Published by McGraw-Hill, a business unit of The McGraw-Hill Companies, Inc., 1221 Avenue of the Americas, New York, NY 10020. Observing voltage and current waveforms from a computer simulation accomplishes some of the same objectives as those.
PREFACE
Most of the topics presented in this text are appropriate for junior or senior level electrical engineering students. Much of this text includes computer simulation using PSpice® to complement analytical circuit solution techniques.
Introduction
- POWER ELECTRONICS
- CONVERTER CLASSIFICATION
- POWER ELECTRONICS CONCEPTS
- ELECTRONIC SWITCHES
- SWITCH SELECTION
- SPICE, PSPICE, AND CAPTURE
- SWITCHES IN PSPICE
- BIBLIOGRAPHY
The device type is selected from the on and off requirements, the voltage and current requirements of the switch for on and off states, and the required switching speed. Load Resistor Voltage Switch Control Voltage 10.0 V. Standard BJT QbreakN can be used instead of a detailed transistor model for a rudimentary simulation.
Problems
Emanuel, “Powers in Nonsinusoidal Situations, A Review of Definitions and Physical Meaning”, IEEE Transactions on Power Delivery, vëll.
Power Computations
INTRODUCTION
POWER AND ENERGY
The average power absorbed by a DC voltage source v(t)Vdc having a periodic current i(t) is derived from the basic definition of average power in Eq. Therefore, average power absorbed by a DC voltage source is the product of the voltage and the average current.
INDUCTORS AND CAPACITORS
The voltage across the inductor is calculated from v(t) L(di/dt) and shown in Fig. When p(t) is positive, the inductor is the power sink and when p(t) is negative, the inductor is the power supply.
ENERGY RECOVERY
The inductor voltage is equal to the source voltage when the transistors are on (Figure 2-5b):. The average power absorbed by the resistor is given by Eq. c) The source current is equal to the inductor current when the switch is closed and is zero when the switch is open.
EFFECTIVE VALUES: RMS
Determine the rms value of the periodic pulse waveform with a duty ratio of Das, shown in figure. The rms value of the shifted triangular waveform can be determined by using the result of part (a).
APPARENT POWER AND POWER FACTOR
In alternating current circuits (linear circuits with sinusoidal sources), apparent power is the magnitude of complex power. In sinusoidal AC circuits, the above calculation results in pf ⫽coswhere is the phase angle between the voltage and current sinusoids.
POWER COMPUTATIONS FOR SINUSOIDAL AC CIRCUITS
The second term of the integration is the constant cos( ), which has an average value of cos(. -52) and power factor of cos (– ) for sinusoidal AC circuits are special cases and are not applicable to non-sinusoidal voltages and currents.
POWER COMPUTATIONS FOR NONSINUSOIDAL PERIODIC WAVEFORMS
The power relationships for these circuits can be expressed in terms of the components of the Fourier series. Note that the total average power is the sum of the powers at the frequencies in the Fourier series.
POWER COMPUTATIONS USING PSPICE
PSpice is used to determine the steady-rms current and the Fourier components of the current. The rms value of the load current can be obtained from the Fourier series in the output file from Eq. Upon entering the variable to be displayed, the spectrum of the Fourier series will appear.
Summary
For numerical calculations in Probe to be accurate, the simulation must represent steady-state voltages and currents. Fourier series expressions are available in PSpice by using Fourier Analysis in Simulation Settings or by using the FFT option in Probe.
Bibliography
Determine (a) the instantaneous power, (b) the average power and (c) the energy absorbed by the device in each period. An inductor is energized as shown in the circuit of Fig. a) Determine the peak energy stored in the inductor. Determine (a) the power absorbed by the load, (b) the power factor of the load, (c) the THD of the load current, (d) the distortion factor of the load current, and (e) the crest factor of the load current.
Half-Wave Rectifiers
INTRODUCTION
RESISTIVE LOAD
Assuming the diode is ideal, the voltage across a forward-biased diode is zero and the current is positive. For the negative half cycle of the source, the diode is biased in the opposite direction, making the current zero. In a real diode, the voltage drop of the diode will ensure that the load voltage and current are equal.
RESISTIVE-INDUCTIVE LOAD
This steady-state current can be found by phasor analysis, resulting in (3-6) where Z2R2(L)2 and tan 1aLR b. The natural response is a transient that occurs when the load is activated. Note also that the inductor voltage is negative when the current decreases (vL L dot). Also P can be calculated from the definition of average power: e) The power factor is calculated from the definition pf PS, and the power Pis supplied by the source, which must be the same as that absorbed by the load.
PSPICE SIMULATION
The time axis is changed to angle by using the x variable and entering Time*60*360; (b) Determination of rms value of current in probe. For example, the objective may be to design a half-wave rectifier circuit to produce a specified value of average current by choosing the correct value of Lin and RLload. Since there is no closed-form solution, a trial-and-error iterative method must be used.
RL-SOURCE LOAD
-26) Assuming the diode and inductor are ideal, there is no average power absorbed by either. The power delivered by the AC source is equal to the sum of the power absorbed by the resistor and the DC source. Determine (a) an expression for the current in the circuit, (b) the power absorbed by the resistor, (c) the power absorbed by the DC source, and (d) the power delivered by the AC source and the power factor of the circuit.
INDUCTOR-SOURCE LOAD
A distinctive feature of this circuit is that the power supplied by the source is the same as that absorbed by the dc source, minus the losses associated with a non-ideal diode and inductor. If the objective is to transfer power from an ac source to a dc source, losses are kept to a minimum by using this circuit. Determine (a) an expression for the current, (b) the power absorbed by the dc source, and (c) the power factor. a).
THE FREEWHEELING DIODE
Since the voltage across the load RL is the same as the source voltage when the source is positive and is zero when the source is negative, the load voltage is a half-wave rectified sine wave. The current reaches the periodic steady state after several periods (depending on the constant L/Rtime), which means that the current at the end of a period is the same as the current at the beginning of the period, as shown in Fig. The Fourier series for the half-wave rectified sine wave for the voltage across the load is.
VFigure 3-9 Steady-state load voltage and current
The average current in the RLload is a function of the applied voltage and the resistance, but not the inductance. If the inductance is infinite, the impedance of the load to alternating current in the Fourier series is infinite, and the load current is purely direct current. The zero-to-peak fluctuation in the load current can be estimated to be equal to the amplitude of the first AC term in the Fourier series.
HALF-WAVE RECTIFIER WITH A CAPACITOR FILTER
The current in the resistor is calculated from iR voR. The current in the capacitor is calculated from. which can also be expressed using the variable tas as. The output voltage is expressed by Eq. b) The peak-to-peak output voltage is described by Eq. c) The capacitor current is determined from Eq. d) The peak diode current is determined from Eq. For an efficient filter capacitor, the output voltage is essentially the same as the peak input voltage.
THE CONTROLLED HALF-WAVE RECTIFIER
The turn-off angle is defined as the angle at which the current returns to zero, as in the case of an uncontrolled rectifier. The analysis of this circuit is very similar to the analysis of the uncontrolled half-wave rectifier discussed earlier in this chapter. Determine (a) the expression for the current, (b) the power absorbed by the resistor, and (c) the power absorbed by the DC source in the load.
PSPICE SOLUTIONS FOR CONTROLLED RECTIFIERS
Design a circuit that will deliver 150 W to the dc voltage source from a 120-Vrms 60-Hz ac source. An uncontrolled rectifier with this source and load will have an average current of 2.25 A and an average power at the dc source of 225 W, as calculated in Example 3-5 earlier. Another option that has been chosen for this application is the half-wave controlled rectifier of Fig.
COMMUTATION
Summary
A freewheeling diode forces the voltage across an RLload to be a half-wave rectified sine wave. An SCR in place of the diode in a half-wave rectifier provides a way to control the output current and voltage. The parametric sweep in PSpice allows different values of a circuit parameter to be tried and aids in circuit design.
Bibliography
Determine (a) the expression for the load current, (b) the average current, (c) the power absorbed by the resistor, and (d) the power factor. Determine (a) the power absorbed by the dc voltage source, (b) the power absorbed by the resistor, and (c) the power factor. Determine (a) the expression for the load current, (b) the average load current, and (c) the power absorbed by the load.
Full-Wave Rectifiers
INTRODUCTION
SINGLE-PHASE FULL-WAVE RECTIFIERS
The direct (average) component of the current in this circuit is 4-7) Sine terms in Fourier analysis are unchanged by a DC source provided the current is continuous. The half-wave rectifier in Fig. a) Determine the peak-to-peak variation of the output voltage. Capacitor C1 is charged to Vm through D1 when. the source is positive and C2 charges to Vm through D4 when the source is negative.
CONTROLLED FULL-WAVE RECTIFIERS
The direct current expression is calculated from Eq. c) The amplitudes of ac terms are calculated from Eqs. The control rectifier in Fig. a) Determine the delay angle so that the power absorbed by the DC source is 1000 W. b). Determine the inductance value that will limit the peak-to-peak load current change to 2 A. The required output voltage is given by Eq.
THREE-PHASE RECTIFIERS
The output voltage across the load is one of the line-to-line voltages of the source. The fundamental frequency of the output voltage is 6, where is the frequency of the three-phase source. -44) where Vm,LLis the peak line-to-line voltage of the three-phase source, which is
CONTROLLED THREE-PHASE RECTIFIERS
The three-phase six-pulse bridge rectifier shows a marked improvement in the quality of the DC output over the single-phase rectifier. The above discussion focused on circuits that act as rectifiers, meaning that the current flow is from the AC side of the converter to the DC side. The DC current must be in the direction shown due to the SCRs in the bridge.
DC POWER TRANSMISSION
The two AC systems each have their own generators, and the purpose of the dc line is to enable power exchange between the AC systems. The directions of the SCRs are such that the current will be positive as shown in the line. Determine the required current carrying capacity of the dc line and calculate the power loss in the line.
COMMUTATION: THE EFFECT OF SOURCE INDUCTANCE
For the uncontrolled three-phase bridge rectifier with source reactance (Fig. 4-27a), assume that diodes D1 and D2 are conducting and the load current is a constant Io. The next transition has transferred the load current from D1 to D3 in the upper half of the bridge.
Summary
A filter capacitor at the output of a rectifier can produce an output voltage that is nearly dc. An LC output filter can further improve the quality of the dc output and reduce the peak current in the diode. Switches such as SCRs can be used to control the output of a single-phase or three-phase rectifier.
Bibliography
Determine (a) the average load current, (b) the power absorbed by the load, and (c) the power factor. Determine (a) the power absorbed by the dc source, (b) the power absorbed by the resistor, and (c) the power factor. Determine (a) the average and rms load current, (b) the average and rms diode current, (c) the rms source current, and (d) the power factor.
AC Voltage Controllers
INTRODUCTION
THE SINGLE-PHASE AC VOLTAGE CONTROLLER
The operating principle for a single-phase AC voltage controller using phase control is very similar to that of the controlled half-wave rectifier of Sec. Determine (a) an expression for load current for the first half period, (b) the rms load current, (c) the rms current, (d) the average SR current, (e) the power delivered to the load , and (f) the labor factor. a) The current is expressed as in Eq. The rms load current is determined from Eq. c) The rms current in each SCR is determined from Eq. d) Average SCR current is obtained from Eq. e) Power absorbed by the load is. f) Power factor is determined from P/S.
THREE-PHASE VOLTAGE CONTROLLERS
