physics_semiconductor
 Semiconductor Classification of metals, conductors and semiconductors on the basis of conductivity(i) Metals: possess very low resistivity (or high conductivity).(ii) Semiconductors: They have resistivity or conductivity intermediate to metals and insulators.(iii) Insulators: They have high resistivity (or low conductivity). Elemental semiconductors: Si , Ge Compound semiconductorsInorganic: CdS, GaAs, CdSe, InP, etc. Organic: anthracene, doped pthalocyanines, etc. Organic polymers: polypyrrole, polyaniline, polythiophene, etc. Resistivity of semiconductors decreases with an increase in temperature. Energy Band:When a huge number of atoms are combined to form a solid however, the discrete energy levels are replaced by discrete ranges of energy, or energy bands. Valence band:The energy band which includes the energy levels of the valence electrons is called the valence band. Conduction band:The energy band above the valence band is called the conduction band.This band contains conduction electrons. This band is either empty or partially filled with electrons.Electrons present in this band take part in the conduction of current.This is the band energy where positive or negative mobile charge carriers exist.Electrons are negative mobile charge carriers.Holes are considered positive mobile charge carriers. Forbidden band or forbidden gapThis is the energy gap which is present between the valence band and conduction band. The separation between these two energy bands is called as forbidden band or forbidden gap. Holes means lack of an electron in the conduction band. Hole is a place where an electron can exist (ie. negative mobile charge carrier), and yet the electron ceases to exist at that particular location. Because the electron has the potential to be there and yet isn't there, it is referred to as positive mobile charge carrier. Energy band gap: The gap between the top of the valence band and bottom of the conduction band is called the energy band gap. When a small amount, (a few parts per million (ppm)) of a suitable impurity is added to the pure semiconductor, the conductivity of the semiconductor is increased manifold. Such materials are known as extrinsic semiconductors or impurity semiconductors. Metallic Conductors:The energy band above the valence band is called the conduction band. With no external energy, all the valence electrons will reside in the valence band. If the lowest level in the conduction band happens to be lower than the highest level of the valence band, the electrons from the valence band can easily move into the conduction band. Normally the conduction band is empty. But whenit overlaps on the valence band electrons can move freely into it. This is the case with metallic conductors. Intrinsic semiconductorExtremely pure semiconductor is called intrinsic semiconductor. Pure silicon, germanium are intrinsic semiconductor.a. An intrinsic semiconductor will behave like an insulator at T = 0 Kb. The conductivity of an intrinsic semiconductor depends on its temperature.c. In intrinsic semiconductors, the number of free electrons electrons (ne${n}_{e}$) is equal to the number of holes (nh)$\phantom{\rule{0.22em}{0ex}}\left({n}_{h}\right)$ .ne=nh=ni${n}_{e}={n}_{h}={n}_{i}$ni${n}_{i}$ is called apparent free particles. Extrinsic semiconductorsa. When a small amount, say, a few parts per million (ppm), of a suitable impurity is added to the pure semiconductor, the conductivity of the semiconductor is increased manifold. Such materials are known as extrinsic semiconductors or impurity semiconductors. b. The deliberate addition of a desirable impurity is called dopingc. The impurity atoms are called dopants. Such a material is also called a doped semiconductor. n-type semiconductora. An extrinsic semiconductor doped with pentavalent impurity, electrons become the majority carriers and holes the minority carriers. These semiconductors are known as n-type semiconductors. b. Pentavalent dopant is donating one extra electron for conduction and hence is known as donor impurity.c. With proper level of doping the number of conduction electrons can be made much larger than the number of holes. d. An extrinsic semiconductor doped with pentavalent impurity, electrons become the majority carriers and holes the minority carriers.ne≫nh${n}_{e}\gg {n}_{h}$wherene=total number of conduction electrons${n}_{e}=total\phantom{\rule{0.22em}{0ex}}number\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}conduction\phantom{\rule{0.22em}{0ex}}electrons$nh=total number of holes${n}_{h}=total\phantom{\rule{0.22em}{0ex}}number\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}holes$ p-type semiconductorp-type semiconductor is obtained when Si or Ge is doped with a trivalent impurity like Al, B, In, etc. Doping:The mixing of a small amount of pentavalent (e.g., phosphorus) or trivalent (e.g., aluminium) substance as impurity in a pure semiconductor (e.g., Ge, Si) is called doping. The deliberate addition of a desirable impurity is called doping and the impurity atoms are called dopants. Such a material is also called a doped semiconductor. Doping increases the conductivity of a semiconductor. Types of dopants a. Pentavalent (valency 5) : Arsenic (As), Antimony (Sb), Phosphorous (P), etc.b. Trivalent (valency 3) : Indium (In), Boron (B), Aluminium (Al), etc. Junction : The border where p-region meets n-region is called junction.Two important processes occur during the formation of a p-n junction: diffusion and drift. Depletion layer: At junction electron from n-side diffuse across junction and recombine with holes on p-side.When electrons leave n-region, it creates +ve icon on n-region, and on recombining with holes it created -ve icon in p-region.So the free charges disappear from the region near junction. So the region near the junction is depleted of free charges. This is called depletion layer. Diffusion current at p-n junction:During the formation of p-n junction, and due to the concentration gradient across p-sides , and n-sides, holes diffuse from p-side to n-side (p→n) and electrons diffuse from n-side to p-side (n→p). This motion of charge carries gives rise to diffusion current across the junction. Drift CurrentWhen an electron diffuses from N to P , it leaves behind an ionised donor on n-side. This ionised donor (positive charge) is immobile as it is bonded to the surrounding atoms. As the electrons continue to diffuse from N to P , a layer of positive charge (or positive space-charge region) on n-side of the junction is developed. When a hole diffuses from P to N due to the concentration gradient, it leaves behind an ionised acceptor (negative charge) which is immobile. As the holes continue to diffuse, a layer of negative charge (or negative space-charge region) on the p-side of the junction is developed. This space-charge region on either side of the junction together is known as depletion region as the electrons and holes taking part in the initial movement across the junction depleted the region of its Due to the positive space-charge region on n-side of the junction and negative space charge region on p-side of the junction, an electric field directed from positive charge towards negative charge develops. Due to this field, an electron on P-side of the junction moves to n-side and a hole on N-side of the junction moves to p-side. The motion of charge carriers due to the electric field is called drift current. Barrier PotentialThe loss of electrons from the n-region and the gain of electron by the p-region causes a difference of potential across the junction of the two regions. This potential tends to prevent the movement of electron from the n region into the p region. This is called a barrier potential. Forward biased : When an external voltage V is applied across a semiconductor diode such that p-side is connected to the positive terminal of the battery and n-side to the negative terminal it is said to be forward biased. Reverse-biased: When positive terminal of the external source is connected to n-side and negative terminal to p-side, the diode is said to be reversed biased. Breakdown of diode:When reverse bias is increased, at particular high value , the reverse current increases suddenly. This is called breakdown of diode.If reverse current is not controlled then p-n junction gets damaged.. The voltage at which diode breakdown occurs is called breakdown voltage. Threshold voltageIn forward bias , the current first increases very slowly, almost negligibly, till the voltage across the diode crosses a certain value. After the characteristic voltage, the diode current increases significantly (exponentially), even for a very small increase in the diode bias voltage. This voltage is called the threshold voltage or cut-in voltage . Special purpose p-n junction Diodesa) Zener diode b) Optoelectronic junction devices Optoelectronic junction devices(i) Photodiode(ii) Light emitting diode(iii) Solar cell Photo DiodeIt is special purpose semiconductor diode. while fabricating, a transparent window is kept to expose its junction to light radiations. It operates in reverse bias. Solar cellA solar cell is basically a p-n junction which generates emf when solar radiation falls on the p-n junction. Light emitting diodeIt is a heavily doped p-n junction which under forward bias emits spontaneous radiation. The diode is encapsulated with a transparent cover so that emitted light can come out. The wave length and intensity of the light emitted by LED depends upon forward bias and energy gap of the semiconductor. A lattice is an ordered array of points describing the arrangement of particles that form a crystal. Zener diode Zener diode as a voltage regulator. Zener diode is operated in in reverse biased conditions. This diode does not get damaged even when the voltage across this exceeds the reverse breakdown voltage. Zener diodes are heavily doped than ordinary diodes. They have extra thin depletion region. When voltage is applied more than the Zener breakdown voltage (range from 1.2 volts to 200 volts), the depletion region vanishes, and large current starts to flow through the junction. There is a crucial difference between an ordinary diode and a Zener diode. The depletion region regains its original position after removal of the reverse voltage in Zener diode whereas in regular diodes don’t, and hence regular diode gets destroyed. The value of R (resistance in series) should be such that the current through the Zener diode is much larger than the load current. This is needed to have good load regulation. So choose Zener current as five times the load current Transistor: structure and actionA transistor has three doped regions forming two p-n junctions between them. Three doped regions are:a. Emitter: It is heavily dopedb. Base: It is thin and lightly dopedc. Collector : It is largest of the three regions and doping level is between doping level of emitter and base. There are two junctions:a) E-B junction (emitter base junction)b) C-B junction (collector base junction) The emitter emits majority of the charge carriers into the base and base passes most of these charges into collector R-R Bias: If both junction are reversed bias then it called R-R bias.F-F Bias: If both junction are forward bias then it called F-F bias.F-R Bias: E-B junction is forward bias and C-B junction is reversed bias, then it called F-R bias. n-p-n transistor : It consists of very thin slice of p-type semiconductor sandwiched between two small blocks of n-type semiconductor. Emitter is given negative potential and Collector is given positive potential with respect to base. p-n-p transistor : It consists of very thin slice of n-type semiconductor sandwiched between two small blocks of p-type semiconductor. Base (B) : The center slice is called baseEmitter (E) : Left block is called emitter.Collector (C) : Right block is called collector. Emitter is given positive potential and Collector is given negative potential with respect to base. Action of transistor F-R biasedWhen a. E-B junction is forward biased and b. C-B junction is reverse biasd, then it is called F-R biased Herea. Large number of electrons enters the base. b. There emitter ejected electrons find reversed biased C-B junction and enter collector and less than 5% recombine with the holes of the base.c. So current in collector is much more than current in base. IE=IB+IC${I}_{E}={I}_{B}+{I}_{C}$ Transistor Action and its charateristics1. C-E Mode (Common Emitter Mode) : (a) Here emitter is common to both input and output.(b) Input is between base and emitter.(c) Output is between collector and emitter. 2. C-B Mode (Common base mode) 3. C-C Mode (Common Collector mode) Input Characteristics (C-E Mode)a) Here graph is plotted between base current (IB)$\left({I}_{B}\right)$ and base emitter voltage VBE${V}_{BE}$ when output voltage (VCE)$\left({V}_{CE}\right)$ is kept constant. b) For VBE${V}_{BE}$ less than barrier potential (0.6V) , IB${I}_{B}$ is nearly zero. IB${I}_{B}$ increases more rapidly when when VBE${V}_{BE}$ crosses barrier potential.c) Dynamic input resistance ri=(𝛥VBE𝛥IB)VCE=Constant${r}_{i}=\left(\frac{𝛥{V}_{BE}}{𝛥{I}_{B}}{\right)}_{{V}_{CE}=Constant}$ Output Characteristics (C-E Mode)a) It is a graph of output current versus output voltage. It is the graph of collector current (IC)$\left({I}_{C}\right)$ versus collector voltage (VCE)$\left({V}_{CE}\right)$ for fixed value of base current IB${I}_{B}$.b) When IB${I}_{B}$ is zero then IC${I}_{C}$ is nearly zero and we say that the transistor is in cut off region.c) For VCE<0.3V${V}_{CE}<0.3V$ , IC${I}_{C}$ increases with VCE${V}_{CE}$ and can be controlled by VCE${V}_{CE}$d) When VCE>VBE${V}_{CE}>{V}_{BE}$ (0.7V), C-E junction gets reversed biased and IC${I}_{C}$ gets saturated. It means that IC${I}_{C}$ does not depend upon VCE${V}_{CE}$ , but can be controlled by IB${I}_{B}$ . In this case we say that transistor is in active region.c) Dynamic output resistance ro=(𝛥VCE𝛥IC)IB=Constant${r}_{o}=\left(\frac{𝛥{V}_{CE}}{𝛥{I}_{C}}{\right)}_{{I}_{B}=Constant}$ Current Ratio in Transistor: 𝛼dc=(alpha)=ICIE${𝛼}_{dc}=\left(alpha\right)=\frac{{I}_{C}}{{I}_{E}}$ = Ratio of collector current to emitter current 𝛽dc=(beta, currrent gain)=ICIB${𝛽}_{dc}=\left(beta,\phantom{\rule{0.22em}{0ex}}currrent\phantom{\rule{0.22em}{0ex}}gain\right)=\frac{{I}_{C}}{{I}_{B}}$ = Ratio of collector current to base current Emitter: It is of moderate size and heavily doped. It supplies a large number of majority carriers for the current flow through the transistor. Base: This is the central segment. It is very thin and lightly doped. Collector: This segment collects a major portion of the majority carriers supplied by the emitter. The collector side is moderately doped and larger in size as compared to the emitter. base–emitter junction is forward-biased : It means that the p-doped side of the junction is at a more positive potential than the n-doped side, and the base–collector junction is reverse-biased. When the transistor is used in the cutoff or saturation state it acts as a switch. Transistor as an amplifierThe transistor works as an amplifier, with its emitter-base junction forward biased and the base-collector junction reverse biased. In the active state of the transistor the emitter-base junction acts as a low resistance while the base collector acts as a high resistance. OscillatorsOscillator is an electronic device that generated AC signal from DC source. Feedback Transistor BiasingBiasing is controlling the operation of the circuit by providing power supply.The biasing of the transistor is done differently for different uses. Thermionic emissionIn thermionic emission electrons are emitted from a metal due to its high temperature.
 Common question and answers 1. In the forward bias arrangement of a PN-junction diode the P-end is connected to the positive terminal of the battery 2. In a PN-junction diode the current in the reverse biased condition is generally very small 3. The cut-in voltage for silicon diode is approximately 0.6 V (approx.) 4. The electrical circuit used to get smooth dc output from a rectifier circuit is filter 5. PN-junction diode works as a insulator, if connected in reverse bias 6. The reverse biasing in a PN junction diode increases the potential barrier 7. The electrical resistance of depletion layer is large because it has no charge carriers. 8. If the forward voltage in a semiconductor diode is doubled, the width of the depletion layer will become half 9. The PN junction diode is used as a rectifier 10. At absolute zero, Si acts as insulator 11. By increasing the temperature, the specific resistance of a conductor increases and a semiconductor decreases 12. The energy band gap is maximum in insulators. 13. The part of a transistor which is most heavily doped to produce large number of majority carriers is emitter 14. Formation of covalent bonds in compounds exhibits wave nature of electron 15. A strip of copper and another germanium are cooled from room temperature to 80 K. The resistance of copper strip decreases and that of germanium increases. 16. The difference in the variation of resistance with temperature in a metal and a semiconductor arises essentially due to the difference in the variation of the number of charge carriers with temperature. 17. In the middle of the depletion layer of a reverse­ biased p-­n junction, the electric field is zero 18. What is the importance of energy gap in a semiconductor?Answer:a. Energy band gap is the difference between the conduction band and the valence band.b. When an electron of the semiconductor gets enough energy to cross this energy gap to reach in the conduction band, then the semiconductor can conduct electricity.c. An electron in the valence band cannot move to the conduction band until it is provided the sufficient amount of energy needed for the electron to cross the energy barrier between the two bands.d. Energy band gap differentiates the metals into conductor, insulator, semiconductor. 19On which factors does the electrical conductivity of a pure semiconductor depend at a given temperature?Answer:a. Forbidden energy gap : If forbidden energy gapis more then electrical conductivity of semiconductor will be less. If forbidden energy gap will be less then electrical conductivity will be more.b. Intrinsic charge carrier concentration: A pure (intrinsic) semiconductor is one where the number of holes is equal to the number of free electrons. Here the hole and electron concentrations are equal and is called the intrinsic concentration. The current flow in intrinsic semiconductor is due to the movement of electrons and holes in opposite directions. Their charges are of opposite sign, the current due to each is in the same direction. 20 (choose the correct option)A reverse biased diode, is equivalent to:(A) an off switch(B) an on switch(C) a low resistance(D) none of the above Answer: A 21 (choose the correct option) The potential barrier in p-n diode is due to:(A) depletion of positive charges near the junction(B) accumulation of positive charges near the junction(C) depletion of negative charges near the junction,(D) accumulation of positive and negative charges near the junction Answer : D 22 n-type semiconductor: When silicon or germanium crystal is doped with a pentavalent impurity such as phosphorus, arsenic, or antimony we get n-type semiconductor. 23What causes a larger current through a p-n junction diode when forward biased?Answer: Forward biased diode. Initially, the current is very low and then there is a sudden rise in the current. The point at which current rises sharply is shown as the ‘knee’ point on the I-Vcharacteristic curve. A diode effectively becomes a short circuit above this knee point and conduct a very large current. 24On which factors does the electrical conductivity of a pure semiconductordepend at a given temperature?AnswerElectrical conductivity of a pure semiconductor depends upon: a. The width of the forbidden bandb. Intrinsic charge carrier concentration Q.25Why is the conductivity of a n-type semiconductor greater than that of p-type semiconductor even when both of these have same level of doping?Answera. In n-type semiconductor electron are the majority carrier.b. In p-type semiconductor holes are the majority carrier.c. Under a given electric field, the mobility of electron is higher than that of hole so the conductivity of a n-type semiconductor greater than that of p-type semiconductor. 26Explain the importance of the depletion region in a p-n junction diode.AnswerThe importance of depletion layer in p-n junction diode1. In reverse bias the width of the depletion region increases and the p-n junction behaves like ahigh resistance. So depletion layer acts as a high resistence to charge carriers in reverse bias condition. 2. In forward bias the width of the depletion region decreases and the p-n junction offers a low resistance path allowing a high current to flow across the junction. 27Order of magnitude of energy gap in an intrinsic semiconductor is : 0.2 eV to 3.0 eV 28In an unbiased p-n junction, holes diffuse from the p-region to the n-region because the concentration of holes in the p-region is much higher than their concentration in the n-region. 29In a half wave rectifier, the a.c. input source of frequency 50 Hz is used. What is the frequencyof the output ? What will be the output frequency if a full wave rectifier was used for the same input frequency ?AnswerFrequency of input voltage is n = 50 HzOutput frequency of half wave rectifier = n = 50 HzOutput frequency of full wave rectifier = 2n = 2x50=100 Hz
 Questions Q.A.14.1In an n-type silicon : Holes are minority carriers and pentavalent atoms are thedopants. Q.A.14.2For p-type semiconductos : Holes are majority carriers and trivalent atoms are the dopants. Q.A.14.3Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (Eg)c$\left({E}_{g}{\right)}_{c}$, (Eg)Si$\left({E}_{g}{\right)}_{Si}$and (Eg)Ge$\left({E}_{g}{\right)}_{Ge}$ , then (Eg)c > (Eg)Si > (Eg)Ge $\left({E}_{g}{\right)}_{c}\phantom{\rule{0.44em}{0ex}}>\phantom{\rule{0.22em}{0ex}}\left({E}_{g}{\right)}_{Si}\phantom{\rule{0.22em}{0ex}}>\phantom{\rule{0.22em}{0ex}}\left({E}_{g}{\right)}_{Ge}\phantom{\rule{0.22em}{0ex}}$ Q.A.14.4In an unbiased p-n junction, holes diffuse from the p-region to n-region because hole concentration in p-region is more as compared to n-region. Q.A.14.5When a forward bias is applied to a p-n junction, it lowers the potential barrier. Q.A.14.6For transistor action:i). The base region must be very thin and lightly doped.ii). The emitter junction is forward biased and collector junction is reverse biased. Q.A.14.7For a transistor amplifier, the voltage gain is low at high and low frequencies and constant at midfrequencies. Q.14.8In half-wave rectification, what is the output frequency if the input frequency is 50 Hz. What is the output frequency of a full-wave rectifier for the same input frequency. A.14.8Input frequency = 50 Hz In half-wave rectification Output frequency = Input frequencyOutput frequency = 50 Hz In full-wave rectification Output frequency = 2 X Input frequencyOutput frequency = 2 X 50 Hz = 100 Hz Q.14.9For a CE-transistor amplifier, the audio signal voltage across the collected resistance of 2 kΩ $2\phantom{\rule{0.22em}{0ex}}k\Omega \phantom{\rule{0.22em}{0ex}}$ is 2V. Suppose the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1 kΩ$1\phantom{\rule{0.22em}{0ex}}k\Omega$ A.14.9GivenRC=2 kΩ ${R}_{C}=2\phantom{\rule{0.22em}{0ex}}k\Omega \phantom{\rule{0.22em}{0ex}}$RB=1 kΩ ${R}_{B}=1\phantom{\rule{0.22em}{0ex}}k\Omega \phantom{\rule{0.22em}{0ex}}$Vo=2 V ${V}_{o}=2\phantom{\rule{0.22em}{0ex}}V\phantom{\rule{0.22em}{0ex}}$RC=2 kΩ ${R}_{C}=2\phantom{\rule{0.22em}{0ex}}k\Omega \phantom{\rule{0.22em}{0ex}}$ICIB=100 $\frac{{I}_{C}}{{I}_{B}}=100\phantom{\rule{0.22em}{0ex}}$ To find : Vi=?${V}_{i}=?$ V0=IcRc${V}_{0}={I}_{c}{R}_{c}$Ic=V0Rc=22000=0.001 A${I}_{c}=\frac{{V}_{0}}{{R}_{c}}=\frac{2}{2000}=0.001\phantom{\rule{0.22em}{0ex}}A$ IB=IC100=0.001100=0.00001 A ${I}_{B}=\frac{{I}_{C}}{100}=\frac{0.001}{100}=0.00001\phantom{\rule{0.22em}{0ex}}A\phantom{\rule{0.22em}{0ex}}$ Vi=IB×RB=0.00001×1000 ${V}_{i}={I}_{B}×{R}_{B}=0.00001×1000\phantom{\rule{0.22em}{0ex}}$ ⟹Vi=0.001 V $⟹{V}_{i}=0.001\phantom{\rule{0.22em}{0ex}}V\phantom{\rule{0.44em}{0ex}}$ Q.14.10Two amplifiers are connected one after the other in series (cascaded). The first amplifier has a voltage gain of 10 and the second has a voltage gain of 20. If the input signal is 0.01 volt, calculate the output ac signal.A.14.10Input signal Voltage = Vi=0.01 V ${V}_{i}=0.01\phantom{\rule{0.22em}{0ex}}V\phantom{\rule{0.22em}{0ex}}$Voltage gain factor of first amplifier = V1=10${V}_{1}=10$Voltage gain factor of second amplifier = V2=20${V}_{2}=20$ Output signal Voltage = V0=Vi×V1×V2${V}_{0}={V}_{i}×{V}_{1}×V2$V0=0.01×10×20=2 Volt${V}_{0}=0.01×10×20=2\phantom{\rule{0.22em}{0ex}}Volt$V0=2 Volt${V}_{0}=2\phantom{\rule{0.22em}{0ex}}Volt$ Q.14.11A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm? A.14.11Band gap = 2.8 eVwavelength = 𝜆=6000 nm$𝜆=6000\phantom{\rule{0.22em}{0ex}}nm$ Energy associated with signalE=hc𝜆 $E=\frac{hc}{𝜆}\phantom{\rule{0.22em}{0ex}}$E=6.626×10-34×3×1086000×10-9 Joule$E=\frac{6.626×{10}^{-34}×3×{10}^{8}}{6000×{10}^{-9}}\phantom{\rule{0.22em}{0ex}}Joule$ E=6.626×10-34×3×1086000×10-9 Joule$E=\frac{6.626×{10}^{-34}×3×{10}^{8}}{6000×{10}^{-9}}\phantom{\rule{0.22em}{0ex}}Joule$ E=3.313×10-20J$E=3.313×{10}^{-20}J$ E=3.313×10-201.6×10-19 eV$E=\frac{3.313×{10}^{-20}}{1.6×{10}^{-19}}\phantom{\rule{0.22em}{0ex}}eV$E=0.20706 eV $E=0.20706\phantom{\rule{0.22em}{0ex}}eV\phantom{\rule{0.22em}{0ex}}$ As 0.20706 eV < 2.8 eV $0.20706\phantom{\rule{0.22em}{0ex}}eV\phantom{\rule{0.22em}{0ex}}<\phantom{\rule{0.22em}{0ex}}2.8\phantom{\rule{0.22em}{0ex}}eV\phantom{\rule{0.22em}{0ex}}$ so it cannot detect a wavelength of 6000 nm. Q.14.12The number of silicon atoms per m3${m}^{3}$ is 5×1028 $5×{10}^{28\phantom{\rule{0.22em}{0ex}}}$. This is doped simultaneously with 5×1022 $5×{10}^{22\phantom{\rule{0.22em}{0ex}}}$atoms per m3${m}^{3}$ of Arsenic and 5×1020 $5×{10}^{20\phantom{\rule{0.22em}{0ex}}}$ per m3${m}^{3}$ atoms of Indium. Calculate the number of electrons and holes.Given that ni=1.5×1016 m3 ${n}_{i}=1.5×{10}^{16}\phantom{\rule{0.22em}{0ex}}{m}^{3}\phantom{\rule{0.22em}{0ex}}$. Is the material n-type or p-type? A.14.12Arsenic is pentavalent Indium is trivalent total number of electronsne=5×1022-1.5×1016=4.9999985×1022 ${n}_{e}=5×{10}^{22}-1.5×{10}^{16}=4.9999985×{10}^{22}\phantom{\rule{0.22em}{0ex}}$ total number of holes=nh$={n}_{h}$ In thermal equilibriumnh×ne=n2i ${n}_{h}×{n}_{e}={n}_{i}^{2}\phantom{\rule{0.22em}{0ex}}$nh×4.9999985×1022=(1.5×1016)2 ${n}_{h}×4.9999985×{10}^{22}=\left(1.5×{10}^{16}{\right)}^{2}\phantom{\rule{0.22em}{0ex}}$nh=0.45×1010=4.5×109${n}_{h}=0.45×{10}^{10}=4.5×{10}^{9}$ As nh