Using a balance of thorough analysis and insight, readers are shown how to work with electronic circuits and apply the techniques they have learned. All mathematics is kept to a suitable level, and there are several exercises throughout the book. Password-protected solutions for instructors, together with eight laboratory exercises that parallel the text, are available.
Preface 1. Basic concepts and resistor circuits 2. AC circuits 3. Band theory and diode circuits 4. Bipolar junction transistors 5. Field-effect transistors 6. Operational amplifiers 7. Oscillators 8. Digital circuits and devices Appendices Index. If the variable device has two leads, it is called a rheostat. One must also select the proper power rating for a resistor.
The power rating of common carbon resistors is indicated by the size of the device. Typical values are 1 1 1 8 , 4 , 2 , 1, and 2 watts. As noted in Eq. The following are some equivalent circuit laws for resistors.
To see this, consider the circuit shown in Fig. We would like to replace the circuit on the left by the equivalent circuit on the right.
The circuit on the right will be equivalent if the current supplied by the battery is the same. By KCL, the current in each resistor is the same. The components are often drawn in parallel lines, hence the name. Again, we would like to replace the circuit on the left by the equivalent circuit on the right.
Solving these for the three currents and substituting in Eq. A frequent task is to analyze two resistors in parallel. Of course, for this special case of Eq. It is often more illuminating to write this 1 as an equation for Req rather than Req. Example For the circuit shown in Fig.
What must its power rating be? Solution As we will see, there is more than one way to solve this problem. Here we use a method that relies on basic electronics reasoning and our resistor equivalent circuit laws.
Thus, if we can get the current supplied by the battery we can solve the problem. To get the battery current, we combine all our resistors into one equivalent resistor. The implementation of this strategy goes as follows. Combining these we get using Eq. This 6. As a check, it is comforting to note that this current is less than the total battery current, as it must be. This is too much for a 18 W resistor, so we must use at least a 14 W resistor.
The theorem states that any two-terminal network of sources and resistors can be replaced by a series combination of a single resistor Rth and voltage source Vth. This is represented by the example in Fig. The sources can include both voltage and current sources the current source is described below. A more general version of the theorem replaces the word resistor with impedance, a concept we will develop in Chapter 2.
Vth is the voltage across the circuit terminals when nothing is connected to the terminals. The voltage across the terminals is thus the same as Vth. In practice, the voltage across the terminals must be calculated by analyzing the original circuit. There are two methods for calculating Rth ; you can use whichever is easiest. This means that you replace the voltage sources by a wire and disconnect the current sources.
Now only resistors are left in the circuit. These are then combined into one resistor using the resistor equivalent circuit laws. This one resistor then gives the value of Rth. Call this the short circuit current Isc. This equivalent circuit is shown in Fig.
The current source is usually less familiar that the voltage source, but the two can be viewed as complements of one another. An ideal voltage source will maintain a constant voltage across it and will provide whatever current is required by the rest of the circuit. Similarly, an ideal current source will maintain a constant current through it while the voltage across it will be set by the rest of the circuit.
If we short the terminals, it is clear from the Norton equivalent circuit that all of Inor will pass through the shorting wire. We have seen previously that the voltage across the terminals when nothing is connected is equal to Vth. Here are a few examples. Current limiting. For example, the ubiquitous LED light emitting diode typically operates with a voltage drop of 1.
Suppose you have a 9 V battery and wish to light the LED. How can you operate the 1. By the discriminating use of a resistor! Consider the circuit in Fig. Without it, the LED would burn out immediately. Voltage divider. Another very common resistor circuit is shown in Fig. Some voltage Vin is applied to the input and the circuit provides a lower voltage at the output. The analysis is simple. This equation is used so frequently it is worth memorizing.
The current divider circuit is shown in Fig. A current source is applied to two resistors in parallel and we would like to obtain an expression that tells us how the current is divided between the two. Since the two resistors are in parallel, the voltage across them must be the same.
In electronics, one frequently has the need to measure voltage and current. The instrument of choice for many experi- mentalists is the multimeter, which can measure voltage, current, and resistance. The analog version of the multimeter uses a simple meter as a display. Both of these can be accomplished by judicious use of resistors. The circuit in Fig. The physical meter within the dotted lines is represented by an ideal current measuring meter in series with a resistor Rm.
When a current I is applied to the terminals, part goes through the meter and part through the shunt. The circuit is simply a current divider, so we have cf. Another simple addition will allow us to use our meter to measure voltage.
This circuit is then seen to be a voltage divider. Inverting Eq. Our goal is to solve for the current through resistor R4 in Fig. In Fig. We thus have four equations relating the three unknown currents I0 , I1 , and I2 and need to solve for I1.
In practice we need only three independent equations to solve for the currents, but we have given all four here to illustrate the method. Solving Eq. The mesh loop method Our second method for solving circuit problems is the mesh loop method. In this method, currents are assigned to the circuit loops rather than the actual physical branches of the circuit. This is shown in Fig. We then move around these loops, applying KVL, but including contributions from both loop currents.
Note that the resulting set of equations is simpler in this method: two equations in two unknowns I1 and I2. For this reason the mesh loop method is usually preferable for more complicated circuits. Furthermore, our equations can be rearranged into the conventional form of a system of linear algebraic equations. Thus Eq. The usual brute force method also works: solving Eq. We form the required two terminal network by removing R4 and taking the two terminals at the points where R4 was attached.
The remaining circuit should look familiar — if we combine R2 and R3 it is the previously considered voltage divider. Rth Vth R4 Figure 1. An important fact to keep in mind when making such measurements is that the measuring instrument becomes part of the circuit. The act of measuring thus inevitably changes the thing we are trying to measure because we are adding cir- cuitry to the original circuit.
To help us cope with this problem, test instrument manufacturers specify a quantity called the input resistance Rin or, as we will see later, the input impedance.
The effect of attaching the instrument is the same as attaching a resistor with value Rin. To see how this helps, suppose we are mea- suring the voltage across some resistor R0 in a complicated circuit, as depicted in Fig.
As an example of what happens when Rin is not large, consider the circuit in Fig. Ignoring the meter for a moment we see that the original circuit is a voltage divider, and application of Eq. Using this in Eq. So, unless we are aware of the effect of input resistance, we run the danger of making a false measurement. On the other hand, if we are aware of this effect, we can analyze the effect and determine the true value of our voltage when the meter is unattached.
How does one determine the value of the input resistance for a given instrument? Here are some common ways. Look in the instrument manual under input resistance or input impedance. The value should be in ohms. This is usually printed on the face of the meter itself, as shown in Fig. To get Rin , multiply this number by the full scale voltage selected.
You may have to analyze the instrument circuitry itself. Constant voltages and currents are described as DC quantities in electronics. For future reference, we list here some of the most common AC signals. Sinusoidal signals. This is probably the most fundamental signal in electronics since, as we will see later, any signal can be constructed from sinusoidal signals. A typical sinusoidal voltage is shown in Fig.
There are several ways to specify the amplitude of a sinusoidal signal that are in common use. These include the following. This is useful for power calculations involving sinusoidal waves. For example, suppose we want the power dissipated in a resistor given the sinusoidally varying voltage across it.
We cannot simply use Eq. The same argument applies to Eq. Some other typical waveforms of electronics are shown in Figs. Square wave. Sawtooth wave. Triangle wave. V A 0 t tramp Figure 1. Pulse train. These are random signals of thermal origin or simply unwanted signals coupled into the circuit. Noise is usually described by its frequency content, but that is a more advanced topic.
V 0 t Figure 1. What is the resistance of a nichrome wire 1 mm in diameter and 1 m in length? Compute the current through R3 of Fig. Compute the current through R1 and R2 of Fig. The output of the voltage divider of Fig. What voltage will each indicate? A real battery can be modeled as an ideal voltage source in series with a resistor the internal resistance.
What is the internal resistance of the battery? What is the resistance across the terminals of Fig. Suppose that a 25 V battery is connected to the terminals of Fig. Compute the current through R2 and R3 of Fig. Find the Thevenin voltage and Thevenin resistance of the circuit shown in Fig. The two terminals for this problem are the points where R5 was connected. Exercises 25 In the circuit of Fig.
Compute all the currents labeled in the circuit of Fig. Suggestion: use the mesh loop method. Do the same for the Norton equivalent circuit.
Alexander and Matthew N. Anderson and W. James J. James Diefenderfer and Brian E. Robert E. In this chapter we introduce two other basic circuit components, the capacitor and the inductor. The treatment of these components depends on the details of how things are changing in time, and this will require the development of some new analysis techniques.
A capacitor is formed by any pair of conductors, but the usual form is two parallel plates. Note that, like the resistance, the capacitance depends only on the physical characteristics of the device. The unit of capacitance is coulombs per volt or farads, abbreviated F. When purchasing a capacitor, you must specify its voltage rating in addition to its capacitance value. This rating tells you the maximum voltage you can apply across the capacitor before there is electrical breakdown through the dielectric material.
So what does a capacitor do? One answer is that it is a charge storage device. When a voltage V is applied to a capacitor, a charge of magnitude Q will be stored on each plate. If we take the time derivative of Eq. We wish to combine the capacitors to form the equivalent circuit on the right. Let Q1 be the charge on capacitor C1 and so on. Applying KVL and Eq. Again, let Q1 be the charge on capacitor C1 and so on.
Comparing this with Eq. Putting these together means that time-varying currents in a circuit induce voltages. While any circuit loop has inductance, we usually ignore this like we ignore the small resistance of connecting wires and, if inductance is required in a circuit, add discrete inductors made of coils of wire. For a long coil i.
The unit of inductance is volts times seconds per amp or henries abbreviated H. The derivation of the equivalent circuit laws for inductors in series and parallel is similar to that for resistors, and we leave the details to the reader. R V0 C Figure 2. We next take the derivative of Eq. Rearranging Eq. Equation 2. We will see how to determine the constant I0 in a moment.
The voltage across the resistor is just IR. Employing Eq. Suppose we assemble the circuit with an uncharged capacitor so that Vc is initially zero. Using this information in Eq. Using this information in Eqs. It is worth noting some of the key features. The product RC, which has units of time, is called the time constant of the decay. We now throw the switch to the down position. As the capacitor discharges, the capacitor voltage decays exponentially with a time constant RC and approaches zero the voltage level it is now attached to.
R Vin C Vout Figure 2. We can represent this case with the circuit diagram of Fig. Some representative cases are shown in Fig. We thus see the same shape waveform as we saw in the switch problem. Note that for this case, the output voltage looks like a square wave with rounded leading edges.
It now tries to charge to this input voltage, but, again, does not have the time to get there. Now the output voltage is quite different from a square wave. In this case, the waveform is a series of rising and falling lines forming a triangle wave. Suppose we apply our square wave to the rearranged circuit of Fig. Now our output voltage is the voltage across the resistor.
The current and thus the output voltage starts at its maximum value and decays exponentially toward zero. When the input voltage switches polarity, the current goes negative and again decays toward zero. The output voltage resembles a series of positive and negative spikes.
While it may appear that we are just making a minor change to the input waveform, the change is actually much more profound. For the switching case or the square wave , the input voltage was, at each instant of time, constant. This allowed us to solve the differential equation Eq. This approach is called a time domain analysis. For more complicated input voltages, time domain analysis is not always possible because we cannot solve the resulting differential equation. In these cases, it is sometimes useful to analyze the circuit in terms of its sine wave response, which we will call a frequency domain analysis.
The relevant RC circuit is shown in Fig. Plugging into Eq. But we prefer to obtain general algebraic results, so we use the following trick. A right triangle satisfying Eq. Thus our circuit is a positive phase shifter.
We examine the easily calculable extreme limits to get some idea of the behavior. Plotting Eq. Higher frequencies, on the other hand, are relatively unattenuated and pass through the circuit with little change in their amplitude.
Using our former result for Ip Eq. This circuit is thus a negative phase shifter. The full plot of Eq. In this case the plot shows that lower frequencies are unattenuated while the higher frequencies are suppressed.
Now from Eq. Lastly, we note that these conclusions are consistent with the response of the RC circuit to a square-wave input voltage that we studied earlier. The triangle is indeed the integral of the square wave since the integral of a constant is a linear function rising or falling according to the sign of the constant.
This transforms the differential equation into an algebraic equation which we can then solve. This, however, involves considerable work and requires that we know certain trig identities. We can solve these problems more easily by employing complex numbers. This approach also has the advantage of producing a broader conceptual understanding of resistors, capacitors, and inductors.
Each of these types has rules that apply to the manipulation of the member numbers: addition, multiplication, exponentiation, etc. A complex number is simply another type of number system with its own set of rules for manipulation. Such a representation is shown in Fig.
We now use these complex voltages and currents to solve the circuit equation. This, of course, is a cheat. The voltage we apply to the circuit is a real number, not a complex one. It remains to massage the answer into a nicer form. We use the fact, noted above, that any complex number can be written in complex exponential form with the magnitude and angle given by Eqs. Applying this to the denominator of Eq. At a minimum, we now have another method that allows us to solve the differential equations arising from the analysis of LRC circuits with sinusoidal drive voltages.
To my taste, the complex exponential method is easier because it does not require the trig identities and algebraic tricks of our former method. The price, of course, is learning to use and manipulate complex numbers. There is, however, an additional advantage and a new conceptual insight that comes from the use of complex numbers.
Look at Eq. A similar result occurs when we analyze circuits that involve inductors. To do this, we introduce the concept of impedance. Impedance is a generalization of resistance that applies to resistors, capacitors, and inductors alike.
Finally, some vocabulary. The real part of a complex impedance is called the resistive impedance or simply the resistance, while the imaginary part is called the reactive impedance or simply the reactance. In either case, we can plug into Eq. This can be obtained by multiplying the current times the impedance of the component. For example, suppose we want the voltage across the resisitor as the output.
The entire curve is shown in Fig. Suppose now that we use the same LR circuit but take the output voltage across the inductor. Thus, from Eq. So for very low 1 and very high frequencies we get no current. Somewhere in between there must be a maximum. Systems exhibiting a large response at a particular drive frequency are called resonant systems, and the frequency at which the response peaks is called the resonant frequency.
Apparently, at the resonant frequency, the effect of the inductive and capacitive impedance cancels. Thus, the smaller R is, the narrower the curve and the higher the peak current.
Such circuits are routinely used to tune-in a selected communication channel while suppressing the neighboring transmission frequencies.
R V0 C L Figure 2. As an example, consider the circuit in Fig. It will be useful, therefore, to solve for the charge Q directly. The general solution to such an equation is the sum of 1 the general solution to the homogeneous equation i.
That leaves the homogeneous equation to solve. For the general solution we must use both. To complete the solution we must specify V3 by applying the initial conditions. This behavior is known as ringing since it is reminiscent of a ringing bell sound. This behavior is shown in Fig.
Applying these conditions to Eq. Combining this with the former result then gives the full behavior of the switched underdamped circuit as shown in Fig. This underdamped behavior is seen in other physical systems as well. For exam- ple, a lightly damped harmonic oscillator that has its equilibrium position suddenly changed will oscillate around that new equilibrium position until the oscillations are damped out.
Vc V0 Figure 2. A plot of these functions is shown in Fig. There are no oscillations in this case. Now 2 the square root in Eq. In such cases the methods of ordinary differential equations can be applied to yield a second independent solution. This critically damped case, shown in Fig. But what about other types of drive voltage e. As we will now show, our analysis is applicable to any periodic signal as a consequence of a remarkable theorem named for Joseph Fourier.
The theorem is easily stated. Note that, as a consequence of Eq. Some powerful insights result from consideration of this theorem. But Eq. The output would be roughly the sum of the lower frequency portions of the original signal. This is not usually required in electronics since the typical periodic functions used have their Fourier series already tabulated, but for the curious we do one example.
Suppose our periodic function is the sawtooth wave shown in Fig. Since the integral in Eq. Similar series can be written for other common waveforms. At power generating plants, huge transformers step up the voltage so that a large amount of power can be transmitted without necessitating very large transmission wires. Since the high voltages are dangerous, at the consumer end of the power grid transformers step down the voltage to a safe level.
The required voltage supplies inside most consumer electronics are even lower, so another step-down transformer is found inside these devices. A typical transformer is shown schematically in Fig.
A length of wire is wrapped around one portion of the transformer core n1 times. These windings are called the primary windings. Another wire is wrapped around the core n2 times to form the secondary windings. Using Eq. There is no free lunch in physics. Equations 2. Another use for the transformer is in impedance matching. Suppose we have a voltage source with a voltage V0 and an internal resistance R0.
This is connected to a load resistor RL as shown in Fig. The question is, how should R0 and RL be related if we want the maximum power transferred to the load resistor?
Somewhere between these limits we will get a maximum in the power. This general result shows that the load resistance or, more generally, the load impedance must match the internal resistance of the source if we want to transfer the most power to the load. While this is desirable, we are sometimes faced with situations where this is not the case. To achieve the desired impedance matching, we can employ a transformer. Consider the circuit shown in Fig. Using Eqs. By choosing a transformer with appropriate turns ratio, we can match Reff to whatever source resistance we have, thereby insuring maximum power transfer.
In effect, the transformer has matched the source impedance to the load impedance. Further details on analyzing circuits involving transformers are given in Appendix C. Find the equivalent capacitance across the terminals of the circuit in Fig. C1 C2 2. Sketch Vout versus time for the circuit shown in Fig.
Include appropriate numeric scales on the axes. R V0 C Vout Figure 2. Referring to Fig. Suppose we use the circuit of the previous problem and wait a long time after the switch is closed so that the capacitor is fully charged. How long will it take for the voltage across the capacitor to drop to 1. Calculate the peak, the rms, and the average currents though the resistor. What power rating should the resistor have? Calculate the magnitude and the phase of the total impedance for the circuit shown in Fig.
Suppose we change the frequency of the signal generator in Fig. Of the voltage across the resistor? Find the magnitude and phase of the impedance of the circuit of Fig. Derive the following expression for the circuit of Fig. Consider the circuit of Fig. Assume the signal generator outputs a Vrms sine wave with a frequency of 60 Hz. Refering to Fig. What is the necessary turns ratio for such a transformer?
A step-down transformer with no markings is measured to have a 1. Which side is the primary and which side is the secondary? If the input voltage is Vrms , what will the output voltage be? Write down the Fourier series for the signal shown in Fig. Hint: it is not necessary to do any calculation. Think about how this signal is related to one for which we already know the Fourier series. V V0 0 t Figure 2. Given a sawtooth signal with a period of 1 ms, design a circuit including all component values that will take this sawtooth as input and give a 3 kHz sine wave as output.
If the peak voltage of the sawtooth is 1 V, what will the amplitude of the resulting sine wave be? Each of these energy levels can accept up to two electrons. We also know that if an atom absorbs energy from the outside for example, by absorbing a photon , an electron can be promoted to a higher energy level.
Conversely, an electron that falls from a higher to a lower energy level emits a photon. What happens to this energy level model when we assemble many atoms together into a solid? As the atoms get closer together, we must start to talk about the energy levels of the solid as a whole rather than those of the individual atoms. Rather than doing quantum mechanics for an isolated potential the atom , we do it for a periodic array of atoms that exhibits a periodic potential.
The net result of this is that, during the assembly of N atoms, the individual atomic levels split into N levels. This is shown schematically in Fig. Just as electrons in individual atoms cannot have energies between the atomic energy levels, so electrons in a solid are forbidden to have energies between the allowed bands. We now start to add electrons to the solid, starting with the lowest energy level. Here a is the distance between atoms. The light gray shading represents the large number of closely spaced energy levels produced by the splitting of the original atomic energy level.
The dotted line represents the equilibrium spacing. At this spacing the solid is characterized by energy a bands. The character of the resulting band structure determines whether a material is a conductor, an insulator, or a semiconductor. Consider, for example, the case shown in Fig. In order to produce a current, electrons in a material must move and thus must increase their energy slightly.
They must, therefore, be able to move to a slightly higher energy level. This is possible for this material because there are lots of empty energy levels in the top-most band. This material will thus be a good conductor. This situation should be contrasted with the band structure shown in Fig. Here we have a full, allowed band, followed by a wide forbidden band, followed by an empty allowed band.
This is the characteristic band structure of an insulator. At this point we should quantify some of these statements. The energy gained by an electron contributing to a current in a material is quite small. The voltage applied to produce the current accelerates the electrons, but before they gain much velocity they undergo a collision that changes their direction and slows them down.
Thus there is no way the electrons in an insulator can gain enough energy from conduction to cross the forbidden band. There is, however, another source of energy that plays an important role in semiconductor physics: thermal energy. A basic insight from thermal physics tells us that temperature is a macroscopic measure of the microscopic motion of particles. Thus particles in our material can have energies beyond those set by their position on the band structure.
It is thus possible if the temperature is high enough and the band gap is small enough for an electron to jump to the next allowed band. Once it is there, it can produce a current because now there are lots of the open energy levels required for conduction. If our material is indeed an insulator, the forbidden band must be wide enough to prevent this from happening, i.
Our third type of material, the semiconductor, is characterized by the opposite situation: the forbidden band is narrow enough to allow thermal transitions to the next allowed band. When an electron in a semiconductor moves from the valance band to the conduction band, it leaves behind a vacancy in the valance band. This vacancy is called a hole see Fig. The hole behaves as if it were a positive particle. To see this, imagine that a voltage is applied across our material, with the negative terminal of the voltage on the left and the positive terminal of the voltage on the right.
If an electron is promoted to the conduction band, it will also move toward the right, toward the positive voltage. While moving up in energy, this electron will also move to the right, trying to get to the positive terminal.
Thus the vacancy it leaves will move to the left. It is thus possible to view the process in terms of the motion of the hole rather than the motion of the electrons. In this view, the hole a positive conduction particle moves toward the negative terminal as it increases in energy here we must switch signs on our energy scale and let the hole energy increase downwards.
We include the hole picture for completeness, since it is often encountered in the electronics literature. This behavior can be altered, however, by adding impurities to the pure semiconductor.
This is called doping the semiconductor. For our purposes, there are two ways of doping a semiconductor. Suppose, for example, that we have a pure germa- nium semiconductor. The location of these addi- tional levels depends on the type of material, type of impurity, and other factors.
A desirable location is shown in Fig. The new levels are near the bottom of the conduction band. It is thus relatively easy for these electrons to be promoted into the conduction band compared to those in the valance band of the pure semicon- ductor.
Thus a donor impurity makes it easier for the semiconductor to conduct. Because the charges that produce this current the so-called charge carriers are electrons, this type of doped semiconductor is called an n-type semiconductor n for negative.
Conversely, suppose we dope our pure germanium semiconductor with gallium outermost shell 4s2 4p1. Such an impurity is called an acceptor. The effect of an acceptor impurity is to add empty, localized energy levels to the band structure. In this case, it is desirable to have these additional energy levels located just above the valance band, as shown in Fig.
It is then easy for electrons from the valance band to move up into these empty levels. But since these levels are localized, the promoted electrons cannot move through the material as they must in order to contribute to the current.
The holes left behind in the valance band, on the other hand, can move throughout the material since the valance band energy levels are not localized. We thus can again have an enhanced level of current due to the impurity, but now the current is produced by the positive hole motion.
This type of doped semiconductor is thus called p-type. The electrons in the n-type material are called majority charge carriers, while the holes are called minority charge carriers.
Conversely, in p-type material, the current is predominately carried by the hole motion, but a few electrons promoted from the valance band to the conduction band also contribute, so the holes in a p-type material are the majority charge carriers while the electrons are the minority charge carriers. We represent our two pieces as shown in Fig. For the n-type material, the number of electrons in the conduction band decreases with energy because it is harder for the electrons to be promoted to these higher levels we will quantify this in a moment.
Similarly, there are fewer holes at the higher hole energy levels remember, hole energy increases downward in this diagram. When the two materials are brought together, the higher density electrons in the n-material will diffuse into the p-material where the density of electrons in the con- duction band is lower. The opposite happens for the high density holes in the p-material; they diffuse into the n-material.
Because the behavior of electrons and holes is analogous, from now on we will focus our attention on the electrons alone. This diffusion of charges leads to a charge imbalance, with excess electrons accu- mulating in the p-material.
The region near the p-n junction where this diffusion takes place is called the depletion region because the density of the charge carriers electrons in the n-material and holes in the p-material is markedly reduced. What does all this do to the band structure?
Thus, in equilibrium, the band structure of the p-n junction will appear as shown in Fig. Originally, f1 is greater than f2 because there are more electrons on the n-side than on the p-side.
To quantify this, we note that the electrons in the conduction band have been promoted to that level by thermal energy. A more rigorous approach would use the Fermi—Dirac distribution for F.
If the negative side of our voltage is applied to the p-material and the positive side to the n-material, we get the situation shown in Fig. The energy levels on the p-side have been raised by an amount eV0 , where e is the charge of an electron. Also, since the energy step in the band structure is larger than in the equilibrium case, more charge carriers are required to support it and the depletion layer is widened. Now we apply an external voltage V0 to our p-n junction such that the negative side of our voltage is applied to the n-material and the positive side to the p-material.
This is called the forward biased case. This, again, is consistent with the notion that electrons will tend to be attracted to the positive bias on the p-side.
Since the energy step in the band structure is smaller than in the equilibrium case, the depletion layer is narrowed. A graph of this result is shown in Fig. I0 is usually very small compared to a typical forward bias current and is thus often approximated as zero.
The electronic symbol for this device is shown in Fig. We have also indicated which end of the diode corresponds to the n- and p-material.
Avalanche breakdown. In this type of breakdown, electrons from the p-side are accelerated to high enough kinetic energy to ionize other atoms in the depletion layer, thus producing a new electron-hole pair.
The new electron is also accelerated and can produce more pairs, and so on. The resulting chain reaction adds many electrons to the conduction band and thus rapidly increases the current. Zener breakdown. This process again produces copious electron-hole pairs and rapidly increases the current. Note that while the breakdown current increases rapidly, the voltage stays fairly constant.
The magnitude of this voltage is called the breakdown voltage. Despite the name, both types of breakdown are non-destructive. That is, the diode can be operated in breakdown mode without destroying it, and, as we shall see later, certain circuits deliberately use the steep rise of the reverse current to achieve useful results.
Recall that, for an atom, a photon is emitted when an electron moves from a higher energy state to a lower one. A similar phenomenon can occur in our p-n junction. Since the holes represent vacancies in a lower energy level, it is possible for the electron to jump to one of these lower energy levels while a photon is emitted to conserve energy. If the diode is constructed so that these photons can exit the material, we have a light source.
This is the basis of the ubiquitous light emitting diode or LED, which is used as an indicator light on many modern electronic devices. Photons can also be absorbed by our p-n junction. In this case, an incoming photon promotes an electron from the valance band to the conduction band, thus producing a new electron-hole pair.
As we will see, this unusual behavior allows us to use the diode for many purposes, but it also complicates the analysis of diode circuits. To see why this is true, consider the simple diode circuit shown in Fig. If we knew this, we could plug into any of the other equations and obtain I and we would be done. There are two standard ways of dealing with transcendental equations.
A second way is to solve the equation graphically, and this will help us to see some of the key features of the solution. To proceed, we note that Eq. In electronics, this solution is called the operating point. This procedure is shown in Fig. The diode I—V characteristic Eq. This latter equation is referred to as the load line, and this graphical solution is often called the load line method in electronics. We can thus imagine what would happen to the solution if V0 or RL was varied.
If V0 increases, the operating point will move up the diode characteristic and the current will rapidly increase. Also note that the analysis is not restricted to positive V0. While the load line method offers some insights into the detailed behavior of the circuit, it is cumbersome to use for routine circuit analysis.
Assuming the diode is forward biased, Eq. Since we know V0 and RL , we can obtain the circuit current I directly without the use of a graphical or numerical method. Several caveats are now in order. Two other common approximations are often seen in textbooks. Since the actual diode voltage depends on the current, it is hard to argue persuasively for either value, so we simply note the difference in convention. While this makes analysis even easier, in this case the approximation loses important information: a forward biased diode has a non-zero voltage drop.
Without this understanding, some laboratory observations will be puzzling and some electronic circuits will seem without merit we shall see some examples later.
Another problem lies in our assumption of the diode being forward biased. When we are doing circuit analysis, how do we know at the start if the diode will be forward biased? What do we do if it is not? To answer these questions, we return to the load line analysis. Figure 3. This is consistent with our general voltage loop law notions: if the current in the circuit is zero, there is no voltage drop across the resistor so the voltage across the diode must be equal to the battery voltage.
An alternative tactic to use to avoid mistakes is to check your answer for con- sistency. We proceed as before and get Eq. Since this is impossible with a positive voltage source and, furthermore, is inconsistent with our assumption of a forward biased diode, this assumption must be incorrect.
Before discussing the circuit, we note the introduction of a new circuit symbol, the common or ground symbol. This is shown connected to the right end of the resistor. Ground serves as a common reference point for all other voltages referred to on the circuit diagram recall that voltage is always between two points. Thus, for example, V0 in this circuit is the voltage relative to ground.
This might be provided by a battery of voltage V0 connected between ground and the point labeled V0. This might be used in a circuit where you have available one voltage say, a 9 V battery and need a slightly lower voltage. Each diode in the chain drops the voltage by approximately 0.
The next example is the limiter or clipper circuit shown in Fig. Vin Vout Vb Vb Figure 3. In order to forward bias a diode we must have the input voltage magnitude greater than 0.
When this happens one of the diodes begins to conduct depending on the polarity of the input voltage and holds the voltage across itself to 0. A variable limiting level can be achieved by adding a battery to the circuit as shown in Fig.
The clamp circuit shown in Fig. The output voltage is thus shifted up by a constant amount. An example case for a sinusoidal input is shown in Fig. Diodes are often used to protect switches in inductive circuits as shown in Fig. The diode does nothing at this point because it is reverse biased.
Since the voltage across the inductor is L dI dt , the abrupt change in current produces a very large voltage, often large enough to produce an arc across the opening switch. If the switch is used frequently, this arcing will damage the switch.
To prevent this, a diode is placed across the inductor. The polarity of the large induced voltage is such as to forward bias the diode and cause it to conduct, thus shorting out the inductor and protecting the switch. Diodes can also be used to make logic circuits. An example of such usage is shown in Fig. If any switch is connected to ground, a circuit is completed which forward biases the diode connected to that switch.
The voltage drop across that diode is then 0. By choosing an appropriate value for R, we can cause the light to illuminate under these conditions. Except for the voltage source, this circuit is identical to that considered before cf. Recall that the current through the diode will be zero until the voltage source exceeds 0.
As a consequence, the current and, thus, the voltage across the load resistor now has a non-zero average. If we think of the goal of creating a constant voltage supply, this is a step in the right direction.
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