Line-frequency diode rectifiers .................................................................................................. 1
Introduction ............................................................................................................................ 1
Basic rectifier concepts .......................................................................................................... 2
Pure resistive load .............................................................................................................. 2
Inductive load ..................................................................................................................... 2
Load with an internal DC voltage ...................................................................................... 4
Single-phase diode bridge rectifiers ....................................................................................... 5
Idealized circuit with LS = 0 ............................................................................................... 5
Practical diode bridge rectifiers ........................................................................................ 16
Three-phase, full-bridge rectifiers ........................................................................................ 19
Idealized circuit with LS=0 ............................................................................................... 19
Constant DC-side voltage vd(t)=Vd .................................................................................. 25
Comparison of single-phase and three-phase rectifiers ........................................................ 27
Line-frequency diode rectifiers
In most power electronic applications, the power input is in the form of a 50- or 60-Hz sine
wave ac voltage provided by the electric utility, that is first converted to a dc voltage.
Increasingly, the trend is to use the inexpensive rectifiers with diodes to convert the input ac
into dc in an uncontrolled manner, using rectifiers with diodes, as illustrated by the block
diagram of Fig. 5- 1. In such diode rectifiers, the power flow can only be from the utility ac
side to the dc side. A majority of the power electronics applications such as switching dc
power supplies, ac motor drives, dc servo drives, and so on, use such uncontrolled rectifiers.
In most of these applications, the rectifiers are supplied directly from the utility source
without a 60-Hz transformer. The avoidance of this costly and bulky 60-Hz transformer is
important in most modem power electronic systems.
The dc output voltage of a rectifier should be as ripple free as possible. Therefore, a large
capacitor is connected as a filter on the dc side. As will be shown in this chapter, this
capacitor gets charged to a value close to the peak of the ac input voltage. As a consequence,
the current through the rectifier is very large near the peak of the 60-Hz ac input voltage and it
does not flow continuously; that is, it becomes zero for finite durations during each half-cycle
of the line frequency. These rectifiers draw highly distorted current from the utility. Now and
even more so in the future, harmonic standards and guidelines will limit the amount of current
distortion allowed into the utility, and the simple diode rectifiers may not be allowed.
Fig. 5.1 Block diagram of a rectifier.
Rectifiers with single-phase and three-phase inputs are discussed in this chapter. The diodes
are assumed to be ideal in the analysis of rectifiers. In a similar manner, the electromagnetic
interference (EMI) filter at the ac input to the rectifiers is ignored, since it does not influence
the basic operation of the rectifier.
Basic rectifier concepts
Rectification of ac voltages and currents is accomplished by means of diodes. Several simple
circuits are considered to illustrate the basic concepts.
Pure resistive load
Consider the circuit of Fig. 5-2a, with a sinusoidal voltage source vs. The waveforms in Fig.
5-2b show that both the load voltage Vd and the current i have an average (dc) component.
Because of the large ripple in Vd and i, this circuit is of little practical significance.
Fig.5.2 Basic rectifier with a load resistance.
Let us consider the load to be inductive, with an inductor in series with a resistor, as shown in
Fig. 5-3a. Prior to t = 0, the voltage vs is negative and the current in the circuit is zero.
Subsequent to t = 0, the diode becomes forward biased and a current begins to flow. Then,
the diode can be replaced by a short, as shown in the equivalent circuit of Fig. 5-3e. The
current in this circuit is governed by the following differential equation:
v s Ri L (5-1)
where the voltage across the inductor vL = L di/dt. The resulting voltages and current are
shown in Figs. 5-3b and c. Until t1, vs> vR (hence vL = vs - vR is positive), the current builds
up, and the inductor stored energy increases. Beyond t1, vL becomes negative, and the current
begins to decrease. After t2, the input voltage vs becomes negative but the current is still
positive and the diode must conduct because of the inductor stored energy.
Fig. 5.3 Basic rectifier with an inductive load.
The instant t3, when the current goes to zero and the diode stops conducting, can be obtained
as follows: The inductor equation vL = L di/dt can be rearranged as
v L dt di (5-2)
Integrating both sides of the above equation between zero and t3 and recognizing that i(0)
and i(t3) are both zero give
t3 i ( t3 ) t3
v L dt di i (t 3 ) i (0) 0 (5-3), we can observe that L dt 0 (5-4)
0 i ( 0) 0
A graphical interpretation of the above equation is as follows: Equation 5-4 can be written as
v dt v dt 0
which in terms of the volt-second areas A and B of Fig. 5-3c is
Area A- Area B=0 (5-6)
Therefore, the current goes to zero at t3 when area A = B in Fig. 5-3c.
Beyond t3, the voltages across both R and L are zero and a reverse polarity voltage (=-vs)
appears across the diode, as shown in Fig. 5-3d. These waveforms repeat with the time period
T = 1/f.
The load voltage Vd becomes negative during the interval from t2 to t3. Therefore, in
comparison to the case of purely resistive load of Fig. 5-2a, the average load voltage is less.
Load with an internal DC voltage
Next, we will consider the circuit of Fig. 5-4a where the load consists of an inductor L and a
dc voltage Ed. The diode begins to conduct at t1 when vs, exceeds Ed. The current reaches its
peak at t2 (when vs is again equal to Ed) and decays to zero at t3, with t3 determined by the
requirement that the volt-second area A be equal to area B in the plot Of vL shown in Fig. 5-
4c. The voltage across the diode is shown in Fig. 5-4d.
Fig. 5.4 Basic rectifier with an internal dc voltage.
Single-phase diode bridge rectifiers
A commonly used single-phase diode bridge rectifier is shown in Fig. 5-5. A large filter
capacitor is connected on the dc side. The utility supply is modeled as a sinusoidal voltage
source vs in series with its internal impedance, which in practice is primarily inductive.
Therefore, it is represented by Ls. To improve the line-current waveform, an inductor may be
added in series on the ac side, which in effect will increase the value of Ls. The objective of
this chapter is to thoroughly analyze the operation of this circuit. Although the circuit appears
simple, the procedure to obtain the associated voltage and current waveforms in a closed form
is quite tedious. Therefore, we will simulate this circuit using PSpice and MATLAB.
However, we will next analyze many simpler and hypothetical circuits in order to gain insight
into the operation of the circuit in Fig. 5-5.
Idealized circuit with LS = 0
As a first approximation to the circuit of Fig. 5-5, we will assume Ls to be zero and replace
the dc side of the rectifier by a resistance R or a constant dc current source Id, as shown in
Figs. 5-6a and b, respectively. It should be noted in the circuit of Fig. 5-6a that although it is
very unlikely that a pure resistive load will be supplied through a diode rectifier, this circuit
models power-factor-corrected rectifiers.
Fig. 5.5 Single-phase diode bridge rectifier.
Fig. 5.6 Idealized diode bridge rectifiers with Ls=0
Similarly, the representation of the load by a constant dc current in the circuit of Fig. 5-6b is
an approximation to a situation where a large inductor may be connected in series at the dc
output of the rectifier for filtering in Fig. 5-5. This is commonly done in phase-controlled
The circuits in Fig. 5-6 are redrawn in Fig. 5-7, which shows that this circuit consists of two
groups of diodes: the top group with diodes 1 and 3 and the bottom group with diodes 2 and 4.
With Ls = 0, it is easy to see the operation of each group of diodes. The current id flows
continuously through one diode of the top group and one diode of the bottom group.
In the top group, the cathodes of the two diodes are at a common potential. Therefore, the
diode with its anode at the highest potential will conduct id. That is, when vs, is positive,
diode 1 will conduct id and vs, will appear as a reverse-bias voltage across diode 3. When vs
goes negative, the current id shifts (commutates) instantaneously to diode 3 since Ls = 0. A
reverse-bias voltage appears across diode 1.
In the bottom group, the anodes of the two diodes are at a common potential. Therefore, the
diode with its cathode at the lowest potential will conduct id. That is, when vs is positive,
diode 2 will carry id and vs will appear as a reverse-bias voltage across diode 4. When vs goes
negative, the current id instantaneously commutates to diode 4 and a reverse-bias voltage
appears across diode 2.
The voltage and current waveforms in the circuits of Fig. 5-6 are shown in Figs. 5-8a and b.
There are several items worth noting. In both circuits, when vs is positive, diodes
Fig. 5.7 Redrawn rectifiers of Fig. 5.6.
1 and 2 conduct and vd = vs, and is = id. When vs goes negative, diodes 3 and 4 conduct and,
therefore, vd = -vs, and is = -id. Therefore, at any time, the dc-side output voltage of the diode
rectifier can be expressed as
v d (t ) v s (5-7)
Similarly, the ac-side current can be expressed as
i if v s 0
is d (5-8)
id if v s 0
and the transition between the two values is instantaneous due to the assumption of zero Ls.
The average value Vdo, (where the subscript o stands for the idealized case with Ls = 0) of the
dc output voltage in both circuits can be obtained by assigning an arbitrary time origin t=0 in
Fig. 5-8 and then integrating v s 2Vs sin t over one-half time period (where =2f and
2V cost T 0/ 2
1 1 2
Vdo 2Vs sin t dt 2Vs (5-9)
T / 2
(T / 2)
Fig. 5.8 Waveforms in the rectifiers of (a) Fig. 5.6a and (b) Fig. 5.6b.
Vdo 2V s 0.9V s (5-10)
where Vs, is the rms value of the input voltage.
With id(t) = Id, vs and is waveforms are shown in Fig. 5-9a along with the fundamental-
frequency component is1. Applying the basic definition of the rms value to the is waveform in
this idealized case yields
By Fourier analysis of is, the fundamental and the harmonic components have the following
rms values in this idealized case:
I s1 2 I d 0. 9 I d (5-12)
0 for even values of h and
I sh (5-13)
I s1 / h for odd values of h
The harmonic components in is are shown in Fig. 5-9b. The total harmonic distortion can
be calculated by Eq. 3-36 to be
Fig. 5.9 Line current is in the idealized case.
Fig. 5.10 Single-phase rectifier with Ls.
By visual inspection of is waveform in Fig. 5-9a, it is apparent that is1 is in phase with the vs
waveform. Therefore from Fig. 5-9a,
PF=DPF Is1/Is=0.9 (5-16)
Effect of Ls on current commutation
Next, we will look at the effect of a finite ac-side inductance Ls on the circuit operation. We
will assume that the dc side can be represented by a constant dc current Id shown in Fig. 5-10.
Due to a finite Ls, the transition of the ac-side current is from a value of +Id to -Id (or vice
versa) will not be instantaneous. The finite time interval required for such a transition is
called the current commutation time, or the commutation interval u, and this process where
the current conduction shifts from one diode (or a set of diodes) to the other is called the
current commutation process.
In order to understand this process fully, let us first consider a simple hypothetical circuit of
Fig. 5-11a with two diodes supplied by a sinusoidal voltage source v s 2Vs sin t . The
output is represented by a constant dc current source Id. For comparison purposes, Fig. 5-11b
shows vs, vd, and is, waveforms with Ls = 0.
Prior to time t = 0, the voltage vs is negative and the current Id is circulating through D2 with
vd = 0 and is = 0. When vs becomes positive at t = 0, a forward-bias voltage appears across
D1 and it begins to conduct. With a finite Ls, the buildup of is can be obtained from the circuit
redrawn as in Fig. 5-12a (valid only for 0 < is < Id). Since D2 is conducting, it provides a
short-circuit (with vd = 0, assuming an ideal diode) path through which is can build up.
Fig. 5.11 Basic circuit to illustrate current commutation. Waveforms assume Ls=0
Fig. 5.12 (a) Circuit during the commutation. (b) Circuit after the current commutation is
The two mesh currents shown are Id and is. In terms of these mesh currents, the diode current
iD2 = Id – is. Therefore, as is builds up to a value Id during the commutation interval t = u, iD2
is positive and D2 conducts in the circuit of Fig. 5-12a. The current is cannot exceed Id since it
will result in a negative value of iD2 that is not possible. As a consequence, the diode D2 stops
conducting at t = u, resulting in the circuit shown in Fig. 5-12b. The waveforms are plotted
in Fig. 5-13 as a function of t.
It is clear from the above introduction that the current is through the inductor starts with a
value of zero at the beginning of the commutation interval and ends up with a value of Id at
the end. Therefore, to obtain the length of the commutation interval u, we should consider the
inductor equation. During the commutation interval, the input ac voltage appears as a current
commutation voltage across the inductor in Fig. 5-12a:
v L 2V s sin t Ls 0 t u (5-17)
The right side of the above equation can be written as Lsdis/d(t). Therefore
2Vs sin t d (t ) Ls dis (5-18)
Fig. 5.13 Waveforms in the basic circuit of Fig. 5.11. Note that a large value of Ls is used to
clearly show the commutation interval.
Integrating both sides of Eq. 5-18 and recognizing that is goes from zero to Id during the
commutation interval from zero to u, we get
2Vs sin t d (t ) Ls dis Ls I d
In Eq. 5-19, the left side is the integral of the inductor voltage vL during the commutation
interval. The above voltage integral is the same as the volt-radian area Au in Fig. 5-13:
2Vs sin t d (t ) 2Vs (1 cos u) (5-20)
Combining Eqs. 5-19 and 5-20 yields
Au 2Vs (1 cos u ) Ls I d (5-21)
The important observation from Eq. 5-21 is that the integral of the commutation voltage over
the commutation interval can always be calculated by the product of , Ls, and the change in
the current through Ls, during commutation. From Eq. 5-21,
Ls I d
cos u 1 (5-22)
Equation 5-22 confirms that if Ls=0, cos u=1 and the current commutation will be
instantaneous with u=0. For a given frequency , the commutation interval u increases with
Ls, and Id and decreases with increasing voltage Vs.
The finite commutation interval reduces the average value of the output voltage. In Fig. 5-
11b with Ls =0, the average value Vdo of vd is
1 2 2
Vdo 2Vs sin t d (t ) Vs 0.45Vs (5-23)
With a finite Ls and hence a nonzero u in Fig. 5-13, vd=0 during the interval u. Therefore,
Vd 2Vs sin t d (t ) (5-24)
which can be written as
Vd 2Vs sin t d (t ) 2Vs sin t d (t ) (5-25)
Substituing Eqs. 5-23 and 5-19 into Eq. 5-25 yields
area Au L s
Vd 0.45V s 0.45V s Id (5-26)
where the reduction in the average output voltage by Vd from Vdo is
area Au Ls
V d Id (5-27)
We will now extend this analysis to the circuit of Fig. 5- 1 0, redrawn in Fig. 5-14a.
The waveforms are shown in Fig. 5-14b.
Fig. 5.14 (a) Single-phase diode rectifier with Ls. (b) Waveforms.
Once again, we need to consider the current commutation process. Prior to t=0 in Fig. 5-14,
diodes 3 and 4 are conducting Id (as in the circuit of Fig. 5-6b with Ls=0) and is = -Id. The
circuit of Fig. 5-14a is carefully redrawn in Fig. 5-15 to show the current commutation
process during 0 < t < u. Subsequent to t = 0, vs becomes positive and diodes 1 and 2
become forward biased because of the short-circuit path provided by the conducting diodes 3
and 4. The three mesh currents are shown in Fig. 5-15, where the two commutation currents iu
are equal, based on the assumption of identical diodes. All four diodes conduct during the
commutation interval, and therefore, vd = 0. In terms of these mesh currents, we can express
diode currents and the line current is during the commutation interval as
iD1=iD2=iu iD3=iD4=Id-iu (5-28)
where iu builds up from zero at the beginning to Id at the end of the commutation interval.
Therefore, at t = u, iD1 = iD2 = Id and is = Id. During this commutation of current from diodes
3 and 4 to diodes 1 and 2, the current through inductor Ls changes from -Id to Id.
Fig. 5.15 Redrawn circuit of Fig. 5.14a during current commutation
Following the analysis previously carried out on the hypothetical circuit of Fig. 5-11a, volt-
radian area Au in the waveforms of Fig. 5-14b and c can be written from Eqs. 5-19 to 5-21 as
volt radian area Au
2Vs sin t d (t ) Ls di
s 2Ls I d (5-30)
where the lower limit of integration now is is(0) = -Id. Therefore,
Au 2Vs (1 cos u ) 2Ls I d (5-31)
cos u 1 Id (5-32)
A similar commutation takes place one-half cycle later when is, goes from Id to -Id.
In this circuit, the average value of vd in the idealized case (with Ls= 0) was calculated in Eq.
5-10 as Vdo = 0.9Vs. Therefore in the presence of Ls, the average value Vd can be calculated,
following the procedure outlined previously by Eqs. 5-23 through 5-26. Alternatively, we can
calculate Vd by inspecting Fig. 5-14b, where compared to the idealized case, the area Au is
"lost" every half-cycle from the integral of voltage Vd. Therefore,
area Au 2Ls I d
Vd Vdo 0.9V s (5-33)
Constant DC-side voltage vd(t) = Vd
Next, we will consider the circuit shown in Fig. 5-16a, where the assumption is that the dc-
side voltage is constant. It is an approximation to the circuit of Fig. 5-5 with a large value of
C. Another assumption here is that the circuit conditions are such that the current Id is zero
during the zero crossing of vs, as shown by the waveforms in Fig. 5-16c. Under these
conditions, the equivalent circuit is drawn in Fig. 5-16b. Consider the waveforms in Fig. 5-
16c. When vs exceeds Vd at b, diodes 1 and 2 begin to conduct. The current reaches its peak
at p, beyond which vL becomes negative. The current becomes zero at f when the volt-
second areas A and B become equal and negative of each other. The current remains zero
until +b. With a given value of Vd, the average value Id of the dc
current can be calculated by the following procedure:
1. The angle b can be calculated from the equation
Vd 2V s sin b (5-34)
Fig. 5.16 (a) Rectifier with a constant dc-side voltage. (b) Equivalent circuit. (c) Waveforms.
2. As shown in Fig. 5-16c, the inductor voltage starts at zero at b, and becomes zero at p
prior to becoming negative. From symmetry in Fig. 5-16c,
p b (5-35)
3. When the current is flowing, the inductor voltage VL is given by
v L Ls sV s sin(t ) Vd (5-36)
and its integral with respect to t can be written as
Ls did ( 2Vs sin t Vd ) d (t )
where > b. Recognizing that id at b is zero, Eq. 5-37 results in
id ( ) 2Vs sin t Vd dt (5-38)
4. The angle f at which id goes to zero can be obtained from Eq. 5-38 as
2V s sin t Vd d (t ) (5-39)
It corresponds to area A = B in Fig. 5-16c.
5. The average value Id Of the dc current can be obtained by integrating id() from b to f and
then dividing by :
d ( ) d
It is intuitively obvious that for given circuit parameters, Id will depend on the value of Vd and
vice versa. To present the relationship between the two in a general manner, we will
normalize Vd by Vdo, and Id by lshort circuit, where
I sh ort circuit (5-41)
is the rms current that will flow if the ac voltage source vs was short circuited through Ls.
Following the above procedure results in the plot shown in Fig. 5-17, where the current
reaches zero as Vd approaches the peak value of the ac input voltage.
Fig. 5.17 Normalized Id versus Vd in the rectifier of Fig. 5.16a with a constant dc side voltage.
The approximation of a constant dc voltage vd(t) = Vd may be reasonable if the capacitance in
the practical circuit of Fig. 5-5 is large. That is, if the load were replaced by an equivalent
resistance Rload as in Fig. 5-20 later, the time constant CdRload is much larger than the line-
frequency cycle time, resulting in a very small ripple in vd. This approximation allows us to
present the characteristics of the rectifier in a general fashion. In Figs. 5-18 and 5-19, various
quantities are plotted as a function of the dc current Id, normalized with respect to the short-
circuit current Ishort circuit (given by Eq. 5-4 1 ) in order to combine the effects of Ls and
frequency in the same plots. (See Problem 5-17 for justification that allows such a
Fig. 5.18 Total harmonic distortion, DPF and PF in the rectifier of Fig. 5.16a with a constant
Fig. 5.19 Normalized Vd and the crest factor in the rectifier of Fig. 5.16a with a constant dc-
For a given value of ld, increasing Ls results in a smaller Ishort circuit and hence a larger Id/lshort
circuit. Therefore, Figs. 5-18 and 5-19 show that increasing Ls results in improved is waveform
with a lower THD, a better power factor, and a lower (improved) crest factor.
Practical diode bridge rectifiers
Having considered the simplified circuits in the previous sections, we are now ready to
consider the practical circuit of Fig. 5-5, which is redrawn in Fig. 5-20. The load is
represented by an equivalent resistance Rload. In this circuit, there will be some ripple in the
capacitor voltage, and therefore, it must be analyzed differently than the circuit of Fig. 5-16a.
A circuit such as shown in Fig. 5-20 can be easily analyzed by a circuit simulation program
such as PSpice. However, for educational purposes we will first analytically calculate the
Analytical Calculations under a Highly Discontinuous Current
The circuit operating conditions are assumed to result in a highly discontinuous id, similar to
the waveform in Fig. 5-16c, where id goes to zero prior to the zero crossing of vs every half-
cycle. Then the equivalent circuit of Fig. 5-21 can be used to calculate the voltages and
currents in Fig. 5-20. If the above condition is not met, the current commutation discussed in
the earlier sections must be included, which makes analytical calculations difficult.
In order to describe the system in Fig. 5-2 1, the state variables chosen are the inductor current
id and the capacitor voltage vd. During each half-cycle of line frequency, there are two
distinct intervals, similar to those shown earlier in Fig. 5-16c (where tb = b/ and tf = f/):
Fig. 5.20 Practical diode-bridge rectifier with a filter capacitor.
Fig. 5.21 Equivalent circuit of Fig. 5.20.
(a) tb < t < tf. When the current is flowing during tb < t < tf, where tb is the beginning of
conduction and tf is the final conduction time, the following equations describe the circuit
every half-cycle of line frequency:
v s R s i d Ls vd (u sin g KVL ) (5-42)
id C d d (u sin g KCL ) (5-43)
where KVL and KCL are the Kirchhoff voltage and current laws. Rearranging the above
equations in the state variable form during tb < t < tf yields
dt L Ls id 1 / Ls
dv 1s 1
v d 0
C d Rload
The state variable vector x consists of id and vd. The state transition matrix is
A s (5-45)
C d Rload
1 / Ls
Using Eqs. 4-11 through 4-13 for the trapezoidal rule of integration yields
x(t ) Mx (t t ) N v s (t ) v s (t t ) (5-47)
M I A I A
N I A b
(b) t f t t b 1 T . During the interval if t f t t b 1 T , when the diode bridge is not
dt C d Rload
The solution to Eq. 5-50 can be expressed as
( t t f ) /( Cd Rload )
v d (t ) v d (t f )e (5-51)
In the solution of Eqs. 5-47 and 5-51, we need the time instant tb at which the current
conduction starts. Since it is not known prior to the solution, we will use an estimated value.
An exact value of tb will result in the beginning of current conduction exactly one-half cycle
later. Therefore, we will use this condition to check for the accuracy of our choice of tb and
slowly increment tb from its initially chosen value until we reach the exact value within a
Distorted currents drawn by loads such as the diode bridge rectifiers can result in distortion in
the utility voltage waveform. For example, consider the circuit of Fig. 5-20 which is redrawn
in Fig. 5-25. Here Ls1 represents the internal impedance of the utility source and Ls2 may be
intentionally added as a part of the power electronics equipment. A resistance Rs is included
that also can be used to represent the diode resistances.
The voltage across other equipment at the point of common coupling (PCC) is
v PCC v s Ls1 (5-52)
where vs is assumed to be sinusoidal.
Expressing is in Eq. 5-52 in terms of its fundamental and harmonic components yields
v PCC v s Ls1 s1 Ls1 (5-53)
dt h 1
( v PCC )1 v s Ls1 (5-54)
and the voltage distortion component due to the harmonics is
(v PCC ) dis Ls1 (5-55)
Fig. 5.25 Line-voltage notching and distortion.
Three-phase, full-bridge rectifiers
In industrial applications where three-phase ac voltages are available, it is preferable to use
three-phase rectifier circuits, compared to single-phase rectifiers, because of their lower ripple
content in the waveforms and a higher power-handling capability. The three-phase, six-pulse,
full-bridge diode rectifier shown in Fig. 5-30 is a commonly used circuit arrangement. A filter
capacitor is connected at the dc side of the rectifier.
Similar to the analysis of single-phase, full-bridge rectifiers, we will begin with simplified
circuits prior to the discussion of the circuit in Fig. 5-30.
Idealized circuit with LS=0
In the circuit of Fig. 5-31a, the ac-side inductance Ls is assumed to be zero and the dc side is
replaced by a constant dc current Id. We will see later that replacing the dc current Id by a load
resistance Rload makes little difference in the circuit operation.
Fig. 5-30 Three-phase, full-bridge rectifier.
Fig 5-31 Three-phase rectifier with a constant dc current.
The rectifier in Fig. 5-31a can be redrawn as in Fig. 5-31b. With Ls= 0, the current Id flows
through one diode from the top group and one from the bottom group. Similar to the
discussion in the case of single-phase rectifiers, in the top group, the diode with its anode at
the highest potential with conduct and the other two become reversed biased. In the bottom
group, the diode with its cathode at the lowest potential will conduct and the other two
become reverse biased.
The voltage waveforms in the circuits of Fig. 5-31 are shown in Fig. 5-32a, where vPn is the
voltage at the point P with respect to the ac voltage neutral point n. Similarly, vNn is the
voltage at the negative dc terminal N. Since Id flows continuously, at any time, vPn and vNn
can be obtained in terms of one of the ac input voltages van, vbn, and vcn. Applying KVL in the
circuit of Fig. 5-31 on an instantaneous basis, the dc-side voltage is
Fig 5-32 Waveforms in the circuit of Fig. 5-31.
The instantaneous waveform of vd consists of six segments per cycle of line frequency.
Hence, this rectifier is often termed a six-pulse rectifier. Each segment belongs to one of the
six line-to-line voltage combinations, as shown in Fig. 5-32b. Each diode conducts for 120°.
Considering the phase a current waveform in Fig. 5-32c,
I d when diode 1 is conducting
ia I d when diode 4 is conducting (5-65)
0 when neither diode 1 or 4 is conducting
The commutation of current from one diode to the next is instantaneous, based on the
assumption of Ls = 0. The diodes are numbered in Fig. 5-31 in such a way that they conduct in
the sequence 1, 2, 3, . . ,. Next, we will compute the average value of the output dc voltage
and rms values of the line currents, where the subscript o is added due to the assumption of
To obtain the average value of the output de voltage, it is sufficient to consider only one of the
six segments and obtain its average over a 60° or /3-rad interval. Arbitrarily, the time origin
t= 0 is chosen in Fig. 5-32a when thc line-to-line voltage vab is at its maximum. Therefore,
vd vab 2VLL cost t (5-66)
where VLL is the rms value of line-to-line voltages.
By integrating vab, the volt-second area A is given by
2VLL cost d (t ) 2VLL (5-67)
and therefore dividing A by the /3 interval yields
Vdo / 6 2VLL cost d (t ) 2VLL 1.35VLL
One of the phase voltages and the corresponding phase current (labeled vs and is ) are redrawn
in Fig. 5-33a. Using the definition of rms current in the phase current waveform of Fig. 5-33a,
the rms value of the line current is in this idealized case is
Is I d 0.816 I d (5-69)
By means of Fourier analysis of is in this idealized case, the fundamental-frequency
component is1 shown in Fig. 5-33a has an rms value
I s1 6 I d 0.78I d (5-70)
The harmonic components Ish can be expressed in terms of the fundamental-frequency
Ish=Is1/h (5-71 )
where h = 5, 7, 11, 13, . . . . The even and triplen harmonics are zero, as shown in Fig. 5-33b.
Since is1 is in phase with its utility phase voltage,
DPF = 1.0 (5-72)
Fig. 5-33 Line current in a three-phase rectifier in the idealized case with Ls=0 and a constant
The voltage waveforms will be identical if the load on the dc side is represented by a
resistance Rload instead of a current source Id. The phase currents will also flow during
identical intervals, as in Fig. 5-32. The only difference will be that the current waveforms will
not have a flat top, as in Fig. 5-32.
Effect of Ls on current commutation
Next, we will include Ls on the ac side and represent the dc side by a current source id = Id, as
shown in Fig. 5-34. Now the current commutations will not be instantaneous. We will look at
only one of the current commutations because all others are identical in a balanced circuit.
Consider the commutation of current from diode 5 to diode 1, beginning at t or t = 0 (the
time origin is chosen arbitrarily). Prior to this, the current id is flowing through diodes 5 and 6.
Figure 5-35a shows the subcircuit pertinent to this current commutation.
Fig. 5-34 Three-phase rectifier with a finite Ls and a constant dc current
Fig. 5-35 Current commutation process.
The current commutation only involves phases a and c, and the commutation voltage
responsible is vcomm = van-vcn. The two mesh currents iu and Id are labeled in Fig. 5-35a. The
commutation current iu flows due to a short-circuit path provided by the conducting diode 5.
In terms of the mesh currents, the phase currents are
ia = iu
These are plotted in Fig. 5-35b, where iu builds up from zero to Id at the end of the
commutation interval tu = u. In the circuit of Fig. 5-35a,
v La Ls a Ls diu / dt (5-75)
v Lc Ls c Ls u (5-76)
noting that ic= Id – iu and therefore dic/dt = d(Id – iu)/dt = - diu/dt. Applying KVL in the upper
loop in the circuit of Fig. 5-35a and using the above equations yield
vcomm van vcn vLa vLc 2 Ls (5-77)
Therefore from the above equation,
di v v
Ls u an cn (5-78)
The commutation interval u can be obtained by multiplying both sides of Eq. 5-78 by and
Ls diu an cn d ( t ) (5-79)
where the time origin is assumed to be at the beginning of the current commutation. With this
choice of time origin, we can express the line-to-line voltage (van – vcn) as
van vcn 2VLL sin t (5-80)
Using Eq. 5-80 in Eq. 5-79 yields
2VLL (1 cosu )
Ls diu Ls I d (5-81)
2Ls I d
cos u 1 (5-82)
If the current commutation was instantaneous due to zero Ls, then the voltage vPn will be
equal to van beginning with t = 0, as in Fig. 5-32c. However, with a finite Ls, during 0 < t <
tu in Fig. 5-35c
di v v
vPn van Ls u an cn (using Eq. 5-78) (5-83)
where the voltage across Ls[=Ls(diu/dt)] is the drop in the voltage vPn during the commutation
interval shown in Fig. 5-35c. The integral of this voltage drop is the area Au, which according
to Eq. 5-81 is
Au Ls I d (5-84)
This area is "lost" every 60° (/3 rad) interval, as shown in Fig. 5-32c. Therefore, the average
dc voltage output is reduced from its Vdo value, and the voltage drop due to commutation is
L I 3
Vd s d Ls I d (5-85)
Therefore, the average dc voltage in the presence of a finite commutation interval is
Vd Vdo Vd 1.35VLL Ls I d (5-86)
where Vdo is the average voltage with an instantaneous commutation due to Ls= 0, given by
Constant DC-side voltage vd(t)=Vd
Next, we will consider the circuit shown in Fig. 5-36a, where the assumption is that the de-
side voltage is a constant dc. It is an approximation to the circuit of Fig. 5-30 with a large
value of C. In order to simplify our analysis, we will make an assumption that the current id
on the dc side of the rectifier flows discontinuously, and therefore only two diodes-one from
the top group and one from the bottom group-conduct at any given time. This assumption
allows the equivalent circuit shown in Fig. 5-36b, where the input voltage is made up of the
line-to-line voltage segments shown in Fig. 5-36c. The diode DP corresponds to one of the
diodes D1, D3, and D5 from the top group. Similarly, the diode DN corresponds to one of the
diodes D2, D4 and D6. The resulting phase current waveform is shown in Fig. 5-36c.
Distortion in the Line-Current Waveforms
It is useful to know the power factor, total harmonic distortion, and the dc output voltage in
the practical circuit of Fig. 5-30. However, to present these results in a generalized manner
requires the assumption of a constant dc output voltage, as made in this section.
Fig. 5-36 (a) Three-phase rectifier with a finite Ls and a constant dc voltage. (b) Equivalent
circuit. (c) Waveforms.
The dc-side current Id is normalized by the per-phase short-circuit curtent. This per-phase
short-circuit current can be obtained in terms of the line-to-line input ac voltages:
V / 3
I sh ort circuit LL (5-87)
Fig. 5-37 Total harmonic distortion, DPF, and PF in the rectifier of Fig. 5-36 with a constant
Fig. 5-38 Normalized Vd and crest factor in the rectifier of Fig. 5-36 with a constant dc
Figures 5-37 and 5-38 show the plots of PF, DPF, THD, crest factor (CF), and Vd/Vdo as a
function of Id normalized by Ishort circuit.
Comparison of single-phase and three-phase rectifiers
The line current in a single-phase rectifier contains significantly more distortion compared to
a three-phase rectifier. This results in a much poorer power factor in a single-phase rectifier
compared to a three-phase rectifier. This is confirmed by comparing Figs. 5-18 and 5-19 for
single-phase rectifiers with the results for three-phase rectifiers in Figs. 5-37 and 5-38. The
displacement power factor (cos 1) is high in both rectifiers.
The ripple in the dc cucrent is smaller in a three-phase rectifier in comparison to a single-
phase rectifier. The ripple current, which flows through the filter capacitor, dictates the
capacitance and the current-handling capability required of the filter capacitor. Therefore, in
some applications, the filter capacitance required may be much smaller in a three-phase
rectifier compared to a single-phase rectifier.
In a three-phase rectifier, the maximum regulation in the dc voltage Vd from no-load to a full-
load condition will generally be less than 5%, as seen from Fig. 5-38. This regulation is often
much larger in single-phase rectifiers.
Based on the foregoing discussion and the fact that the use of single-phase rectifiers in three-
phase, four-wire systems introduces large currents in the neutral (even in a balanced system),
it is always preferable to use a three-phase rectifier over a single-phase rectifier.