ABSTRACT
Consider a hypothetical experiment in which we have two crystals of a semiconductor: one is n-type and the other is p-type The band diagrams for these before the materials are brought together are shown in Figure 51
When we bring the two semiconductors together to form a p-n junction, electrons start flowing from the n-side to the p-side to minimize the concentration gradient This will continue until a new common Fermi energy level (EF) is established Since electrons are negatively charged, the direction of the current associated with their motion is opposite to the direction of their actual motion The direction of the current associated with the diffusion of electrons is from the p-side to the n-side of the p-n junction (Figure 52)
A p-doped material has a much higher concentration of holes on the p-side than that on the n-side Thus, when a p-n junction is formed, the holes diffuse from the p-side to the n-side Since holes are positively charged particles, the current induced due to their movement is in the same direction as their motion (Figure 52)
Note that a concentration gradient of dopant atoms exists because there is a significant concentration of donor atoms on the n-side and almost no donor atoms on the p-side However, there is no diffusion of donor atoms or ions, as their mobility is very low at or near room temperature Thus, here we consider the diffusion of holes and electrons only
Since diffusion is concentration gradient driven, we expect that the flow of electrons and holes induced will continue until the concentrations of electrons and holes become equal on both sides of the p-n junction For example, if we join a crystal of copper (Cu) and a crystal of nickel (Ni) and then heat it to a high temperature (eg, 500°C) to promote diffusion (Figure 53), then after a few hours, the copper atoms will diffuse into the nickel crystal and vice versa This process of interdiffusion will continue until the concentrations of copper and nickel atoms are equal on both sides of the original interface
We will now discuss the difference between the interdiffusion of copper and nickel atoms (Figure 53) and the diffusion of holes and electrons in the formation of a p-n junction (Figure 52) When electrons in the n-type material begin to diffuse out into the p-type material, they leave behind positively charged donor ions For example, when a donor such as phosphorus (P) is added as a neutral atom, it donates an electron and becomes ionized, turning into a P1+ ion Similarly, when holes from the p-side begin to diffuse onto the n-side of the p-n junction, they leave behind negatively charged acceptor ions (eg, boron; B1− ion)
Thus, as electrons diffuse from the n-type into the p-material, they leave behind a region of positively charged donor ions, shown with + signs in Figure 51b Similarly, as holes diffuse from the p-side to the n-side, they create a region comprised of negatively charged acceptor ions (Figure 51b) Thus, at the p-n junction, there is a region known as the depletion region, so named because it is depleted of electrons and holes It is also called the space-charge region or space-charge layer, which refers to the positive and negative charges present in this region
Across the width of the depletion region (w), an internal or built-in electric field (E) is set up because of the presence of positively charged donor ions and negatively charged acceptor ions (Figure 51) This electric field (E) is directed from the positively charged donor ions to the negatively charged acceptor ions If an electron tries to diffuse from the n-side to the p-side, it begins to “see” the acceptor ions The motion of such electrons toward the p-side is opposed by the negatively charged acceptor ions (Na−) Similarly, if a hole tries to diffuse from the p-side to the n-side, it begins to experience the repelling force of the positively charged donor ions ( Nd+ ) We will use the symbols Nd and Na for the concentrations of donor and acceptor atoms or ions, respectively Thus, the built-in electric field stops the diffusion of electrons from the n-side to the p-side and the diffusion of holes from the p-side to the n-side
The voltage corresponding to the internal electric field is known as the contact potential or builtin potential (V0) and is expressed in volts This potential difference can appear when a p-n junction between different materials is formed As we can see from Figure 51, the built-in electric field is directed toward the −x-direction In the depletion region, the electric field is related to the potential as follows:
E x dV x
dx ( )
( )= − (51)
The electrostatic potential (V) is higher on the n-side, which is a part of the depletion layer with a net positive charge (Figure 51b) The contact potential (V0) is the difference between Vn and Vp, where Vn and Vp are the electrostatic potentials in the n-and p-neutral regions, respectively Thus,
V V V0 = −n p (52)
We plot the electron energy, which is related to the electric potential by a factor of −q, on the band diagram Therefore, Ec,n, EF,n and Ev,n, that is, different energy levels on the n-side, appear lower on the p-n junction band diagram than the corresponding levels on the p-side (Figure 51b)
The development of the contact potential, also known as the diffusion potential, makes the p-n junction interesting and useful for device applications We will see in Section 510 that by applying a forward bias, that is, by connecting the positive terminal of an external voltage supply to the p-side, this built-in potential can be overcome If this happens, the p-n junction begins to conduct as the diffusion of electrons and holes resumes Conversely, if we apply a reverse bias by connecting the negative terminal of an external voltage supply to the p-side, then we add to the built-in potential barrier (see Section 59) Thus, the p-n junction will be able to carry very little current A p-n junction is a “tunable device” and is used to create diodes, transistors, and so on
The built-in or internal electric field (E) created in the p-n junction stops further diffusion of electrons and holes It also plays another important role by setting up drift currents (Figure 52) The term “drift” refers to the motion of charge carriers under the influence of an internal or external electric field When thermally generated electrons from the p-side (where they are minority carriers) experience the internal electric field, they are driven toward the positively charged region of donor ions The internal electric field causes electrons on the p-side to drift toward the n-side (Figure 52) Similarly, thermally generated holes on the n-side drift toward the negatively charged space-charge region on the p-side The internal electric field causes these drift currents The total drift current in a p-n junction is known as the generation current (see Section 511) It is an important part of the current-voltage (I-V) characteristics of a p-n junction Note that the directions of the diffusion of electrons and the drift of electrons are opposite The current due to electron diffusion is from p-side to n-side The electron drift current is from the n-side to the p-side The directions of the diffusion of majority carriers and the drift of minority carriers on the n-side and p-side are shown in Figure 52
For a p-n junction at equilibrium (Figure 52), the drift and diffusion current densities (J) cancel out, since a p-n junction at the equilibrium carries no electrical current
Therefore,
J Jp pdiffusion drift( ) ( )+ = 0 (53)
and
J Jn ndiffusion drift( ) ( )+ = 0 (54)
Subscripts p and n refer to the motion of holes and electrons, respectively
In Chapter 4, we learned how to calculate the relative position of the Fermi energy level for a semiconductor (Examples 42 and 43) and the invariance of the Fermi energy level (Section 410) We will now draw the band diagram for a p-n junction using this information and graphically calculate the value of qV0, and hence V0 This is illustrated in Example 51
SOLUTION a. As we have seen in Chapter 4,
If the doping level on the n-side increases, then EF,n will be farther away from EF,i,n (Figure 55) and the Ec,n and Ev,n will move down to center themselves around EF,i,n If nothing changes on the p-side, then the value of qV0 will be expected to increase Similarly, if the doping level on the p-side is increased, then EF,i,p will move up relative to EF,p (Figure 54) Then, Ev,p and Ec,p will also move up to center around EF,i,p If nothing changes on the n-side, qV0 will increase
We will now derive an equation that quantitatively describes how the value of V0 changes with the dopant concentrations on the n-and p-sides of the p-n junction
According to Fick’s first law of diffusion, the flux of diffusing species is proportional to the negative of the concentration gradient The negative sign means that there will be a movement of species from a region of higher concentration to a region of lower concentration In this case, we want to calculate the electrical charge flowing per unit time, not just the number of electrons or holes Therefore, we will multiply the flux of species by q, the magnitude of the charge of an electron or a hole
Thus, the diffusion current density (J) due to the motion of the holes is
J q D dp x
dxp,diffusion p = − × × ( ) (510)
We have taken the direction from p to n as the positive x-direction (Figure 52) In this equation, Dp is the diffusion coefficient for holes
The drift current due to the movement of the holes from the n-side, where they are the minority carriers, to the p-side is given by
J q p x E xp,drift p= × × ×µ ( ) ( ) (511)
In Equation 511, p(x) is the hole concentration along x-direction and E(x) is the built-in electric field We know the hole concentration on both the p-side and the n-side, where the holes are minority carriers
The current induced by the diffusion of majority carriers and the current caused by the drift of minority carriers are in opposite directions and cancel each other out (Equation 53) Therefore, we get
q D dp x
dx q p x E x× × = × ×p p
( ) ( ) ( )µ (512)
Simplifying,
dp x
dx p x D E x
( )
( ) ( )
1 = µp
(513)
Since we want to calculate the value of V0, the contact potential, we change the electric field (E) to electrostatic potential (V) by substituting for E(x) from Equation 51 into Equation 513:
dp x
dx p x D
dV x
dx
( )
( )
( )1 = −
(514)
Using the so-called Einstein relation (not derived here) and applying it to holes,
µp
q
k T = (515)
From Equations 514 and 515, we get
dp x
dx p x
q
k T
dV x
dx
( )
( )
( )
= −
(516)
We know the concentrations of holes on both the p-side and the n-side in the neutral regions, that is, the regions away from the p-n junction that do not have any built-up net charge We assume a one-dimensional model; that is, we assume that the carriers will diffuse and drift along x-direction (+ or −) only We now integrate Equation 516 from the p-side to the n-side:
− =∫ ∫qk T dV dp
pV
(517)
In Equation 517, Vp and Vn are the electrostatic potentials on the p-side and n-side of the neutral regions, where there is no built-up net charge The hole concentrations in the neutral regions on the p-side and n-side are pp and pn, respectively The subscripts indicate the side of the p-n junction Simplifying Equation 517,
− − = − =
q
k T V V p p
p
( ) ln ( ) ln ( ) ln (518)
Note that the potential difference Vn − Vp is V0, the contact potential
−
=
q
k T V
p
pB
p 0 ln (519)
Eliminating the negative sign, we get
V k T
q
p
p0 =
ln (520)
For a step junction, we move abruptly from the p-side, with Na acceptors per cubic centimeter, to the n-side with Nd donors per cubic centimeter We rewrite Equation 520 as
V k T
q
N N
n0 2 =
ln (521)
To get Equation 521, we used pp = Na and pn · Nd = ni2 This form is useful in calculating the contact potential (V0) associated with a p-n junction
We can rewrite Equation 520 as
p
p
qV
k T p
=
exp 0 (522)
Since pp × np = pn × nn = ni2, we can write
p
p
n
n
qV
k T p
= =
exp 0 (523)
We can see from Equation 523 that if the dopant concentration on the p-side increases, then qV0 will increase We also saw this in the calculation of V0 from the p-n junction band diagram in Section 53 and Figure 56
From Equation 522, we substitute for pp and pn in terms of the density of states (N) and the difference between EF relative to the valence bandedge on each side:
p
p
N E E
k T
N E E
= −
−
− −
exp
exp n
k T
qV
k T
=
exp 0 (524)
In Equation 524, the additional subscript for E refers to the side of the junction Thus, EF,p is the Fermi energy level on the p-side, Ev,n is the valence bandedge on the n-side, and so on Rearranging Equation 524,
exp
exp
exp
qV
k T
E E
k T
E 0
B
=
− −
−
E
k T
Note that the Fermi energy is constant for a p-n junction under equilibrium; that is, EF,n − EF,p = 0, so we get
exp exp exp
qV
k T
E E
k T
E0
v,
=
−
−
=
−E k T
E E
k T exp
Therefore,
qV E E0 = −v,p v,n (525)
The contact potential barrier energy (qV0) is the difference between the valence and conduction bandedges on each side of the p-n junction This was shown in Figure 56, the band diagram for a p-n junction Example 52 examines the application of these equations
SOLUTION From Equation 5.21,
The depletion region has an extremely small concentration of electrons or holes The charge density for the depletion region on the n-side of the junction is q × Nd, where Nd is the concentration of donor ions Assume that the cross-sectional area of the p-n junction is A and the width of the depletion region on the n-side is xn,0 The subscript n refers to the n-side, and the subscript 0 stands for a p-n junction under equilibrium, that is, no external voltage or bias is applied Thus, the volume of the depletion region on the n-side is A × xn,0 In this region, the total positive electrical charge is
Q q A x N+ = × × ×n d,0 (526)
Similarly, if xp,0 is the width of the penetration of the depletion region into the p-side and the crosssectional area is A, then the magnitude of the negative charge on the p-side of the depletion region is
Q q A x N= × × ×p,0 a (527)
The donor and acceptor ion charges accumulated on the n-and p-sides must be equal for an electrical neutrality of the entire p-n junction, that is, Q+ = Q− or
q A x N q A x N× × × = × × ×n d p a, ,0 0 (528)
Therefore,
x
N
x
N n
The width of penetration of the depletion region varies inversely to the dopant concentration For example, if the p-side of the semiconductor is doped more heavily than the n-side (as in Example 51), the penetration of the depletion region on the p-side will be smaller If one side is heavily doped, it needs a lesser volume of that material to compensate for the charge on the other
side (Figure 58) If one side is lightly doped (in this case, the n-side), then we need more volume of that material to compensate for the charge on the other side This means a higher penetration depth in the one-dimensional model
Using Poisson’s equation, we can calculate the electric field and the electron energy variation across the depletion region According to the one-dimensional form of this equation, the gradient of the electric field is related to the charge density and the dielectric permittivity of the material (ε):
d x
dx
x dE x
dx
φ ρ ε
( ) ( ) ( )= − = − Poisson s equatio’ n (530)
where ϕ is the electric potential, ρ is the charge density, ε is the dielectric permittivity of the material, E is the electric field, and x is the distance Note that ε ε ε= ×0 r, where εr is the dielectric constant and ε0 is the permittivity of the free space The dielectric constant (εr) is a measure of the ability of a dielectric material to store an electrical charge
We assume that the doping on the n-and p-sides is uniform, and hence the charge density is as shown in Figure 58a
Applying Poisson’s equation to the n-side region in the depletion zone (x = 0 to x = xn,0),
dE
dx
qN x x= ≤ ≤d n,0forε
( )0 (531)
We can obtain the electric field (E) variation by integrating Equation 531
E qN
dx qN
x C= = +∫ d dε ε 1 (532) Note that when x = xn,0, E = 0, that is, outside the depletion layer, there is no net built-up charge and the electric field becomes zero Using this condition in Equation 532, we get
0 1
= +
∴ = −
qN x C
C qN
x
ε
ε
(533)
+
Using this value of the constant of integration (C1) in Equation 532, the electric field variation across the n-side of the depletion region is given by
E x qN
x x x x( ) ( )= − − ≤ ≤
0 (534)
The maximum in the electric field occurs at x = 0 Its magnitude is given by
E qN
xmax =
(535)
Similarly, we can show that on the p-side of the junction, in the depletion region,
E x qN
x x x x( ) ( )= − + − ≤ ≤
0 (536)
The variation in the electric field across the depletion layer width (w), that is, from −xp,0 to xn,0, is shown in Figure 58
As we can see from Poisson’s equation, the electric field variation is in the form of a straight line for a uniform charge density With a linear variation in the electric field, the electric potential will have a parabolic change
We use Poisson’s equation to compute the electrostatic potential (ϕ) across the p-n junction We can obtain the electrostatic potential by integrating the electric field across the distance over which the potential appears
φ( ) ( )x E x dx= −∫ (537) For the p-side of the depletion region, substituting for E(x) from Equation 536 into Equation 537, we get
φ ε
( ) ( )x qN
x x dx= − − +∫ a p,0 (538)
Therefore,
φ ε
( )x qN x
x x C= + +
22 (539)
Since the electrostatic potential is zero at x = −xp,o, we can calculate the integration constant C2 in Equation 539 as follows:
C qN
x2 22 = a p,0ε
(540)
Substituting the value of C2 in Equation 539, we get
φ ε ( ) ( ) ( )x
qN x x x x= + − ≤ ≤a p,0 p,02
02 (541)
We will now calculate the electrostatic potential (ϕ) on the n-side of the depletion layer by integrating the electric field from Equation 534 as follows:
φ ε ( ) ( )x
qN x x dx= − −
−∫ d n,0 (542)
Therefore,
φ ε
( )x qN
x x x
C= −
+
32 (543)
The electrostatic potential is continuous across the p-n junction, that is, at x = 0, the value of potential (ϕ) can also be calculated using Equation 541 and is equal to the value given by Equation 543 Therefore, we evaluate the integration constant C3 using the following equation:
C qN
x3 22 =
(544)
Substituting this value of C3 into Equation 543, we get
φ ε ε
( ) (x qN
x x x qN
x= −
+
2 2 0 ≤ ≤x xn,0) (545)
This variation in the electrostatic potential across the p-n junction is shown in Figure 59 (Neaman 2006)
The value of the contact potential (V0) can be calculated by evaluating this equation at x = xn,0 Therefore,
V x x q
N x N x0 2 22 = = = + φ ε
Note that the electrostatic potential (ϕ; unit is volts) and the electron energy (unit is electron volts) shown on the band diagram are related by the following equation:
Electron energy = − ×q φ (547)
The electrostatic potential is higher on the n-side (Figure 59), which means that the electron energy is lower on the n-side One way to visualize the higher electrostatic potential on the n-side is to note that this is the side of the p-n junction with positively charged donor ions left behind (Figure 51)
The semiconductor p-n junction is electrically neutral as a whole (Equations 528 and 529) Substituting for xp from Equation 529 into Equation 546 and solving for xn, we get
x V
q
N
N N Nn a
=
+
2 10ε
(548)
Note that we dropped the subscript 0 in xn,0, and that the p-n junction is not biased Similarly, we can solve for xp by substituting for xn from Equation 529 into Equation 546:
x V
q
N
N N Np d
=
+
2 10ε
(549)
Now, the width of the depletion layer is
w x x= +p n (550)
Substituting for xp and xn,
w V
q
N N
N N =
+
(551)
Thus, for a p-n junction with known doping levels, we can calculate the built-in potential (V0) using Equation 521 or 546 We can calculate the depletion layer width using Equation 551, and the maximum electric field in a p-n junction at x = 0 using Equation 535 Examples 53 and 54 illustrate the calculation of the contact potential (V0), the depletion layer width (w), and the maximum electric field
SOLUTION a. For this p-n junction, the depletion layer penetration depth on the p-side will be smaller
SOLUTION a. From Equation 5.21,
The real utility of the p-n junction is that its ability to conduct can be changed with doping and by application of an external voltage Consider a reverse-biased p-n junction This means that we apply the n-side of the junction to the positive terminal of an external direct current (DC) voltage supply (Figure 510)
Applying a reverse bias, that is, connecting the n-side (which has the positive space-charge region) to the positive terminal, causes the electrostatic potential on the n-side (Vn) to increase (Figure 510b)
Recall that electron energy is related to electrostatic potential by −q Thus, with a reverse bias, the Fermi energy level on the n-side is decreased The band diagram for a reverse-biased p-n junction is shown in Figure 510c
The internal electric field and the external electric field follow the same direction, from the n-side to the p-side The depletion layer width (w) for a reversebiased junction is higher than that for a p-n junction with no bias (Figure 511a)
Electrons attempting to diffuse from the n-side to the p-side under a concentration gradient will encounter a larger energy barrier of height q(V0 + VR), which can be visualized easily on a band diagram (Figure 510c) The magnitude of the diffusion current due to the diffusion of electrons from the n-side to the p-side becomes very small Similarly, holes attempting to diffuse from the p-side to the n-side, again under a concentration gradient, will encounter a larger potential barrier that can be easily visualized on the electrostatic potential diagram (Figure 510b) The magnitude of the total diffusion current due to the movement of holes and electrons is negligible because of the increased energy barrier (Figure 511c)
We can calculate the width of the depletion region (w) and the penetration depths in the p-and n-regions of a p-n junction by substituting (VR + V0) for V0 in Equations 548, 549, and 551 derived in Section 58
Therefore, for a reverse-biased p-n junction,
x V V
q
N
N N Nn R a
= +
+
2 10ε( )
(553)
x V V
q
N
N N Np = +
+
(554)
The depletion layer width under the reverse bias can be calculated by adding xn and xp values or by using the following equation:
w V V
q
N N
N N = +
+ +
(555)
In a forward-biased p-n junction, the p-side is connected to the positive terminal of an external power supply This increases the electrostatic potential on the p-side Due to the built-in potential, the applied electric field will now oppose the internal electric field (Figure 511a) This means that the total height of the potential barrier will be reduced (Figure 512b)
Electrons, the majority carriers on the n-side, can now diffuse more easily to the p-side In the p-region, these injected electrons move toward the positive terminal, where they are collected While making their way through the p-region, some of the electrons will recombine with the holes The positive terminal of the power supply compensates for the holes lost as a result of the recombination The current due to the electron diffusion is thus maintained by the electrons from the n-side The negative terminal of the battery provides these electrons
Similarly, we can see on the electrostatic potential diagram (Figure 512c) that the energy barrier for hole diffusion is reduced This means that more holes, the majority carriers on the p-side, diffuse onto the n-side The holes injected in the n-side diffuse toward the negative terminal of the power supply Some holes will end up recombining with the electrons on the n-side The negative terminal of the power supply replaces the electrons lost to the recombination The current is sustained as more holes continue to diffuse over to the n-side (Figure 511b) The positive terminal of the power supply supplies these holes
The motion of the majority carriers produces a forward current (IF) Under a sufficient forward bias, the total diffusion current increases and the p-n junction begins to conduct The depletion layer width (w) decreases under a forward bias because the net electric field in the depletion layer is decreased (Figure 511b) We can obtain the new values of the penetration depths and the depletion layer width by substituting V0 by (V0 − Vf) in Equations 548, 549, and 551 for these quantities
In a p-n junction at room temperature, there are thermally-created electron-hole pairs (EHPs) in both the neutral regions and the depletion regions of the p-n junction The thermally generated carriers in the neutral region first diffuse to the depletion region under a concentration gradient For example, there are thermally generated holes in the n-side neutral region (x > xn) However, the concentration of holes near the space-charge region is essentially zero at x = xn Thus, the holes flow from the neutral region to the boundary of the depletion layer due to the concentration gradient These holes then drift toward the n-side under the influence of the electric field present in the depletion region The hole diffusion length (Lp) is the distance that thermally generated holes on the neutral n-side can travel from the transition region to the n-side through the electric field in the depletion region Similarly, the electron diffusion length (Ln) is the average distance that an electron can travel before recombining with a hole Carriers generated within the length of the diffusion distance will successfully pass to the depletion layer without recombination Holes or electrons generated at a distance larger than the diffusion distance (for that carrier) will most likely recombine and thus will not contribute to this so-called “reverse current” All thermally generated carriers therefore do not end up contributing to generation current
The first source for the drift current (IShockley) is induced by the diffusion of thermally generated minority carriers in the neutral region, followed by their drift across the depletion region This current is given by the Shockley equation:
I qD
L N
qD
L NShockley p
=
+
ni2 (556)
In the Shockley equation, q is the magnitude of the charge on the electron or hole, D is the diffusion coefficient, L is the diffusion length, and subscripts p and n refer to the holes and electrons, respectively Recall from Chapter 4 that the intrinsic carrier concentration (ni) increases exponentially with the temperature and is inversely related to the bandgap (Eg)
The second source for the drift current is from the thermally generated carriers inside the depletion region The internal electric field separates these carriers, and they drift toward the neutral regions This aspect of the reverse current is given by
I qWn
= τ
(557)
In Equation 557, τg is the mean thermal generation time This is the average time needed to thermally create an EHP The total current due to the drift of thermally generated minority carriers, whether created in the neutral region or in the space-charge region, is called the generation current (I0)
Thus, the total generation current (I0) is given by combining Equations 556 and 557:
I qD
L N
qD
L N n0 =
+
i p
2 + qW ni g
iτ (558)
From Equation 558, we can see that either the first term with ni2 or the second, thermal-generation term with ni is controlling the generation current In Figure 513, the reverse current for a germanium (Ge) diode is shown In this case, we show a dark current to avoid any current due to the photogeneration of EHPs
The total generation current is controlled at low temperatures (< 240 K) by the second term, thermal generation At lower temperatures, the slope of the line is proportional to Eg/2 because ni ∝ exp(−Eg/2kT) At higher temperatures, the diffusion term dominates, and the current is
controlled by the ni2 term The slope of the line shown in Figure 513 is nearly proportional to the bandgap of the semiconductor
The generation current (I0) is independent of the applied bias; this is because the electric field in the depletion region affects how rapidly the carriers will overcome the potential barrier, but not the total amount of electrical charge moving across per unit of time Although the forward current (IF), which is due to the diffusion of majority carriers, is strongly affected by the type and amount of bias, the magnitude of the generation current (I0) is independent of the applied voltage (Figure 511)
The generation current depends upon the temperature because the carrier concentration due to thermal excitation also depends on the temperature Similarly, the absorption of light energy leads to the creation of EHPs This also causes an increase in the generation current
Thus, the total current (I) flowing through the p-n junction is given by
I I I= +generation diff (559)
When there is no applied voltage (Figure 511a), the total drift current and the current due to the diffusion of majority carriers are equal and opposite, that is,
I I Vdiff generation for= = 0 (560)
When the p-n junction is reverse-biased (Figures 511c and 514c), the diffusion of majority carriers is almost negligible, and the total current is equal to the generation current
I Igeneration for reverse bias= 0 (561)
Therefore, the generation current (I0) is also known as the reverse-bias saturation current (Is) The generation current is usually in the range of 10−14−10−12 A The magnitude of the generation current depends upon its doping levels, the size of the cross-sectional area of the junction, temperature, and so on (Equation 558) It does not depend on the reverse-applied voltage
For a p-n junction under a forward bias (Figures 511b and 514b), the diffusion current (Idiffusion) changes exponentially with the applied voltage and is given by
I I qV
kTdiffusion =
0 (562)
Thus, the total current under a forward bias is given by the difference between the diffusion currents:
I I qV
k T =
−
0 1exp
(563)
Equation 563 is also known as the ideal diode equation The current-voltage (I-V) curve for a p-n junction is shown in Figure 517
If the applied bias (V) is large compared to the magnitude of (kBT/q), then the exponential term will be much larger than 1, and Equation 563 can be rewritten as
I I V
q k T ≈
0
exp ( / )B
(564)
At room temperature (T = 300 K), the value of (q/kBT) is ∼0026 V, hence
I I V
T≈ =0 0 026
exp ( )
( . )
in V
V at 300 K (565)
The p-n junction behaves like a one-way electrical valve It functions as a rectifier by allowing current to flow through under a forward bias but not under a reverse bias
A summary of p-n junction band diagrams and changes in the electric field in the depletion layer width, as well as electrostatic potential variation under zero, forward, and reverse bias, is shown in Figure 514
The electrical symbol for a diode is shown in Figure 515 A diode can serve several useful functions, including rectification For an alternating current
voltage (AC) input, the diode will produce the output voltage shown in Figure 516 Two diodes can be used to create a full-wave rectification of
AC voltage The p-n junction is also used as the basic building block for solar cells and devices known as transistors Millions of transistors are connected to form miniature circuits, which are then used to create semiconductor chips that are used in computers and other electronic equipment
An actual I-V characteristic curve for a p-n junction is shown in Figure 517 Note that under a forward bias, the current (I) is in
milliamperes (mA) Under a reverse bias, the current is in microamperes (μA) When the forwardbias voltage is less than this knee voltage (a voltage close to the built-in potential V0), a small current is set up across the p-n junction As the voltage increases, the current also increases However, this increase is nonlinear; that is, it does not follow Ohm’s law (Figure 518)
For silicon (Si), the knee voltage is ∼07 V Recall that this is the typical built-in potential (V0) for silicon p-n junctions (Example 51) As the temperature increases from 25°C to 25 + ΔT, the knee voltage decreases slightly At an elevated temperature the forward current (IF) at any given voltage is higher than that at 25°C As expected, more minority carriers are generated on both sides of the p-n junction at higher temperatures There is thus a very slight increase in the reverse-bias current at higher temperatures The sketch in Figure 517 is not to scale and exaggerates this difference in the change in the magnitude of the generation current
For germanium-based p-n junctions, the knee voltage is ∼03 V A schematic of I-V curves expected for p-n junctions for different semiconductors are shown in Figure 519
SOLUTION a. When the applied voltage is 0.5 V, the new height of the barrier seen on the electrostatic
The p-n junction under forward bias has a dynamic resistance (rd′) that changes with the applied voltage (Figure 520) Most materials (eg metals, alloys, semiconductors, and so on) show a resistance that is constant and is not voltage-dependent
The dynamic resistance of a p-n junction can be calculated from the slope of the I-V curve at a given value of voltage:
′ =r dV
(566)
A p-n junction circuit uses a resistor (RL), known as the limiting resistor, to limit the total current and protect the p-n junction from damage due to an excessive current When such a resistor is used in series with the diode, the forward current is given by
I V V
= −( )
(567)
If a limiting resistor is not used, the p-n junction can be damaged by Joule heating If the dynamic resistance of the p-n junction is accounted for, then the forward current is given by
I V V
= − + ′
( )
( ) (568)
The application of this equation is illustrated in Example 56
SOLUTION a. From Equation 5.67,
A reverse bias is sometimes so high that it causes a reverse-bias dielectric breakdown of the p-n junction (Figure 521) When this happens, the current flowing through the p-n junction increases rapidly Critical breakdown voltages (Vbr) for silicon diodes generally start at ∼60 V The term “breakdown” is a bit misleading; if the p-n junction breaks down electrically, this does not mean that the device is permanently damaged The breakdown of a p-n junction, which should be distinguished from a broken or defective p-n junction, is useful in some applications
A limiting resistor (RL) is used to limit the flow of current during a reverse-bias breakdown If such a current-limiting resistor is not used, the p-n junction will be damaged because of overheating caused by the excessive current This will occur in both forward and reverse bias
The reverse-bias breakdown can occur by two mechanisms In the avalanche breakdown mechanism, a high-energy conduction electron collides with a silicon-silicon bond and knocks off a
valence electron from that bond, which makes this electron free This process is known as impact ionization (Figure 522)
From a band diagram viewpoint, the impact ionization process sends an electron from the valence band to the conduction band, creating a hole in the valence band The first high-energy electron continues, knocking off another electron from one more silicon-silicon covalent bond and continuing the process The impact ionization process is shown on the band diagram in Figure 523
In the mechanism known as an avalanche breakdown, multiple collisions from a single high-energy electron can create many EHPs
A p-n junction with one side heavily doped is known as a one-sided junction If the p-side is heavily doped, we refer to this as a p+-n junction The depletion layer of a p+-n is mainly on the n-side The charge density distribution for this junction is shown in Figure 524 When this junction breaks down electrically, the breakdown will occur on the n-side at a location where the electric field is at a maximum
Figure 524 shows the critical breakdown field as a function of background doping for a onesided p-n junction, in which one of the sides is very heavily doped If the p-side is heavily doped, the junction is referred to as p+-n; if the n-side is heavily doped, the junction is shown as p-n+ The charge density expected for a p+-n junction is shown in Figure 525
The breakdown voltage for a p+-n junction is given by
V E
qNBD = ε critical
2 (569)
where Nd is the donor dopant concentration and Ecritical is the critical field that causes breakdown (Singh 2001) The breakdown field for a p+-n junction will depend upon the level of doping This change for the silicon and gallium arsenide (GaAs) p-n junctions is shown in Figure 524
Example 57 shows the calculation of the breakdown voltage by avalanche breakdown mechanism
As the doping level increases, another mechanism of breakdown can be applied In the Zener tunneling mechanism, electrons tunnel across the p-n junction instead of climbing over the energy barrier This is commonly seen in heavily doped p-n junctions The breakdown occurs across very thin depletion regions and thus typically occurs at low voltages (∼01−5 V) Heavier doping means a smaller depletion region width (Equations 548 and 549) With Zener tunneling, electrons from the p-side valence band can tunnel across into the conduction band on the n-side because the reverse bias pushes down the Ec on the n-side and aligns it with the valence bandedge (Ev) on the p-side The Zener tunneling mechanism is also known as the Zener effect or band-to-band tunneling (Figure 526)
The probability of Zener tunneling (T) is given by
T m E
q E ≈ −
exp
* /4 2
(570)
where me* is the reduced effective mass of the electron, Eg is the bandgap, and E is electric field (Singh 2001) The Zener breakdown is important in heavily doped junctions and in narrow-bandgap materials The value of probability (T) generally needs to be ∼10−6 for tunneling to initiate this breakdown process Example 58 illustrates the calculation of the tunneling probability
SOLUTION From Equation 5.70, the probability of tunneling is
The breakdown by either Zener or avalanche mechanism is the basis for the Zener diode (Figure 527a), a device that is typically operated under a reverse bias The I-V curve for a Zener diode including the reverse-breakdown region is shown in Figure 527b The normal operating region for a regular p-n junction diode is also shown in this figure for comparison (Figure 527c)
Since the Zener diode has an avalanche or Zener breakdown after a certain applied voltage is reached, the voltage across this diode remains essentially constant Therefore, Zener diodes are used as voltage regulators (Example 59) Under a forward bias, the Zener diode will function like a regular diode, with the typical knee voltage of ∼07 V for silicon diodes
The voltage at which a Zener diode will begin to conduct current under a reverse bias is known as the Zener voltage Diodes can be designed to have a Zener voltage ranging from a few volts to a few hundred volts
In practice, the Zener I-V curve is not completely vertical (Figure 528) We define ΔVZ as the change in voltage across the reverse-biased Zener diode as the current changes from a Zener test current (IZT) to a higher value, up to the Zener maximum current (IZM) The Zener diode impedance or resistance (ZZ) is given by
Z V
= ∆ ∆ (571)
In Equation 571, ΔIZ = IZM − IZT Example 59 shows how a Zener diode under a reverse bias functions as a voltage regulator
51 A p-n junction is formed in Ge so that the donor and acceptor concentrations are 1016 and 1017 atoms/cm3, respectively Ge has a small bandgap (Eg ∼ 067 eV) Based on this, do you expect the built-in potential to be smaller or greater than Si (Eg ∼ 11 eV)? What is the built-in potential (V0) for this p-n junction? Use the intrinsic carrier concentrations for Ge from Figure 37 and assume T = 300 K
52 A p-n junction is formed in GaAs so that the donor and acceptor concentrations are 1016 and 1017 atoms/cm3, respectively Show that the built-in potential for this p-n junction is 122 V Use the intrinsic carrier concentrations for GaAs from Figure 37 and assume that T = 300 K
53 A p-n junction is such that the acceptor and donor dopant levels are 016 eV from the nearest bandedge The doping level on both sides is 1016 atoms/cm3 What is the built-in potential for this junction?
