The PN Junction Diode
2.4 The Depletion Approximation
Now,
(Ef−Ev)p−side=kT ln Nv
pp
=(0.026 eV) ln
1.06×1019cm−3 1017cm−3
=0.12 eV
p-side n-side
Ec
Ef Ev
(Ef –Ev)p–side = 0.12 eV
Eg = 1.11 eV
E0 = qV0 = 0.817 eV
To understand this depletion it is necessary to consider two separate semiconductors, one n-type and one p-type. If they are brought together, electrons in the n-type material and holes in the p-type material close to the junction will quickly diffuse across the junction and annihilate each other by recombination leaving a deficit of holes and electrons. This diffusion and recombination will be complete after a very short time, and then equilibrium conditions will be established. Electrons and holes further away from the junction will also have a chance to diffuse across the junction; however, the electric field that is built up near the junction opposes this and also causes drift currents and the resulting band diagram now is correctly described by Figure 2.2. The equilibrium currents that continue to flow are described by Equation 2.2.
A simplification called the depletion approximation is used to model the depletion of charges. We assume complete depletion of charge carriers in a depletion region of width W0at the junction, and then assume that the carrier concentrations abruptly return to their equilibrium levels on either side of the depletion region as shown in Figure 2.10. Carrier concentrations do not make abrupt concentration changes in real materials or devices and instead gradients in carrier concentrations exist; however, the depletion approximation dramatically simplifies the quantitative model for the p-n junction and the results are highly relevant to real diodes. In Section 2.5 we will consider a more detailed view of these carrier gradients.
Normally a doped semiconductor consists of both mobile carriers and non-mobile ionized dopants as discussed in Section 1.13. The net charge density in the semiconductor is zero since the dopant ions have a charge density that is equal in magnitude and opposite in sign to the mobile charges that they provide.
Depletion region
p n
Ionized donors x
Ionized acceptors –qNa qNd
W0
Wp0
Wn0 ρ
0
Figure 2.10 A depletion region of widthW0is assumed at the junction. Charge densityρ is zero outside of the depletion region. Inside the depletion region a net charge density due to ionized dopants is established. The origin of the x-axis is placed at the junction for convenience
The semiconductor material within the depletion region of a p-n junction is depleted of carriers but it is still doped either p-type or n-type and dopant ions are therefore present in this region. This means that there is a net fixed (non-mobile) charge density in the depletion region from the ionized dopants.
In the p-side the ions are negative. For example, in silicon with p-type aluminium doping the aluminium ions are negatively charged having accepted an electron to yield a valence band hole. In the n-side the ions are positive. Conversely, in silicon with n-type phosphorus doping the phosphorus ions would be positively charged having donated an electron to the conduction band. Once these holes and electrons recombine the depletion region will be left with Al−ions and P+ions having concentrations of Naand Ndrespectively.
Charge densities−q Na and+q Nd (coulombs per cm3) will reside on the p-side and n-side of the depletion region respectively, as indicated in Figure 2.10. The magnitude of depletion charge on either side of the junction must be the same since one negative and one positive ion results from each recombination of a hole and an electron. Hence the magnitude of the charge present on either side of the junction is
Q=q NaWp0 A=q NdWn0 A (2.9)
where Wp0 and Wn0 are the widths of the depletion regions on the p- and n-sides of the junction in equilibrium respectively, and A is the cross-sectional area of the diode. From Equation 2.9 we can write
Wp0
Wn0 = Nd
Na (2.10)
The charged regions in the depletion region give rise to an electric field. This is the same field that causes the energy bands to tilt in Figure 2.2, and we can now calculate it. Using Gauss’s law we can enclose the negative charge on the p-side of the depletion region with a Gaussian surface having surface area A, as shown in Figure 2.11.
p n
0 x
Gaussian surface
–Q A
Wn0 Wp0
W0
Figure 2.11 A Gaussian surface having volume AWp0(shaded) encloses the negative charge of magnitude Q on the p-side of the depletion region
Gauss’s law relates the surface integral of electric field ε for a closed surface to the enclosed charge Q or
S
ε· ds= Q
(2.11)
If we assume our Gaussian volume has a relatively large cross-sectional surface area A, and a small depth Wp0, then we can approximate the total surface area as 2 A. Since there is symmetry and the two cross-sectional areas A are equivalent, we can use Equation 2.11 to write
2εpA= Q 0r
whereεpis the magnitude of the electric field caused by the depletion charge on the p-side andεris the relative permittivity of the semiconductor, and therefore
εp= Q
20rA (2.11a)
The same reasoning may be applied to a Gaussian surface that encloses depletion charge on the n-side. This will give rise to a field of magnitude
εn= Q
20rA (2.11b)
Since electric field is a vector quantity the relevant electric field directions are shown in Figure 2.12, and it is clear that the total field in equilibrium at the junction is the vector sum of−→εp and−→εn yielding the equilibrium electric field at the semiconductor junction
ε0= − Q 0rA which may be rewritten using Equation 2.9 as
ε0= −q NdWn0 0r
= −q NaWp0
0r
(2.12)
–Q +Q
–Wp0 0 Wn0 x
Figure 2.12 The electric field directions for the two parts of the depletion region showing that the fields add at the junction
The minus sign indicates that the field points in the negative x direction. The field at other points along the x-axis may also be evaluated. We have assumed that the cross-sectional area of the junction is very large compared to the depletion width. Since the electric field due to an infinite plane of charge is independent of the distance from the plane, the vector quantities in Figure 2.12 will cancel out and the net field will be zero at x= −Wp0 and at x= −Wn0. At other x values, the electric field will vary linearly as a function of x and may be calculated by appropriately applying Gauss’s law (see Problem 2.5). The resulting electric field will also give rise to a potential
V0(x)= −
Wn0
−Wp0
ε(x)dx (2.13)
Bothε(x) and V (x) are shown in Figure 2.13, which also shows that a contact potential V0results directly from the depletion model. This contact potential V0is the same quantity that we introduced in Section 2.3.
p n
–qNa Wn0 –Wp0
ρ
x
x
x ε
ε0
V
V0
Figure 2.13 The equilibrium electric fieldε(x) and potential V(x) for the p-n junction follow from the application of Gauss’s law to the fixed depletion charge. Note that V on the n-side is higher compared to the p-side, whereas in Figure 2.2 the energy levels on the n-side are lower.
This is the case because the energy scale in Figure 2.2 is forelectronenergy levels; however, the voltage scale in Figure 2.13 is established for apositivecharge by convention
The integral (Equation 2.13) is the area under theεversus x curve in Figure 2.13, which may be calculated using the area of a triangle, and we obtain
V0= 1 2bh= 1
2W0ε0
and using Equation 2.12 to expressε0, V0= q NdWn0
20r
W0= q NaWp0 20r
W0
Since W0=Wn0+Wp0,
V0= q 20r
NaNd
Na+Nd
W02 (2.14)
or
W0 =
20rV0
q 1
Na+ 1 Nd
(2.15a) with
Wp0= W0Nd
Na+Nd (2.15b)
and
Wn0= W0Na
Na+Nd
(2.15c) It is interesting to note that the depletion approximation is consistent with an externally applied voltage falling across the depletion region. Since the depletion region has high resistivity, and the neutral regions on either side of it have high conductivities, we are justified in stating that applied reverse-bias voltages will drop across the depletion region.
Example 2.2
(a) Find the depletion layer width in both the n-side and the p-side of the abrupt silicon p-n junction diode of Example 2.1 doped with Na=1017 cm−3 on the p-side and Nd=1017cm−3on the n-side. Find the equilibrium electric field at the semiconductor junctionε0. Sketch the electric field and the potential as a function of position across the junction.
(b) Repeat if the doping in the p-side is increased to Na=1×1018cm−3and the doping in the n-side is decreased to Nd=1×1016cm−3. This is called a p+-n junction to indicate the heavy doping on the p-side.
Solution
(a) W0=
20rV0
q 1
Na+ 1 Nd
=
2
8.85×10−14×11.8 F cm−1
(0.817 V) 1.6×10−19C
1
1017cm−3 + 1 1017cm−3
=1.51×10−5cm=0.15μm On p-side:
Wp0 = W0Nd
Na+Nd = 0.15μm×1017cm−3
1017cm−3+1017cm−3 =0.075μm On n-side:
Wn0= W0Na
Na+Nd = 0.15μm×1017cm−3
1017cm−3+1017cm−3 =0.075μm ε0= −q NdWn0
0r
=1.6×10−19C×1017cm−3×0.75×10−5cm 8.85×10−14F cm−1×11.8
=1.15×105V cm−1
p-side n-side
V0
V
x x
ε0
ε
(b)
V0=kT/q ln
(NaNd) n2i
=(0.026 V) ln
1018cm−3×1016cm−3
1.5×1010cm−32
=0.817 V
W0=
20rV0
q 1
Na+ 1 Nd
=
2
8.85×10−14×11.8 F cm−1
(0.817 V) 1.6×10−19C
1
1018cm−3 + 1 1016cm−3
=3.4×10−5cm=0.34μm On p-side:
Wp0= W0Nd
Na+Nd = 0.34μm×1016cm−3 1018cm−3+1016cm−3
∼=0.33×10−2μm=3.3 nm On n-side:
Wn0 = W0Na
Na+Nd = 0.34×1018cm−3
1018cm−3+1016cm−3 ∼=0.33μm ε0 = −q NdWn0
0r
= 1.6×10−19C×1016cm−3×3.3×10−5cm 8.85×10−14F cm−1×11.8
=5.1×104V cm−1
Note that the depletion region is almost entirely in the n-side of the p+-n junction.
Almost all the built-in potential V0appears in the n-side. The reverse is true for an n+-p junction.
p-side n-side
V0
V
x x
ε0
ε