holes.

The spatial variations of the quasi-Fermi levels are shown, along with the spatial variations of the energy bands, in Figure 12-9(b) for the case of forward bias and in Figure 12-9(c) for the case of reverse bias.

The largest splitting of $$E_\text{Fc}$$ and $$E_\text{Fv}$$ occurs in the depletion layer. The largest gradients in $$E_\text{Fc}$$ and $$E_\text{Fv}$$ exist in the diffusion regions just outside the depletion layer. The quasi-Fermi levels gradually merge into $$E_\text{Fp}$$ in the homogenous p region and into $$E_\text{Fn}$$ in the homogenous n region.

In the presence of a bias voltage, the Fermi levels in the homogenous p and n regions are not aligned any more:

$\tag{12-93}E_\text{Fn}-E_\text{Fp}=eV$

In the case of forward bias, $$E_\text{Fn}\gt{E}_\text{Fp}$$ and $$E_\text{Fc}\gt{E}_\text{Fv}$$, as shown in Figure 12-9(b). In the case of reverse bias, $$E_\text{Fn}\lt{E}_\text{Fp}$$ and $$E_\text{Fc}\lt{E}_\text{Fv}$$, as shown in Figure 12-9(c).

The relation in (12-93) and the characteristics of the quasi-Fermi levels shown in Figure 12-9 are valid for p-N and P-n heterojunctions as well.

Example 12-10

Find the bias voltage that lines up the band edges of the p and n regions on the two sides of the p-n homojunction considered in Example 12-8. What is the bias voltage that is needed to line up the conduction-band edges of the p and n regions on the two sides of the P-n heterojunction considered in Example 12-9? What is the bias voltage needed to line up the valence-band edges?

Because the bandgap is the same on both sides of a homojunction, a bias voltage that lines up the conduction-band edge also lines up the valence-band edge. From (12-88), we find that $$V=V_0$$ for $$E_\text{cp}=E_\text{cn}$$ and $$E_\text{vp}=E_\text{vn}$$ across a homojunction. Therefore, we need a forward bias voltage of $$V=V_0=1.209\text{ V}$$ to line up the band edges of the p and n regions on the two sides of the p-n homojunction considered in Example 12-8.

For a heterojunction, a single bias voltage does not line up conduction-band edges and valence-band edges simultaneously because the bandgaps are different on the two sides of the junction. For the P-n heterojunction considered in Example 12-9, we find from (12-91) that the forward bias voltage required for $$E_\text{cp}=E_\text{cn}$$ to line up the conduction-band edges is

$V=V_0+\frac{\Delta{E}_\text{c}}{e}=1.333\text{ V}+0.65\times0.374\text{ V}=1.576\text{ V}$

We then find from (12-92) that the forward bias voltage required for $$E_\text{vp}=E_\text{vn}$$ to line up the valence-band edges is

$V=V_0-\frac{\Delta{E}_\text{v}}{e}=1.333\text{ V}-0.35\times0.374\text{ V}=1.202\text{ V}$

We find that these two bias voltages are quite different because of the existence of conduction-band and valence-band offsets caused by the bandgap difference on the two sides of the heterojunction.

Depletion Layer

The depletion layer is created by the diffusion of holes from the p side, where the hole concentration is high, to the n side, where the hole concentration is low, and the diffusion of electrons from the n side, where the electron concentration is high, to the p side, where the electron concentration is low.

The depletion layer has a width of

$\tag{12-94}W=x_\text{p}+x_\text{n}$

where $$x_\text{p}$$ and $$x_\text{n}$$ are the penetration depths of the depletion layer into the p and n regions, respectively.

The depletion layer is also known as the space-charge region because depletion of the majority of carriers in this region leaves the immobile negatively charged acceptor ions on the p side and the immobile positively charged donor ions on the n side as space charges in this region.

As shown in Figure 12-10, the p side has a negative space charge density of $$-eN_\text{a}$$ over a penetration depth of $$x_\text{p}$$, and the n side has a positive space charge density of $$eN_\text{d}$$ over a penetration depth of $$x_\text{n}$$.

Because of the overall neutrality of the semiconductor, the total negative space charges on the p side must be equal to the total positive space charges on the n side. Therefore, we have

$\tag{12-95}N_\text{a}x_\text{p}=N_\text{d}x_\text{n}$

By combining (12-94) and (12-95), we find that

$\tag{12-96}x_\text{p}=\frac{N_\text{d}}{N_\text{a}+N_\text{d}}W\qquad\text{and}\qquad{}x_\text{n}=\frac{N_\text{a}}{N_\text{a}+N_\text{d}}W$

Clearly, the depletion layer can penetrate p and n regions unevenly, depending on the relative doping on the two sides. The depletion layer penetrates deeper into the region that has a lighter doping concentration. This characteristic is demonstrated in Figure 12-10.

The electric field associated with the gradient of the band edges is created by the space charges in the depletion layer. Therefore, most of the band-edge gradient exists in the depletion layer and most of the potential difference between the n and p regions is distributed across this layer.

Because $$\pmb{E}=-\boldsymbol{\nabla}V$$ and $$\boldsymbol{\nabla}\cdot\pmb{E}=\rho(x)/\epsilon(x)$$, where $$\epsilon(x)$$ is the spatially varying electric permittivity of the semiconductor, we have the following Poisson equation to described the potential variations across the depletion layer:

$\tag{12-97}\nabla^2V=\frac{\text{d}^2V}{\text{d}V^2}=-\frac{\rho(x)}{\epsilon(x)}=\begin{cases}\frac{eN_\text{a}}{\epsilon_\text{p}},\qquad\text{ for }-x_\text{p}\lt{x}\lt0\\-\frac{eN_\text{d}}{\epsilon_\text{n}},\qquad\text{for }0\lt{x}\lt{x}_\text{n}\end{cases}$

where $$\epsilon_\text{p}$$ and $$\epsilon_\text{n}$$ are the electric permittivities of the p and n regions, respectively.

The boundary conditions are $$\pmb{E}(-x_\text{p})=\pmb{E}(x_\text{n})=0$$ and $$\epsilon_\text{p}\pmb{E}(0_{-})=\epsilon_\text{n}\pmb{E}(0_{+})$$, meaning that $$\text{d}V/\text{d}x=0$$ at $$x=-x_\text{p}$$ and $$x=x_\text{n}$$, and $$\epsilon_\text{p}\text{d}V/\text{d}x|_{x=0_{-}}=\epsilon_\text{n}\text{d}V/\text{d}x|_{x=0_{+}}$$.

By integrating (12-97) through the depletion layer and applying these boundary conditions, we find that, in the presence of a bias voltage $$V$$,

$\tag{12-98}V_0-V=V(x_\text{n})-V(-x_\text{p})=\frac{e}{2\epsilon}\frac{N_\text{a}N_\text{d}}{N_\text{a}+N_\text{d}}W^2$

where $$\epsilon$$ is an effective electric permittivity defined as

$\tag{12-99}\epsilon=\frac{\epsilon_\text{p}\epsilon_\text{n}(N_\text{a}+N_\text{d})}{\epsilon_\text{p}N_\text{a}+\epsilon_\text{n}N_\text{d}}$

Therefore, the width of the depletion layer can be expressed as a function of the applied bias voltage $$V$$ in the following form:

$\tag{12-100}W=\left[\frac{2\epsilon}{e}\left(\frac{N_\text{a}+N_\text{d}}{N_\text{a}N_\text{d}}\right)(V_0-V)\right]^{1/2}$

When a junction is in thermal equilibrium without bias, $$x_\text{p}=x_\text{p0}$$, $$x_\text{n}=x_\text{n0}$$, and $$W=W_0$$, as shown in Figure 12-10(a). From (12-100), we see that $$W\lt{W_0}$$ for $$V\gt0$$ and $$W\gt{W}_0$$ for $$V\lt0$$. Therefore, the depletion layer narrows with forward bias, as shown in Figure 12-10(b), and broadens with reverse bias, as shown in Figure 12-10(c).

Example 12-11

The static dielectric constant of GaAs is $$\epsilon/\epsilon_0=13.18$$. Find the width of the depletion layer, $$W_0$$, and the penetration depths, $$x_\text{p0}$$ and $$x_\text{n0}$$, for the GaAs p-n homojunction described in Example 12-8 when it is in thermal equilibrium without bias at $$300\text{ K}$$.

The width of the depletion layer in thermal equilibrium without bias can be found from (12-100) with $$V=0$$. With $$\epsilon=13.18\epsilon_0$$, $$\epsilon_0=8.854\times10^{-12}\text{ F m}^{-1}$$, $$N_\text{a}=1\times10^{23}\text{ m}^{-3}$$, $$N_\text{d}=1\times10^{22}\text{ m}^{-3}$$, and $$V_0=1.209\text{ V}$$ from Example 12-8, we find that

\begin{align}W_0&=\left[\frac{2\times13.18\times8.854\times10^{-12}}{1.6\times10^{-19}}\times\left(\frac{1\times10^{23}+1\times10^{22}}{1\times10^{23}\times1\times10^{22}}\right)\times1.209\right]^{1/2}\text{ m}\\&=440\text{ nm}\end{align}

From (12-96), we find the following penetration depths:

$x_\text{p0}=\frac{N_\text{d}}{N_\text{a}+N_\text{d}}W_0=\frac{1\times10^{22}}{1\times10^{23}+1\times10^{22}}\times440\text{ nm}=40\text{ nm}$

$x_\text{n0}=\frac{N_\text{a}}{N_\text{a}+N_\text{d}}W_0=\frac{1\times10^{23}}{1\times10^{23}+1\times10^{22}}\times440\text{ nm}=400\text{ nm}$

We see that $$x_\text{n0}=10x_\text{p0}$$ for this junction because $$N_\text{a}=10N_\text{d}$$.

Carrier Distribution

As mentioned above, the majority and minority carrier concentrations are $$p_\text{p0}$$ and $$n_\text{p0}$$, respectively, in the homogeneous p region and are $$n_\text{n0}$$ and $$p_\text{n0}$$, respectively, in the homogeneous n region.

The depletion layer is not completely devoid of free carriers. In thermal equilibrium, the electron and hole concentrations in the depletion layer are determined by (12-22) and (12-23) [refer to the electron and hole concentrations tutorial], respectively, with spatially varying band edges, $$E_\text{c}(x)$$ and $$E_\text{v}(x)$$; therefore, $$n_\text{p0}\ll{n}_0(x)\ll{n}_\text{n0}$$ and $$p_\text{n0}\ll{p}_0(x)\ll{p}_\text{p0}$$ for $$-x_\text{p0}\lt{x}\lt{x}_\text{n0}$$. The distributions of the electron and hole concentrations across a p-n junction in thermal equilibrium are illustrated in Figure 12-11(a).

A bias voltage can cause substantial changes in the minority carrier concentrations at $$x=-x_\text{p}$$ and $$x=x_\text{n}$$, where the edges of the depletion layer are located.

From Figures 12-9(b) and (c), we find that $$E_\text{Fc}(-x_\text{p})-E_\text{Fp}\approx{E}_\text{Fn}-E_\text{Fp}=eV$$ and $$E_\text{c}(-x_\text{p})\approx{E}_\text{cp}$$ at $$x=-x_\text{p}$$, and $$E_\text{Fn}-E_\text{Fv}(x_\text{n})\approx{E}_\text{Fn}-E_\text{Fp}=eV$$ and $$E_\text{v}(x_\text{n})\approx{E}_\text{vn}$$ at $$x=x_\text{n}$$. Therefore, using (12-41) and (12-42) [refer to the electron and hole concentrations tutorial], we find that the minority carrier concentrations at the edges of the depletion layer are

$\tag{12-101}n_\text{p}(-x_\text{p})=n_\text{p0}\text{e}^{eV/k_\text{B}T}$

$\tag{12-102}p_\text{n}(x_\text{n})=p_\text{n0}\text{e}^{eV/k_\text{B}T}$

At the edges of the depletion layer, the bias voltage creates the following changes in the minority carrier concentrations from their equilibrium values:

$\tag{12-103}\Delta{n}_\text{p}=n_\text{p}(-x_\text{p})-n_\text{p0}=n_\text{p0}(\text{e}^{eV/k_\text{B}T}-1)$

$\tag{12-104}\Delta{p}_\text{n}=p_\text{n}(x_\text{n})-p_\text{n0}=p_\text{n0}(\text{e}^{eV/k_\text{B}T}-1)$

As shown in Figure 12-11(b), a forward bias creates excess minority carriers on both p and n sides, $$\Delta{n}_\text{p}\gt0$$ and $$\Delta{p}_\text{n}\gt0$$ for $$V\gt0$$, as a result of minority carrier injection.

In contrast, as shown in Figure 12-11(c), a reverse bias depletes minority carriers on both p and n sides, $$\Delta{n}_\text{p}\lt0$$ and $$\Delta{p}_\text{n}\lt0$$ for $$V\lt0$$, as a result of minority carrier extraction.

Because of the diffusion of minority carriers, changes in the minority carrier concentrations caused by a bias voltage are not localized at the edges of the depletion layer. Instead, the minority carrier concentrations have the following spatially dependent variations across the diffusion regions:

\tag{12-105}\begin{align}n_\text{p}(x)-n_\text{p0}&=\Delta{n}_\text{p}\text{e}^{(x+x_\text{p})/L_\text{e}}\\&=n_\text{p0}(\text{e}^{eV/k_\text{B}T}-1)\text{e}^{(x+x_\text{p})/L_\text{e}},\qquad\text{for }x\lt-x_\text{p}\end{align}

\tag{12-106}\begin{align}p_\text{n}(x)-p_\text{n0}&=\Delta{p}_\text{n}\text{e}^{-(x-x_\text{n})/L_\text{h}}\\&=p_\text{n0}(\text{e}^{eV/k_\text{B}T}-1)\text{e}^{-(x-x_\text{n})/L_\text{h}},\qquad\text{for }x\gt{x}_\text{n}\end{align}

where $$L_\text{e}=(D_\text{e}\tau_\text{e})^{1/2}$$ is the electron diffusion length in the p region and $$L_\text{h}=(D_\text{h}\tau_\text{h})^{1/2}$$ is the hole diffusion length in the n region. Here $$\tau_\text{e}$$ is the lifetime of the minority electrons in the p region, and $$\tau_\text{h}$$ is the lifetime of the minority holes in the n region. They are not subject to the condition given in (12-54) [refer to the carrier recombination tutorial] for excess electrons and holes at the same location because they are the minority carrier lifetimes in separate regions on the two opposite sides of the junction. Therefore, $$\tau_\text{e}$$ and $$\tau_\text{h}$$ that define $$L_\text{e}$$ and $$L_\text{h}$$ here are independent of each other.

Because of charge neutrality in the diffusion regions, the concentrations of majority carrier also vary in space correspondingly:

$\tag{12-107}p_\text{p}(x)-p_\text{p0}=n_\text{p}(x)-n_\text{p0},\qquad\text{for }x\lt-x_\text{p}$

$\tag{12-108}n_\text{n}(x)-n_\text{n0}=p_\text{n}(x)-p_\text{n0},\qquad\text{for }x\gt{x}_\text{n}$

The relations in (12-105) - (12-108) that describe the carrier distributions are valid for p-N and P-n heterojunctions as well as for p-n homojunctions.

The distributions of majority and minority carrier concentrations for the cases when a junction is subject to a forward bias and when it is subject to a reverse bias are shown in Figures 12-11(b) and 12-11(c), respectively.

Both diffusion regions on p and n sides are clearly defined by the diffusion lengths of minority carriers because the spatial distributions of both minority and majority carriers are determined by the diffusion lengths of the minority carriers.

In the case when $$p_\text{p0}\gg{n}_\text{p0}$$ and $$n_\text{n0}\gg{p}_\text{p0}$$, as is the situation in many practical  junction devices, the equal amount of local changes in the majority and minority carrier concentrations in the diffusion regions is relatively insignificant for the total majority carrier concentration, but it can be substantial for the total minority carrier concentration.

Example 12-12

In this example, we consider the minority carrier concentrations at $$300\text{ K}$$ for the GaAs p-n homostructure described in Example 12-8. Find $$n_\text{p0}$$ and $$p_\text{n0}$$ first. Then, find the changes in the minority carrier concentrations, $$\Delta{n}_\text{p}$$ and $$\Delta{p}_\text{n}$$, at the two edges of the depletion layer for two different forward bias voltages of $$V=1\text{ V}$$ and $$V=V_0$$, respectively.

From Example 12-8, we know that $$p_\text{p0}\approx{N}_\text{a}=1\times10^{23}\text{ m}^{-3}$$, $$n_\text{n0}\approx{N}_\text{d}=1\times10^{22}\text{ m}^{-3}$$, and $$n_\text{i}=2.33\times10^{12}\text{ m}^{-3}$$. Because both p and n regions are nondegenerate, we can use the law of mass action given in (12-31) [refer to the electron and hole concentrations tutorial] for nondegenerate semiconductors to find that

$n_\text{p0}=\frac{n_\text{i}^2}{p_\text{p0}}=\frac{(2.33\times10^{12})^2}{1\times10^{23}}\text{ m}^{-3}=54.3\text{ m}^{-3}$

$p_\text{n0}=\frac{n_\text{i}^2}{n_\text{n0}}=\frac{(2.33\times10^{12})^2}{1\times10^{22}}\text{ m}^{-3}=543\text{ m}^{-3}$

At $$T=300\text{ K}$$, $$k_\text{B}T=0.0259\text{ eV}$$. For a forward bias voltage of $$V=1\text{ V}$$, we find from (12-103) and (12-104) that

$\Delta{n}_\text{p}=54.3\text{ m}^{-3}\times\left(\text{e}^{1/0.0259}-1\right)=3.18\times10^{18}\text{ m}^{-3}$

$\Delta{p}_\text{n}=543\text{ m}^{-3}\times\left(\text{e}^{1/0.0259}-1\right)=3.18\times10^{19}\text{ m}^{-3}$

For a forward bias voltage of $$V=V_0=1.209\text{ V}$$, we have

$\Delta{n}_\text{p}=54.3\text{ m}^{-3}\times\left(\text{e}^{1.209/0.0259}-1\right)=1\times10^{22}\text{ m}^{-3}=n_\text{n0}$

$\Delta{p}_\text{n}=543\text{ m}^{-3}\times\left(\text{e}^{1.209/0.0259}-1\right)=1\times10^{23}\text{ m}^{-3}=p_\text{p0}$

We find that through $$n_\text{p0}$$ and $$p_\text{n0}$$ are extremely small in this example, $$\Delta{n}_\text{p}$$ and $$\Delta{p}_\text{n}$$ are still quite substantial at a reasonable forward bias voltage because of their exponential dependence on bias voltage. We also find that $$\Delta{n}_\text{p}=n_\text{n0}$$ and $$\Delta{p}_\text{n}=p_\text{p0}$$ when the junction is forward biased at $$V=V_0$$.

Current-Voltage Characteristics

The electric current flowing in a semiconductor under bias consists of an electron current and a hole current each having both drift and diffusion components.

The total current is the vectoral sum of the individual current components, which may flow in different directions.

The total current is constant throughout the semiconductor under a constant bias voltage, but the electron and hole currents, as well as their drift and diffusion components, vary from one location to another because of spatial variations in the carrier distribution and in the electric field distribution.

In the depletion layer, there are drift and diffusion currents for both electrons and holes because in this layer a large electric field exists and the carrier concentration gradients for both electrons and holes are large.

In the diffusion regions, both majority and minority carrier diffusion currents are significant because both majority and minority carriers have large concentration gradients here. Furthermore, there is an appreciable majority carrier drift current because the majority carrier concentration is high through the electric field is small.

In homogeneous regions, almost the entire current is carried by majority carrier diffusion because both the carrier distribution gradients and the minority carrier concentration are negligibly small in these regions.

There are negligible generation and recombination of carriers in the depletion layer because the large electric field in the depletion layer sweeps the carriers across this layer very swiftly.

In this situation, the total electron current density, $$J_\text{e}(x)$$, and the total hole current density, $$J_\text{h}(x)$$, are constant for $$-x_\text{p}\lt{x}\lt{x}_\text{n}$$ across the depletion layer.

Consequently, the total current density in a semiconductor can be evaluated as

$\tag{12-109}J=J_\text{e}(-x_\text{p})+J_\text{h}(x_\text{n})$

where $$J_\text{e}(-x_\text{p})$$ is the minority carrier current density at the boundary between the depletion layer and the diffusion region on the p side, and $$J_\text{h}(x_\text{n})$$ is the minority carrier current density at the boundary between the depletion layer and the diffusion region on the n side.

Because the minority carrier currents in the diffusion regions are purely diffusive, we have, using (12-105) and (12-106),

$\tag{12-110}J_\text{e}(-x_\text{p})=eD_\text{e}\frac{\text{d}n_\text{p}}{\text{d}x}\rvert_{x=-x_\text{p}}=\frac{eD_\text{e}}{L_\text{e}}n_\text{p0}\left(\text{e}^{eV/k_\text{B}T}-1\right)$

$\tag{12-111}J_\text{h}(x_\text{n})=-eD_\text{e}\frac{\text{d}p_\text{n}}{\text{d}x}\rvert_{x=x_\text{n}}=\frac{eD_\text{h}}{L_\text{h}}p_\text{n0}\left(\text{e}^{eV/k_\text{B}T}-1\right)$

Consequently, the total current density varies with the bias voltage $$V$$ as

$\tag{12-112}J=J_\text{sat}(\text{e}^{eV/k_\text{B}T}-1)$

where $$J_\text{sat}$$ is the saturation current density given by

$\tag{12-113}J_\text{sat}=\frac{eD_\text{e}}{L_\text{e}}n_\text{p0}+\frac{eD_\text{h}}{L_\text{h}}p_\text{n0}$

The minority carrier currents given in (12-110) and (12-111) are contributed by the minority carrier injection across the depletion layer to the diffusion regions. They have the following ratio:

$\tag{12-114}\frac{J_\text{e}}{J_\text{h}}=\frac{D_\text{e}L_\text{h}}{D_\text{h}L_\text{e}}\frac{n_\text{p0}}{p_\text{n0}}=\left(\frac{D_\text{e}\tau_\text{h}}{D_\text{h}\tau_\text{e}}\right)^{1/2}\frac{n_\text{p0}}{p_\text{n0}}$

By using (12-79) for a p-N junction and (12-85) for a P-n junction, this ratio can be expressed as

$\tag{12-115}\frac{J_\text{e}}{J_\text{h}}=\frac{D_\text{e}L_\text{h}}{D_\text{h}L_\text{e}}\frac{n_\text{n0}}{p_\text{p0}}\frac{N_\text{cp}N_\text{vp}}{N_\text{cn}N_\text{vn}}\text{e}^{(E_\text{gn}-E_\text{gp})/k_\text{B}T}=\frac{D_\text{e}L_\text{h}}{D_\text{h}L_\text{e}}\frac{N_\text{d}}{N_\text{a}}\frac{N_\text{cp}N_\text{vp}}{N_\text{cn}N_\text{vn}}\text{e}^{(E_\text{gn}-E_\text{gp})/k_\text{B}T}$

This relation is valid for p-n homojunctions as well as for p-N and P-n heterojunctions.

In the case of a homojunction, $$J_\text{e}/J_\text{h}=D_\text{e}L_\text{h}N_\text{d}/D_\text{h}L_\text{e}N_\text{a}$$ because $$N_\text{cp}=N_\text{cn}$$, $$N_\text{vp}=N_\text{vn}$$, and $$E_\text{gn}=E_\text{gp}$$. Therefore, the relative importance of electron and hole injection is determined by the diffusion parameters of the minority carriers and the doping concentrations in the n and p regions, respectively.

It can be seen by examining the numerical values listed in Table 12-2 [refer to the current density in semiconductors tutorial] that $$D_\text{e}\gt{D}_\text{h}$$ because electrons have a higher mobility than holes in the same semiconductor. We thus reach the following important conclusion: unless the p side is much more heavily doped than the n side, the injection current through a homojunction is predominantly carried by the electrons injected from the n side into the p side.

In the case of a heterojunction, the exponential dependence on bandgap difference in (12-115) can be significant if $$\Delta{E}_\text{g}\gt{k}_\text{B}T$$. Because $$k_\text{B}T=25.9\text{ meV}$$ for $$T=300\text{ K}$$, this exponential dependence dominates in most practical heterojunctions where the value of $$\Delta{E}_\text{g}$$ is many times this value.

Consequently, $$J_\text{e}\gg{J}_\text{h}$$ for a p-N junction where $$E_\text{gn}\gt{E}_\text{gp}$$, whereas $$J_\text{h}\gg{J}_\text{e}$$ for a P-n junction where $$E_\text{gp}\gt{E}_\text{gn}$$.

An important conclusion is reached for heterojunctions: in a the case of a p-N junction the diffusion current is mainly contributed by the injection of electrons from the wide-gap n-type semiconductor to the narrow-gap p-type semiconductor, whereas in the case of a P-n junction it is mainly contributed by the injection of holes from the wide-gap p-type semiconductor to the narrow-gap n-type semiconductor.

This important characteristic can be understood from the observation in earlier discussions on energy bands that the energy barrier across a heterojunction is lower for majority carriers of the wide-gap semiconductor than for majority carriers of the narrow-gap semiconductor by the amount of the bandgap difference $$\Delta{E}_\text{g}$$ between the two semiconductors.

For a junction that has a cross-sectional area $$\mathcal{A}$$, the total current is $$I=J\mathcal{A}$$. Therefore, we have the following diode equation between the current and the bias voltage for an ideal p-n junction:

$\tag{12-116}I=I_\text{sat}(\text{e}^{eV/k_\text{B}T}-1)$

where $$I_\text{sat}=J_\text{sat}\mathcal{A}$$ is the saturation current.

This relation is valid for both forward and reverse bias conditions. It also applies to p-N and P-n heterojunctions.

Figure 12-12(a) shows the current-voltage characteristics, also known simply as the $$I-V$$ characteristics, described by (12-116) for an ideal diode.

Deviations from these ideal characteristics are found in all realistic p-n, p-N, and P-n junctions. Some important characteristics of realistic junctions are shown in Figure 12-12(b) and are summarized in the following.

1. At a certain critical reverse bias voltage, known as the breakdown voltage, $$V_\text{br}$$, a realistic junction breaks down with a sharp increase of reverse breakdown current.
2. The ideal diode equation is obtained by ignoring carrier recombination and generation in the depletion layer. When the effects of carrier recombination and generation in the depletion layer are considered, the reverse current does not saturate at $$-I_\text{sat}$$ but slightly increases in magnitude with the reverse bias voltage before a sudden change takes place at the breakdown voltage. Correspondingly, the forward current depends on the bias voltage with a modified factor in the exponent. Therefore, a realistic diode ahs the following current-voltage relation for $$V\gt{V}_\text{br}$$: $\tag{12-117}I=I_0(\text{e}^{eV/ak_\text{B}T}-1)$ where $$I_0$$ is a constant current different from $$I_\text{sat}$$ and $$a$$ is a factor that has a value between 1 and 2.
3. At high injection levels when the minority carrier concentrations in the diffusion regions become comparable to the majority carrier concentrations, (12-105) and (12-106) are not valid. A detailed analysis of this situation results in the current-voltage characteristic described by (12-117) with $$a=2$$ at these high injection levels.
4. The exponential rise of the current with forward bias voltage does not continue at high current levels as the voltage drop associated with finite resistivity in the neutral regions becomes significant at high currents.

Example 12-13

For the GaAs p-n homostructure described in Example 12-8, the p region is doped 10 times more heavily than the n region. Take the minority carrier lifetimes to be $$\tau_\text{e}=10\text{ ns}$$ for electrons in the p region and $$\tau_\text{h}=100\text{ ns}$$ for holes in the n region.

(a) By using the electron and hole diffusion coefficients for GaAs listed in Table 12-2 [refer to the current density in semiconductors tutorial], find the diffusion lengths that define the diffusion regions on the p and n sides, respectively.

(b) Find the saturation current density.

(c) Compare the relative importance of electron and hole injection currents.

(d) Consider a junction that has a 100 μm x 100 μm cross section. Find the saturation current in reverse bias. Use the ideal diode equation to find the current under forward bias voltages of $$V=1\text{ V}$$ and $$V=V_0$$, respectively.

(a)

From Table 12-2 [refer to the current density in semiconductors tutorial], we find that $$D_\text{e}=220\text{ cm}^2\text{ s}^{-1}=2.2\times10^{-2}\text{ m}^2\text{ s}^{-1}$$ and $$D_\text{h}=10\text{ cm}^2\text{ s}^{-1}=1\times10^{-3}\text{ m}^2\text{ s}^{-1}$$. Therefore,

$L_\text{e}=(D_\text{e}\tau_\text{e})^{1/2}=(2.2\times10^{-2}\times10\times10^{-9})^{1/2}\text{ m}=14.8\text{ μm}$

$L_\text{h}=(D_\text{h}\tau_\text{h})^{1/2}=(1\times10^{-3}\times100\times10^{-9})^{1/2}\text{ m}=10\text{ μm}$

(b)

From Example 12-12, we find that $$n_\text{p0}=54.3\text{ m}^{-3}$$ and $$p_\text{n0}=543\text{ m}^{-3}$$. Therefore, from (12-113), the saturation current density is

\begin{align}J_\text{sat}&=\left(\frac{1.6\times10^{-19}\times2.2\times10^{-2}}{14.8\times10^{-6}}\times54.3+\frac{1.6\times10^{-19}\times1\times10^{-3}}{10\times10^{-6}}\times543\right)\text{ A m}^{-2}\\&=2.16\times10^{-14}\text{ A m}^{-2}\end{align}

(c)

By using the relation in (12-114), we find that

$\frac{J_\text{e}}{J_\text{h}}=\frac{D_\text{e}L_\text{h}}{D_\text{h}L_\text{e}}\frac{n_\text{p0}}{p_\text{n0}}=\frac{2.2\times10^{-2}\times10}{1\times10^{-3}\times14.8}\times\frac{54.3}{543}=1.49$

Therefore, despite the fact that the doping concentrations in the p region is 10 times that in the n region and $$\tau_\text{h}$$ is 10 times $$\tau_\text{e}$$, we still find that the electron injection current is 1.49 times the hole injection current. The electron current would be even more important if the p region were not so heavily doped compared to the n region or if the hole lifetime were not so much longer than the electron lifetime.

(d)

We have a cross-sectional area of $$\mathcal{A}=100\text{ μm}\times100\text{ μm}=1\times10^{-8}\text{ m}^2$$. Thus the saturation current is $$I_\text{sat}=J_\text{sat}\mathcal{A}=2.16\times10^{-22}\text{ A}$$. At a sufficiently high reverse bias voltage of $$V\ll-k_\text{B}T/e=-25.9\text{ mV}$$ at $$T=300\text{ K}$$, $$I=-I_\text{sat}=-2.16\times10^{-22}\text{ A}$$. At a forward bias voltage of $$V=1\text{ V}$$, we find that

$I=2.16\times10^{-22}\text{ A}\times(\text{e}^{1/0.0259}-1)=12.7\text{ μA}$

At a forward bias voltage of $$V=V_0=1.209\text{ V}$$, the current is

$I=2.16\times10^{-22}\text{ A}\times(\text{e}^{1.209/0.0259}-1)=40.5\text{ mA}$

Comparing these results, we find that the current saturates with voltage in reverse bias but increases very quickly with voltage in forward bias.

Capacitance

There are two types of capacitance associated with a p-n, p-N, or P-n junction:

1. The junction capacitance, $$C_\text{j}$$, also known as the depletion-layer capacitance
2. The diffusion capacitance, $$C_\text{d}$$, also known as the charge-storage capacitance.

In a junction under reverse bias, only the junction capacitance is important. In a junction under forward bias, however, the diffusion capacitance dominates.

The depletion layer acts as a capacitor by holding negative space charges on the p side and positive space charges on the n side of the following magnitude:

$\tag{12-118}Q=eN_\text{a}x_\text{p}\mathcal{A}=eN_\text{d}x_\text{n}\mathcal{A}=e\frac{N_\text{a}N_\text{d}}{N_\text{a}+N_\text{d}}W\mathcal{A}$

where $$\mathcal{A}$$ is the cross-sectional area of the junction and $$W$$ is the width of the depletion layer.

By using (12-100) and (12-118), we find that the junction capacitance associated with the depletion layer is given by

$\tag{12-119}C_\text{j}=|\frac{\text{d}Q}{\text{d}V}|=\frac{\epsilon\mathcal{A}}{W}$

Because the width of the depletion layer $$W$$ decreases with forward bias but increases with reverse bias, the junction capacitance increases when the junction is subject to a forward bias voltage but decreases when it is subject to a reverse bias voltage.

Because the diffusion capacitance, $$C_\text{d}$$, is associated with the storage of minority carrier charges in the diffusion region, it exists only when a junction is under forward bias.

This capacitance is a complicated function of the minority carrier lifetime and the modulation frequency of the bias voltage, but it is directly proportional to the injection current.

When a junction is under forward bias, $$C_\text{d}$$ can be significantly larger than $$C_\text{j}$$ at high injection currents though $$C_\text{j}$$ can be already large in this situation.

When a junction is under reverse bias, $$C_\text{j}$$ is the only capacitance of significance though it can be small.

Consequently, the capacitance of a junction can be substantially smaller when it is under reverse bias than when it is under forward bias.

Example 12-14

A GaAs p-n homojunction as described in Example 12-8 and considered in Examples 12-11 to 12-13 has a 100 μm x 100 μm cross section. Consider the junction in thermal equilibrium without bias at $$300\text{ K}$$. Find the amount of the positive and negative space charges stored in the depletion layer. Find the junction capacitance.

The cross-sectional area of the junction is $$\mathcal{A}=100\text{ μm}\times100\text{ μm}=1\times10^{-8}\text{ m}^2$$. An equal amount of positive and negative space charges is stored on the n and p sides, respectively, of the junction in the depletion layer. By using $$N_\text{a}=1\times10^{23}\text{ m}^{-3}$$ given in Example 12-8 and $$x_\text{p0}=40\text{ nm}$$ found in Example 12-11, we find from (12-118) that

$Q=eN_\text{a}x_\text{p0}\mathcal{A}=1.6\times10^{-19}\times1\times10^{23}\times40\times10^{-9}\times1\times10^{-8}\text{ C}=6.4\text{ pC}$

From Example 12-11, we know that $$\epsilon=13.18\epsilon_0$$ for GaAs and $$W_0=440\text{ nm}$$ for the junction under consideration. Therefore, from (12-119), the junction capacitance is

$C_\text{j}=\frac{13.18\times8.854\times10^{-12}\times1\times10^{-8}}{440\times10^{-9}}\text{ F}=2.65\text{ pF}$

The next tutorial covers the topic of radiative recombination