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Файл:Steady electric current. Tutorial
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71
where e is the electron charge, N is the number of electrons emitted by the
cathode in one second. Therefore, an increase in the anode voltage after
reaching the saturation current does not cause a change in the anode current.
The magnitude of the saturation current Isat depends on the temperature T of the cathode, its area S and the work function W of the electron. The sat-
uration current density is expressed by the Richardson1–Dushman2 law:
. (4.34)
4.5. P r o b l e m s
Problem 1. Find how the specific electrical conductivity of copper
and pure silicon changes with increasing temperature from 0 to 100 °C. The
temperature coefficient of resistance of copper is 0.004 °C−1, the activation
energy of silicon electrical conductivity is 1.1 eV.
Given
SI
Solution
t1 = 0 °C
t2 = 100 °C
α = 0.004 C−1
ε
A
= 1.1 eV
273 K
373 K
0,004 K−1
1.76∙10
−19
J
We convert C and eV to SI units:
T = t + 273; 1 eV = 1.6∙10
−19
J.
Resistance R of a copper conductor
changes with temperature according to the
law (2.48)
)( αt1RR
0
+=)1
,
σ1/σ
2
– ?
where R0 is resistance at 0 °C; α is temperature coefficient of resistance.
Taking into account (2.2), the specific resistance of the conductor will
change according to the same law
)( αt1
0
+= ρρ)2
.
1
Owen Willans Richardson (1879–1959) was a British physicist who won the Nobel
Prize in Physics in 1928 for his work on thermionic emission, which led to Richardson's
law. He also researched the photoelectric effect, the gyromagnetic effect, the emission
of electrons by chemical reactions, soft X-rays, and the spectrum of hydrogen.
2
Saul Dushman (1883–1954) was a Russian-American physical chemist. His main research interests were quantum mechanics, electromotive force, atomic structure, electron emission.
72
Having written this expression twice for t1 and t2, after division we get
2
1
2
1
αt1
αt1
ρ
ρ
+
+
=)3
.
Taking into account the relationship (2.3) between ρ and σ we obtain
2
1
1
2
αt1
αt1
+
+
=
σ
σ
)4
.
The conductivity of pure silicon varies to an exponential law (4.1)
)exp() /kTε
A
−=
0
5 σσ
.
By performing similar actions, we obtain
)6
12 2A1A
/kTε /kTε −=exp(/) σσ
.
Substitute numerical data.
Answer: (σ2/σ1)
Сu
= 0.714; (σ2/σ1)
Si
= 525.
Problem 2. During the electrolysis of copper sulfate, 0.4 g of copper
was released at the cathode in 1 hour. The area of each electrode is 70 cm2.
Find the current density in the electrolyte. What electrical energy will be consumed if the electrolysis is carried out at a voltage of 5 V.
Given
SI
Solution
t = 1 h
М = 0.4 g
S = 70 cm2
U = 5 V
3600 s
4∙10−4 kg
7∙10−3 m2
According to the first law of electrolysis
(4.13) the mass of the substance released at the
electrodes is proportional to the current I and
the time of its flow t
tIkkQM ==)1
,
j ‒ ? W ‒ ?
where the proportionality coefficient k is called the electrochemical equivalent
(4.14), it depends on the molar (atomic) mass of the ion μ and its valence Z:
Z
μ
F
1
k =)2
,
F = 96485.33 C/mol is the Faraday constant.
The atomic mass of copper is 63.5∙10‒3 kg/mol, the valence is 2.
The current density (2.4) is defined as
S
I
j =)3
.
73
Substituting expressions (1) and (2) into formula (3), we obtain
tSμ
ZM
Fj =)4
.
The expended electrical energy W, equal to the work of the current,
ultimately goes to heating the electrolyte:
5) W = IUt = jSUt.
We perform the substitution of numerical data.
Answer: j = 48 A/m2; W = 6 kJ.
Problem 3. A potential difference of 5 V is applied to the electrodes
of the discharge tube, the distance between them is 10 cm. The gas in the
tube is ionized once. The number of ions of each sign in a unit volume of gas
is 108 m−3; the mobility of positive ions is equal to 3∙10−2 m2/(V∙s) and negative ions is 3∙102 m2/(V∙s). In the weak field approximation, find the current
density created by ions of each sign and the total current density in the tube.
What fraction of the total current is carried by positive ions?
Given
SI
Solution
U = 5 V
l = 10 cm
n = 100 cm−3
b+ = 3∙10−2 m2/(V∙s)
b‒ = 3∙102 m2/(V∙s)
0,1 m
1∙108 m−3
The current densities created by
positive and negative ions, as well as
the total current density in the gas in
a case of weak currents, are determined
by formulas (4.26)‒(4.28):
Eqnbj
++
=)1
,
Eqnbj
−−
=)2
,
Ebbqnjjj )()3
−+−+
+=+=
.
j+ ‒ ?
j− ‒ ?
j ‒ ?
I+/I ‒ ?
In the case of weak currents, the electric field in the gas-discharge gap
can be considered uniform, then the following is fulfilled
4)
.
According to the problem statement, the gas is singly ionized, therefore,
the charges of the ions are equal to the elementary charge e = 1,6∙10
−19
C.
The fraction of current carried by positive ions to the total current is
5)
.
74
We perform the substitution of numerical data.
Answer: j+ = 2.4.10
−11
A/m2; j– = j.10−7 A/m2; j+/j = 10−4 = 0,01 %.
Problem 4. A potential difference of 20 V is applied to the electrodes
of the ionization chamber, the area of each electrode is 100 cm2, the distance
between them is 8 cm. In a unit of volume per unit of time, 1015 m−3s−1 pairs
of ions are formed. The resulting singly charged ions have the same mobility
of 10−4 m2/(V∙s), the recombination coefficient is 10
−12 m3
/s. Find the current
in the chamber for given parameters and the saturation current.
Given
SI
Solution
U = 20 V
S = 100 cm2
l = 8 cm
Δn
i
= 1015 m−3s−1
b+ = b‒ = 10−4 m2/(V∙s)
r = 1∙10
–12 m3
/s
0.01 m2
0.08 m
The ionizer produces in a unit
of volume per unit of time Δni pairs of
ions of both signs, some of ions Δnr recombine and form neutral molecules.
As a result, a dynamic equilibrium is
established with an ion concentration
1)
Δ
.
I – ?
I
sat
– ?
These ions create a current in the gas in the weak electric field mode.
The current density of a gas discharge under the condition of a weak
field (4.28) is equal to
2)
Δ
,
where the electric field strength E can be found as in the case of a uniform
field of a flat capacitor
3)
.
Finally, the discharge current under weak field conditions is equal to
4)
Δ
.
In the case of a strong field, the current does not depend on the volta-
ge (4.30), and the saturation phenomenon occurs
5) I
sat
= j
sat
.
S = qΔnilS.
Substitution of numerical data.
Answer: I = 2.5 nA; I
sat
= 0.13 μA.
75
Problem 5. How many times will the specific thermionic emission of
tungsten at a temperature of 2000 K change, if:
a) the temperature of tungsten is increased to 2500 K;
b) it is replaced with thoriated tungsten?
The emission constant of pure tungsten is 0.6 106 A/(m2K4), thoriated
tungsten is 0.3∙107 A/(m2K4).
Given
Solution
T1 = 2000 K
T2 = 2500 K
Ct = 0.6∙106 A/(m2K4)
C
t-t
= 0.3∙107 A/(m2K4)
Specific thermo-electronic emission, that is,
emission from a unit surface of the cathode is
equal to the saturation current density and is expressed by Richardson–Dashman law (4.34)
1)
.
The required values of the electron work function W for pure tungsten
and thoriated tungsten can be found in the reference book: Wt = 4.5 eV;
W
t−t
= 2.63 eV (1 eV = 1.6∙10
−19
J).
Using the Richardson–Dashman law, we find how many times the specific thermo-electronic emission of tungsten will change when the temperature increases from T1 to T2:
2)
.
In the second part of the problem, cathodes made of different materials
(tungsten and thoriated tungsten) are used, but thermionic emission is observed at the same temperature of T1 = 2000 K, therefore, also based on the
Richardson–Dashman law, we can write
3)
.
We perform the substitution of numerical data.
Answer:
S e l f - c h e c k q u e s t i o n s
1. Describe the mechanism of formation of current carriers in pure
(without impurities) semiconductors.
2. Describe the processes of generation and the recombination of elec-
tron-hole pairs in semiconductors.
3. How is the temperature dependence of electrical conductivity of
pure semiconductors explained?
4. Describe the conduction mechanism of n-type semiconductor with
donor impurity.
5. Describe the conductivity mechanism of a p-type semiconductor
with acceptor impurity.
6. What are electrolytes, what are the current carriers in electrolytes?
7. What is called the mobility of electrolyte ions, what does it depend on?
8. Explain why and how the mobility of electrolyte ions depends on
temperature.
9. Write down and explain the expression for current density in elec-
trolytes.
10. Write down Faraday's first law and explain its physical meaning.
11. Write down Faraday's second law. What is the electrochemical
equivalent?
12. What is called the Faraday number?
13. What is called gas discharge?
14. What is called self-sustaining and non-self-sustaining gas dis-
charges?
15. What is the current density of a non-self-sustaining discharge un-
der conditions of a weak electric field?
16. Write down and explain the expression for the current density of
a non-self-sustaining discharge in the saturation state.
17. Draw an approximate volt-ampere characteristic of a gas discharge
and indicate the limits of applicability of Ohm's law.
18. What is the phenomenon of thermionic emission, how does it de-
pend on temperature?
19. Write down and explain the Richardson–Dushman law.
76
5 . E L E M E N T S O F T H E B A N D T H E O R Y
λne
V2m
2
~
~
==σρ
1
(5.1)
πm
8kT
V =
~
OF S O L I D S
The classical electron theory of Drude–Lorentz made it possible to ex-
plain Ohm’s and Joule–Lenz’s laws, but could not explain a number of ex-
perimentally established regularities.
According to classical theory, the resistance of metals (3.12), taking
into account formula (3.1) for the arithmetic mean velocity of electrons,
should increase proportionally to the square root of the temperature
However, it has been experimentally established that in a wide range of temperatures, the resistance of metals increases (3.48) proportionally to the temperature to the first power.
According to classical theory, the molar heat capacity C of a metal
should be equal to the sum of the heat capacities of the crystal lattice 3 R and
the electron gas (3/2) R. However, the experiment gives only 3 R, which
means that there is no heat capacity of the electron gas?
These and other shortcomings of the classical theory were overcome
in the quantum theory of electrical conductivity of metals, according to which
electrons, both in an atom and in a solid, obey the laws of wave, quantum
mechanics.
Below we will consider the main elements of the quantum band theory
of solids, which made it possible to explain from a unified position the properties of metals, semiconductors and dielectrics, although their specific resistances differ by 20 orders of magnitude. The foundations of the band theory of solids were developed in 1928 by F. Bloch1 and L. Brillouin2. According to the band theory, the energy spectrum of electrons in solids consists of
alternating zones of allowed and forbidden energies.
1
Felix Bloch (1905–1983) was an American physicist. Work in the field of solid state
physics, magnetism, quantum electrodynamics, nuclear physics. Nobel Prize laureate in
physics (1952) for the discovery of nuclear magnetic resonance.
2
Leon Brillouin (1889–1968) was a French physicist. Works in the field of quantum
mechanics, solid state physics, radiophysics.
77
78
5.1. E n e r g y b a n d s
The band theory of solids uses quantum concepts about the motion of
electrons and their energy spectrum in solids (crystalline) bodies. According
to the band theory, the energy spectrum of electrons in solids consists of alternating bands of allowed and forbidden energies.
Assumptions used in the band theory:
1) atomic nuclei are located at the nodes of an ideal crystal lattice;
2) atomic nuclei, due to their large mass compared to electrons, are
considered motionless;
3) the interaction of electrons with the atoms (ions) of the lattice and
with each other is described by some effective intracrystalline force field.
Let us consider the process of formation of energy bands. The unification of isolated atoms into a solid means the merging of energy levels of
individual atoms into a single system. However, according to the Pauli principle, at each energy level, both in an isolated atom and in a solid, there can
be no more than two electrons with oppositely oriented spins (proper angular
momenta). Consequently, each energy level of an isolated atom in a crystal
will be represented by a system of N levels, where N is the number of atoms
in the crystal.
The set of closely spaced energy levels formed by the “merging” of the
energy levels of isolated atoms is called an allowed energy band. Allowed
energy bands are separated by forbidden bands or band gaps. The width of the
allowed or forbidden energy bands is equal to several electron-volts1 (eV), and
the distance between adjacent levels within the allowed band is of the order
of 10
−22
eV. Energy bands (fig. 5.1) are formed by the merging of energy
levels of isolated atoms when they come closer together.
The outer electrons form the widest energy bands, while the inner
electron bands are narrow. It is obvious that the width of the bands is
determined by the intensity of the effective intracrystalline field on the
corresponding electrons. The inner electrons interact mainly with the nuclei
of atoms; the electrons on the upper shells screen them from the field of
neighboring atoms.
1
An electron-volt (eV) is a non-systemic unit of energy used in atomic and nuclear physics, and in elementary particle physics. 1 eV is equal to the kinetic energy acquired by
an electron under the influence of an electric field when moving between points with
a potential difference of one volt in a vacuum. One electron-volt is equal to 1.6 ‧ 10
−19
J.
79
Fig. 5.1. Scheme of formation of energy bands: levels 1s, 2s, 2p of isolated
atoms (r = r∞) when they approach the equilibrium distance (r = r0)
in a solid are split, forming the allowed energy bands
Even with the above assumptions, the calculation of energy bands performed by quantum mechanics methods is a complex task. This problem is
not the subject of this tutorial. Below, only the issues of the structure of
energy bands formed by external (valence) electrons will be considered.
The degree of filling of valence bands and adjacent bands with electrons, the
width of forbidden bands, and the presence of impurity levels determine the
electrical conductivity of solids.
5.2. M e t a l s i n b a nd t h e o r y
Metals and dielectrics have different widths of forbidden bands and
different filling of allowed bands with electrons. The last partially or completely filled allowed band is formed by external valence electrons and is
called the valence band.
A solid will be a current conductor in two cases: 1) if the valence band
is not completely filled with electrons (fig. 5.2a), 2) a completely filled valence band overlaps with the upper empty or partially filled band, forming
the so-called hybrid band (fig. 5.2b). In both cases, there are free levels to
which electrons can move, increasing their energy under the influence of an
electric field.
80
а b
Fig. 5.2. Scheme of formation of conduction bands in metals
An incompletely filled valence band (in the first case) or a hybrid band
(in the second case) are called conduction bands. The degree of filling of the
conduction band with electrons, i. e., the ratio of the number of electrons in
the conduction band to the total number of levels in this band, determines the
electrical conductivity of the metal. The specific electrical conductivity of
metals σ ~ 107 S∙m−1.
5.3. D i e l e c t r i c s i n b a n d t h e o r y
In dielectrics:
a) the valence band is completely filled;
b) the valence band does not overlap with the empty upper band—a hybrid band is not formed;
c) the valence band is separated from the upper empty band by a wide
forbidden band.
Typical dielectrics have a band gap width ΔΕfb > 3 eV, for example:
diamond → ΔΕfb = 5,2 eV, aluminum oxide ceramics Al2O3 → ΔΕfb = 7 eV.
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