Semiconductor Devices-Intro PDF

SEMICONDUCTOR DEVICES DR. U.P.SINGH Senior Professor School of Electronics Engineering E-mail : [email protected] 8/13/2024 DR.UDAI P. SINGH 1 Introduction to Semiconductor and P-N Junction...

SEMICONDUCTOR DEVICES DR. U.P.SINGH Senior Professor School of Electronics Engineering E-mail : [email protected] 8/13/2024 DR.UDAI P. SINGH 1 Introduction to Semiconductor and P-N Junction theory Energy band diagram and classification, Semiconductor under equilibrium Condition (n0, p0 and intrinsic carrier concentration), Intrinsic and extrinsic semiconductor, Density states and Fermi level, Drift and diffusion current, Carrier generation and recombination, Excess carrier concentration, Continuity equation, Carrier lifetime, Qualitative description of charge flow in a p-n Junction, Junction Theory (concept of potential barrier, built in electric field, depletion layer width, and junction capacitance qualitative only), PN Junction operation under bias, and I~V relationship. 8/13/2024 DR.UDAI P. SINGH 2 SOLID : A solid consists of atoms, ions or molecules packed closely together, and the forces that hold them together in place give rise to the distinctive properties of the various kinds of solids. 8/13/2024 DR.UDAI P. SINGH 3  Crystalline Amorphous Polycrystalline 8/13/2024 DR.UDAI P. SINGH 4 8/13/2024 DR.UDAI P. SINGH 5  What is Semiconductor ? The conductivity of semiconductors occupies the area between conductors and insulators. The conductivity can range over many orders of magnitude. The conductivity of semiconductors can be adjusted through a number of means, each related to the physical properties of semiconductors. Typical methods for adjusting the conductivity of a semiconductor are: a. Temperature b. Purity (Doping) c. Optical excitation d. Electrical excitation. 8/13/2024 DR.UDAI P. SINGH 6  What is a Semiconductor?– contd. Low resistivity => “conductor” High resistivity => “insulator” Intermediate resistivity => “semiconductor” – conductivity lies between that of conductors and insulators Large variation in conductivity ?? Concentration of Electrons 8/13/2024 DR.UDAI P. SINGH 7  conductivity σ (S/cm)=1/(resistivity ρ (Ω -cm)) The conductivity of a semiconductor is generally sensitive to temperature, illumination, magnetic field and minute amounts of impurity atoms. 8/13/2024 DR.UDAI P. SINGH 8 Element semiconductors Germanium – In the early 1950s, was the major semiconductor material Silicon – Better device properties – High quality silicon dioxide – Cheaper – Abundant 9  The other important criteria for semiconductors are : – Energy Band Gap (eV) – Carrier concentration or density (cm-3) 8/13/2024 DR.UDAI P. SINGH 10  Energy Band Theory in Solids The physical basis of energy bands is quantum mechanics Atoms have … oquantized electron states odiscrete energies oa specific spatial distribution of electrons that is, electron orbitals have a certain shape s-type orbitals have spherical symmetry p-type orbitals have a lobe-shaped symmetry 8/13/2024 DR.UDAI P. SINGH 11 8/13/2024 DR.UDAI P. SINGH 12 8/13/2024 DR.UDAI P. SINGH 13  Energy Band Formation in Solids Energies of the electrons in an atom can’t have arbitrary value, but only some definite values given by quantum mechanical laws. The electrons have a well defined energy levels for a free atom. If the atom belongs to a crystal where it is surrounded by neighboring atoms, then the energy levels are modified. Pauli exclusion principle, which states that no two electrons or other fermions can have the same quantum numbers. 8/13/2024 DR.UDAI P. SINGH 14  Band formation Si: 1s2,2s2,2p6,3s2,3p2 d c b a Interatomic spacing r Conductor: Valence band partially filled (half full) Cu. or Conduction band overlaps the valence band Na: 1s2,2s2,2p6 3s1 Mg: 1s2,2s2,2p6 3s2 8/13/2024 DR.UDAI P. SINGH 16 8/13/2024 DR.UDAI P. SINGH 17  Band Overlap Example: Magnesium (Mg; Z =12): 1s22s22p63s2 – Might expect to be insulator; however, it is a metal – 3s-band overlaps the 3p-band, so now the conduction band contains 8N energy levels, while only have 2N electrons – Other examples: Zn, Be, Ca, Bi  Energy Band Diagram Conduction band Ec Band gap Eg Ev Valence band Energy band diagram shows the bottom edge of conduction band, Ec , and top edge of valence band, Ev.  Ec and Ev are separated by the band gap energy, Eg.  Semiconductor  Elemental : Si and Ge  Compound : GaAs, CdTe, InP etc.  Direct band gap and indirect band gap One distinguished feature of semiconductor is the location of the conduction band energy minimum w.r.t. the valence band maximum on the E-k diagram Slide 1-20  Measuring the Band Gap Energy by Light Absorption electron Ec photons Eg photon energy: hv > Eg Ev hole Eg can be determined from the minimum energy (hn) of photons that are absorbed by the semiconductor. Bandgap energies of selected semiconductors Semi- conductor InSb Ge Si GaAs GaP ZnSe Diamond Eg (eV) 0.18 0.67 1.12 1.42 2.25 2.7 6  The main characteristic of a semiconductor element is that it has four electrons in its outer or valence orbit. The unique capability of semiconductor atoms is their ability to link together to form a physical structure called a crystal lattice. 8/13/2024 DR.UDAI P. SINGH 22 8/13/2024 DR.UDAI P. SINGH 23 8/13/2024 DR.UDAI P. SINGH 24  The longest wavelength thatcan be absorbed by Silicon, which has the bandgap of 1.12 eV, is 1 mm. If the longest wavelength that can be absorbed by another material is 0.87 mm, then the bandgap of the material is ?? 8/13/2024 DR.UDAI P. SINGH 25 Direct Band gap Materials Actual Band structure are more complex. Plot between energy(E) and momentum(k=ħk) called E-K diagram. For Direct band materials, minimum of CB and Max of VB are same for k=0,so min energy transition without change in ‘k’. When electron fall from CB to VB give off energy difference Eg as photon of light. In direct semiconductor such as GaAs,GaP, an electron in the conduction band can fall to an empty state in the valence band, giving off the energy difference Eg as a photon of light. Example:GaAs,GaP  Indirect Band gap Materials In an indirect semiconductor such as Si, the electron in the conduction band cannot fall directly to the valence band but must undergo a momentum change as well as changing its energy.When the electron falls it must go some changes in energy as well as momentum. The electron may go to defect state (Ef) with in the band gap and involves a change in ‘k’. The energy given as heat to the lattice rather than photon. It is important for deciding which semiconductors can be used in devices requiring light output. Example: Silicon, Germanium  Intrinsic and Extrinsic Semiconductor  The semiconductor has a filled V.B. and empty C.B. at 0 oK  As the temperature is raised – increase in C.B. electrons – thermal excitation across the band gap  After the electrons are excited to the C.B., the empty states left in the V.B. can contribute to the conduction process.  The introduction of impurities has an important effect on the energy band structure and on the availability of charge carriers.  The concentration( cm-3) of conduction electrons and holes in a semiconductor can be modulated in several ways  By adding special impurity atoms  By applying an electric field  By changing the temperature (thermal excitation)  By irradiation (optical excitation) Bond Model Band Model Intrinsic carrier conc. (@RT) GaAs 2x106 cm-3 Si 1x1010 cm-3 Ge 2x1013 cm-3 Why the carrier conc. Is different for different semi- conductor? 8/13/2024 DR.UDAI P. SINGH 30 8/13/2024 DR.UDAI P. SINGH 31 8/13/2024 DR.UDAI P. SINGH 32 The extra electron on the phosphorous atom is easily removed and becomes a free electron without generating a hole. The phosphorous atom becomes positively charged (ionized). 8/13/2024 DR.UDAI P. SINGH 33 8/13/2024 DR.UDAI P. SINGH 34 8/13/2024 DR.UDAI P. SINGH 35  The boron atom ‘steals’ an electron from a neighboring Si atom to complete the four bonds with the surrounding Si atoms, generating a hole at the neighboring Si atom. The boron atom becomes negatively charged (ionized). All the dopants are ionized at room temperature!! 8/13/2024 DR.UDAI P. SINGH 36  P-type Semiconductor 8/13/2024 DR.UDAI P. SINGH 37 8/13/2024 DR.UDAI P. SINGH 38  The situation for the group III-V semiconductor is different Group VI impurities (S, Se,Te) acts as donors in GaAs, these impurities substitute the Gr V atoms (As) Group II impurities (Be,Zn,Cd) substitute for Gr III (Ga) in GaAs An interesting case arises when III-V compound is doped with Gr IV (e.g. Si) If the impurity resides on Gr III  Donors (more common) If the impurity resides on Gr V  Acceptors Such impurities are called Amphoteric Impurities. Semiconductor that can be made both n or p type are said to be Ambipolar 8/13/2024 DR.UDAI P. SINGH 39  FERMI ENERGY OR FERMI LEVEL In dealing with large number of particles, we are interested only in statistical behaviour of the group as a whole rather than the behaviour of each individual particle Likewise in crystal, the electrical characteristics will be determined by the statistical behaviour of a large number of electrons. Statistical Laws  Maxwell Boltzmann Probability Function - particles are considered to be distinguishable - with no limit to the numbers of particles allowed in each energy state ( behaviour of gas molecules in a container at very low pressure)  Bose Einstein Function - Particles are indistinguishable - no limit to the number of particles permitted in each quantum states ( e.g. behaviour of photons, black body radiation ) 8/13/2024 DR.UDAI P. SINGH 40  Fermi Dirac Distribution Function - particles are indistinguishable, identical - one particle is permitted in each quantum states - These particles must obey Pauli’s exclusion principle (e.g. electrons in a crystal, such particles are called FERMIONS Pauli’s Exclusion Principle: No two electron in an atom can be in the same quantum state The function f(E) specifies, under equilibrium conditions, the probability that an available 1 state at an energy E will be occupied by an f (E)  electron. 1  e ( E  EF ) / kT EF is called the Fermi energy or the Fermi 8/13/2024 level DR.UDAI P. SINGH 41  To understand the meaning of the distribution function and the Fermi Energy we can plot the distribution function vs Energy Consider T  0K For E < Ef 1 For E > Ef 0 8/13/2024 DR.UDAI P. SINGH 42 T>0K If E = Ef , then f(E) = 1/2  At 0K, the distribution function f(E) takes a simple rectangular shape. It implies that all states upto Ef are filled with electrons and states above Ef are empty  At temperature greater then 0K, some probability exists for states above Fermi Level to get filled with electrons  At any temperature (T1) , there is a probability f(E) that states above Ef are filled. A corresponding probability 1-f(E) exists that states below Ef are empty 8/13/2024 DR.UDAI P. SINGH 43  Fermi energy defines as that energy for which the probability of occupancy f(E) equals ½ OR An energy state at the Fermi Level has a probability of ½ being occupied. 1-f(E) is the probability that a state at energy E is unoccupied/empty. 8/13/2024 DR.UDAI P. SINGH 44  Intrinsic Semiconductor An insulator/intinsic semiconductor at T = 0 K has a filled valence band and an empty conduction band. The Fermi level lies midway between these bands. The Fermi–Dirac probability that an energy state E is occupied at T > 0 K is shown to the left.  Charge carriers in Semiconductor The distribution(w.r.t. energy) of the electrons in the CB is given by density of allowed quantum states times the probability that a state is occupied by an electron. n(E) = N(E) F(E) Density of quantum states in CB Fermi Dirac probability function The total electron concentration per unit volume in the CB is found by integrating the above equation over the entire CB energy n(E) =∫ N(E) F(E) dE The lower limit of the integration is Ec and the upper limit is taken as infinity 8/13/2024 DR.UDAI P. SINGH 47 The integral is complicated. For ease of calculation, one assume (E-Ef) > 3kT, with this limitation the F.D. distribution function can be approx. by Boltzman distribution 1 f (E)  1  e ( E  EF ) / kT For E > Ef The reason for restricting the energy levels to values greater than Ef is that an energy level E < Ef in above equation make the probability greater than 1, which is meaningless. N(E) is given as For E > Ec 8/13/2024 DR.UDAI P. SINGH 48  n or no= Nc exp (Ef-Ec/kT) Nc is effective density of states in the CB 8/13/2024 DR.UDAI P. SINGH 49 Similarly, the distribution (w.r.t. energy) of holes in the VB is the density of allowed quantum states in the VB multiplied by the probability that a state is not occupied by an electron The total hole conc. per unit volume is found by integrating this function over the entire VB energy Where, Commonly p(E), is written as po (equilibrium hole conc.) 8/13/2024 DR.UDAI P. SINGH 50  The Intrinsic carrier conc. 8/13/2024 DR.UDAI P. SINGH 51 8/13/2024 DR.UDAI P. SINGH 52 8/13/2024 DR.UDAI P. SINGH 53 8/13/2024 DR.UDAI P. SINGH 54 8/13/2024 DR.UDAI P. SINGH 55   The distance between the bottom of the CB and the Fermi energy is a logarithmic function of donor concentration  As the donor concentration increases, the F.L. moves closer to the CB.  Conversely, if the F.L. moves closer to the CB then the electron concentration in the CB is increasing 8/13/2024 DR.UDAI P. SINGH 56 8/13/2024 DR.UDAI P. SINGH 57 It is useful to express electron and hole densities in terms of the intrinsic carrier concentration ni and the intrinsic Fermi level Ei Also, 8/13/2024 DR.UDAI P. SINGH 58  From (1) Substituting in (3) Similarly, 8/13/2024 DR.UDAI P. SINGH 59  PROBLEMS Si at RT (300 K) contains an acceptor impurity concentration of NA =1016 cm-3. Determine the concentration of donor impurity atoms that must be added so that the Si is n-type and the Fermi energy is 0.2 eV below the CB edge. Given NC==2.8x1019 cm-3 = 2.8x1019 exp (-7.7221) = 2.8x1019 x 0.000443 = 1.24x1016 cm-3 ND= 1.24X1016 +NA =1.24X1016 + 1016 =2.24X1016 cm-3 Solve Prob. 15, Chap 1 8/13/2024 DR.UDAI P. SINGH 60 The intrinsic carrier concentration of Si sample at 300 K is 1.5x1016 m-3. If after doping, the number of majority carriers is 5x1020m-3, the minority carrier density is 4.50 x1011 m-3 Silicon bar is doped with donor impurities ND = 2.25x1015 atoms/cm3. Given the intrinsic carrier conc. of Si at RT is 5x1010 cm-3. Assuming the complete ionization, the equilibrium electron and hole concentrations are no= 2.25x1015 cm-3 po= 1x105 cm-3 8/13/2024 DR.UDAI P. SINGH 61 Electron density as a function of temperature for a Si sample with a donor concentration of 1015 cm-3. 8/13/2024 DR.UDAI P. SINGH 62  Figure (previous slide) shows electron density in Si as a function of temperature for a donor concentration of ND=1015cm-3.  At low temperatures, the thermal energy in the crystal is not sufficient to ionize all the donor impurities present. Some electrons are “frozen” at the donor level and the electron density is less than the donor concentration.  As the temperature is increased, the condition of complete ionization is reached,  nn=ND  As the temperature is further increased, the electron concentration remains essentially the same over a wide temperature range. This is the extrinsic region.  However, as the temperature is increased even further, one reach a point where the intrinsic carrier concentration becomes comparable to the donor concentration.  Beyond this point, the semiconductor becomes intrinsic.  The temperature at which the semiconductor becomes intrinsic depends on the impurity concentrations and the bandgap value 8/13/2024 DR.UDAI P. SINGH 63 8/13/2024 DR.UDAI P. SINGH 64  A Semiconductor device requires n-type material, it is to be operated at 400 K. Would Si doped with 1015 atoms/cm3 of As be useful in this application? Could Ge doped with 1015 cm-3 Sb be used? At 400K Si  1013 cm-3 Ge  5x1015 cm-3 8/13/2024 DR.UDAI P. SINGH 65  Drift and Diffusion Current Carrier motion is caused by two conditions :  The application of an electric field whose force accelerates the carriers and, second a difference of carrier concentration between two points causes carriers to move by diffusion from region of high conc. to regions of low conc.  Electrons and holes in semiconductors are in constant motion because of the thermal energy they receive. Since they are in motion, they are not associated with any particular lattice position.  At any one time, the electrons and holes move in random directions with a mean random velocity. Because of the motion in random directions, the current resulting from the motion of all the carriers in any one direction is zero.  The average distance between each collision is called the mean free path. 8/13/2024 DR.UDAI P. SINGH 66  Diffusion Current It is possible to have a non-uniform concentration of particles in a semiconductor. 8/13/2024 DR.UDAI P. SINGH 67 8/13/2024 DR.UDAI P. SINGH 68  Similarly for electrons 8/13/2024 DR.UDAI P. SINGH 69 Assume that, in an n-type semiconductor at T = 300K, the electron concentration varies linearly from 1 × 1018 to 7 × 1017 cm-3 over a distance of 0.1 cm. Calculate the diffusion current density if the electron diffusion coefficient is Dn = 22.5 cm2/s. The diffusion current density is given by =- - - = 8/13/2024 DR.UDAI P. SINGH 70  Drift Current 8/13/2024 DR.UDAI P. SINGH 71  The above discussion shows, a steady state drift speed has been superimposed upon the random thermal motion of the electrons. Such a directed flow of electrons constitutes a current. 8/13/2024 DR.UDAI P. SINGH 72  Current Density  N no of electrons are contained in a length L of a conductor. If it takes an electron a time T (sec) to travel a distance L meter in the conductor, the total no. of electrons passing through any cross section of wire in unit time is N/T  Thus the total charge per second passing any area is I(current) = Nq/T = Nq/L { Since L/T is the average or drift speed v(m/s) of the electron.}  The current density (J), current per unit area, J = I/A = Nq/LA  From figure, one can say that LA is the volume containing N electrons, and so N/LA is the electron concentration density ‘n’ 8/13/2024 DR.UDAI P. SINGH 73   J = nq = ρ where ρ = nq is the charge density.  Now  = μE and J = nq  J = nqμE = σE where σ = nqμ is the conductivity 8/13/2024 DR.UDAI P. SINGH 74  For semiconductor, we have electron and hole as carriers, having mobility μn and μp.  These particles move in opposite directions in an electric field E, but they are of opposite sign, the current of each is in the same direction.  Hence, the current density, J = (nμn + pμp)qE = σE where σ = (nμn + pμp)q 8/13/2024 DR.UDAI P. SINGH 75  Total Current 8/13/2024 DR.UDAI P. SINGH 76  Carrier Mobility and Velocity Mobility - the ease at which a carrier (electron or hole) moves in a semiconductor – Symbol: mn for electrons and mp for holes Drift velocity – the speed at which a carrier moves in a crystal when an electric field is present – For electrons: vd = mn E – For holes: vd = mp E  The electron in the semiconductor are moving rapidly in all directions due to thermal energy. The thermal motion of an individual electron may be visualized as a succession of random scattering from collisions with lattice atoms, impurity atoms etc. The random motion of electrons leads to zero net displacement of an electron over a sufficiently long period of time. The average distance between collisions is called the mean free path, and the average time between collisions is called the mean free time t. 8/13/2024 DR.UDAI P. SINGH 78 8/13/2024 DR.UDAI P. SINGH 79 8/13/2024 DR.UDAI P. SINGH 80  In Eq. 1 the mobility is related directly to the mean free time between collisions, which in turn is determined by the various scattering mechanisms. The two most important mechanisms are lattice scattering and impurity scattering. Lattice scattering results from thermal vibrations of the lattice atoms at any temperature above absolute zero. These vibrations disturb the lattice periodic potential and allow energy to be transferred between the carriers and the lattice. Since lattice vibration increases with increasing temperature, lattice scattering becomes dominant at high temperatures; hence the mobility decreases with increasing temperature. 8/13/2024 DR.UDAI P. SINGH 81  Theoretical analysis shows that the mobility due to lattice scattering μL will decrease in proportion to T–3/2. Impurity scattering results when a charge carrier travels past an ionized dopant impurity (donor or acceptor). The charge carrier path will be deflected because of Coulomb force interaction. The probability of impurity scattering depends on the total concentration of ionized impurities (the sum of the concentration of negatively and positively charged ions). However, unlike lattice scattering, impurity scattering becomes less significant at higher temperatures. At higher temperatures, the carriers move faster; they remain near the impurity atom for a shorter time and are therefore scattered less effectively. 8/13/2024 DR.UDAI P. SINGH 82  The mobility due to impurity scattering μI can be shown to vary as T3/2/NT, where NT is the total impurity conc. 1. The mobility depends very strongly on the temperature and on irregularities in the crystal, which may be due to the growing of the crystal or the impurities. 2. So long as the drift velocity is less than the thermal velocity, the mobility varies directly with, and is proportional to, the time between collisions. 3. If the drift velocity approaches the saturation velocity through an increase of the electric field intensity, the mobility decreases with an increase of the electric field. 8/13/2024 DR.UDAI P. SINGH 83  Electron mobility in silicon versus temperature for various donor concentrations. Inset shows the theoretical temperature dependence of electron mobility. 8/13/2024 DR.UDAI P. SINGH 84  Resistivity and Conductivity Fundamental material properties 1 1   q m n no  m p po  q m n  m p ni 1    Resistivity n-type semiconductor p-type semiconductor Generally, in extrinsic semiconductors, only one of the components is significant because of the many orders-of-magnitude difference between the two carrier densities  Find the room-temperature resistivity of an n-type silicon doped with 1016 phosphorus atoms/cm3 ρ = 0.48 Ω-cm 8/13/2024 DR.UDAI P. SINGH 89 Problem: 8/13/2024 DR.UDAI P. SINGH 90 8/13/2024 DR.UDAI P. SINGH 91  Current Flow Va Va I   R L 1    A  q  m n no  m p po     Aq  m n no  m p po  Va I  L Va E L I  Aq  m n no  m p po  E Conductivity of semiconductor  Conductivity of intrinsic semiconductor: σi = ni e( μe + μh)  Conductivity of Extrinsic semiconductor: For n – type semiconductor: σn = n e μe = ND e μe For p – type semiconductor: σp = p e μh = NA e μh Where ND and NA are the concentration of donor and acceptor impurity.  Generation and Recombination of charge carriers  The photon transition is a direct, band-to-band, generation/recombination process.  An electron from the conduction band falls back to the valence-band and releases its energy in the form of a photon (light)  Recombination  The reverse process, the generation of an electron-hole pair, is triggered by a sufficiently energetic photon which transfers its energy to a valence band electron which is excited to the conduction band leaving a hole behind.  The photon energy for this process has to be at least of the magnitude of the band- gap energy Eg.  The recombination in semiconductors eliminates free electrons and holes, setting up equilibrium in steady state.  Recombination in semiconductors can be categorized into two groups: radiative and non-radiative recombination.  Non-radiative recombination mechanisms in semiconductors are : I. auger recombination, II. recombination by defects, and III. surface recombination.  Radiative recombination occurs when an electron from the conduction band recombines with a hole in the valence band.  As a result of electron-hole recombination, a photon is emitted.  Radiative recombination is the radiative transition of the electron from the conduction band to the valence band, which involves optical processes: a)spontaneous emission, b)absorption or gain, and c) stimulated emission 8/13/2024 DR.UDAI P. SINGH 95   In a pure semiconductor the number of holes is equal to the number of free electrons.  Thermal agitation, however, continues to generate g new hole- electron pairs per unit volume per second, while other hole-electron pairs disappear as a result of recombination;  Free electrons fall into empty covalent bonds, resulting in the loss of a pair of mobile carriers.  On an average, a hole (an electron) will exist for tp(tn) sec before recombination.  This time is called the mean lifetime of the hole and electron, respectively. 8/13/2024 DR.UDAI P. SINGH 96 8/13/2024 DR.UDAI P. SINGH 97  The increase in hole concentration p equals that for the electron density n, the percentage increase for electrons in an n-type semiconductor (where electrons are plentiful) is very small.  On the other hand, the percentage increase in holes may be tremendous, because holes are scarce in an n-type crystal.  The radiation affects the majority concentration hardly at all.  The discussion is limited to the behavior of the minority carriers.  After a steady state is reached, at t = 0 , the radiation (external source) is removed.  One can show that the excess carrier density returns to zero exponentially with time.  For this, we must derive the differential equation which governs the hole concentration as a function of time for t > 0 (with no external excitation). 8/13/2024 DR.UDAI P. SINGH 98 8/13/2024 DR.UDAI P. SINGH 99 the excess, or injected, carrier density p’ is defined as the increase in minority concentration above the equilibrium value. Since p’ is a function of time, then 8/13/2024 DR.UDAI P. SINGH 100  the differential equation controlling p’ is The minus sign indicates that the change is a decrease in the case of recombination and an increase when the concentration is recovering from a temporary depletion Since the radiation results in an initial (at t > 0 is from From Sze and Lee book The excess carrier decreases exponentially 8/13/2024 DR.UDAI P. SINGH 101  Solution: lnp’ = - t/tp+c1 at t=0, c1 = lnp’(0) lnp’= -t/tp +lnp’(0) or p’/p’(0)= e –t/tp or p’ = p’(0) e –t/tp 8/13/2024 DR.UDAI P. SINGH 102 An N-type silicon sample is uniformly illuminated with light which generates 1020 electron-hole pairs per cm3 per second. The minority carrier life time in the sample is 1 microsecond. In the life steady state, the hole concentration in the sample is approximately 10x, where x is an integer. The value of x is ? Answer, we know Generation of carriers (g) = 1020 electron pairs per cm3 per second Under steady state condition Thereby, 8/13/2024 DR.UDAI P. SINGH 103  Continuity Equation If one disturb the equilibrium concentration of carriers in a semiconductor material, the concentration of holes or electrons will vary with time. Carrier concentration in the body of a semiconductor is a function of both time and distance. One can derive the differential equation which governs this functional relationship. This equation is based on the fact that charge can neither be created not destroyed. 8/13/2024 DR.UDAI P. SINGH 104  Consider the infinitesimal element of volume of area A and length dx in (figure below) within which the average hole concentration is p. Assume that the problem is 1D and that the hole current Ip is a function of x. If (figure) , the current entering the volume at x is Ip at time t and leaving at x + dx is Ip +dIp at the same time t, there must be dIp more coulombs per second leaving the volume than entering it (for a positive value of dIp). 8/13/2024 DR.UDAI P. SINGH 105  Hence, the decrease in number of coulombs per second within the volume is dIp. Since the magnitude of the carrier charge is q, then dIp/q equals the decrease in the number of holes per second within the elemental volume A dx. Remembering that the current density Jp = Ip / A, we have Decrease in hole concentration (holes per  unit volume)per second, due to current Ip Since charge can neither be created not destroyed, the increase in holes per unit volume per second, dp/dt, must equal the algebraic sum of all the increases listed above, or 8/13/2024 DR.UDAI P. SINGH 106  (since both p and Jp are functions of both t and x, then partial derivatives are used in this equation) Continuity equation (law of conservation of charge is given as Increase in holes per second = Holes generated per unit volume due to thermal generation - Holes lost per unit volume due to recombination - Holes leaving the bar per unit volume OR Rate of change of holes in dx = rate of generation & recombination + the rate at which holes enter the small piece less the rate at which they enter 8/13/2024 DR.UDAI P. SINGH 107