The resulting voltage and current waveforms assuming a resistive load are shown below. A simple full-wave ac phase angle voltage controller is shown in Figure P However, capacitor C charges up through resistor R, and when the voltage vC t builds up to the breakover voltage of D1, the PNPN diode will start to conduct.
Figure P shows a three-phase full-wave rectifier circuit supplying power to a dc load. The circuit uses SCRs instead of diodes as the rectifying elements. At what phase angle should the SCRs be triggered in order to operate this way?
Sketch or plot the output voltage for this case. The sketch of output voltage is reproduced below, and the ripple is 4. The following table shows which SCRs must conduct in what order to create the output voltage shown below.
The times are expressed as multiples of the period T of the input waveforms, and the firing angle is in degrees relative to time zero. These waveforms are shown below. The inputs to this function are the current phase angle in degrees, the offset angle of the waveform in degrees, and the firing angle in degrees.
What could be done to reduce the harmonic content of the output voltage? It would be easy to modify the function to use any arbitrary dc voltage, if desired. After the voltages are generated, function vout will be used to calculate vout t and the frequency spectrum of v out t.
Finally, the program will plot v in t , v x t and v y t , v out t , and the spectrum of v out t. Note that in order to have a valid spectrum, we need to create several cycles of the 60 Hz output waveform, and we need to sample the data at a fairly high frequency. This problem creates 4 cycles of vout t and samples all data at a 20, Hz rate. Declare arrays. There are two plots here, one showing the entire spectrum, and the other one showing the close-in frequencies those under Hz , which will have the most effect on machinery.
Note that there is a sharp peak at 50 Hz, which is there desired frequency, but there are also strong contaminating signals at about Hz and Hz. If necessary, these components could be filtered out using a low-pass filter. This increase in sidelobe frequency has two major advantages: it makes the harmonics easier to filter, and it also makes it less necessary to filter them at all. Since large machines have their own internal inductances, they form natural low-pass filters. If the contaminating sidelobes are at high enough frequencies, they will never affect the operation of the machine.
Thus, it is a good idea to design PWM modulators with a high frequency reference signal and rapid switching. Calculate the current that would flow through the resistor. How does this number compare to the amount of electric power being generated by the loop? This machine is acting as a generator, converting mechanical power into electrical power. Develop a table showing the speed of magnetic field rotation in ac machines of 2, 4, 6, 8, 10, 12, and 14 poles operating at frequencies of 50, 60, and Hz.
A three-phase four-pole winding is installed in 12 slots on a stator. There are 40 turns of wire in each slot of the windings. The flux per pole in the machine is 0. Since this is a four-pole machine, there are two sets of coils 4 slots associated with each phase. A three-phase Y-connected Hz two-pole synchronous machine has a stator with turns of wire per phase.
What rotor flux would be required to produce a terminal line-to-line voltage of 6 kV? What happens to the resulting net magnetic field? If an ac machine has the rotor and stator magnetic fields shown in Figure P, what is the direction of the induced torque in the machine? Is the machine acting as a motor or generator?
The machine is acting as a generator. In the early days of ac motor development, machine designers had great difficulty controlling the core losses hysteresis and eddy currents in machines. They had not yet developed steels with low hysteresis, and were not making laminations as thin as the ones used today. To help control these losses, early ac motors in the USA were run from a 25 Hz ac power supply, while lighting systems were run from a separate 60 Hz ac power supply.
What was the fastest rotational speed available to these early motors? At a location in Europe, it is necessary to supply kW of Hz power. The only power sources available operate at 50 Hz. It is decided to generate the power by means of a motor-generator set consisting of a synchronous motor driving a synchronous generator.
How many poles should each of the two machines have in order to convert Hz power to Hz power? A V kVA 0. At 60 Hz, its friction and windage losses are 24 kW, and its core losses are 18 kW. The field circuit has a dc voltage of V, and the maximum I F is 10 A.
Column 1 contains field current in amps, and column 2 contains open-circuit terminal voltage in volts. It is 4. The required torque is PIN Assume that the field current of the generator in Problem has been adjusted to a value of 4. What happens to the phasor diagram for the generator? Assume that the field current of the generator in Problem is adjusted to achieve rated voltage V at full load conditions in each of the questions below. Note that the maximum current will be A in any case.
Assume that the field current of the generator in Problem has been adjusted so that it supplies rated voltage when loaded with rated current at unity power factor. You may ignore the effects of R A when answering these questions.
This generator is a very long way from that limit. Since sin This machine can also be paralleled with the normal power supply a very large power system if desired. Show both by means of house diagrams and by means of phasor diagrams what happens to the generator. How much reactive power does the generator supply now? Show this behavior both with phasor diagrams and with house diagrams. The generator must have the same voltage as the power system.
The phase sequence of the oncoming generator must be the same as the phase sequence of the power system. The frequency of the oncoming generator should be slightly higher than the frequency of the running system. The circuit breaker connecting the two systems together should be shut when the above conditions are met and the generator is in phase with the power system.
This generator is operating in parallel with a large power system infinite bus. How much reserve power or torque does this generator have at full load? Sketch the corresponding phasor diagram. Assume I F is still unchanged. The loads supplied by the two generators consist of kW at 0. That action will increase the terminal voltages of the system without changing the power sharing between the generators.
Three physically identical synchronous generators are operating in parallel. They are all rated for a full load of 3 MW at 0. The no-load frequency of generator A is 61 Hz, and its speed droop is 3. The no-load frequency of generator B is The no-load frequency of generator C is At what load does one of the generators exceed its ratings? Which generator exceeds its ratings first? Why or why not?
Generator B is the first to exceed its ratings as load increases. Its rated power is reached at a total load of 6. On the other hand, Generator C gets into trouble as the total load is reduced.
When the total load drops to 2. A paper mill has installed three steam generators boilers to provide process steam and also to use some its waste products as an energy source. Since there is extra capacity, the mill has installed three 5-MW turbine generators to take advantage of the situation.
Each generator is a V kVA 0. Generators 1 and 2 have a characteristic power-frequency slope sP of 2. At what frequency does this limit occur? How much power does each generator supply at that point? Generator 3 will be the first machine to reach that limit.
Generator 3 will supply this power at a frequency of 5. The field currents of the three generators must then be adjusted to get them supplying a power factor of 0. At that point, each generator will be supplying its rated real and reactive power.
Its armature resistance RA is 0. The core losses of this generator at rated conditions are 7 kW, and the friction and windage losses are 8 kW. The open-circuit and short-circuit characteristics are shown in Figure P Column 1 contains field current in amps, and column 2 contains short-circuit terminal current in amps.
It is 0. The open circuit voltage at 0. This is a straight line, so we can determine its value by comparing the ratio of the air-gap voltage to the short-circuit current at any given field current. Each of these files are organized in two columns, where the first column is field current and the second column is either open-circuit terminal voltage or short-circuit current.
A program to read these files and calculate and plot X S is shown below. The saturated synchronous reactance at rated conditions was found to be 0. From the open-circuit characteristic, the required field current would be 0. What is the voltage regulation of this generator at the rated current and power factor? If this generator is operating at the rated conditions and the load is suddenly removed, what will the terminal voltage be? What are the electrical losses in this generator at rated conditions?
If this machine is operating at rated conditions, what input torque must be applied to the shaft of this generator? Express your answer both in newton-meters and in pound-feet. The input power to this generator is equal to the output power plus losses. Assume that the generator field current is adjusted to supply V under rated conditions. What is the static stability limit of this generator? Note: You may ignore R A to make this calculation easier.
How close is the full-load condition of this generator to the static stability limit? Normal generators would have more margin than this. If the field current and the magnitude of the load current are held constant, how will the terminal voltage change as the load power factor varies from 0.
Make a plot of the terminal voltage versus the impedance angle of the load being supplied by this generator. Also, the magnitude of the armature current is constant.
Assume that the generator is connected to a V infinite bus, and that its field current has been adjusted so that it is supplying rated power and power factor to the bus. You may ignore the armature resistance R A when answering the following questions. The reactive power supplied would increase as shown below. A MVA A three-phase Y-connected synchronous generator is rated MVA, Its synchronous reactance is 0.
The phase voltage is Note that this is not quite true, if the armature resistance R A is included, since R A does not scale with frequency in the same fashion as the other terms. Two identical kVA V synchronous generators are connected in parallel to supply a load. The prime movers of the two generators happen to have different speed droop characteristics.
When the field currents of the two generators are equal, one delivers A at 0. A generating station for a power system consists of four MVA kV 0. Three of these generators are each supplying a steady 75 MW at a frequency of 60 Hz, while the fourth generator called the swing generator handles all incremental load changes on the system while maintaining the system's frequency at 60 Hz.
Suppose that you were an engineer planning a new electric co-generation facility for a plant with excess process steam.
You have a choice of either two 10 MW turbine-generators or a single 20 MW turbine generator. What would be the advantages and disadvantages of each choice?
If two 10 MW generators are chosen, one of them could go down for maintenance and some power could still be generated. A MVA three-phase Express the answer both in ohms per phase and in per-unit. The extrapolated air-gap voltage at this point is The generator is connected in parallel with a Hz If the internal generated voltage E A is decreased by 5 percent, what will the new armature current I A be?
A V, 60 Hz, four-pole synchronous motor draws 50 A from the line at unity power factor and full load. Assuming that the motor is lossless, answer the following questions: a What is the output torque of this motor? Express the answer both in newton-meters and in pound-feet.
Explain your answer, using phasor diagrams. This increase in E A changes the angle of the current I A , eventually causing it to reach a power factor of 0. A V, 60 Hz, hp 0. Ignore its friction, windage, and core losses for the purposes of this problem. How near is this value to the maximum possible induced torque of the motor for this field current setting?
The induced torque is POUT A V hp 0. The open-circuit characteristic of this motor is shown in Figure P Answer the following questions about the motor, assuming that it is being supplied by an infinite bus. What would the new power factor be? How much reactive power is being consumed or supplied by the motor?
This voltage would require a field current of 4. Since the power supplied by the motor does not change when I F is changed, this quantity will be a constant. The new power factor is cos 3. At a power factor of 0. The internal generated voltage at 0. Plot the V-curves I A versus I F for the synchronous motor of Problem at no-load, half-load, and full- load conditions.
It may simplify the calculations required by this problem. Also, you may assume that R A is negligible for this calculation. Note that we are assuming that R A is negligible in each case.
The flattening visible to the right of the V-curves is due to magnetic saturation in the machine. If a Hz synchronous motor is to be operated at 50 Hz, will its synchronous reactance be the same as at 60 Hz, or will it change? Hint: Think about the derivation of X S. The armature reaction voltage is caused by the armature magnetic field B S , and the amount of voltage is directly proportional to the speed with which the magnetic field sweeps over the stator surface.
The higher the frequency, the faster B S sweeps over the stator, and the higher the armature reaction voltage Estat is. Therefore, the armature reaction voltage is directly proportional to frequency. Similarly, the reactance of the armature self-inductance is directly proportional to frequency, so the total synchronous reactance X S is directly proportional to frequency. A V kW 0. The rotational losses are also to be ignored. A V Y-connected synchronous motor is drawing 40 A at unity power factor from a V power system.
The field current flowing under these conditions is 2. Assume a linear open-circuit characteristic. Therefore, I A2 0. A synchronous machine has a synchronous reactance of 2. How much power P is this machine consuming from or supplying to the electrical system? How much reactive power Q is this machine consuming from or supplying to the electrical system?
Figure P shows a synchronous motor phasor diagram for a motor operating at a leading power factor with no RA. This generator is supplying power to a V kW 0. The synchronous generator is adjusted to have a terminal voltage of V when the motor is drawing the rated power at unity power factor. What is its new value? To make finding the new conditions easier, we will make the angle of the phasor E A, g the reference during the following calculations.
The resulting phasor diagram is shown below. Note that an increase in machine flux has increased the reactive power supplied by the motor and also raised the terminal voltage of the system. This is consistent with what we learned about reactive power sharing in Chapter 5. A V, kW, Hz, four-pole, Y-connected synchronous motor has a rated power factor of 0.
At full load, the efficiency is 91 percent. The armature resistance is 0. The Y-connected synchronous motor whose nameplate is shown in Figure has a per-unit synchronous reactance of 0. Is it the armature current or the field current that limits the reactive power output? Express the answer both in newton- meters and in pound-feet. We will have to check each one separately, and limit the reactive power to the lesser of the two limits.
The stator apparent power limit defines a maximum safe stator current. This limit is the same as the rated input power for this motor, since the motor is rated at unity power factor. The maximum E A is V or 1. A V three-phase Y-connected synchronous motor has a synchronous reactance of 1. A V, kVA, 0. Ignore all losses. A hp V 0. Its efficiency at full load is 89 percent.
What is the phase current of the motor at rated conditions? To simplify this part of the problem, we will ignore R A. Answer the following questions about the machine of Problem Is it consuming reactive power from or supplying reactive power to the power system? Is the machine operating within its ratings under these circumstances? Why is this so? A V, three-phase, two-pole, Hz induction motor is running at a slip of 5 percent. Answer the questions in Problem for a V, three-phase, four-pole, Hz induction motor running at a slip of 0.
A kW, V, Hz, six-pole induction motor has a slip of 6 percent when operating at full-load conditions. At full-load conditions, the friction and windage losses are W, and the core losses are W. Calculate the slip and the electrical frequency of the rotor at no-load and full-load conditions. What is the speed regulation of this motor [Equation ]? A V, two-pole, Hz Y-connected wound-rotor induction motor is rated at 15 hp. The power consumed by the Thevenin equivalent circuit must be the same as the power consumed by the original circuit.
For the motor in Problem , what is the slip at the pullout torque? What is the pullout torque of this motor? For the motor of Problem , how much additional resistance referred to the stator circuit would it be necessary to add to the rotor circuit to make the maximum torque occur at starting conditions when the shaft is not moving?
Plot the torque-speed characteristic of this motor with the additional resistance inserted. If the motor in Problem is to be operated on a Hz power system, what must be done to its supply voltage? What will the equivalent circuit component values be at 50 Hz? Answer the questions in Problem for operation at 50 Hz with a slip of 0. Figure a shows a simple circuit consisting of a voltage source, a resistor, and two reactances.
Find the Thevenin equivalent voltage and impedance of this circuit at the terminals. Figure P shows a simple circuit consisting of a voltage source, two resistors, and two reactances in parallel with each other. If the resistor RL is allowed to vary but all the other components are constant, at what value of RL will the maximum possible power be supplied to it?
Prove your answer. Use this result to derive the expression for the pullout torque [Equation 7- 54 ]. A V Hz two-pole Y-connected induction motor is rated at 75 kW. For the motor in Problem , what is the pullout torque? What is the slip at the pullout torque? What is the rotor speed at the pullout torque?
SOLUTION The slip at pullout torque is found by calculating the Thevenin equivalent of the input circuit from the rotor back to the power supply, and then using that with the rotor circuit model. If the motor in Problem is to be driven from a V Hz power supply, what will the pullout torque be?
What will the slip be at pullout? The resulting equivalent circuit is shown below. At what slip does Pout equal the rated power of the machine? An appropriate program is shown below. It follows the calculations performed for Problem , but repeats them at many values of slip, and then plots the results.
It produces an output power of 75 kW at 3. A V, 60 Hz, six-pole Y-connected hp design class B induction motor is tested in the laboratory, with the following results: No load: V, A V, four-pole, hp, Hz, Y-connected three-phase induction motor develops its full-load induced torque at 3.
At full load, the output power of this motor is 50 hp and its slip is 3. The two curves are plotted below.
As you can see, only the 0. The Thevenin equivalent of the input circuit was calculate in part a. The easiest way to find the line current or armature current at starting is to get the equivalent impedance Z F of the rotor circuit in parallel with jX M at starting conditions, and then calculate the starting current as the phase voltage divided by the sum of the series impedances, as shown below.
Answer the following questions about the motor in Problem What will the voltage be at the motor end of the transmission line during starting? Note that this voltage sagged by In this chapter, we learned that a step-down autotransformer could be used to reduce the starting current drawn by an induction motor. While this technique works, an autotransformer is relatively expensive.
Answer the following questions about this type of starter. At start-up, the motor develops 1. This motor is to be started with an autotransformer reduced voltage starter. A wound-rotor induction motor is operating at rated voltage and frequency with its slip rings shorted and with a load of about 25 percent of the rated value for the machine.
For most loads, the induced torque will decrease. This reduces the phase current and line current in the motor and on the secondary side of the transformer by a factor of 0. However, the current on the primary of the autotransformer will be reduced by another factor of 0. When it is necessary to stop an induction motor very rapidly, many induction motor controllers reverse the direction of rotation of the magnetic fields by switching any two stator leads. When the direction of rotation of the magnetic fields is reversed, the motor develops an induced torque opposite to the current direction of rotation, so it quickly stops and tries to start turning in the opposite direction.
If power is removed from the stator circuit at the moment when the rotor speed goes through zero, then the motor has been stopped very rapidly. This technique for rapidly stopping an induction motor is called plugging. The motor of Problem is running at rated conditions and is to be stopped by plugging. What is the power flowing into or out of the machine?
When the loop goes beyond the pole faces, eind will momentarily fall to 0 V, and the current flow will momentarily reverse. Therefore, the average current flow over a complete cycle will be somewhat less than 2. Refer to the simple two-pole eight-coil machine shown in Figure P Consider the internal resistance of the machine in determining the current flow. SOLUTION a This winding is progressive, since the ends of each coil are connected to the commutator segments ahead of the segments that the beginnings of the coils are connected to.
There are 16 conductors in this machine, and about 12 of them are under the pole faces at any given time. A dc machine has 8 poles and a rated current of A. How much current will flow in each path at rated conditions if the armature is a simplex lap-wound, b duplex lap-wound, c simplex wave-wound? How many parallel current paths will there be in the armature of a pole machine if the armature is a simplex lap-wound, b duplex wave-wound, c triplex lap-wound, d quadruplex wave-wound?
An eight-pole, kW, V DC generator has a duplex lap-wound armature, which has 64 coils with 16 turns per coil. How wide must each one be? Since it is duplex-wound, each brush must be wide enough to stretch across 2 complete commutator segments. Since there are 16 parallel paths through the machine, the armature resistance of the generator is 0. Figure P shows a small two-pole dc motor with eight rotor coils and four turns per coil. The flux per pole in this machine is 0.
Ignore any internal resistance in the motor. If K is known, then the speed of the motor can be found. Refer to the machine winding shown in Figure P How wide should they be? At the time shown, those windings are 1, 2, 9, and Therefore, the brushes should be connected to short out commutator segments b-c-d and j-k-l at the instant shown in the figure.
Each brush should be two commutator segments wide, since this is a duplex winding. Of that number, an average of 14 of them would be under the pole faces at any one time. Therefore, there are 28 conductors divided among 4 parallel paths, which produces 7 conductors per path.
Describe in detail the winding of the machine shown in Figure P If a positive voltage is applied to the brush under the North pole face, which way will this motor rotate? If a positive voltage is applied to the brush under the North pole face, the rotor will rotate in a counterclockwise direction.
Magnetization curve as shown in Figure P Column 1 contains field current in amps, and column 2 contains the internal generated voltage EA in volts. In Problems through , assume that the motor described above can be connected in shunt.
The equivalent circuit of the shunt motor is shown in Figure P The correct values are given in the text, but shown incorrectly on the figure. This will be corrected at the second printing. Assuming no armature reaction, what is the speed of the motor at full load? What is the speed regulation of the motor? Therefore, EA Assume no armature reaction, as in the previous problem. How does it compare to the result for Problem ? What is the starting current of this machine if it is started by connecting it directly to the power supply VT?
How does this starting current compare to the full-load current of the motor? This much current is extremely likely to damage the motor. For the separately excited motor of Problem a What is the maximum no-load speed attainable by varying both V A and Radj?
What is its full-load speed? What is its speed regulation? Calculate and plot the torque- speed characteristic for this motor. Neglect armature effects in this problem. Both curves are plotted on the same scale to facilitate comparison. The motor is connected cumulatively compounded and is operating at full load. How does the new speed compared to the full-load speed calculated in Problem ? The motor is now connected differentially compounded.
Note that this plot has a larger vertical scale to accommodate the speed runaway of the differentially-compounded motor. The core losses are W, and the mechanical losses are W at full load.
Assume that the mechanical losses vary as the cube of the speed of the motor and that the core losses are constant. Its armature resistance is 0. Column 1 contains magnetomotive force in ampere-turns, and column 2 contains the internal generated voltage EA in volts. Armature reaction is negligible in this machine. Neglect rotational losses. Therefore the speed of the motor at these conditions is EA The motor described above is connected in shunt.
Derive the shape of its torque- speed characteristic. Compare the two curves. A series motor is now constructed from this machine by leaving the shunt field out entirely. Derive the torque-speed characteristic of the resulting motor.
To make a practical series motor out of this machine, it would be necessary to include 20 to 30 series turns instead of An automatic starter circuit is to be designed for a shunt motor rated at 15 hp, V, and 60 A.
The armature resistance of the motor is 0. The motor is to start with no more than percent of its rated armature current, and as soon as the current falls to rated value, a starting resistor stage is to be cut out. How many stages of starting resistance are needed, and how big should each one be? The maximum desired starting current is 2. The three stages of starting resistance can be found from the resistance in the circuit at each state during starting.
Armature reaction may be ignored in this machine. What is the output torque of the motor? What would happen to the motor if its field circuit were to open? Ignoring armature reaction, what would the final steady- state speed of the motor be under those conditions? The magnetization curve for a separately excited dc generator is shown in Figure P Its field circuit is rated at 5A.
The machine in Problem is reconnected as a shunt generator and is shown in Figure P The magnetization curve and the field resistance line are plotted below. As you can see, they intersect at a terminal voltage of V. As shown in the figure below, there is a difference of 3. This program created the plot shown above. Note that there are actually two places where the difference between the E A and VT lines is 3. The code shown in bold face below prevents the program from reporting that first unstable point.
Tell user. As shown in the figure below, there is a difference of 7. The figure below shows that a triangle consisting of 3. There is no point where a triangle consisting of 3. Assume no armature reaction. The point where the distance between the E A and VT curves is exactly 4. The new point where the distance between the E A and VT curves is exactly 4. Note that decreasing the field resistance of the shunt generator increases the terminal voltage. Its equivalent circuit is shown in Figure P Answer the following questions about this machine, assuming no armature reaction.
Compare it to the terminal characteristics of the shunt dc generators in Problem d. If the machine described in Problem is reconnected as a differentially compounded dc generator, what will its terminal characteristic look like? Derive it in the same fashion as in Problem Compare it to the terminal characteristics of the cumulatively compounded dc generator in Problem and the shunt dc generators in Problem d.
A cumulatively compounded dc generator is operating properly as a flat-compounded dc generator. The machine is then shut down, and its shunt field connections are reversed. SOLUTION a The output voltage will not build up, because the residual flux now induces a voltage in the opposite direction, which causes a field current to flow that tends to further reduce the residual flux. A three-phase synchronous machine is mechanically connected to a shunt dc machine, forming a motor- generator set, as shown in Figure P The dc machine is connected to a dc power system supplying a constant V, and the ac machine is connected to a V Hz infinite bus.
The dc machine has four poles and is rated at 50 kW and V. It has a per-unit armature resistance of 0. The ac machine has four poles and is Y-connected. It is rated at 50 kVA, V, and 0. Assume that the magnetization curves of both machines are linear. How much power is being supplied to the dc motor from the dc power system?
How large is the internal generated voltage E A of the dc machine? How large is the internal generated voltage E A of the ac machine? What effect does this change have on the real power supplied by the motor-generator set? On the reactive power supplied by the motor- generator set? Calculate the real and reactive power supplied or consumed by the ac machine under these conditions. On the reactive power supplied by the motor-generator set?
How can the real power flow through an ac-dc motor-generator set be controlled? How can the reactive power supplied or consumed by the ac machine be controlled without affecting the real power flow? This is also the power converted from electrical to mechanical form in the dc machine, since all other losses are neglected. The internal generated voltage E A of the dc machine is The phasor diagram illustrating this change is shown below. Therefore, E A on the dc machine will decrease to 0.
This is also the output power of the dc machine, the input power of the ac machine, and the output power of the ac machine, since losses are being neglected.
Note that changes in power flow also have some effect on the reactive power of the ac machine: in this problem, Q dropped from 35 kvar to 30 kvar when the real power flow was adjusted. This adjustment has basically no effect on the real power flow through the MG set. The rotational losses may be assumed constant over the normal operating range of the motor. If the slip is 0. Repeat Problem for a rotor slip of 0. Assuming that the rotational losses are still 51 W, will this motor continue accelerating or will it slow down again?
Pconv is The motor will slow down5. Therefore, the losses will really be less than 51 W, and this motor might just be able to keep on accelerating slowly—it is a close thing either way. Note that this program shows the torque-speed curve for both positive and negative directions of rotation. Using Eq. Because the second term on the right-hand side of the Eq.
From Fig. However, the mobilities and resistivities for these two samples are different. Ev Assuming complete ionization, the Fenni level measm ed from the intrinsic Fenni level is 0. Then from Eq. The value is comparable to the thermal velocity, the linear relationship between drift velocity and the electric field is not valid. The impmity profile is, x J. We use the Poisson 's equation for calculation of the electric field E x.
The int1insic caniers density in Si at different temperatures can be obtained by using Fig. The interceptions give the built-in potential of the p-n junctions. ILpdX x. The hole diffusion length is larger than the length of neutral region. We can use following equations to determine the parameters of the diode. As the temperature increases, the total reverse current also increases.
That is, the total electron current increases. The impact ionization takes place when the electron gains enough energy from the electrical field to create an electron-hole pair. When the temperature increases, total number of electron increases resulting in easy to lose their energy by collision with other electron before breaking the lattice bonds.
This need higher breakdown voltage. The obtained electrostatic potentials are 1. The depletion widths are 3. The total depletion width will be reduced when the heterojunction is forward-biased from the thermal equilibrium condition. On the other hand, when the heterojunction is reverse-biased, the total depletion width will be increased. We can sketch p n x p n 0 curves by using a computer program: 1. By multiplying this 2 value by q and the cross-sectional area A, we can obtain the same expression as Q B.
In Problem 3, 1. In Eq. W2 Therefore, the collector current is directly proportional to the minority carrier charge stored in the base. And the collector current components are given by 1. The emitter efficiency can be obtained by I En The value is very close to unity. The mobility of an average impurity concentration of 6. Comparing the equations with Eq. Referring Eq. The neutral base width should be 0. The impmity concentration of the n1 region is cm Qot q d Co The bandgap in degenerately doped Si is around 1eV due to bandgap-narrowing effect.
Pros: 1. Higher operation speed. High device density Cons: 1. More complicated fabrication flow. High manufacturing cost. Therefore, From Eq. The pinch-off voltage is qN D d12 1. The pinch-off voltage is 1. By neglecting the second term in Eq. For same energy but a width of 8 meV, we use the same well thickness of 6. The resonant-tunneling current is related to the integrated flux of electrons whose energy is in the range where the transmission coefficient is large. From Eqs. The Fresnel transmission coefficient from Eq.
We assume a conventional p-n junction laser and a stripe DH laser have the same active area. It is impractical. Take the solution which is the only practical one, i.
The threshold current in Fig. The photons with 1. However, since the cells are in series, the top cell wiI1 block the current generated by the bottom cell once the current is larger than the top cell's dark saturation current va1ue.
We can assume that absorption and hence photogeneration occurs over the entire i layer. The earth is hotter. The efficiencies are The segregation coefficient of boron in silicon is 0. It is smaller than unity, so the solubility of B in Si under solid phase is smaller than that of the melt.
Therefore, the excess B atoms will be thrown-off into the melt, then the concentration of B in the melt will be increased. The tail-end of the crystal is the last to solidify. Therefore, the concentration of B in the tail-end of grown c1ystal will be higher than that of seed-end. The reason is that the solubility in the melt is prop01tional to the temperature, and the temperature is higher in the center prut than at the perimeter.
Therefore, the solubility is higher in the center prui, causing a higher impmity concentration there. The segregation coefficient of Ga in Si is 8 x 1o-3 FromEq. We have from Eq. The corresponding doping concentration varies from 2. From the Fig. Therefore, the As content will be lost when the temperature is increased. Thus the composition of liquid GaAs always becomes gallium rich.
We divide the wafer into four symmetrical patts for convenient dicing, and discard the perimeter parts of the wafer. Usually the quality of the perimeter patts is the worst due to the edge effects.
Where f. For close-packing arrange, there are 3 pie shaped sections in the equilateral triangle. The molecular weight is The x value is about 0. The time required to grow 0. For the field oxide with an original thickness 0. A1 A2 8.
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