5
RECTIFICATION OF UTILITY INPUT USING DIODE RECTIFIERS

As discussed in the introduction to Chapter 1, the role of power electronics is to facilitate power flow, often in a controlled manner, between two systems shown in Figure 5.1: one of them a “source” and the other a “load.” Typically, power is provided by a single-phase or a three-phase utility source, for example, in adjustable-speed motor drives. (Of course, there are exceptions, for example, in wind turbines, where the wind-turbine generator is the source of power to the utility grid that acts like a “load.”)

FIGURE 5.1 Block diagram of power electronic systems.

Such power-electronic interfaces often consist of a voltage-link structure, discussed in Section 1.5.1, where the input from the AC source is first rectified into a DC voltage across a large capacitor. If reversing power flow is not an objective, it is possible to rectify the AC input, single-phase or three-phase, by means of diode rectifiers discussed in this chapter. The knowledge of such systems is essential for learning about thyristor converters, discussed in Chapter 13, which are used in important applications such as high-voltage DC transmission (HVDC) systems.

5.1 INTRODUCTION

In diode rectifiers, unless corrective action is taken as described in the next chapter, power is drawn by means of highly distorted currents, which have a deleterious effect on the power quality of the utility source. This issue and the basic principles of diode-rectifier operation are examined in this chapter.

5.2 DISTORTION AND POWER FACTOR

To quantify distortion in the current drawn by power electronic systems, it is necessary to define certain indices. As a base case, consider the linear upper R minus upper L load shown in Figure 5.2a, which is supplied by a sinusoidal source in steady state. The voltage and current phasors are shown in Figure 5.2b, where phi is the angle by which the current lags the voltage. Using RMS values for the voltage and current magnitudes, the average power supplied by the source is

FIGURE 5.2 Voltage and current phasors in a simple R-L circuit.

upper P equals upper V Subscript s Baseline upper I Subscript s Baseline cosine phi (5.1)

The power factor left-parenthesis upper P upper F right-parenthesis at which power is drawn is defined as the ratio of the real average power P to the product of the RMS voltage and the RMS current:

StartLayout 1st Row 1st Column upper P upper F equals StartFraction upper P Over upper V Subscript s Baseline upper I Subscript s Baseline EndFraction equals cosine phi 2nd Column left-parenthesis using Equation 5 zero width space .1 right-parenthesis EndLayout (5.2)

where upper V Subscript s Baseline upper I Subscript s is the apparent power. For a given voltage, from Equation (5.2), the RMS current drawn is

upper I Subscript s Baseline equals StartFraction upper P Over upper V Subscript s Baseline dot upper P upper F EndFraction period (5.3)

This shows that the power factor PF and the current upper I Subscript s are inversely proportional. This current flows through the utility distribution lines, transformers, and so on, causing losses in their resistances. This is the reason why utilities prefer unity power factor loads that draw power at the minimum value of the RMS current.

5.2.1 RMS Value of Distorted Current and the Total Harmonic Distortion (THD) [1]

The sinusoidal current drawn by the linear load in Figure 5.2 has zero distortion. However, power electronic systems with diode rectifiers as the front-end draw currents with a distorted waveform such as that shown by i Subscript s Baseline left-parenthesis t right-parenthesis in Figure 5.3a. The utility voltage v Subscript s Baseline left-parenthesis t right-parenthesis is assumed sinusoidal. The following analysis is general, applying to the utility supply that is either single-phase or three-phase, in which case the analysis is on a per-phase basis.

FIGURE 5.3 Current drawn by power electronics equipment with diode-bridge front-end.

The current waveform i Subscript s Baseline left-parenthesis t right-parenthesis in Figure 5.3a repeats with a time period upper T 1. By Fourier analysis of this repetitive waveform, we can compute its fundamental frequency (equals 1 slash upper T 1) component i Subscript s Baseline 1 Baseline left-parenthesis t right-parenthesis, shown dotted in Figure 5.3a. The distortion component i Subscript d i s t o r t i o n Baseline left-parenthesis t right-parenthesis in the input current is the difference between i Subscript s Baseline left-parenthesis t right-parenthesis and the fundamental-frequency component i Subscript s Baseline 1 Baseline left-parenthesis t right-parenthesis:

i Subscript d i s t o r t i o n Baseline left-parenthesis t right-parenthesis equals i Subscript s Baseline left-parenthesis t right-parenthesis minus i Subscript s Baseline 1 Baseline left-parenthesis t right-parenthesis comma (5.4)

where i Subscript d i s t o r t i o n Baseline left-parenthesis t right-parenthesis using Equation (5.4) is plotted in Figure 5.3b. This distortion component consists of components at frequencies that are the multiples of the fundamental frequency.

To obtain the RMS value of i Subscript s Baseline left-parenthesis t right-parenthesis in Figure 5.3a, we will apply the basic definition of RMS:

upper I Subscript s Baseline equals StartRoot StartFraction 1 Over upper T 1 EndFraction integral Underscript upper T 1 Endscripts zero width space i Subscript s Superscript 2 Baseline left-parenthesis t right-parenthesis dot d t EndRoot period (5.5)

Using Equation (5.4),

i Subscript s Superscript 2 Baseline left-parenthesis t right-parenthesis equals i Subscript s Baseline 1 Superscript 2 Baseline left-parenthesis t right-parenthesis plus i Subscript d i s t o r t i o n Superscript 2 Baseline left-parenthesis t right-parenthesis plus 2 i Subscript s Baseline 1 Baseline left-parenthesis t right-parenthesis times i Subscript d i s t o r t i o n Baseline left-parenthesis t right-parenthesis period (5.6)

In a repetitive waveform, the integral of the products of the two harmonic components (including the fundamental) at unequal frequencies, over the repetition time-period, equals zero:

StartLayout 1st Row 1st Column integral Underscript upper T 1 Endscripts g Subscript h 1 Baseline zero width space left-parenthesis t right-parenthesis dot g Subscript h 2 Baseline zero width space left-parenthesis t right-parenthesis dot d t equals 0 2nd Column h 1 not-equals h 2 EndLayout period (5.7)

Therefore, substituting Equation (5.6) into Equation (5.5) and making use of Equation (5.7) that implies that the integral of the third term on the right side of Equation (5.6) equals zero (assuming the average component to be zero),

upper I Subscript s Baseline equals StartRoot ModifyingBelow StartFraction 1 Over upper T 1 EndFraction zero width space integral Underscript upper T 1 Endscripts zero width space i Subscript s Baseline 1 Superscript 2 Baseline left-parenthesis t right-parenthesis dot d t With presentation form for vertical right-brace Underscript upper I Subscript s Baseline 1 Superscript 2 Baseline Endscripts plus ModifyingBelow StartFraction 1 Over upper T 1 EndFraction integral Underscript upper T 1 Endscripts zero width space i Subscript d i s t o r t i o n Superscript 2 Baseline left-parenthesis t right-parenthesis dot d t With presentation form for vertical right-brace Underscript upper I Subscript d i s t o r t i o n Superscript 2 Baseline Endscripts plus 0 EndRoot comma (5.8)

or

upper I Subscript s Baseline equals StartRoot upper I Subscript s Baseline 1 Superscript 2 Baseline plus upper I Subscript d i s t o r t i o n Superscript 2 Baseline EndRoot comma (5.9)

where the RMS values of the fundamental-frequency component and the distortion component are as follows:

upper I Subscript s Baseline 1 Baseline equals StartRoot StartFraction 1 Over upper T 1 EndFraction integral Underscript upper T 1 Endscripts i Subscript s Baseline 1 Superscript 2 Baseline left-parenthesis t right-parenthesis dot d t EndRoot comma (5.10)

and

upper I Subscript d i s t o r t i o n Baseline equals StartRoot StartFraction 1 Over upper T 1 EndFraction integral Underscript upper T 1 Endscripts i Subscript d i s t o r t i o n Superscript 2 Baseline left-parenthesis t right-parenthesis dot d t EndRoot period (5.11)

Based on the RMS values of the fundamental and the distortion components in the input current i Subscript s Baseline left-parenthesis t right-parenthesis, a distortion index called the total harmonic distortion (THD) is defined in percentage as follows:

percent-sign upper T upper H upper D equals 100 times StartFraction upper I Subscript d i s t o r t i o n Baseline Over upper I Subscript s Baseline 1 Baseline EndFraction period (5.12)

Using Equation (5.9) into Equation (5.12),

percent-sign upper T upper H upper D equals 100 times StartFraction StartRoot upper I Subscript s Superscript 2 Baseline minus upper I Subscript s Baseline 1 Superscript 2 Baseline EndRoot Over upper I Subscript s Baseline 1 Baseline EndFraction period (5.13)

The RMS value of the distortion component can be obtained based on the harmonic components (except the fundamental) as follows using Equation (5.7):

upper I Subscript d i s t o r t i o n Baseline equals StartRoot sigma-summation Underscript h equals 2 Overscript infinity Endscripts upper I Subscript s h Superscript 2 Baseline EndRoot comma (5.14)

where upper I Subscript s h is the RMS value of the harmonic component “h.”

5.2.1.1 Obtaining Harmonic Components by Fourier Analysis

By Fourier analysis, any distorted (non-sinusoidal) waveform g left-parenthesis t right-parenthesis that is repetitive with a fundamental frequency f 1, for example, i Subscript s in Figure 5.3a, can be expressed as a sum of sinusoidal components at the fundamental frequency and its multiples (harmonic frequencies):

g left-parenthesis t right-parenthesis equals upper G 0 plus sigma-summation Underscript h equals 1 Overscript infinity Endscripts g Subscript h Baseline left-parenthesis t right-parenthesis equals upper G 0 plus sigma-summation Underscript h equals 1 Overscript infinity Endscripts left-brace a Subscript h Baseline cosine left-parenthesis h omega t right-parenthesis plus b Subscript h Baseline sine left-parenthesis h omega t right-parenthesis right-brace comma (5.15)

where the average value upper G 0 is DC,

upper G 0 equals StartFraction 1 Over 2 pi EndFraction integral Subscript 0 Superscript 2 pi Baseline g left-parenthesis t right-parenthesis dot d left-parenthesis omega t right-parenthesis period (5.16)

The sinusoidal waveforms in Equation (5.15) at the fundamental frequency f 1 (h equals 1) and the harmonic components at frequencies h times f 1 can be expressed as the sum of their cosine and sine components,

a Subscript h Baseline equals StartFraction 1 Over pi EndFraction integral Subscript 0 Superscript 2 pi Baseline g left-parenthesis t right-parenthesis cosine left-parenthesis h omega t right-parenthesis d left-parenthesis omega t right-parenthesis h equals 1 comma 2 comma period period period comma infinity (5.17)
b Subscript h Baseline equals StartFraction 1 Over pi EndFraction integral Subscript 0 Superscript 2 pi Baseline g left-parenthesis t right-parenthesis sine left-parenthesis h omega t right-parenthesis d left-parenthesis omega t right-parenthesis h equals 1 comma 2 comma period period period comma infinity period (5.18)

The cosine and the sine components above, given by Equations (5.17) and (5.18), can be combined and written as a phasor in terms of its RMS value,

upper G overbar Subscript h Baseline equals upper G Subscript h Baseline angle zero width space phi Subscript h Baseline comma (5.19)

where the RMS magnitude in terms of the peak values a Subscript h and b Subscript h equals

upper G Subscript h Baseline equals StartFraction StartRoot a Subscript h Superscript 2 Baseline plus b Subscript h Superscript 2 Baseline EndRoot Over StartRoot 2 EndRoot EndFraction comma (5.20)

and the phase phi Subscript h can be expressed as

tangent phi Subscript h Baseline equals StartFraction minus b Subscript h Baseline Over a Subscript h Baseline EndFraction period (5.21)

It can be shown that the RMS value of the distorted function g left-parenthesis t right-parenthesis can be expressed in terms of its average and the sinusoidal components as

upper G equals StartRoot upper G 0 squared plus sigma-summation Underscript h equals 1 Overscript infinity Endscripts upper G Subscript h Superscript 2 Baseline EndRoot period (5.22)

In Fourier analysis, by appropriate selection of the time origin, it is often possible to make the sine or the cosine components in Equation (5.15) to be zero, thus considerably simplifying the analysis, as illustrated by a simple example below.

Example 5.1

A current i Subscript s of a square waveform is shown in Figure 5.4a. Calculate and plot its fundamental frequency component and its distortion component. What is the %THD associated with this waveform?

FIGURE 5.4 Example 5.1.

Solution From Fourier analysis, by choosing the time origin as shown in Figure 5.4a, i Subscript s Baseline left-parenthesis t right-parenthesis in Figure 5.4a can be expressed as

i Subscript s Baseline equals StartFraction 4 Over pi EndFraction upper I left-parenthesis sine omega 1 t plus one-third sine 3 omega 1 t plus one-fifth sine 5 omega 1 t plus one-seventh sine 7 omega 1 t plus zero width space ellipsis zero width space right-parenthesis period (5.23)

The fundamental frequency component and the distortion component are plotted in Figures 5.4b and 5.4c.

From Figure 5.4a, it is obvious that the RMS value upper I Subscript s of the square waveform is equal to upper I. In the Fourier expression of Equation (5.23), the RMS value of the fundamental-frequency component is

upper I Subscript s Baseline 1 Baseline equals StartFraction left-parenthesis 4 slash pi right-parenthesis Over StartRoot 2 EndRoot EndFraction upper I equals 0.9 upper I

Therefore, the distortion component can be calculated from Equation (5.9) as

upper I Subscript d i s t o r t i o n Baseline equals StartRoot upper I Subscript s Superscript 2 Baseline minus upper I Subscript s Baseline 1 Superscript 2 Baseline EndRoot equals StartRoot upper I squared minus left-parenthesis 0.9 upper I right-parenthesis squared EndRoot equals 0.436 upper I period

Therefore, using the definition of THD,

percent-sign upper T upper H upper D equals 100 times StartFraction upper I Subscript d i s t o r t i o n Baseline Over upper I Subscript s Baseline 1 Baseline EndFraction equals 100 times StartFraction 0.436 upper I Over 0.9 upper I EndFraction equals 48.4 percent-sign period

5.2.2 The Displacement Power Factor (DPF) and Power Factor (PF)

Next, we will consider the power factor at which power is drawn by a load with a distorted current waveform such as that shown in Figure 5.3a. As before, it is reasonable to assume that the utility-supplied line-frequency voltage v Subscript s Baseline left-parenthesis t right-parenthesis is sinusoidal, with an RMS value of upper V Subscript s and a frequency f 1 left-parenthesis equals StartFraction omega 1 Over 2 pi EndFraction right-parenthesis. Based on Equation (5.7), which states that the product of the cross-frequency terms has a zero average, the average power P drawn by the load in Figure 5.3a is due only to the fundamental-frequency component of the current:

upper P equals StartFraction 1 Over upper T 1 EndFraction integral Underscript upper T 1 Endscripts v Subscript s Baseline left-parenthesis t right-parenthesis dot i Subscript s Baseline left-parenthesis t right-parenthesis dot d t equals StartFraction 1 Over upper T 1 EndFraction integral Underscript upper T 1 Endscripts v Subscript s Baseline left-parenthesis t right-parenthesis dot i Subscript s Baseline 1 Baseline left-parenthesis t right-parenthesis dot d t period (5.24)

Therefore, in contrast to Equation (5.1) for a linear load, in a load that draws distorted current, similar to Equation (5.1),

upper P equals upper V Subscript s Baseline upper I Subscript s Baseline 1 Baseline cosine phi 1 comma (5.25)

where phi 1 is the angle by which the fundamental-frequency current component i Subscript s Baseline 1 Baseline left-parenthesis t right-parenthesis lags behind the voltage, as shown in Figure 5.3a.

At this point, another term called the displacement power factor (DPF) needs to be introduced, where,

upper D upper P upper F equals cosine phi 1 period (5.26)

Therefore, using the DPF in Equation (5.25),

upper P equals upper V Subscript s Baseline upper I Subscript s Baseline 1 Baseline zero width space left-parenthesis upper D upper P upper F right-parenthesis period (5.27)

In the presence of distortion in the current, the meaning and therefore the definition of the power factor, at which the real average power P is drawn, remains the same as in Equation (5.2), that is, the ratio of the real power to the product of the RMS voltage and the RMS current:

upper P upper F equals StartFraction upper P Over upper V Subscript s Baseline upper I Subscript s Baseline EndFraction period(5.28)

Substituting Equation (5.27) for P into Equation (5.28),

upper P upper F equals left-parenthesis StartFraction upper I Subscript s Baseline 1 Baseline Over upper I Subscript s Baseline EndFraction right-parenthesis left-parenthesis upper D upper P upper F right-parenthesis period (5.29)

In linear loads that draw sinusoidal currents, the current-ratio left-parenthesis upper I Subscript s Baseline 1 Baseline slash upper I Subscript s Baseline right-parenthesis in Equation (5.29) is unity, hence upper P upper F equals upper D upper P upper F. Equation (5.29) shows the following: a high distortion in the current waveform leads to a low power factor, even if the DPF is high. Using Equation (5.13), the ratio left-parenthesis upper I Subscript s Baseline 1 Baseline zero width space slash zero width space upper I Subscript s Baseline right-parenthesis in Equation (5.29) can be expressed in terms of the total harmonic distortion as

StartFraction upper I Subscript s Baseline 1 Baseline Over upper I Subscript s Baseline EndFraction equals StartFraction 1 Over StartRoot 1 plus left-parenthesis StartFraction percent-sign upper T upper H upper D Over 100 EndFraction right-parenthesis squared EndRoot EndFraction period (5.30)

Therefore, in Equation (5.29),

upper P upper F equals StartFraction 1 Over StartRoot 1 plus left-parenthesis StartFraction percent-sign upper T upper H upper D Over 100 EndFraction right-parenthesis squared EndRoot EndFraction dot upper D upper P upper F (5.31)

The effect of THD on the power factor is shown in Figure 5.5 by plotting left-parenthesis upper P upper F slash upper D upper P upper F right-parenthesis versus THD. It shows that even if the displacement power factor is unity, a total harmonic distortion of 100% (which is possible in power electronic systems unless corrective measures are taken) can reduce the power factor to approximately 0.7 (or StartFraction 1 Over StartRoot 2 EndRoot EndFraction equals 0.707, to three decimal places, to be exact) which is unacceptably low.

FIGURE 5.5 Relation between PF/DPF and THD.

5.2.3 Deleterious Effects of Harmonic Distortion and a Poor Power Factor

There are several deleterious effects of high distortion in the current waveform and the poor power factor that results due to it. These are as follows:

  • Power loss in utility equipment such as distribution and transmission lines, transformers, and generators increases, possibly to the point of overloading them.
  • Harmonic currents can overload the shunt capacitors used by utilities for voltage support and may cause resonance conditions between the capacitive reactance of these capacitors and the inductive reactance of the distribution and transmission lines.
  • The utility voltage waveform will also become distorted, adversely affecting other linear loads, if a significant portion of the load supplied by the utility draws power by means of distorted currents.

5.2.3.1 Harmonic Guidelines

In order to prevent degradation in power quality, recommended guidelines (in the form of the IEEE-519) have been suggested by the IEEE (Institute of Electrical and Electronics Engineers). These guidelines place the responsibilities of maintaining power quality on the consumers and the utilities as follows: (1) on the power consumers, such as the users of power electronic systems, to limit the distortion in the current drawn, and (2) on the utilities to ensure that the voltage supply is sinusoidal with less than a specified amount of distortion.

The limits on current distortion placed by the IEEE-519 are shown in Table 5.1, where the limits on harmonic currents, as a ratio of the fundamental component, are specified for various harmonic frequencies. Also, the limits on the THD are specified. These limits are selected to prevent distortion in the voltage waveform of the utility supply. Therefore, the limits on distortion in Table 5.1 depend on the “stiffness” of the utility supply, which is shown in Figure 5.6a by a voltage source upper V overbar s in series with an internal impedance upper Z Subscript s. An ideal voltage supply has zero internal impedance. In contrast, the voltage supply at the end of a long distribution line, for example, will have a large internal impedance.

FIGURE 5.6 (a) Utility supply; (b) short-circuit current.

TABLE 5.1  Harmonic current distortion (upper I Subscript h Baseline slash upper I Subscript l)

Odd Harmonic Order h (in %)Total Harmonic Distortion (%)
Isc/I1 h < 1111 ≤ h < 1717 ≤ h < 2323 ≤ h <3535 ≤ h
<204.02.01.50.60.35.0
20–507.03.52.51.00.58.0
50–10010.04.54.01.50.712.0
100–100012.05.55.02.01.015.0
>100015.07.06.02.51.420.0

To define the “stiffness” of the supply, the short-circuit current upper I Subscript s c is calculated by hypothetically placing a short circuit at the supply terminals, as shown in Figure 5.6b. The stiffness of the supply must be calculated in relation to the load current. Therefore, the stiffness is defined by a ratio called the short-circuit ratio (SCR):

StartLayout 1st Row 1st Column Short hyphen Circuit hyphen Ratio 2nd Column upper S upper C upper R equals StartFraction upper I Subscript s c Baseline Over upper I Subscript s Baseline 1 Baseline EndFraction EndLayout comma (5.32)

where upper I Subscript s Baseline 1 is the fundamental-frequency component of the load current. Table 5.1 shows that a smaller short-circuit ratio corresponds to lower limits on the allowed distortion in the current drawn. For the short-circuit ratio of less than 20, the total harmonic distortion in the current must be less than 5%. Power electronic systems that meet this limit would also meet the limits of more stiff supplies.

It should be noted that the IEEE-519 does not propose harmonic guidelines for individual pieces of equipment but rather for the aggregate of loads (such as in an industrial plant) seen from the service entrance, which is also the point of common coupling (PCC) with other customers. However, the IEEE-519 is frequently interpreted as the harmonic guidelines for specifying individual pieces of equipment such as motor drives. There are other harmonic standards, such as the IEC-1000, which apply to individual pieces of equipment.

5.3 CLASSIFYING THE “FRONT-END” OF POWER ELECTRONIC SYSTEMS

Interaction between the utility supply and power electronic systems depends on the “front-ends” (within the power-processing units), which convert line-frequency AC into DC. These front-ends can be broadly classified as follows:

  • Diode-bridge rectifiers (shown in Figure 5.7a) in which power flows only in one direction.
  • Switch-mode converters (shown in Figure 5.7b) in which the power flow can reverse and the line currents are sinusoidal at the unity power factor.
  • Thyristor converters (shown in Figure 5.7c) in which the power flow can be made bi-directional.

FIGURE 5.7 Front-end of power electronics equipment.

All of these front-ends can be designed to interface with single-phase or three-phase utility systems. In the following discussion, a brief description of the diode interface shown in Figure 5.7a is provided, supplemented by an analysis of results obtained through computer simulations. Interfaces using switch-mode converters in Figure 5.7b and thyristor converters in Figure 5.7c are discussed later in this book.

5.4 DIODE-RECTIFIER BRIDGE “FRONT-END”

Most power electronic systems use diode-bridge rectifiers, such as the one shown in Figure 5.7a, even though they draw currents with highly distorted waveforms and the power through them can flow only in one direction. In switch-mode DC power supplies, these diode-bridge rectifiers are supplemented by a power-factor-correction circuit to meet current harmonic limits, as discussed in the next chapter.

Diode rectifiers rectify line-frequency AC into DC across the DC-bus capacitor without any control over the DC-bus voltage. For analyzing the interaction between the utility and the power electronic systems, the switch-mode converter and the load can be represented by an equivalent resistance upper R Subscript e q across the DC-bus capacitor. In our theoretical discussion, it is adequate to assume the diodes are ideal.

In the following subsections, we will consider single-phase as well as three-phase diode rectifiers operating in steady state, where waveforms repeat from one line-frequency cycle upper T 1 left-parenthesis equals 1 slash f 1 right-parenthesis to the next.

5.4.1 Single-Phase Diode-Rectifier Bridge

At power levels below a few kW, for example, in residential applications, power electronic systems are supplied by a single-phase utility source. A commonly used full-bridge rectifier circuit is shown in Figure 5.8a, in which upper L Subscript s is the sum of the inductance internal to the utility supply and an external inductance, which may be intentionally added in series. Losses on the AC side can be represented by the series resistance upper R Subscript s.

FIGURE 5.8 Full-bridge diode rectifier.

To understand the circuit operation, the rectifier circuit can be drawn as in Figure 5.8b, where the AC-side inductance and resistance have been ignored. The circuit consists of a top group and a bottom group of diodes. If the DC-side current i Subscript d r is to flow, one diode from each group must conduct to facilitate the flow of this current. In the top group, both diodes have their cathodes connected together. Therefore, the diode connected to the most positive voltage will conduct; the other will be reverse biased. In the bottom group, both diodes have their anodes connected together. Therefore, the diode connected to the most negative voltage will conduct; the other will be reverse biased.

As an example, resistance upper R Subscript d is connected across the terminals on the DC-side, as shown in Figure 5.9a. The circuit waveforms are shown Figure 5.9b. During the positive half-cycle of the input voltage v Subscript s, diodes 1 and 2 conduct the DC-side current i Subscript d r, equal to v Subscript s Baseline slash upper R Subscript d, and the DC-side voltage is v Subscript d Baseline equals v Subscript s. During the negative half-cycle of the input voltage v Subscript s, diodes 3 and 4 conduct the DC-side current i Subscript d r, equal to Math bar pipe bar symblom v Subscript s Math bar pipe bar symblom slash upper R Subscript d, and the DC-side voltage is v Subscript d Baseline equals Math bar pipe bar symblom v Subscript s Baseline Math bar pipe bar symblom period

FIGURE 5.9 Full-bridge diode rectifier with resistive load.

The average value upper V Subscript d of the voltage across the DC-side of the converter can be obtained by averaging the v Subscript d Baseline left-parenthesis t right-parenthesis waveform in Figure 5.9b over only one half-cycle (by symmetry):

upper V Subscript d Baseline equals StartFraction 1 Over pi EndFraction integral Subscript 0 Superscript pi Baseline zero width space ModifyingAbove upper V With ˆ Subscript s Baseline sine omega t dot d left-parenthesis omega t right-parenthesis equals StartFraction 2 Over pi EndFraction ModifyingAbove upper V With ˆ Subscript s Baseline equals 0.9 upper V Subscript s Baseline left-parenthesis rms right-parenthesis comma (5.33)

and

upper I Subscript d r Baseline equals StartFraction upper V Subscript d Baseline Over upper R Subscript d Baseline EndFraction equals StartFraction 0.9 upper V Subscript s Baseline left-parenthesis rms right-parenthesis Over upper R Subscript d Baseline EndFraction (5.34)

As another example, consider the load on the DC-side to have a large inductance, as shown in Figure 5.10a, such that the DC-side current is essentially DC. Assuming that i Subscript d r Baseline left-parenthesis t right-parenthesis asymptotically-equals upper I Subscript d r to be purely DC, the waveforms are as shown in Figure 5.10b.

FIGURE 5.10 Full-bridge diode rectifier with an inductive load where i Subscript d r Baseline left-parenthesis t right-parenthesis asymptotically-equals upper I Subscript d r (DC).

Note that the waveform of v Subscript d Baseline left-parenthesis t right-parenthesis in Figure 5.10b is identical to that in Figure 5.9b for a resistive load. Therefore, the average value upper V Subscript d of the voltage across the DC-side of the converter in Figure 5.10a is the same as in Equation (5.33), that is upper V Subscript d Baseline equals 0.9 upper V Subscript s Baseline left-parenthesis rm s right-parenthesis. Similarly, the average value of the DC-side current is related to upper V Subscript d by upper R Subscript d, as upper I Subscript d r Baseline equals upper V Subscript d Baseline slash upper R Subscript d.

5.4.1.1 DC-Bus Capacitor to Achieve a Low-Ripple in the DC-Side Voltage

Figures 5.9b and 5.10b show that the DC-side voltage has a large ripple. To eliminate this, in order to achieve a voltage waveform that is fairly DC, a large filter capacitor is connected on the DC-side, as shown in Figure 5.8a. As shown in Figure 5.11, at the beginning of the positive half-cycle of the input voltage v Subscript s, the capacitor is already charged to a DC voltage v Subscript d. So long as v Subscript d exceeds the input voltage magnitude, all diodes remain reverse biased, and the input current is zero. Power to the equivalent resistance upper R Subscript e q is supplied by the energy stored in the capacitor up to t 1.

FIGURE 5.11 Waveforms for the full-bridge diode rectifier with a DC-bus capacitor.

Beyond t 1, the input current i Subscript s Baseline left-parenthesis equals i Subscript d r Baseline right-parenthesis increases, flowing through diodes upper D 1 and upper D 2. Beyond t 2, the input voltage becomes smaller than the capacitor voltage, and the input current begins to decline, falling to zero at t 3. Beyond t 3, until one half-cycle later than t 1, the input current remains zero, and the power to upper R Subscript e q is supplied by the energy stored in the capacitor. At left-parenthesis t 1 plus upper T 1 slash 2 right-parenthesis during the negative half-cycle of the input voltage, the input current flows through diodes upper D 3 and upper D 4. The rectifier DC-side current i Subscript d r continues to flow in the same direction as during the positive half-cycle; however, the input current i Subscript s Baseline equals minus i Subscript d r, as shown in Figure 5.11. Figure 5.12 shows waveforms obtained by LTspice simulations for two values of the AC-side inductance, with current THD of 86% and 62%, respectively (higher inductance reduces THD, as discussed in the next section).

FIGURE 5.12 Single-phase diode-bridge rectification for two values of Ls.

The fact that i Subscript d r flows in the same direction during both the positive and the negative half-cycles represents the rectification process. In the circuit of Figure 5.8a in steady state, all waveforms repeat from one cycle to the next. Therefore, the average value of the capacitor current over a line-frequency cycle must be zero so that the DC-bus voltage is in steady state. As a consequence, the average current through the equivalent load-resistance upper R Subscript e q equals the average of the rectifier DC-side current; that is, upper I Subscript d Baseline equals upper I Subscript d r.

5.4.1.2 Effects of Ls and Cd on the Waveforms and the THD

As Figures 5.11 and 5.12 show, power is drawn from the utility supply by means of a pulse of current every half-cycle. The larger the “base” of this pulse during which the current flows, the lower its peak value and the total harmonic distortion (THD). This pulse widening can be accomplished by increasing the AC-side inductance upper L Subscript s. Another parameter under the designer’s control is the value of the DC-bus capacitor upper C Subscript d. At its minimum, it should be able to carry the ripple current in i Subscript d r and in i Subscript d (which in practice is the input DC-side current, with a pulsating waveform, of a switch-mode converter) and keep the peak-to-peak ripple in the DC-bus voltage to some acceptable value, for example, less than 5% of the DC-bus average value. Assuming that these constraints are met, the lower the value of upper C Subscript d, the lower the THD in current and the higher the ripple in the DC-bus voltage.

In practice, it is almost impossible to meet the harmonic limits specified by IEEE-519 by using the above techniques. Rather, the power-factor-correction circuits described in the next chapter are needed to meet the harmonic specifications.

5.4.1.3 Simulation Using LTspice

The simulation of a single-phase diode-bridge rectifier is demonstrated by means of an example:

Example 5.2

In the single-phase diode-bridge rectifier shown in Figure 5.8a, upper L Subscript s Baseline equals 1 m upper H, upper R Subscript s Baseline equals 0.2 upper Omega, upper C Subscript d Baseline equals 1200 mu upper F, and upper R Subscript e q Baseline equals 15 upper Omega. The supply voltage is upper V Subscript s Baseline equals 120 upper V RMS at 60 Hz period Simulate this rectifier using LTspice.

Solution The simulation file used in this example is available on the accompanying website. The LTspice model is shown in Figure 5.13, and the steady-state waveforms from the simulation of this model are shown in Figure 5.14.

FIGURE 5.13 LTspice model.

FIGURE 5.14 LTspice simulation results.

5.4.2 Three-Phase Diode-Rectifier Bridge

It is preferable to use a three-phase utility source, except at a fractional kilowatt, if such a supply is available. A commonly used full-bridge rectifier circuit is shown in Figure 5.15a.

FIGURE 5.15 Three-phase diode bridge rectifier.

To understand the circuit operation, the rectifier circuit can be drawn as in Figure 5.15b. The circuit consists of a top group and a bottom group of diodes. Initially, upper L Subscript s is ignored, and the DC-side current is assumed to flow continuously. At least one diode from each group must conduct to facilitate the flow of i Subscript d r. In the top group, all diodes have their cathodes connected together. Therefore, the diode connected to the most positive voltage will conduct; the other two will be reverse biased. In the bottom group, all diodes have their anodes connected together. Therefore, the diode connected to the most negative voltage will conduct; the other two will be reverse biased.

Ignoring upper L Subscript s and assuming that the DC-side current i Subscript d r is a pure DC, the waveforms are as shown in Figure 5.16. In Figure 5.16a, the waveforms (identified by the dark portions of the curves) show that each diode, based on the principle described above, conducts during 120 degree. The diodes are numbered so that they begin conducting sequentially: 1, 2, 3, and so on. The waveforms for the voltages v Subscript upper P and v Subscript upper N, with respect to the source-neutral, consist of 120 degree-segments of the phase voltages, as shown in Figure 5.16a. The waveform of the DC-side voltage v Subscript d Baseline left-parenthesis equals v Subscript upper P Baseline minus v Subscript upper N Baseline right-parenthesis is shown in Figure 5.16b. It consists of 60 degree-segments of the line-line voltages supplied by the utility. Line currents on the AC-side are as shown in Figure 5.16c. For example, phase-a current flows for 120 degree during each half-cycle of the phase-a input voltage; it flows through diode upper D 1 during the positive half-cycle of v Subscript a and through diode upper D 4 during the negative half-cycle.

FIGURE 5.16 Waveforms in a three-phase rectifier (a constant i Subscript d r).

The average value of the DC-side voltage can be obtained by considering only a 60 degree-segment in the 6-pulse (per line-frequency cycle) waveform shown in Figure 5.16b. Let us consider the instant of the peak in the 60 degree-segment to be the time-origin, with ModifyingAbove upper V With ˆ Subscript upper L upper L as the peak line-line voltage. The average value upper V Subscript d can be obtained by calculating the integral from omega t equals negative pi slash 6 to omega t equals pi slash 6 (the area shown by the hatched area in Figure 5.16b) and then dividing by the interval pi slash 3:

upper V Subscript d Baseline equals StartFraction 1 Over pi slash 3 EndFraction integral Subscript negative pi slash 6 Superscript pi slash 6 Baseline ModifyingAbove upper V With ˆ Subscript upper L upper L Baseline cosine omega t dot d left-parenthesis omega t right-parenthesis equals StartFraction 3 Over pi EndFraction ModifyingAbove upper V With ˆ Subscript upper L upper L Baseline period (5.35)

This average value is plotted as a straight line in Figure 5.16b.

5.4.2.1 Effect of DC-Bus Capacitor

In the three-phase rectifier of Figure 5.15a with the DC-bus capacitor filter, the input current waveforms obtained by computer simulations are shown in Figure 5.17.

FIGURE 5.17 Effect of upper L Subscript s variation (a) upper L Subscript s Baseline equals 0.1 mH; (b) upper L Subscript s Baseline equals 3 mH.

Figure 5.17a shows that the input current waveform within each half-cycle consists of two distinct pulses when upper L Subscript s is small. For example, in the i Subscript a waveform during the positive half-cycle, the first pulse corresponds to the flow of DC-side current through the diode pair left-parenthesis upper D 1 comma upper D 6 right-parenthesis and then through the diode pair left-parenthesis upper D 1 comma upper D 2 right-parenthesis. At larger values of upper L Subscript s, within each half-cycle, the input current between the two pulses does not go to zero, as shown in Figure 5.17b.

The effects of upper L Subscript s and upper C Subscript d on the waveforms can be determined by a parametric analysis, similar to the case of single-phase rectifiers. The THD in the current waveform of Figure 5.17b is much smaller than in Figure 5.17a (23% versus 82%). The AC-side inductance upper L Subscript s is required to provide a line-frequency reactance upper X Subscript upper L Sub Subscript s Baseline left-parenthesis equals 2 pi f 1 upper L Subscript s Baseline right-parenthesis that is typically greater than 2% of the base impedance upper Z Subscript b a s e, which is defined as follows:

StartLayout 1st Row 1st Column Base Impedance 2nd Column upper Z Subscript b a s e Baseline equals 3 StartFraction upper V Subscript s Superscript 2 Baseline Over upper P EndFraction EndLayout comma (5.36)

where upper P is the three-phase power rating of the power electronic system, and upper V Subscript s is the RMS value of the phase voltage. Therefore, typically, the minimum AC-side inductance should be such that

upper X Subscript upper L Sub Subscript s Subscript Baseline greater-than-or-equal-to left-parenthesis 0.02 times upper Z Subscript b a s e Baseline right-parenthesis period (5.37)
5.4.2.2 Simulation Using LTspice

The simulation of a three-phase diode-bridge rectifier is demonstrated by means of an example:

Example 5.3

In the three-phase diode-bridge rectifier shown in Figure 5.15a, upper L Subscript s Baseline equals 0.1 m upper H, upper C Subscript d Baseline equals 470 mu upper F, and upper R Subscript e q Baseline equals 15 upper Omega. The supply voltage is 208 upper Vline-line RMS at 60 upper H z. Simulate this rectifier using LTspice.

Solution The simulation file used in this example is available on the accompanying website. The LTspice model is shown in Figure 5.18, and the steady-state waveforms from the simulation of this model are shown in Figure 5.19.

FIGURE 5.18 LTspice model.

FIGURE 5.19 LTspice simulation results.

5.4.3 Comparison of Single-Phase and Three-Phase Rectifiers

Examination of single-phase and three-phase rectifier waveforms shows the differences in their characteristics. Three-phase rectification results in six identical “pulses” per cycle in the rectified DC-side voltage, whereas single-phase rectification results in two such pulses. Therefore, three-phase rectifiers are superior in terms of minimizing distortion in line currents and ripple across the DC-bus voltage. Consequently, as stated earlier, three-phase rectifiers should be used if a three-phase supply is available. However, three-phase rectifiers, just like single-phase rectifiers, are also unable to meet the harmonic limits specified by IEEE-519 unless corrective actions such as those described in Chapter 12 are taken.

5.5 MEANS TO AVOID TRANSIENT INRUSH CURRENTS AT STARTING

In power electronic systems with rectifier front-ends, it may be necessary to take steps to avoid a large inrush of current at the instant the system is connected to the utility source. In such power electronic systems, the DC-bus capacitor is very large and initially has no voltage across it. Therefore, at the instant the switch in Figure 5.20a is closed to connect the power electronic system to the utility source, a large current flows through the diode-bridge rectifier, charging the DC-bus capacitor.

FIGURE 5.20 Means to avoid inrush current.

This transient current inrush is highly undesirable; fortunately, several means of avoiding it are available. These include using a front-end that consists of thyristors, discussed in Chapter 14, or using a series semiconductor switch as shown in Figure 5.16b. At the instant of starting, the resistance across the switch lets the DC-bus capacitor to get charged without a large inrush current, and subsequently, the semiconductor switch is turned on to bypass the resistance. There are other techniques that can also be employed.

5.6 FRONT-ENDS WITH BI-DIRECTIONAL POWER FLOW

In stop-and-go applications such as elevators, it is cost effective to feed the energy recovered by regenerative braking of the motor drive back into the utility supply. Converter arrangements for such applications are considered in Chapter 12.

REFERENCES

  1. 1. N. Mohan, T.M. Undeland, and W.P. Robbins, Power Electronics: Converters, Applications and Design, 3rd Edition (New York: John Wiley & Sons, 2003).

PROBLEMS

  • 5.1 In a single-phase diode rectifier bridge, upper I Subscript s Baseline equals 10 normal upper A left-parenthesis rms right-parenthesis, upper D upper P upper F equals 0.85 comma and upper D upper P upper F equals 0.85. Calculate upper I Subscript d i s t o r t i o n, percent-sign upper T upper H upper D, and PF.
  • 5.2 In a single-phase diode bridge rectifier circuit, the following operating conditions are given: upper V Subscript s Baseline equals 120 upper V left-parenthesis rms right-parenthesis, upper P equals 0.95 kW, upper I Subscript s Baseline 1 Baseline equals 10 normal upper A, and upper T upper H upper D equals 75 percent-sign. Calculate the following: upper D upper P upper F, upper I Subscript d i s t o r t i o n, upper I Subscript s, and upper P upper F.
  • 5.3 In a single-phase rectifier the input current can be approximated by a triangular waveform every half cycle with a peak of 10 normal upper A and a base of 60 degree. Calculate the RMS current through each diode.
  • 5.4 In the above problem, calculate the ripple component in the DC-side current that will flow through the DC-side capacitor.
  • 5.5 In a three-phase rectifier, if the input currents are of rectangular waveforms, as shown in Figure 5.16c, with an amplitude of 12 A, calculate upper I Subscript s, upper I Subscript s Baseline 1, upper I Subscript d i s t o r t i o n, percent-sign upper T upper H upper D and the power factor. Assume that upper D upper P upper F equals 1.0, as associated with the current waveforms in Figure 5.14c.
  • 5.6 In the above problem, calculate the three-phase power through the rectifier bridge if upper V Subscript upper L upper L Baseline left-parenthesis rms right-parenthesis equals 208 normal upper V.

Simulation Problems

  • 5.7 In the single-phase rectifier in Figure 5.8a, upper L Subscript s Baseline equals 1 mH comma upper R Subscript s Baseline equals 0.25 upper Omega comma upper C Subscript d Baseline equals 1 comma 000 mu normal upper F and upper R Subscript e q Baseline equals 20.0 upper Omega period
    • Obtain the %THD in the input current for the three values of the input inductance 1 mH, 3 mH, and 5 mH. Comment on the input current waveform as a function of the AC-side inductance in part (a).
    • Measure the average and the peak-to-peak ripple in the output voltage for the three values of the input-side inductance in part (a), and comment on the output voltage waveform as a function of the AC-side inductance.
    • Keeping the input inductance value as 1 mH, change the DC-side capacitance values to be 220 mu normal upper F, 470 mu normal upper F and 1 comma 000 mu normal upper F. Measure the peak-to-peak ripple in the output capacitor voltage for these three values of the output capacitance.
  • 5.8 In the three-phase rectifier in Figure 5.15a, upper L Subscript s Baseline equals 0.1 mH comma upper C Subscript d Baseline equals 500 mu normal upper F and upper R Subscript e q Baseline equals 16.5 normal upper Omega Req = 16.5 Ω.
    • Obtain the %THD in the input current for the three values of the input inductance of 0.1 mH, 0.5 mH, and 1.0 mH.
    • Comment on the input current waveform as a function of the AC-side inductance for values.
    • Comment on the output voltage waveform as a function of the AC-side inductance. Measure the average and the peak-to-peak ripple in the output voltage for the three values of the input-side inductance.
    • Measure the peak-to-peak ripple in the output capacitor current for the three values of the input-side inductance. What is the fundamental frequency of this current?
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