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Chapter 1
Conduction in Semiconductors
1.1 Introduction
All solid-state devices, e.g. diodes and transistors, are fabricated from materials known as semi- conductors. In order to understand the operation of these devices, the basic mechanism of how currents flow in them must be understood. This chapter covers the fundamentals of conduction in semiconductors. The chapter is not intended to be an extensive introduction to the area of solid-state physics. Instead, only those topics which lead to a better understanding of the macro- scopic properties of semiconductors are covered. The mechanisms of conduction in a metal and in a semiconductor are compared. The effects of impurities on conduction in semiconductors are dis- cussed. The formation of a p-n semiconductor junction is described and its conduction properties are discussed.
1.2 Classification of Conductors
Figure 1.1 illustrates a two-dimensional view of an atom that is called the Bohr model of the atom. It consists of a positively charged nucleus and a system of negatively charged electrons which rotates around the nucleus. In a neutral atom, the total charge is zero. This means that the positive charge on the nucleus is equal to the total negative charge on the electrons. The electrons are bound to the nucleus by the forces of attraction between oppositely charged particles. They are arranged systematically in layers called shells. The closer a shell is to the nucleus, the more tightly bound are the electrons in that shell to the atom. The shell closest to the nucleus can contain no more than two electrons. The outer shell can have no more than eight. The number in the shells in between is determined by the laws of quantum mechanics. The outermost shell in an atom contains what are called the valence electrons. These govern the nature of chemical reactions of the elements. In addition, they play a large part in determining the electrical behavior of the elements and the crystalline structure of solids. The metallic elements tend to have one, two, or three valence electrons. The nonmetals have five, six, or seven. The inert gases have eight. The class of elements which have four valence electrons is called semiconductors. If a valence electron escapes its parent atom, it becomes free to move about. The parent atom then has a net positive charge and is called an ionized atom or an ion. If an electric field is applied to a material, the free electrons have forces exerted on them which cause them to move. This constitutes the flow of a current in the material that is called a conduction current or a drift current.
1
2 CHAPTER 1. CONDUCTION IN SEMICONDUCTORS
Figure 1.1: Two-dimensional Bohr model of an atom showing the nucleus and three shells.
Depending on the number n of free electrons per unit volume in a solid, the material is classified as being a good conductor, a semiconductor, or an insulator. For a good conductor, n is very large and is independent of temperature. A typical value is n ' 1028 per m^3. For an insulator at ordinary temperatures, n is much smaller and has a typical value n ' 107 per^ m^3. For a semiconductor, it lies between the values for a good conductor and an insulator and is a function of the temperature. Silicon is an important semiconductor for which n ' 1. 5 × 1016 per m^3 at room temperature (T = 300 K).
1.3 Conduction in Metals
Metals are classified as good conductors. The valence electrons are so loosely bound to the atoms that they are free to move about in the conductor. Fig. 1.2 shows a two-dimensional illustration of the atoms in a metal with the free electrons distributed randomly among the immobile ions. The free electrons can be visualized as molecules of a gas that permeate the region between the ions. Analogous to the random motion of molecules in a gas, thermal energy causes the free electrons to be in continuous random motion. Observation of an individual electron would reveal that its direction of motion changes randomly after each collision with an ion. Because the direction of motion of each electron is random, the average number of electrons passing through any area per unit time is zero. Thus the average current flow in the metal is zero.
1.3.1 Drift Velocity
If an electric field
E ( V/ m) is applied to a metal, an electrostatic force is exerted on the free electrons which causes a conduction current to flow. (The arrow indicates a vector quantity.) The force on an individual electron is given by
F = −q
E ( N), where q is the electronic charge (q = 1. 602 × 10 −^19 C). The electrostatic forces cause the electrons to be accelerated in a direction opposite to that of the applied field. Fig. 1.3 illustrates the path that an individual electron might take under the influence of the electric field. If the electron did not collide with the bound ions, its velocity would increase indefinitely. However, energy is lost with each collision so that the average
4 CHAPTER 1. CONDUCTION IN SEMICONDUCTORS
per m^3. Because the charge per electron is −q, it follows that the free charge per unit volume in the metal is given by ρ = −nq (1.2)
1.3.3 Current Density
The current density
J ( A/ m^2 ) in a conductor is defined as the current per unit area flowing in a particular direction. To relate the current density in a conductor to the drift velocity of the moving charges, consider a section of wire of length ∆c in which a current I is flowing. This is illustrated in Fig. 1.4. The charge in the section is ∆Q = ρ∆V = ρS∆c, where ρ is the charge density and S is the cross-sectional area of the wire. Let ∆t be the time required for the charge in the section to move the distance ∆c. The velocity of the charge is −→v = ba∆c/∆t, where ba is a unit vector in the direction of current flow. The current I flowing in the area S is I = ∆Q/∆t. It follows that the current density
J can be related to the drift velocity −→v as follows:
−→ J = ba
I
S
= ba
∆Q
S∆t
= ba ρS∆c S∆t
= ρ
μ ba ∆c ∆t
= ρ−→v (1.3)
Figure 1.4: Section of wire of length ∆c in which a current I flows.
1.3.4 Conductivity
Using Eqs. (1.1) through (1.3), we can relate the current density
J to the electric field
E in a metal as follows: −→ J = ρ−→v = (−nq)
−μe
E
= nqμe
E = σ
E (1.4)
(Note that two minus signs have canceled so that
J is in the direction of
E .) This equation defines the conductivity σ ( Ω−^1 m−^1 ) of the metal. It is given by
σ = nqμe (1.5)
Because n is independent of temperature in a metal, it follows that the decrease in electron mobility μe with temperature causes the conductivity σ to decrease with temperature.
1.4. CONDUCTION IN INTRINSIC SEMICONDUCTORS 5
Example 1 Aluminum has three valence electrons per atom, an atomic weight of 0 .02698 kg/ mol, a density of 2700 kg/ m^3 , and a conductivity of 3. 54 × 107 S/ m−^1. Calculate the electron mobility in aluminum. Assume that all three valence electrons in each atom are free.
Solution. Recall from introductory chemistry that a mole of any substance is a quantity equal to its atomic weight and contains a number of molecules equal to Avogadro’s number which is
02 × 1023. It follows that the number of aluminum atoms per m^3 is
02 × 1023 atoms mol
×
mol kg
× 2700
kg m^3
= 6. 024 × 1028
atoms m^3
Thus the electron density in the aluminum is n = 3 × 6. 024 × 1028 = 1. 807 × 1029 per m^3. From Eq. (1.5), the mobility is given by
μe = σ nq
3. 54 × 107
6. 024 × 1028 × 1. 602 × 10 −^19
= 3. 67 × 10 −^3 m^2 V−^1 s−^1
1.3.5 Resistance
Consider the section of wire illustrated in Fig. 1.4. A conduction current I is flowing in the wire so that the current density is
J = baI/S, where ba is a unit vector in the direction of current flow and S is the cross-sectional area. Let V be the voltage drop across the section of length ∆c so that the electric field in the section is
E = baV /∆c. With the aid of Eq. (1.4), we can write
J = baσV /∆c.
By equating the two relations for
J , we obtain −→ J = ba
I
S
= ba σV ∆c
This equation can be solved for the resistance R of the section of wire to obtain
R =
V
I
∆c σS
Thus the resistance is directly proportional to the length of the wire and inversely proportional to its area. Because the conductivity σ decreases with temperature, it follows from this equation that R increases with temperature. In most metals, the resistance increases linearly with temperature.
Example 2 The conductivity of copper is 5. 8 × 107 S/ m. If a 1 m length of copper wire has a resistance of 1 Ω, what is the thickness of the wire? Assume a circular cross section.
Solution. Let d be the diameter of the wire. Using Eq. (1.7), we can write S = π (d/2)^2 = ∆c/σR = 1/
5. 8 × 107 × 1
. Solution for d yields d = 0.148 mm.
1.4 Conduction in Intrinsic Semiconductors
Semiconductors are the class of elements which have four valence electrons. Two important semi- conductors are germanium (Ge) and silicon (Si). Early solid-state electronic devices were fabricated almost exclusively from germanium, whereas modern devices are fabricated almost exclusively from silicon. Gallium arsenide (GaAs) is a semiconductor compound made up of gallium, which has three
1.4. CONDUCTION IN INTRINSIC SEMICONDUCTORS 7
free electrons to drift to the left. Fig. 1.7 illustrates the drift of holes. In effect, a hole drifts to the right when a bound valence electron shifts to the left from one atom to another. The arrows in the figure point from the new position of a hole to its former position, i.e. in the direction of movement of the bound electrons in the lattice. The movement of holes may be likened to the movement of bubbles of air in water, where the water represents the bound electrons and the bubbles represent the holes. The movement of a bubble in one direction is really the result of a movement of water in the opposite direction. In summary, the flow of current in the semiconductor is the result of the flow of two components. One component is the flow of free electrons in one direction. The other component is the flow of the absence of bound electrons in the other direction. Because of the opposite charge polarities, the electron current and the hole current add to produce the total conduction current.
Figure 1.6: Illustration of the drift of free electrons under the application of an external electric field.
1.4.1 Recombinations
Because hole-electron pairs are continually created by thermal agitation of a semiconductor lattice, it might seem that the number of holes and free electrons would continually increase with time. This does not happen because free electrons are continually recombining with holes. At any temperature, a stable state is reached when the creation rate of hole-electron pairs is equal to the recombination rate. The mean lifetime τ (^) n ( s) of a free electron is the average time that the electron exists in the free state before recombination. The mean lifetime τ (^) p ( s) for the hole is defined similarly. In the intrinsic semiconductor, τ (^) n is equal to τ (^) p because the number of free electrons must be equal to the number of holes. However, the addition of an impurity to the semiconductor lattice can cause the mean lifetimes to be unequal.
1.4.2 Intrinsic Concentration
Denote the number of free electrons per m^3 in a semiconductor by n and the number of holes per m^3 by p. In an intrinsic semiconductor, the hole concentration must equal the electron concentration.
8 CHAPTER 1. CONDUCTION IN SEMICONDUCTORS
Figure 1.7: Illustration of the drift of holes under the application of an external electric field.
In this case we write n = p = ni, where ni is called the intrinsic concentration. It can be shown that ni can be written
ni = n 0 T 3 /^2 exp
μ −VG 2 VT
where n 0 is a constant, T is the absolute temperature, VG is the semiconductor bandgap voltage, and VT is the thermal voltage. (The bandgap voltage multiplied by q represents the minimum energy required to cause a bound valence electron to become a free electron.) For silicon, the bandgap voltage at T = 300 K has the value VG = 1.11 V. The thermal voltage is related to the temperature by
VT = kT q
where k is the Boltzmann constant (k = 1. 381 × 10 −^23 J/ K). At T = 300 K, the thermal voltage has the value VT = 0.0259 V.
Example 3 The initial temperature of a specimen of silicon is T 1 = 300 K. By what factor does the intrinsic concentration ni increase if the temperature increases by 10 ◦C ( 18 ◦F)? Assume the bandgap voltage at both temperatures is VG = 1.11 V.
Solution. Let T 1 = 300 K be the initial temperature and T 2 = 310 K the final temperature. At 300 K, the thermal voltage is VT 1 = 0.0259 V. At 310 K, it is VT 2 = (1. 380 × 10 −^23 × 310)/ 1. 602 × 10 −^19 = 0.0267 V. The factor by which ni increases is calculated from Eq. (1.8) as follows:
ni 2 ni 1
(T 2 )^3 /^2 exp (−VG/ 2 VT 2 ) (T 1 )^3 /^2 exp (−VG/ 2 VT 1 )
μ 310 300
exp
μ − 1
- 0267
1.4.3 Conductivity
When an electric field
E is applied to an intrinsic semiconductor, the free electrons drift with a ve- locity −→v (^) e = −μe
E , where μe is the electron mobility. The holes drift with a velocity −→v (^) h = +μh
E ,
10 CHAPTER 1. CONDUCTION IN SEMICONDUCTORS
Figure 1.8: Two-dimensional illustration of the crystal lattice of an n-type semiconductor
Hole-electron pairs are continually formed by thermal agitation of the lattice in an n-type semiconductor. Because of the large number of donor electrons, there are many more free electrons available for recombination with the holes. This decreases the mean lifetime for the holes which decreases the number of holes in the n-type semiconductor compared to the intrinsic semiconductor. For this reason, the current due to the flow of holes in an n-type semiconductor is often neglected in calculations. It is important to understand that a donor atom is electrically neutral if its fifth valence electron does not become a free electron in the lattice. If the fifth electron becomes a free electron, the number of protons in the atom is greater than the number of electrons by one. In this case, the donor atom becomes a bound positively charged ion.
1.5.2 p-Type Semiconductor
A p-type semiconductor is produced by adding an acceptor impurity such as gallium, boron, or indium to an intrinsic semiconductor. Each acceptor atom has three valence electrons. When an acceptor atom replaces an atom in the crystal lattice, there are only three valence electrons shared with the surrounding atoms. This leaves a hole as illustrated in Fig. 1.9. The number of holes created by the acceptor atoms is much greater than the number of free electrons and holes in the intrinsic semiconductor. This makes the conductivity of the p-type semiconductor much greater that of the intrinsic semiconductor. Because the number of holes is far greater than the number of electrons, the holes are the majority carriers. The semiconductor is called p-type because the majority carriers have a positive charge. Hole-electron pairs are continually formed by thermal agitation of the lattice in a p-type semicon- ductor. Because of the large number of holes, there are many more holes available for recombination with the free electrons. This decreases the mean lifetime for the free electrons which decreases the number of electrons in the p-type semiconductor compared to the intrinsic semiconductor. For this reason, the current due to the flow of free electrons in a p-type semiconductor is often neglected in calculations. It is important to understand that an acceptor atom is electrically neutral if the hole created by the absence of its fourth valence electron is not filled by an electron from an adjacent silicon atom.
1.5. N-TYPE AND P-TYPE SEMICONDUCTORS 11
Figure 1.9: Two-dimensional illustration of the crystal lattice of a p-type semiconductor.
Once an electron fills the hole, the number of electrons in that atom is greater than the number of protons by one. In this case, the acceptor atom becomes a bound negatively charged ion.
1.5.3 Mass-Action Law
In an intrinsic semiconductor, we have noted that the electron concentration and the hole concen- tration are both equal to the intrinsic concentration, i.e. n = p = ni. If this were not true, the material would not be electrically neutral. We have seen that adding an n-type impurity to the semiconductor increases n and decreases p. Similarly, adding a p-type impurity increases p and decreases n. It can be shown that the product of n times p is a constant independent of the doping type and the doping level. The product is given by
np = n^2 i (1.12)
where ni is given by Eq. (1.8). This relation is called the mass-action law. To understand this equation, consider an intrinsic semiconductor in which n = p = ni. Assume that donors with the density DA = ni are added to the semiconductor at t = 0. This initially doubles the total number of free electrons, which causes the recombination rate with the holes to double. This causes the hole density to drop from ni to ni/ 2. The initial free electron density is 2 ni which drops to 2 ni − ni/ 2 after the increase in recombinations. Thus the product of the electron and hole concentrations is (2ni − ni/2) × ni/2 = 3n^2 i / 4. Now suppose that the number of donors is increased by the factor N , where N is large. In this case, the product becomes (N ni − ni/N ) × ni/N = n^2 i
1 − 1 /N 2
' n^2 i. Although this is not an exact proof, it illustrates the basic mechanism. A more detailed proof requires an involved thermodynamic analysis.
1.5.4 Electrical Neutrality
An intrinsic semiconductor is electrically neutral, i.e. there is no net charge stored. The addition of n-type or p-type impurities does not change this. To state this mathematically, let ND be the number of donor atoms per m^3 and NA the number of acceptor atoms per m^3. We assume that all donor atoms and all acceptor atoms are ionized so that there are ND bound positive charges per
1.5. N-TYPE AND P-TYPE SEMICONDUCTORS 13
p '
n^2 i ND − NA
Similarly if NA > ND, the semiconductor becomes a p-type and Eqs. (1.16) and (1.17) hold if NA is replaced by NA − ND. p = n + NA − ND ' NA − ND (1.20)
n '
n^2 i NA − ND
1.5.5 Conductivity
The conduction current density in an intrinsic semiconductor is given by Eq. (1.10). In a doped semiconductor, it is given by −→ J = (nμe + pμh) q
E = σ
E (1.22)
where n is the electron concentration and p is the hole concentration. This equation defines the conductivity σ. It is given by σ = (nμe + pμh) q (1.23)
For an intrinsic semiconductor, n = p = ni and this equation reduces to Eq. (1.11).
1.5.6 Diffusion Current
In an n-type or a p-type semiconductor, it is possible to have a component of current that is not a conduction current. This current is due to the non-uniform density of free electrons or holes and is called a diffusion current. It is not possible to have a diffusion current in a metal. To achieve a non-uniform density of free electrons or holes, the doping concentration in the semiconductor is not constant, i.e. it is a function of position. Such a concentration is called a graded doping concentration. Figure 1.10 illustrates a semiconductor in which the concentration of holes is a function of the coordinate z, i.e. p = p (z). In addition, p (z) is a decreasing function of z. Consider the plane defined by z = z 1. Because the number of holes to the left of the plane is greater than the number to the right, it is reasonable to expect that the random motion of the holes due to thermal energy would cause more holes to migrate from left to right than from right to left. Thus there is a net current flow across the plane from left to right. This current is called a diffusion current. In general, the hole diffusion current density is given by −→ J (^) p = −qDh∇p (1.24)
where Dh is the hole diffusion constant and ∇p is the gradient or directional derivative of p. If bx, by, and bz, respectively, are unit vectors in the x, y, and z directions, ∇p is given by
∇p = bx
∂p ∂x
∂p ∂y
∂p ∂z
For the case illustrated in Fig. 1.10, p is a function of z only so that ∇p = bzdp/dz. (The total derivative is used because p is a function of z only.) Because p (z) is a decreasing function of z in the figure, it follows that dp/dz < 0. This makes the direction of the diffusion current in the +z direction.
14 CHAPTER 1. CONDUCTION IN SEMICONDUCTORS
Figure 1.10: Illustration of a semiconductor in which the hole concentration is a function of the coordinate z.
In an n-type semiconductor that has a non-uniform density of free electrons, the electron diffu- sion current is given by −→ J (^) n = qDe∇n
1.25 where n is the electron concentration function and De is the electron diffusion constant. The diffusion constants are related to the mobility constants by the relation
De μe
Dh μh
= VT (1.26)
where VT is the thermal voltage given by Eq. (1.9). This relation is known as the^ Einstein equation.
1.5.7 Total Current
In general, the total current in a semiconductor is written as the sum of the electron and hole conduction currents and the electron and hole diffusion currents. It is given by
−→ J = σ
E + q(De∇n − Dh∇p) (1.27)
where
E is the electric field intensity and σ is the conductivity given by Eq. (1.23). In an open- circuited semiconductor, the equilibrium current density must be zero. If we set
J = 0, the above equation predicts an electric field in the open-circuited semiconductor given by
−→ E = q σ
(Dh∇p − De∇n) =
VT
nμe + pμh
(μh∇p − μe∇n) (1.28)
where Eqs. (1.23) and (1.26) have been used. As an application of the preceding results, let us calculate the voltage difference between two points in a semiconductor caused by a graded doping concentration. Consider a p-type semicon- ductor in which the hole concentration is a function of the coordinate z, i.e. p = p (z). We assume that there are no external sources connected to the semiconductor so that the equilibrium current density is zero. The semiconductor is illustrated in Fig. 1.11. Let the voltage at z = z 1 be V 1 and the voltage at z = z 2 be V 2. The hole concentrations are labeled p 1 and p 2. We assume that p (z)
16 CHAPTER 1. CONDUCTION IN SEMICONDUCTORS
1.6 The Open-Circuited p-n Junction
A p-n junction is the junction between an n-type semiconductor and a p-type semiconductor. It is fabricated by introducing donor impurities into one side of an intrinsic semiconductor crystal and acceptor impurities into the other side. The transition between the two regions occurs in a very small distance, typically 0. 5 μm. Fig. 1.12 illustrates the cross section of a p-n junction where the donor ions are represented by positive signs and the acceptor ions are represented by negative signs. Initially, we assume that the only charge carriers in the n-type side are free electrons and that the only charge carriers in the p-type side are holes.
Figure 1.12: Diagram of a p-n junction with the width of the depletion region greatly exaggerated.
Because of the unequal electron concentrations and unequal hole concentrations on the two sides of the junction, a diffusion current consisting of both holes and free electrons will flow across the junction. (The diffusion process is similar to the diffusion of different gases between two glass jars joined at the mouths.) Holes diffuse out of the p-type side and into the n-type side and free electrons diffuse out of the n-type side and into the p-type side. This causes the p-type side to become negatively charged and the n-type side to become positively charged. The charges cause an electric field to build up across the junction which is directed from the n-type side to the p-type side. The polarity of the electric field is such that the force it exerts on the holes and free electrons opposes the diffusion process. Equilibrium is reached when the force exerted on the charge carriers by the electric field is equal to the diffusion force. Let us now consider what happens when thermal agitation of the semiconductor lattice produces a hole-electron pair in the region near the junction. The electric field directed from the n-type side to the p-type side exerts a force on the free electron and causes it to be swept to the n-type side. Similarly, the hole is swept to the p-type side. The directions that the charges move are opposite to those due to the diffusion process. When equilibrium is reached, the net number of both electrons and holes crossing the junction is zero.
1.6.1 Depletion Region
Because no free electrons or holes can exist is the region about the junction, there are no mobile charges to neutralize the ions in this region. This is illustrated in Fig. 1.12. The ions on the
1.6. THE OPEN-CIRCUITED P-N JUNCTION 17
n-type side have a positive charge on them and those on the p-type side have a negative charge. These charges are called uncovered charges. The region about the junction in which the uncovered charges exist is called the depletion region. Other names for this are the space-charge region and the transition region. Fig. 1.13a illustrates the plot of the net uncovered charge density in the p-n junction as a function of distance from the junction. The charge distribution is called a dipole distribution because the charge on one side of the junction is the negative of the charge on the other side. The uncovered charges on each side of the junction can be thought of as the charges on the plates of a parallel plate capacitor as shown in Fig. 1.13b.
Figure 1.13: (a) Plot of the charge density as a function of distance from the junction. (b) Parallel plate capacitor analog of the charge distribution.
Because of charge neutrality, the total uncovered charge on the n-type side of the depletion region must be equal to the negative of the total uncovered charge on the p-type side. If the n and p concentrations are equal, it follows that the widths of the uncovered charge regions on the two sides of the junction must be equal. Now, suppose the p concentration is increased while holding the n concentration constant. Charge neutrality requires the width of the p-type side of the depletion region to decrease if the total uncovered charge is to remain constant. Similarly, if the n concentration is increased while holding the p concentration constant, the width of the n-type side must decrease. We conclude, in general, that increasing either p or n or both decreases the total width w of the depletion region illustrated in Fig. 1.13a. This has an important effect on the reverse-bias breakdown characteristics of a junction. This is discussed in the following chapter.
1.6.2 Built-In Potential
Because there is an electric field in the depletion region of a p-n junction that is directed from the n-type side to the p-type side, it follows that there is a difference in potential or voltage across the junction. This voltage difference is called the built-in potential or the contact potential. It can be calculated from either Eq. (1.31) or Eq. (1.32). Let us use Eq. (1.32). In this equation, p 1 is the hole concentration in the p-type side and p 2 is the hole concentration in the n-type side. By Eq. (1.16) the hole concentration in the p-type side is p 1 ' NA, where NA is the acceptor concentration per m^3. By Eq. (1.15) the hole concentration in the n-type side is p 2 ' n^2 i /ND, where ni is the intrinsic concentration per m^3 and ND is the donor concentration per m^3. It follows from Eq.
1.8. THE BIASED P-N JUNCTION 19
the p-n junction, both metal-semiconductor junctions exhibit a built-in potential. The algebraic sum of the three built-in potentials is zero so that there is no net voltage to cause a current to flow in the wire and the second law of thermodynamics is not violated. The characteristics of metal-semiconductor junctions differ primarily from those of p-n junctions in two ways. First, they conduct current well in both directions. Second, the built-in potential does not change when a current flows through the junction. Junctions which have these properties are called ohmic junctions or nonrectifying junctions. We will see in the following that the p-n junction does not have the properties of an ohmic junction. (Not all metal-semiconductor junctions are ohmic. The Schottky-barrier diode is an example which is described in the following chapter.)
1.8 The Biased p-n Junction
1.8.1 Reverse Bias
Figure 1.15 shows a p-n junction with a dc source connected to it. The polarity of the source is chosen so that the positive terminal is connected to the n-type side and the negative terminal is connected to the p-type side. The current that flows across the junction consists of two components, a diffusion current caused by unequal carrier concentrations on the two sides of the junction and a conduction current caused by the electric field across the junction. With VS = 0, these two currents exactly cancel each other so that the net current is zero.
Figure 1.15: Reverse biased p-n junction.
Now let us examine what happens when VS > 0. Because negative charge is attracted by a positive voltage and positive charge is attracted by a negative voltage, both the free electrons in the n-type side and the holes in the p-type side are pulled away from the junction. This causes the width of the depletion region to increase so that there are more uncovered charges on each side of the junction. This is illustrated in Fig. 1.15 compared to Fig. 1.12. The potential across the junction which opposes diffusion is increased by the applied bias to the value VB + VS , where VB
20 CHAPTER 1. CONDUCTION IN SEMICONDUCTORS
is the built-in potential. This is greater than VB so that the electric field across the junction is increased. Because the diffusion force on the charge carriers is opposed by the force exerted by this electric field, it follows that the diffusion current is decreased by the applied voltage. The diffusion current approaches zero as VS is increased. Although the diffusion current goes to zero, the conduction current due to thermally produced hole-electron pairs in the depletion region continues to flow across the junction. When such a hole-electron pair is generated, the electric field across the junction causes the electron to be swept to the n-type side and the hole to be swept to the p-type side. This causes a very small current to flow in the external circuit in the direction indicated in Fig. 1.15. Because the current is so small, the junction is said to be reverse biased.
1.8.2 Forward Bias
Figure 1.16 shows the p-n junction with the polarity of the dc source chosen so that the positive terminal is connected to the p-type side and the negative terminal is connected to the n-type side. Because positive charge is repelled by a positive voltage and negative charge is repelled by a negative voltage, both the free electrons in the n-type side and the holes in the p-type side are forced toward the junction. This causes the width of the depletion region to decrease so that there are fewer uncovered charges on each side of the junction. This is illustrated in Fig. 1.16 compared to Fig. 1.12. The potential across the junction which opposes diffusion is decreased by the applied bias to the value VB − VS , where VB is the built-in potential. This is less than VB so that the electric field across the junction is decreased. This decreases the force which opposes diffusion so that the diffusion current increases rapidly as VS is increased. Because a large current flows, the junction is said to be forward biased.
Figure 1.16: Forward biased p-n junction.
It might seem that the potential across the junction which opposes diffusion could be made to go to zero by increasing VS. Should this happen, the width of the depletion region would go
