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The basic building blocks of electronic circuits are devices that allow for controlled flow of electrons. These devices were mainly vacuum tubes, also known as valves, before the invention of the transistor in 1948. They included the anode (often called plate), and the cathode. The triode has three electrodes: cathode plate grid, cathode plate, and grid. Tetrode and pentode have 4 and 5 electrodes. The electrons are supplied in a vacuum tube by a heated cathode. This allows for controlled flow by changing the voltage between the different electrodes. The inter-electrode area must be vacuum because otherwise, the electrons moving in it may lose their energy due to collisions with air molecules. These devices allow electrons to flow only one direction, from the cathode towards the anode. These devices are commonly referred to simply as valves. These vacuum tube devices can be bulky and require high power.
They also operate at higher voltages (>100 V), and are therefore not reliable and long-lasting. The 1930’s were the year that the first solid-state semiconductor electronics were developed. It was then that it became apparent that certain solidstate semiconductors and their junctions allowed for the control of the flow and direction of charge carriers through them. Simple excitations such as heat, light or small voltage can alter the number of semiconductor mobile charges. The supply Chapter Fourteen SEMICONDUCTOR EELECTRONICS : MATERIALS DEVICES and SIMPLE CIRCUITS (c), NCERT must not be republished Physics 468. The semiconductor devices have charge carriers that are contained within them. In the older vacuum tubes/valves the mobile electrons were extracted from a heated cathode. They were then made to flow in an evacuated or vacuum. The semiconductor devices do not require external heating or large evacuation spaces.
They are small and lightweight, require little power and operate at low voltages. They also have long lives and high reliability. Even Cathode Ray Tubes, which are used in television and computer monitors on the principle vacuum tubes, are being replaced with Liquid Crystal Displays (LCD), monitors that support solid state electronics. Before the full implications of semiconductor devices were fully understood, a naturally occurring galena crystal (Lead sulfuride, PbS), with a metal contact attached, was used to detect radio waves. We will discuss the basics of semiconductor physics in the next sections. We will also describe a few circuits that illustrate their uses. 14.2 CLASSIFICATION METALS, CONDUCTORS and SEMICONDUCTORS Based on conductivity R 10-2 to 10-8 m.s. 102 – 108 S.m-1 (iii) Semiconductors are conductors or resistive materials that are intermediate to insulators and metals. R 10-5 to 106 m.s. 105- 10-6 S.m-1 (iii). Insulators.
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They have high resistance (or low conductivity). r 1011 – 106 m.s. 105 – 10-6 S m-1 (iii)Insulators: They have high resistivity (or low conductivity). The ranges of r and S are only indicative of magnitude. It is possible to go beyond these ranges. The relative resistivity values are not the only way to distinguish metals, semiconductors and insulators from one another. As we move through this chapter, there are other differences that will be apparent. This chapter focuses on semiconductors. Examples include: (i) Elemental and Ge semiconductors; (ii) Complex semiconductors. * Organic: anthracene and doped pthalocyanines * Organic polymers are polypyrrole (polyaniline), polythiophene and polypyrrole. The majority of current semiconductor devices are based upon elemental semiconductors Si and Ge, and compound inorganic (c. NCERT) semiconductors. However, after 1990, a few semiconductor devices using organic semiconductors and semiconducting polymers have been developed signalling the birth of a futuristic technology of polymerelectronics and molecular-electronics.
This chapter will focus on inorganic semiconductors, especially elemental semiconductors Si or Ge. These general concepts, which are used to discuss elemental semiconductors, can be applied to most compound semiconductors. Energy bands according to the Bohr Atomic Model. The orbit around which an atom revolves determines the energy of each electron. However, the atoms that form a solid are very close to one another when they come together. The outer orbits of electrons in neighbouring atoms could overlap or be very close to each other. This would mean that the nature of electron movement in solids is very different to one in an isolated atom. Each electron is unique inside the crystal and each one sees a different pattern of surrounding charges. Each electron will therefore have a different level of energy. These energy levels, which have continuous energy variation, are known as energy bands. The valence band is the energy band that includes the energy levels for the valence electrons.
The conduction band is the energy band that lies above the valence bands. All valence electrons, if there is no external energy, will be found in the valence spectrum. The electrons from the Valence band can move easily into the conduction bands if the lowest level of the conduction spectrum is lower than that of the valence spectrum. The conduction band is usually empty. However, electrons can freely move into the conduction band if it is adjacent to the valence. This is true for metallic conductors. There is a gap between the conduction and valence bands. This means that electrons in the latter band are all bound, and there is no free electron in the conduction. This makes the material an insulation. However, some electrons in the valence bands may be able to gain external energy to cross over the gap between conduction and valence. These electrons will then move into the conduction spectrum. They will also create vacant energy levels within the valence spectrum, where other valence elements can move. This creates conduction both due to electrons within the conduction band and vacancies in valence band.
Let’s look at what happens when Si or Ge crystals contain N atoms. The outermost orbit for Si is the third orbit (n=3), while it is for Ge the fourth orbit (n=4). The outermost orbit has 4 electrons (2s and 2p electrons). The crystal has 4N outer electrons. The outer orbit can hold 8 electrons (2s + 6p electrons). There are 8N energy states available for the 4N valence atoms. These 8N discrete energy levels may form a continuous or separate band depending on the distance between atoms in a crystal. (See box on Band Theory of Solids). The energy band for these 8N states splits into two energy bands depending on the distance between the atoms within the crystal lattices Si and Ge. These energy bands are separated by an E g (Fig. 14.1). The valence band is the lower band, which is (c). NCERT not to republished Physics 470 completely occupied at temperature absolute zero by the 4N valence neutrons. At absolute zero, the conduction band is empty. The other band that consists of 4N energy states is called the conduction bands.
BAND THEORY of SOLIDS Take into account that Si and Ge crystals contain N atoms. Each atom’s electrons will have distinct energies in different orbits. If all the atoms can be separated by a great distance, then the electron energy will remain the same. Crystals have atoms that are very close together (2 to 3 A), and the electrons interact with one another and with their neighbouring atomic cores. The electrons in the outermost orbit will feel the interaction more, while the core electron energies or inner orbit may not be affected. To understand the electron energies of Si and Ge crystals, we must only consider changes in the electron energies within the outermost orbit. The outermost orbit for Si is the third orbit (n=3), while it is the fourth orbit for Ge (n=4). The outermost orbit has 4 electrons (2s and 2p electrons). The crystal has 4N outer electrons. The orbit can hold 8 outer electrons (2s + 6p electrons). Out of the 4N electrons 2N electrons can be found in the 2Ns states (orbital quantum numbers l = 0), and 2N electrons in the 6N p-states.
As shown in Figure’s extreme right, certain p-electron states may be empty. This is true for well-separated or isolated atoms (region A in Figure). These atoms will eventually form a solid if they start to get closer to one another. Due to interaction between electrons from different atoms, the energies of the electrons in their outermost orbits may change (both increase or decrease). The 6N states of l = 1, originally having identical energies in isolated atoms, have now spread out to form an energy band (region B in Figure). The 2N states for the l=0, which had identical energies in isolated atoms, were also split into a second energy band (carefully refer to the region B in Figure), separated from the first by an energetic gap. The bands can merge at a smaller spacing. The lower energy state, which is a splitting from the upper nuclear level, appears to be lower than the higher state that has come down from the lower nuclear level. This region (region C) has no energy gap where the lower and higher energy states mix.
If the distance between atoms decreases further, the energy bands split up and are separated by Eg (region A in Figure). The total number available energy states 8N was re-adjusted between the bands (4N states in each of the lower and higher energy bands). The important point here is that the lower band has exactly the same number of states (4N) than the available valence electrons (four N). This means that the valence band, also known as the upper band, is fully filled and the upper band is empty. The conduction band is the upper band. (c) NCERT not be republished 471 Semiconductor electronics: Materials, Devices, and Simple Circuits. The conduction band’s lowest energy level is shown as EC. The highest energy level is shown in the valence bands as EV. There are many energy levels that are closely spaced between EC and EV, as illustrated in Fig. 14.1. 14.1. It can be small, large, or even zero depending on the material. These situations are shown in Fig. 14.2 is discussed below. Case I: This refers a situation as shown in Fig. 14.2(a). 14.2(a). One can have metal when the conduction band fills partially and the balanced band partially empty, or when the conduction-valance bands overlap.
Electrons can easily cross the conduction band if there is overlap in the valence and conduction bands. This creates a lot of electrons for electrical conduction. Conduction is possible when the valence band has a partial empty. This allows electrons to move from lower levels to higher ones. These materials have low resistance or high conductivity. FIGURE 14.2 The energy bands of (a), metals, (b), insulators, and (c) semiconductors. FIGURE 14.1 Energy band positions in a semiconductor at zero K. The lower band, known as the valence, is composed of tightly spaced, completely filled energy states. (c) Physics 472 Case 2: NCERT must not be republished Physics 472 case II: As shown in Figure. 14.2(b) A large band gap Eg exists (Eg> 3 eV). The conduction band does not contain electrons, so electrical conduction is impossible. The energy gap is too large to allow electrons to be excited by thermal excitation from the valence spectrum to the conduction bands. This is true for insulators. This is the case of insulators. 14.2(c). 14.2(c). The small band gap means that at room temperature, some electrons in the valence band can gain enough energy to cross the energy barrier and enter the conduction bands.
Even though they are small, these electrons can still move in the conduction bands. The resistance of semiconductors is lower than that of insulators. This section has a broad classification for metals, conductors, and semiconductors. The section that follows will explain the conduction process within semiconductors. 14.3 INTRINSIC SMICONDUCTOR Let’s take Ge and Si, whose lattice structures are shown in Fig. 14.3. These structures are known as the diamond-like structures. Each atom surrounds four of its nearest neighbors. We know that Si has four valence electrons and Ge has three. Every Si or Ge atom has a crystalline structure that allows it to share one of its four electrons with each of its nearest four neighbours. Each neighbour also gets one electron. These electron pairs are known as covalent bonds or valence bonds. It can be assumed that the two shared electrons shuttle back and forth between the atoms which hold them together strongly. The 2-dimensional representation of Si and Ge structures shown in Figure 14.4 is schematically illustrated in Figure 14.4. 14.3, which emphasizes the covalent bond.
This picture depicts an idealized scenario in which all bonds are intact and no bonds are broken. This is possible at low temperatures. These electrons become more energetic as the temperature rises. Some of them may also break-away, thereby contributing to conduction. As shown in Figure. 14.5(a). 14.5(a). This is known as a hole. The hole is an apparent free particle that has an effective positive charge. The number of electrons free in intrinsic semiconductors is equal to the number holes, nh. This is ne=nh=ni (14.1), where ni refers to intrinsic carrier concentration. Semiconductors have the unique property of having electrons and holes move. As shown in Figure. 14.5(a). Fig. 14.5(b). 14.5(b). An electron may be released from site 2’s covalent bond and jump to site 1 (hole). After such a jump, the hole at site 2 is now visible and site 1 has an electron. The hole appears to have moved from site 1 into site 2. The electron that was originally freed [Fig. Note that the electron originally set free [Fig. 14.5(a),] is not involved with this process of hole movement. The free electron is independent of the conduction electron, and generates an electron current I e in an applied electric field.
The motion of hole is a convenient way to describe the actual motions of bound electrons. This is true regardless of whether there is an empty bond in the crystal. These holes can move towards negative potential under the action of an electrical field. This creates the hole current Ih. The sum of the electron currents I e and I h is called the total current. I = Ie + Ih (14.2). It should be noted that in addition to the generation of conduction electrons, there is also a process of recombination where the electrons combine with the holes. The rate at which electrons are generated is equal to that of charge carrier recombination. Recombination is caused by an electron colliding to a hole. An intrinsic semiconductor will behave as an insulator at T=0 K, as shown in Figure. 14.6(a). 14.6(a). These thermally excited electrons, T > 0K, partially occupy conduction band. The energy-band diagram for an intrinsic semiconductor will look like the one in Fig. 14.6(b). Some electrons can be seen in the conduction band.
These electrons have come from the conduction band, leaving an equal number of holes. Example 14.1 C and Si have the same lattice structures. C is an insulator, while Si and Ge are intrinsic semiconductors. Solution The fourth, third, and fourth orbits contain the bonding electrons for Si, Ge, and C respectively. It is important to note that the conductivity of an intrinsic semiconductor is dependent on its temperature. However, it is very low at room temperature. These semiconductors are not suitable for the development of important electronic devices. It is therefore necessary to improve their conductivity. Impurities can be used to improve their conductivity. Conductivity is enhanced when a small amount of an impurity, such as a few parts per thousand (ppm), is added to the pure semiconductor. These materials are called extrinsic or impurity semiconductors. Doping is the deliberate addition of desirable impurities to a material. The impurity atoms that are added are called dopants. This material is also known as a doped semiconductor. The dopant must not cause distortion to the original semiconductor lattice. It occupies very few of the original sites for semiconductor atoms in the crystal.
To achieve this, the size of the dopant must be almost equal to the size of the semiconductor atoms. Two types of dopants are used to dop the tetravalent Si/Ge: (i) Pentavalent, valency 5,; like Arsenic/As, Antimony/Sb), Phosphorous/P, etc. FIGURE 14.6 (a). An intrinsic semiconductor behaves as an insulator at T = 0.0 K. (b) Four thermally generated electron/hole pairs are available at T > 0 K. The electrons are represented by the filled circles (or ), while the holes represent them by the empty fields (or ). (ii) Trivalent (valency 3); like Indium (In), Boron (B), Aluminium (Al), etc. Now we will discuss how doping affects the number of charge carriers and hence conductivity of semiconductors. Si and Ge belong to the fourth Periodic table group. Therefore, we select the dopant element of the nearby fifth or third groups, expecting and taking care that their dopant atoms are nearly equal in size to those of Si and Ge. As we will see, the pentavalent (or trivalent) dopants in Si and Ge produce two completely different types of semiconductors. (i) N-type semiconductor Let’s say we dope Si/Ge with a pentavalent component.
The reason is that the five electrons involved in bonding are considered part of the effective core atom. This means that the energy needed to ionize this electron is extremely small. It will move freely in the semiconductor’s lattice even at room temperature. To separate the electron from its atom, it is required to generate a 0.01eV of energy for germanium and 0.05eV for silicon. This contrasts with the energy needed to jump the forbidden spectrum (about 0.72eV for germanium, and about 1.1eV to separate silicon from its atom) at room temperature. The pentavalent dopant, also known as donor impurity, is one electron more for conduction. Dopant atoms make conduction electrons available at different levels depending on their doping level. This is independent from any temperature increase. The temperature does not affect the number of electrons generated by Si atoms. The total number ne of conduction electrons in a doped semiconductor is due to electrons generated intrinsically and donors, while nh refers to holes that are only created from donor electrons. The increase in electrons would cause a greater rate of recombination.
The number of holes would decrease further as a result. With the right level of doping, the number conduction electrons can be increased to exceed the number holes. Extrinsic thereforeDoped with pentavalent impureity, semiconductors make electrons the majority carriers while holes are the minor carriers. These semiconductors are known as n type semiconductors. We have ne >> (13.3) (ii) the p-type of semiconductor. This happens when Si or Ge are doped with a trivalent impurety like Al, B and In. The dopant is one electron less than Si and Ge. This means that it can form covalent bonds between three Si atoms, but not with the fourth Si atom. As shown in Figure. 14.8. 14.8 The hole can therefore be conduced. The trivalent foreign atom is effectively negatively charged when it shares a fourth electron with a Si atom. The dopant atom in p-type materials can therefore be considered core of one negative charge and its associated hole, as shown in Figure. 14.8(b). It is evident that one acceptor atom can give one hole. These holes are not only the intrinsically created holes, but the source of conduction electrons can also be generated.
For such a material, electrons and holes are both majority carriers. Extrinsic semiconductors containing trivalent impurity are referred to as p-type semiconductors. The recombination will decrease the number (ni) of intrinsically generated electrons to make p-type semiconductors. For p-type semiconductors, we have nh >> nu (14.4). Note that the crystal retains an overall charge neutrality because the charge of additional charge carrier is exactly equal to the charge of the ionised inner cores in the lattice. Extrinsic semiconductors have a greater chance of encountering majority current carriers because there are more of them. Doping, which adds a large amount of current carriers of one type to make majority carriers, indirectly reduces the intrinsic concentration of minor carriers. Doping can affect the semiconductor’s energy band structure. Extrinsic semiconductors can also have additional energy states as a result of donor impurities and acceptor impurities.
The energy band diagram for n-type Si semiconductor shows that the donor energy level ED is slightly lower than the bottom EC. This allows electrons to move into the conduction bands with very little energy. At room temperature, most donor atoms are ionized. However, very few Si atoms (10-12), get ionized. As shown in Fig. 14.9(a). Similarly, FIGURE 14.8 (a) Trivalent acceptor atom (In, Al, B etc.) Doped with tetravalent Si, Ge lattice to give p-type semiconductor. (b) A common schematic representation of p type material. It shows only the core of the substituent accepting element with one additional effective negative charge and its associated holes. (c) NCERT to not be republished 477 Semiconductor electronics: Materials, Devices, and Simple Circuits EXAMPLE 14.2 For p-type semiconductors, the acceptor energy level EA slightly exceeds the top EV in the valence band, as shown in Figure. 14.9(b). 14.9(b). Alternately, it is possible to say that a very low energy supply causes the hole at level EA to sink into the valence bands. When they receive external energy, electrons rise and holes sink.
At room temperature, the majority of acceptor atoms are ionized leaving holes in the band’s valence band. At room temperature, the density of holes within the valence band at room temperature is primarily due to impurities in the extrinsic silicon. The electron and hole concentrations in a semiconductor at thermal equilibrium are given by ne = ni 2. (14.5). Although the above description is improbable and hypothetical, it can help you understand the differences between metals, semiconductors, and insulators (extrinsic, intrinsic) in an easy way. The energy gap between the conduction and the valence bands determines the difference in resistivity between C, Si, and Ge. The energy gaps for C (diamond), Si, and Ge are respectively 5.4 eV 1.1 eV 0.7 eV. Sn is also a group IV element, but it is a metallic element because its energy gap is 0 eV. FIGURE 14.9 Energy bands for (a) n type semiconductor at T> 0K, and (b) the p-type semiconductor @ T> 0K. Let’s say that a pure Si crystal contains 5 x 1028 electrons per m-3. It is doped with 1 ppm of pentavalent as. Calculate the number electrons and holes.
Given that ni =1.5×1016 m-3. Solution It is important to note that thermally generated electrons are not as small as those produced by doping. Therefore, ne = ND Because ne = ni 2, the number of holes in a semiconductor device is nh = (2.25×1032 )/(5×1022) 4.5×109 m-3 (c). Physics 478 14.5 P-n JUNCTION Understanding the behaviour of a p-n junction will allow you to analyze the operation of other semiconductor devices. Now we will examine how a junction forms and how it behaves under external voltage (also known as bias). 14.5.1 P-n junction formation Let’s take a thin p type silicon (pSi) semiconductor wafer. A small amount of pentavelent impurety can be added to convert a portion of the pSi wafer into n-Si. A variety of processes can be used to form a semiconductor. The wafer now has a p-region, n-region, and a metalurgical junction between the p- and n regions. Diffusion and drift are two important steps in the formation of a P-N junction. It is known that the concentration (number) of electrons in an n-type semiconductor is higher than the concentrations of holes. In a p type semiconductor, the concentrations of electrons and holes are equal.
The concentration gradient between p and n sides creates a p junction. This causes holes to diffuse from one side to another (p – n), and electrons to diffuse from the other (n – p). The diffusion current across the junction is caused by the motion of charge. An electron that diffuses from n to p leaves behind an ionised donation on the n-side. As it bonds to its surrounding atoms, this ionised donor (positive charges) is immobile. As electrons continue to diffuse between n and p, a layer with a positive charge (or region of positive space-charge) is formed on the n-side. Similar to above, a hole that diffuses from p-n to n because of the concentration gradient leaves behind an immobile ionised acceptor (negative) that is immobile. As the holes diffuse, a layer or negative space-charge area (or negative charge) is formed on the p side of the junction. The space-charge area on each side of the junction is called the depletion zone. This is because the electrons and holes that took part in the initial movement of the junction depleted its free charge (Fig. 14.10).
The depletion area is approximately one-tenth of an inch thick. An electric field is created between the negative space-charge region at p-side and the positive space-charge area at n-side. This field causes an electron on the junction’s p-side to move to the n-side, and a hole at the junction’s n-side to move to the p-side. Drift is the motion of charge carriers caused by the electric field. A drift current is a current that moves in the opposite direction to the diffusion current (Fig. 14.10) starts. FIGURE 14.10 p-n junction formation process. Formation and working of p-n junction diode http://hyperphysics.phy-astr.gsu.edu/hbase/solids/pnjun.html (c) NCERT not to be republished 479 Semiconductor Electronics: Materials, Devices and Simple Circuits EXAMPLE 14.3 Initially, diffusion current is large and drift current is small. The diffusion process will continue, and the space-charge areas on either side of the junction will expand, increasing the electric field strength, and thus the drift current. The process continues until the drift current equals the diffusion current.
This is how a p–n junction is created. Under equilibrium, there is no net current in a p–n junction. A difference in potential between the junctions of the n- and p-regions is caused by the loss of electrons from one region and the gain of electron from the other. This potential’s polarity is so that there is no further flow of carriers, thereby creating an equilibrium condition. Figure 14.11 shows both the potential across the junction and the p-n junction at equilibrium. The n-material has lost electrons and the p material has gained electrons. Therefore, the n material is positive relative to p material. This potential is sometimes called a barrier potential because it tends to block electron movement from the n area into the p. Example 14.3 To get the p-n junction, can we physically join one slab of p type semiconductor to another slab of n-type? Solution: Solution No! The junction will act as a discontinuity to the flow of charge carriers. 14.6 SEMICONDUCTOR DIODE A semiconductor diode, a semiconductor diode. It has two terminals.
The symbol for a p-n junction diode can be seen in . An external voltage V can be applied to the diode to alter the equilibrium barrier potential. p–n junction diode with forward bias. An external voltage V is applied to a semiconductor diode so that the p–side is connected to a positive terminal and the n–side to the negative. [Fig. Forward bias is indicated by 14.13(a). The voltage drop across the junction and the p-side and the n-sides of the junction is negligible. This is due to the fact that the resistance in the depletion area – an area where there are no charged – is much higher than the resistance on the n-side or p-side. The direction of the applied voltage V varies from FIGURE 14.11 (a). Diode under equilibrium (V=0), (b). Barrier potential under no bias. FIGURE 14.12 (a) Semiconductor diode, (b) Symbol for p-n junction diode. (c) Physics 480 built in potential V0 not to be republished. The depletion layer’s width is decreased and the barrier height is lower [Fig. 14.13(b)]. 14.13(b) The effective barrier height for forward bias is V0 -V. The barrier potential will only be slightly reduced if the applied voltage is low.
Only a few carriers in the material, those that are at the highest energy levels, will have enough energy to cross this junction. The current will therefore be very small. The barrier height can be decreased by increasing the applied voltage. This will allow more carriers to have the energy they need. The current thus increases. The applied voltage causes electrons from the n-side to cross the depletion area and reach the p-side (where there are minority carries). Similar to the above, holes from the p-side cross over the junction and reach n-side (where are they minority carries). Minority carrier injection is a process that occurs under forward bias. The junction boundary is where the concentration of minority carriers increases dramatically compared to locations further away from the junction. The concentration gradient causes the injected electrons from p side to diffuse from the junction edge to the other end. The injected holes on the n side diffuse similarly from the junction edge to the other end (Fig. 14.14). The current is created by the motion of charged carriers.
The sum of the hole diffusion current and electron diffusion currents is called total diode forward. This current’s magnitude is typically in the mA range. 14.6.2 p–n junction diode with reverse bias. When an external voltage (V) is applied across the diode, such that n–side is positive and negative, it is called reverse biased [Fig.14.15a] The depletion area is where the applied voltage drops most. The direction of the applied voltage is the same as that of the barrier potential. The change in the electric fields causes the barrier height to increase and the depletion area to widen. [Fig.] The effective barrier height for reverse bias is V0 + V. 14.15(b)]. 14.15(b) The junction’s electric field direction is such that electrons from p-side and holes on n side will drift towards the junction in their random motion. They will then be swept into its majority zone. Current is created by the drift of carriers. The drift current is only a few uA. This is because of the movement of carriers across the junction from their minority side towards their majority side.
Although the drift current can be seen under forward bias, it is much lower (uA), compared to current due to injected carrier currents which are usually in the mA range. The applied voltage is not a significant determinant of the diode’s reverse current. Even a low voltage can sweep minor carriers from one end of the junction to another. FIGURE 14.13 currently shows (a) forward bias p-n junction diode, (b) barrier potential (1) without battery and (2) low battery voltage and (3) high voltage battery. FIGURE 14.14 Forward bias minority carrier injection. (c) NCERT not be republished. FIGURE 14.14 Forward bias minority carrier injection. The magnitude of the applied voltage is not limiting, but it is limited by the concentration of the minor carrier on either side. The current under reverse bias can be voltage independent up to a critical voltage known as Vbr (breakdown voltage). The diode’s reverse current rises sharply when V is greater than Vbr. A slight increase in the bias voltage can cause a significant change in current.
The p-n junction can be destroyed if the reverse current isn’t limited by an external circuit below its rated value. Overheating can cause the diode to melt if it exceeds its rated value. If the forward current exceeds its rated value, this can occur even for diodes under forward bias. Fig. shows the circuit arrangement to study the V-I characteristics (i.e. the variation in current as a function applied voltage). 14.16(a), and (b). 14.16(a) and (b). The current value is indicated for different voltages. As shown in Fig. 14.16(c). 14.16(c). As you can see in Figure. 14.16(c), shows that forward bias causes the current to increase slowly until the voltage across diodes crosses a certain level. The characteristic voltage is reached and the diode current rises significantly (exponentially) even with a small increase in diode bias voltage. This voltage is known as the threshold voltage, or cut-in voltage. It is 0.2V for germanium and 0.7V for silicon diodes. The current for the diode with reverse bias is very small (uA), and almost stays constant with a change in bias. This current is known as reverse saturation current.
The current can suddenly increase in special cases when there is a high level of reverse bias (breakdown voltage). The special diode action is described in Section 14.8. The reverse saturation current area is not the place where the general purpose diode can be used. As you can see, the p-n junction diode primarly permits current to flow only in one direction (forward bias). As compared to reverse bias resistance, the forward bias resistance is lower. This property is useful for rectifying ac voltages, as we will discuss in the next section. We define dynamic resistance for diodes as the ratio of small voltage changes to small current changes 14.7 APPLICATION JUNCTION DIIODE AS A REECTIFIER The V-I characteristic is a junction diode that allows current to flow only when it’s forward biased. If an alternating voltage is applied to a junction diode, current will only flow in the part of the cycle where the diode has been forward biased. This property can be used to rectify alternating currents. The circuit used for this purpose, called a rectifier, is also used.
A series of diodes connected to a load will produce an alternating voltage. However, the load will only be affected by the half-cycles of the ac input when the diode has been biased forward. Fig. 14.18 is also known as a half-wave rectifier. The secondary of a transformer supplies the required ac voltage between terminals A or B. If A is positive, the diode conducts. If A is negative, the diode will conduct if it is reverse-biased. Practically, the reverse saturation current of a diode can be considered to be zero. To protect the diode against reverse breakdown, the reverse breakdown voltage must be sufficient higher than the peak voltage at the secondary transformer. As shown in Fig. 14.18(b) – there is no current in negative halfcycle. The output voltage is again obtained in the next positive cycle. The output voltage is therefore rectified, even though it is still variable. This circuit is known as a half-wave rectifier because the rectified output is only half the input ac wave.
The circuit shown in Fig. 14.19(a) gives the output rectified voltage that corresponds to both the positive and negative halves of the ac cycles. It is also known as full wave rectifier. The p-sides of the diodes are connected with the ends of secondary transformer. The output is measured between the common point of the diodes and midpoint of transformer’s secondary. The secondary of a transformer with full-wave rectifier is equipped with a centre tap. This transformer is also known as a centre-tap transformer. As shown in Fig.14.19 (c), the voltage rectified each diode only half of the total secondary voltage. Each diode only rectifies for half of the cycle. The two other diodes do the same for alternate cycles. The output from their common terminals to the transformer’s centretap becomes a full wave rectifier output. You will also notice that there is a circuit for full-wave rectifier, which doesn’t require a centretap transformer, but does require four diodes. Let’s say that the input voltage to A relative to the centre tap is at any moment positive. As shown in Figure 14.19(b), it is obvious that voltage at B will be negative at that moment because it is out of phase.
Therefore, diode 1 becomes forward biased and conducts while D2 is reverse biased and not conducting. As shown in Figure 14.19(c), we see an output current and a voltage across the load resistor, RL during the positive half cycle. The voltage at B will be positive during the ac cycle, when A’s voltage becomes negative relative to the centre tap. This is the part of the cycle where diode 1 would not conduct, but diode 2 would. It would give an output current and voltage (across Rl) during the negative half of the input AC. This means that we can get the output voltage in both the positive and negative half of the cycle. This circuit is more efficient than the halfwave rectifier for rectified voltage or current. The rectified voltage comes in the form pulses in the shape of half-sinoids. It is not unidirectional, but it has a constant value. A capacitor is usually connected to the output terminals of the pulsating voltage (parallel the load RL) in order to get steady dc output. You can also use an inductor connected in series to RL for the exact same purpose.
These additional circuits are known as filters because they appear to remove the ac ripple from the voltage and provide a pure dc current. We will now discuss the role that capacitors play in filtering. It charges when the voltage across it rises. It will remain charged up to the maximum voltage of the rectified output if there is no load. It is discharged by the load and its voltage begins to drop. It is then charged to its peak value in the second half-cycle of rectified output (Fig. 14.20). The time constant is the rate at which the voltage drops across the capacitor. It is determined by the inverse product C and the effective resistivity RL in the circuit. Large values of C are necessary to make the time constant. So capacitor input filters use large capacitors. Capacitor input filters produce a voltage that is close to the peak voltage of rectified voltage. This filter is used most often in power supplies. This section will discuss devices that are essentially junction diodes, but have been developed for various applications.
Zener diode This special-purpose semiconductor diode is named after its inventor C. Zener. It can operate in reverse bias in the breakdown area and is used as a voltage regulator. Fig. 14.21(a). By heavily doping the junction’s n- and p- sides, Zener diode can be made. This results in a very thin depletion area (10-6m) and an extremely high electric field (5x106V/m even for a small reverse bias voltage (about 5V). Figure. 14.21(b). 14.21(b). A small change in reverse bias voltage can cause a significant change in current after reaching the breakdown voltage Vz. This means that the Zener voltage is constant even though current through Zener diodes fluctuates over a wide range. This property is used to regulate supply voltages, ensuring they remain constant. Let’s look at how reverse current suddenly increases at breakdown voltage. The reverse current is caused by the flow electrons (minority carrier) from p to n and holes between n and p. This is why the electric field at junction becomes more significant.
If the reverse bias voltage Vz is greater than 0, then the electric field strength can pull the valence electrons out of the host atoms on p-side, which are then accelerated to the n-side. These electrons are responsible for the high current seen at the breakdown. Internal field emission, also known as field ionization, is the process of releasing electrons from host atoms. Field ionization requires an electric field of at least 106 V/m. FIGURE 14.20 (a), A full-wave rectifier equipped with a capacitor filter. (b) Input voltage and output voltage for rectifier in (a). FIGURE 14.21 Zener diode, (a) symbol, (b) I-V characteristics. (c) NCERT not republished Physics 486 EXAMPLE 14.5. FIGURE 14.22 Zener Diode as DC Voltage Regulator Zener diode is a voltage regulator. We all know that rectifiers’ rectified output can fluctuate when their ac input voltage changes. A Zener diode is used to get a constant voltage from the unregulated dc output of a rectifier. Fig. 14.22. 14.22 The current through the Zener diode and Rs will increase if the input voltage is higher.
This causes a voltage drop across Rs to increase without any changes in Zener diode voltage. Because the Zener voltage is constant in the breakdown area, even though the current through Zener diode changes, the Zener voltage stays the same. The current through the Zener diode and Rs also drops if the input voltage is lower. The voltage drop across Rs is decreased without any changes in the Zener diode’s voltage. Any increase/ decrease in input voltage causes an increase/ decrease in voltage drop across Rs. The Zener diode acts like a voltage regulator. The output voltage required and the series resistance Rs must be considered when selecting the Zener diode. Example 14.5 A Zener regulated power source uses a Zener diode with a VZ of 6.0 V for regulation. The load current should be 4.0 mA, and the unregulated input should be 10.0 V. What value of the series resistor RS should it be? Solution: The value of RS should not be greater than the load current. This is necessary to ensure load regulation.
Zener current should be five times the load current. I Z = 20mA. The total current through RS therefore is 24 mA. The voltage dr op across RS ranges from 10.0 – 6.0 = 4.0V. This means that RS = 4.0V/(24 x 103) A = 167. 150 is the closest value for a carbon resistor. A series resistor of 150 would be appropriate. It is not important that the resistor’s value is slightly different. What is important is that I Z is sufficient to exceed I L. 14.8.2 Optoelectronic Junction Devices We have seen how a semiconductor junction device behaves under electrical inputs. This section will discuss semiconductor diodes that are generated from photons (photoexcitation). These devices are known as optoelectronics devices. The following optoelectronic gadgets will be examined: (i) Photodetectors are devices that detect optical signals. (iii) Light emitting Diodes (LED), which convert electrical energy to light. (iii). Photovoltaic devices that convert optical radiation into electricity (solar cell). (c) NCERT must not be republished 487 Semiconductor electronics: Materials, Devices, and Simple Circuits EXAMPLE 14.6 (i) Photodiode. A photodiode is a special purpose pn junction diode that has a transparent window to let light fall on it.
It operates under reverse bias. The absorption of photons results in the generation of electron-hole pairs when the photodiode’s surface is illuminated by light (photons), which has an energy (hn) greater that the energy gap (Eg) of the semiconductor. The diode is designed so that the generation and absorption of photons takes place within or near the diode’s depletion region. The junction’s electric field causes electrons and holes to be separated before they can recombine. The direction of the electrical field is such that electrons reach the n-side while holes reach the p-side. An emf is formed when electrons are captured on the n-side while holes are collected at the p-side. Current flows when an external load is connected. Photocurrents are proportional to the intensity of incident sunlight. The photocurrent’s magnitude is dependent on the intensity of the incident light. If a reverse bias is used, it is easier to see the current’s change with light intensity. Photodetector can also be used to detect optical signals.
The circuit diagram for measuring I-V characteristics of a Photodiode. 14.23(a), and a typical I.V characteristics in Fig. 14.23(b). 14.23(b). Why is it necessary to use photodiodes with reverse bias? Take the example of an n-type semiconductor. The majority carrier density (n), is clearly much higher than the minor hole density (p), (i.e., not >> p). Let the excess electrons or holes that are generated by illumination be n and, respectively, p. Here n’ + n p’ = p+ p. The electron and hole concentrations at each illumination are n’, p’, and n, while p is the carriers concentration in the absence of illumination. Keep in mind that n = p, and n >>, p. FIGURE 14.23 (a), An illuminated photodiode with reverse bias, (b), I-V characteristics for a photodiode at different illumination intensities I 4 > 3 > 2 > 1. * It is important to note that in order to create an electron-hole pair, some energy must be expended (photoexcitation or thermal excitation). When an electron and a hole combine, the energy is released as light (radiative or non-radiative) or heat. It all depends on the semiconductor used and the fabrication method used to make the p-n junction. In order to fabricate LEDs, semiconductors such as GaAs and GaAsGaP must be used.
The fractional change in majority carriers (i.e. n/n), would be less than the fractional change in minority carriers (i.e. p/p). We can generally say that the fractional changes due to photo-effects on minority carrier dominated reverse biased current are more easily quantifiable than the fractional changes in the forward bias current. Photodiodes should be used in reverse bias conditions to measure light intensity. (iii) Light emitting Diode This is a highly doped pn junction that emits spontaneous radiation under forward bias. To let the light emitted from the diode out, it is enclosed with a transparent cover. Forward biased diodes send electrons from n to p (where they’re minority carriers), and holes from p to n (where the are majority carriers). The junction boundary is where the concentration of minority carriers rises relative to the equilibrium concentration, i.e. when there is no bias. Excessive minority carriers can be found at the junction border on either side of it, where they recombine with the majority carriers close to the junction. Photons are released as a result of recombination.
Emitted photons have an energy that is equal or slightly lower than the band gap. The intensity of light emitted by a diode with a small forward current is low. The forward current increases, and the intensity of light rises to a maximum. An increase in forward current will result in a decrease in light intensity. The light emitting efficiency of LEDs is maximized because they are biased. A LED’s V-I characteristics are similar to a Si junction diode. However, the threshold voltages for each colour are slightly different and much higher. The reverse voltages of LEDs is very low, usually around 5V. It is important to ensure that they do not have high reverse voltages. Commercially available LEDs can emit red, yellow and orange light. A visible LED-producing semiconductor must have at least a band gap between 1.8 and 0.4 eV. The visible spectrum of visible light ranges from 0.4 um to about 0.7 um, or about 3 eV up to 1.8. For making different colours of LEDs, the compound semiconductor Gallium Arsenide-Phosphide (GaAs1xPx) is used.
Red LEDs are made from GaAs0.6P0.4 (Eg = 1.9 eV). Infrared LEDs are made from GaAs (Eg = 1.4 eV). These LEDs are used extensively in remote controls, burglar alarm system, optical communication, and many other applications. The development of white LEDs that can replace incandescent lamps is a major focus of research. The following benefits of LEDs over traditional incandescent low-power lamps are apparent: (i) They use less power and operate at a lower voltage. (iii) Rapid action with no warm-up required. (iii). The wavelength of the emitted light ranges from 100 A to 500 A, or it is almost (but not quite) monochromatic. (iv) Robustness and long life. (v) Quick on-off switching capabilities. (c) NCERT not be republished 489 Semiconductor electronics: Materials, Devices, and Simple Circuits (iii). A solar cell is essentially a pn junction that generates emf from solar radiation. It operates on the same principle as the photodiode. However, no external bias is used and the junction area for incident solar radiation is kept larger because we are more interested in power.
A p-Si wafer measuring approximately 300 um is removed. Then, a thin layer of n-Si (0.3 um), is grown one-side through diffusion. The back contact is applied to the other side of pSi. A metal finger electrode (or metallic Grid) is placed on top of the n-Si layers. This serves as a front contact. Because the metallic grid takes up only a small portion of the cell’s area (15%), light can pass through the top. When light hits a solar cell, the generation of emf occurs through three processes: separation, generation and collection. (i) The generation of eh pairs by light (with Eg > hn) near the junction. (ii) The separation of electrons from holes due to the electric field in the depletion area. The electrons are swept to the n side and the holes to the p side; (iii), the front contact collects the electrons to the n-side, while the back contact collects the holes to the p-side. Photovoltage is then created when n-side and p-side become positive. As shown in the Fig. 14.25(a), a photocurrent L flows through the load. The Fig. shows typical I-V characteristics for a solar cell.
Because a solar cell doesn’t draw current, but provides the same load, it is not able to draw current. Solar cell fabrication is possible with semiconductors having a band gap of 1.5 eV. You can make solar cells with semiconductors such as Si (Eg=1.1 eV), GaAs(Eg= 1.43 eV), GaAs (1.45 eV), and CdTe (1.45 eV), as well as CuInSe2 (2Eg= 1.04 eV), among others. For the fabrication of solar cells, the most important criteria are: (i) Band gap (1.0-1.8 eV),(ii) high optical absorbtion (104 cm-1), CdTe (Eg = 1.45 eV), CuInSe2 (Eg = 1.04 m-1), (iii), electrical conductivity, (iv), availability of the raw materials, and (v). cost. A solar cell does not necessarily need sunlight. Any light that has photon energies greater then the bandgap can be used. Solar cells can be used to power electronics in satellites and space vehicles, as well as to supply power to calculators. Research is ongoing on the production of inexpensive photovoltaic cells that can produce large amounts of solar energy. FIGURE 14.24 (a). A typical p-n junction solar cells; (b) Cross sectional view. FIGURE 14.25 (a).
A typical illuminated, p-n junction solar cells; (b) I/V characteristics of a sun cell. (c) Physics 490 EXAMPLE 14.7 Example 14.7 NCERT must not be republished Physics 490 EXAMPLE 14.7 Solution. The solar radiation spectrum that we received is shown in Figure. 14.26. FIGURE 14.26. The maxima is close to 1.5 eV. Photo-excitation is achieved by hn > Eg A semiconductor with a band gap of 1.5 eV or less is more likely to have a higher solar conversion efficiency. Eg for silicon is 1.1 eV, while it is 1.53eV for GaAs. Despite having a higher band gap, GaAs has a higher absorption coefficient than Si. CdSe or CdSe can only be used for photo-conversion if they have a high energy component. It begs the question: Why not use PbS (Eg0.4 eV), which meets the conditions hn> Eg for n maxima that correspond to the solar radiation spectrum? This will ensure that most of the sun’s radiation is absorbed by the top layer of the solar cell, and not in the depletion area. We want photo-generation only to take place in the junction area to achieve effective electron-hole separation due to the junction field. 14.9 JUNCTION TRAISTOR J. Bardeen, W.H. Brattain, of Bell Telephone Laboratories U.S.A. invented the transistor.
It was a point contact transistor. William Schockley invented the first junction transistor, consisting of two back to back p-n junctions in 1951. It was simply known as transistor until the invention of the junction transistor. Over the years, new types of transistors have been invented. To distinguish it from the others, it is now known as the Bipolar Junction Transistor. When there is no confusion, transistor (c. NCERT) is still used a lot. 491 Semiconductor electronics: Materials, Devices and Simple Circuits: Semiconductor Electronics: Semiconductor Electronics. Our study will be limited to BJT so we will use the term transistor for BJT. 14.9.1 Transistor structure and action: A transistor is composed of three regions doped that form two p-n junctions. As shown in Fig., there are two types. 14.27. 14.27 (ii) P-n-p transistor : Two segments of p type semiconductor (termed emitter and collector) can be separated by a segment n-type (termed base). Figure shows the schematic representations of an N-p-n configuration and a P-n-p configuration.
The schematic symbols that represent p-n–p and n–p-n transistors are [Fig. 14.27(b), the arrowhead indicates the direction of the conventional current in the transistor. Below is a brief description of each segment of a transistor: * Emitter: This section is on the one side of the transistor as shown in Fig. 14.27(a). It is moderately large and heavily doped. It provides a large amount of majority carriers to the transistor’s current flow. * The base is the central segment. It is thin and lightly doped. * Collector: This section collects the bulk of the carriers supplied by emitters. The collector side is slightly doped and has a larger size than the emitter. As we have seen in the case a p -n junction, there is a depletion area that forms around the junction. Depletion regions in a transistor are formed at the emitter-base junction and the basecollector junction. To understand the actions of a transistor we must consider the nature depletion areas formed at these junctions. When proper voltages are applied to the terminals of the transistor, charge carriers can move through different areas. Different uses require different biasing techniques.
There are two ways to use the transistor. It was originally designed to be an amplifier. This device produces a larger version of a signal. Later, its function as a switch was equally important. These functions are both important and we will examine how the transistor biases to accomplish these mutually exclusive functions. We will first examine what gives the transistor its amplifying abilities. The transistor functions as an amplifier with its emitter-base junction forward biased, and the base-collector Junction reverse biased. Fig. 14.28 shows the situation. VCC and VEE are used to create the biasing. The transistor is considered active when it is biased in this manner. We represent the voltage between emitter & base as VEB, and the voltage between collector & base as VCB. FIGURE 14.27 shows (a) Schematic representations for a n–p–n transistor and (b) Symbols to represent n–p–n and p–n–p transistors. Physics 492 Fig. (c) NCERT to not be republished. 14.28, base represents a common terminal between the two power supplies, whose other terminals are connected respectively to emitter or collector. The VEE and VCC power supplies are therefore represented respectively.
Circuits where emitter acts as the common terminal have the power supply between base and emitter represented by VBB, and collector and emitter represented as VCC. Let’s now see the current carriers that travel in the transistor with emitter/base junction forward biased or base-collector junction reverse biased. A heavily doped emitter contains a large concentration of majority carriers. These will be electrons in an N-p-n transistor and holes in a P-n-p transistor. These majority carriers are abundant in the base region. The base is thin, and light doped. The majority of carriers would therefore be very few. The majority of carriers in a p–n-p transistor are electrons, since the base is an n-type semiconductor. The base emitter has a large amount of holes that overwhelm the few electrons. These holes can cross the junction to enter the collector because the junction is reverse biased. The holes in the base can move towards the junction to combine with electrons from the outside, or they could cross the junction to reach the collector terminal.
The base is thin enough that most holes are located near the reverse-biased junction base-collector junction. They then cross the junction rather than moving towards the base terminal. It is interesting that a lot of current enters the emitter/base junction due to forward bias. However, most of it is diverted into the adjacent reverse-biased bases-collector junction. The current leaving the base is a fraction of that that entered the junction. The forward biased junction is represented by the electron current and the hole current, I h and I e. This means that the emitter current (I E) = I h+ I e and the base current (IB) = Ih+ I e. This is because a large portion of IE flows to the collector and not out of the base terminal. Thus, the base current is only a fraction of the emitter. The emitter current I E is equal to the current entering it from the outside. The current coming from the collector terminal is I E whereas the current from the base terminal I B is the same. This is evident from the above description, as well as from a simple application of Kirchhoff’s law to Fig. 14.28(a), The emitter current is equal to the sum of the collector current and the base current.I E = IC + I B (14.7).
We also see that I I E. The direction of motion for the holes is the same as the conventional current. However, the direction of motion for electrons is opposite that of current. In a p–n-p transistor, the current flows from emitter to base. However, in an n–p–n transistor, it flows from emitter to emitter. The direction of the conventional current is indicated by the arrowhead at the emitter. The descriptions of the paths taken by majority and minor carriers in a N-P-N transistor are identical to those for a P-N-P transistor. However, the current paths are completely opposite as illustrated in Fig. 14.28. Fig. 14.28(b), the electrons are the majority carriers provided by the n type emitter region. They can cross the thin p–base region and reach the collector to provide the collector current, IC. The above description shows that the transistor’s active state has the emitter-base junction acting as a low resistance and the base collector acting as a high resist. 14.9.2 Basic transistor circuit layouts and transistor characteristics. A transistor has only three terminals, namely the emitter (E), collector (C), and base (B). The input/output connections in a circuit must be so that the input and output connections are the same (emitter, base, or collector).
You can connect the transistor in any of the three configurations below: Common Emitter, Common Base (CB), or Common Collector (CC). The most common configuration is the CE and we will only be discussing it. We will only be discussing n-p–n Si transistors as they are more common. For p-n–p transistors, the polarities for external power supplies must be inverted. Common emitter transistor characteristics A transistor that is used in a CE configuration has the input between the base (emitter) and the output between the collector (emitter). The input characteristic is the variation of the base voltage VBE and the base current IB. The output characteristic is the same. It refers to the variation in the collector current I B with the collector-emitter Volt VCE. The input characteristics control the output characteristics. This means that the base current and collector current change with each other. The circuit in Fig. 2 can be used to study the input and output characteristics of n-p–n transistors.
A curve is drawn between the base voltage VBE and the base current IB to study the input characteristics for the transistor in CE configuration. FIGURE 14.29 Circuit arrangement to study the input and output characteristics of an n-p -n transistor in CE configuration. Physics 494 collector-emitter Voltage VCE must not be republished. This allows for the study of VBE’s dependence on I B. The input characteristic of a transistor in active mode is what we are looking for. The collector-emitter voltage VCE must be large enough to reverse bias the base collector junction. VCE = VBE + VCB and Si transistor VBE ranges from 0.6 to 0.75 V, so VCE must be sufficient to cause the base collector junction reverse biased. The transistor acts as an amplifier with a large VCE range, so the reverse bias across this junction is usually high. The input characteristics for VCE can be found in the range 3 V to 20V. This is because the effect of VCE on IB is negligible. Therefore, input characteristics for different values of VCE will produce almost identical curves. It is sufficient to only determine one input characteristic.
The output characteristic can be determined by watching the variation in IC as VCE changes while keeping I B constant. VBE can be increased by a small amount. This will cause both the hole current from emitter region and electron current from base region to increase. In turn, I B and C will both increase in proportion. As IB increases, I C also rises. One output characteristic is shown by the plot of I C and VCE for different fixed IB values. As shown in Figure, there will be different output characteristics for different IB values. 14.30(b). 14.30(b). (i) Input resistance, (ri): This is the ratio of the change in baseemitter current (VBE) to its resulting change (I B) at constant collector–emitter (VCE) voltage. This resistance is dynamic (ac resist) and, as you can see from the input characteristic it varies with the operating voltage in the transistor. CE BE i V V r = (14.8). The value of ri may be anywhere from a few hundred to several thousand ohms. FIGURE 14.30 (a), Typical input characteristics and (b), Typical output characteristics.
Semiconductor Electronics Materials Devices and Simple Circuits EXAMPLE 14.8 (ii) Output Resistance (ro): This refers to the change of collector-emitter voltages (VCE) and the change in collector currents (I C) at constant base currents IB. B CE oC I V r I = (14.9). The output characteristics indicate that IC rises almost linearly initially, even for very low VCE values. This is because the base-collector junction of the transistor is not reverse biased, and it is not in an active state. The transistor is actually in the saturation state, and the supply voltage VCC (=VCE in this section of the characteristic) controls the current. If VCE is greater than the voltage required to reverse bias base-collector junctions, I C will increase very little with VCE. The values of ro are determined by the reciprocal slope of the linear portion of the output characteristic. The bias of the basecollector junction controls the output resistance of the transistor. This diode is in reverse bias, which results in a high output resistance of around 100 k.
This is also why the resistance at its initial portion, when the transistor has reached saturation, is extremely low. (iii). Current amplification factor b: This is the ratio between the change of collector current and the change in base voltage at a constant collector/emitter voltage. It is also called small signal current gain. Simply by finding the ratio between IC and IB, we can calculate what we call dc b for the transistor. C dc I I b = (14.11). Since I C increases almost linearly with IB and I C = 0. when I B is 0, both the bdc/bac values are almost equal. For most calculations, bdc is sufficient. Both bac (or bdc) can be used to calculate VCE, I B (or C ) variations. Example 14.8 The output characteristics as shown in Figure. 14.30(b) Calculate the output characteristics of the transistor from Fig. Solution CE C Ac B V I I I B =, C Dc B I I I B = To determine bac and/or bdc for the given values of VCE/IC, one can follow these steps: Consider the following two characteristics to determine the IB values that are above or below the I C value. Here, IC = 4.0mA. (Choose characteristics to IB = 20 uA and 30 mA. We read the two values for IC from the graph at VCE = 10V.
This depends on the configuration (CC, CE, and CC), the biasing of E-B/B-C junction, and the operation area, cutoff, active, region, saturation We have not covered the CE configuration. Instead, we will concentrate on the biasing as well as the operation region. The transistor acts as a switch when it is in the cutoff state or saturation. The transistor must operate in the active area to be used as an amplifier. (i) The transistor as a switching device We will analyse the behavior of the base-biased transistor under CE configuration to better understand its operation as a switching device. 14.31(a). Kirchhoff’s voltage rule is applied to both the input and outgoing sides of this circuit. VBB = IBRB+ VBE (14.12) andVCE = VCC-I C RC (14.12). (14.13) We will treat VBB as Vi, the dc input voltage Vi, and VCE the dc output voltage VO. We have Vi = VBE + IBRB and Vo = VCC – ICRC. Let’s see how Vo changes when Vi goes up from zero. The case of a Si transistor is that as long as the input Vi is below 0.6 V, the transistor will be in cut-off state. Current IC will also be zero. Vo = VCC If Vi is greater than 0.6V, the transistor will be in an active state with some current in the output path.
The output Vo decreases as FIGURE 14.31 (a. Base-biased transistor in CE layout, (b. Transfer characteristic). (c) Not to be republished in 497 Semiconductor Electronics Materials Devices and Simple Circuits. As Vi increases, I C rises almost linearly, and Vo drops linearly until it reaches 1.0 V. Above this point, the change is non-linear and the transistor enters saturation. The output voltage decreases towards zero as Vi increases, but it could never reach zero. The Vo vs Vi curve (also known as the transfer characteristics of a base-biased transistor) can be plotted (Fig. 14.31(b), we can see that there are areas of non-linearity between cut off and active states, as well as between active and saturation states. This shows that the transitions from active to cutoff state and active to saturation state is not clearly defined. Let’s now see how the transistor works as a switch. Vo is high as long as Vi remains low enough to prevent the transistor from being forward-biased. If Vi is sufficient to drive the transistor to saturation, Vo is very close to zero.
The transistor is switched off when it isn’t conducting and switched on when it is saturated. If we consider low and high voltage levels as corresponding to cutoff or saturation of the transistors, we can then say that a low input turns the transistor off and that a high input turns it on. Or, you can also say that a low input to a transistor produces a high output while a high input results in a low output. The switching circuits are designed so that the transistor is not in an active state. (iii) The transistor can be used as an amplifier. We will use the active area of the Vo versus VI curve to make this possible. The rate at which the output changes with respect to the input is represented by the slope of the linear portion of the curve. It is negative because VCC – IC RC is the output and not I C RC. This is why, as the input voltage of CE amplifier increases, its output voltage decreases. The output is then said to be out-of-phase with the input. Vo/Vi can be described as the amplifier’s small signal voltage gain (AV) if we think of Vo and Vi simply as small variations in the input and output voltages.
The circuit will behave like a CE amplifier with voltage gains Vo/ Vi if the VBB voltage is fixed at the midpoint of the active area. The voltage gain AV can be expressed in terms of both the resistors in a circuit and the transistor’s current gain as follows. Vo = VCC-ICRC Vo = 0 = RCIC IBRB + Vi = IBRB I B + ViBE VBE in this circuit is negligible in comparison to IBRB. The voltage gain of the CE amplifier (Fig. 14.32) is calculated as AV = – RC IC / RB IB= -bac (RC /RB ), (14.14) where bac equals IC/IB in Eq. (14.10). The linear portion of active region can thus be used for amplifiers. The next section will discuss the transistor as an amplifier (CE configuration). Physics 498 14.9.4 NCERT not to republished Physics 498 14.9.4 The transistor can be used as an amplifier by setting its operating point somewhere within its active area. If VBB is fixed to the point at the center of the transfer curve’s linear portion, then the dc collector current I C and the base current VBB will remain constant. VCE = VCC + I C RC would remain constant.
The operating values VCE and IB determine the operating point of the amplifier. A small sinusoidal voltage of amplitude Vs can be superposed on the DC base bias by connecting that signal source in series with VBB supply. The base current will then have sinusoidal variations overlaid on the value I B. The collector current will also have sinusoidal variations that are superimposed on the value IC. This in turn will produce a change in the value VO. By blocking the dc voltages with large capacitors, we can measure the ac variations between the output and input terminals. We haven’t included any ac signals in the description of the amplifier. Amplifiers are generally used to amplify alternating signals. Let’s now superimpose the ac input signal vi on the bias VBB(dc), as shown in Figure. 14.32. The output is the difference between ground and collector. If we assume that vi is zero, the working of an amplifier is easy to understand. Kirchhoff’s law is applied to the output loop to get Vcc = ICE + Ic RL (14.15) and VBB = IB RB + VBE (14.16) If vi is not zero, VBE + vi = IBE + B RB + VBE (R B + Ri ) The change of VBE can be attributed to the input resistance [see Eq. (14.8) and the change of I B Vi = IB (RB + ri) = rIB A change in IB causes an increase in I c
A change in I B results in a change of I c. This causes a change VCE and a voltage drop across resistor RL. VCC is fixed. FIGURE 14.32 An example of a simple circuit for a CE-transistor amp. ( These modifications can be given by Eq. (14.15) VCC = VCE+RLI C = 0, or VCE =-RLI C. The VCE change is the output voltage, v0. Eq. Eq. You can see that the CE configuration has a current gain of bac. We also see the voltage gain Av in this discussion. The power gain Ap is the sum of the voltage gain and the current gain. Mathematically, Ap = bacx Av (14.19) Because bac and Av both exceed 1, we get an ac power increase. It should be noted that a transistor is not a power-generating device. The battery supplies the energy needed to produce the higher AC power at the output. Consider that the transistor will become saturated when VCE is 0V and VBE is 0.8V. Calculate (a), the minimum base current at which the transistor will reach saturation. Therefore, (b), determine V1 at the time that the transistor is “switched on”. (c) Find the V1 ranges for which the transistor has been’switched off” and’switched back’. Answer Given that VCE is 0V, VBE is 0.8V VCE is VCC-ICRC IC = VCC/RC= 5.0V/1.0k= 5.0mA I B = I /b = 20uA/250 = 20uA VIH = VIBB = IBRB+VBE = 20uAx100k + 0.8V = 2.28V.
The voltage below which the transistor is cutoff is VIL = 0.6%V and VIH = 2.8V. The transistor will be in the switched off state. It will switch on when it is between 2.8V to 5.0V. The transistor will be active when I B is between 0.0mA and 20mA. In this range, the IC = bIB value is valid. In the saturation range, IC = bIB is valid. (c) Physics 500 EXAMPLE 14:10 Calculate the dc drop across collector resistance. Refer to Fig. 14.33). Solution The output voltage of the AC is 2.0 V. Therefore, the ac collector current (iC) = 2.0/2000 =1.0 mA. Therefore, the signal current through a base is iB = IC/b = 2.0 V/100 = 0.0.010 mA. The dc base current must be 10x 0.0.010 = 0.10mA. Eq.14.16 shows that RB = (VBB-VBE )/IB. VBE = 0.6 V. RB = (2.0-0.6 )/0.10 =14 k. The dc collector current, I C, is 100×0.10 = 10mA. 14.9.5 Feedback amplifier and transistor oscillator An external input is required to maintain ac in an amplifier’s output. An oscillator produces ac output with no external input signal. An oscillator’s output is self-sustaining.
An amplifier is needed to achieve this. An amplifier is used to achieve this. A portion of the output power is then returned (feedback) in phase with the input (this is called positive feedback). 14.33(a). Inductive coupling, through mutual inductance or LC/RC networks can provide feedback. There are many types of oscillators. They use different methods to connect the output to the input (feedback networks), and they also use different resonant circuits for oscillation at a specific frequency. The circuit in Fig. 2 is a good example of the oscillator’s action. 14.33(b), where feedback is achieved by inductive coupling between one coil winding T1 and another coil winding T2. The coils T2 & T1 are connected by mutual inductance. The base-emitter junction in an amplifier is forward biased, while the base collector junction is reverse biased. For simplicity, we have omitted details about the biasing circuits that were actually used. Let’s look at how oscillations work. Let’s say switch S1 is turned on.