chapter 18 pulse-width modulated rectifiers
TRANSCRIPT
Fundamentals of Power Electronics 1 Chapter 18: PWM Rectifiers
Chapter 18
Pulse-Width Modulated Rectifiers
18.1 Properties of the ideal rectifier
18.2 Realization of a near-ideal rectifier
18.3 Control of the current waveform
18.4 Single-phase converter systems employing ideal rectifiers
18.5 RMS values of rectifier waveforms
18.6 Modeling losses and efficiency in CCM high-quality rectifiers
18.7 Ideal three-phase rectifiers
Fundamentals of Power Electronics 2 Chapter 18: PWM Rectifiers
18.1 Properties of the ideal rectifier
It is desired that the rectifier present a resistive load to the ac powersystem. This leads to
• unity power factor
• ac line current has same waveshape as voltage
iac(t) =vac(t)
Re
Re is called the emulated resistance Re
+
–
vac(t)
iac(t)
Fundamentals of Power Electronics 3 Chapter 18: PWM Rectifiers
Control of power throughput
Re(vcontrol)
+
–
vac(t)
iac(t)
vcontrol
Pav =V ac,rms
2
Re(vcontrol)
Power apparently “consumed” by Re
is actually transferred to rectifier dcoutput port. To control the amountof output power, it must be possibleto adjust the value of Re.
Fundamentals of Power Electronics 4 Chapter 18: PWM Rectifiers
Output port model
Re(vcontrol)
+
–
vac(t)
iac(t)
vcontrol
v(t)
i(t)
+
–
p(t) = vac2/Re
Ideal rectifier (LFR)
acinput
dcoutput
The ideal rectifier islossless and containsno internal energystorage. Hence, theinstantaneous inputpower equals theinstantaneous outputpower. Since theinstantaneous power isindependent of the dcload characteristics, theoutput port obeys apower sourcecharacteristic.
p(t) =vac
2 (t)Re(vcontrol(t))
v(t)i(t) = p(t) =vac
2 (t)Re
Fundamentals of Power Electronics 5 Chapter 18: PWM Rectifiers
The dependent power source
p(t)
+
v(t)
–
i(t)
p(t)
+
v(t)
–
i(t) v(t)i(t) = p(t)
v(t)
i(t)
powersource
powersink
i-v characteristic
Fundamentals of Power Electronics 6 Chapter 18: PWM Rectifiers
Equations of the ideal rectifier / LFR
iac(t) =vac(t)
Re(vcontrol)
v(t)i(t) = p(t)
p(t) =vac
2 (t)Re(vcontrol(t))
Vrms
Vac,rms= R
Re
Iac,rms
Irms= R
Re
Defining equations of theideal rectifier:
When connected to aresistive load of value R, theinput and output rms voltagesand currents are related asfollows:
Fundamentals of Power Electronics 7 Chapter 18: PWM Rectifiers
18.2 Realization of a near-ideal rectifier
1 : M(d(t))
dc–dc converter
Controller
d(t)
Rvac(t)
iac(t)+
vg(t)
–
ig(t)
ig
vg
+
v(t)
–
i(t)
C
Control the duty cycle of a dc-dcconverter, such that the input currentis proportional to the input voltage:
Fundamentals of Power Electronics 8 Chapter 18: PWM Rectifiers
Waveforms
t
vac(t)
t
iac(t)
VM
vg(t) VM
VM /Re
v(t)
ig(t)
V
M(t)
Mmin
vac(t) = VM sin (ωt)
vg(t) = VM sin (ωt)
M(d(t)) =v(t)vg(t)
= VVM sin (ωt)
Mmin = VVM
Fundamentals of Power Electronics 9 Chapter 18: PWM Rectifiers
Output-side current
1 : M(d(t))
dc–dc converter
Controller
d(t)
Rvac(t)
iac(t)+
vg(t)
–
ig(t)
ig
vg
+
v(t)
–
i(t)
C
i(t) =vg(t)ig(t)
V =vg
2(t)VRe
i(t) =V M
2
VResin2(ωt)
=V M
2
2VRe1 – cos (2ωt)
I = i(t)TL
=V M
2
2VRe
P =V M
2
2Re
Averagedoverswitchingperiod
Averagedover ac lineperiod
Fundamentals of Power Electronics 10 Chapter 18: PWM Rectifiers
Choice of converter
M(d(t)) =v(t)vg(t)
= VVM sin (ωt)
• To avoid distortion near line voltage zero crossings, converter shouldbe capable of producing M(d(t)) approaching infinity
• Above expression neglects converter dynamics
• Boost, buck-boost, Cuk, SEPIC, and other converters with similarconversion ratios are suitable
• We will see that the boost converter exhibits lowest transistorstresses. For this reason, it is most often chosen
M(t)
Mmin
Fundamentals of Power Electronics 11 Chapter 18: PWM Rectifiers
18.2.1 CCM Boost converterwith controller to cause input current to follow input voltage
Boost converter
Controller
Rvac(t)
iac(t)+
vg(t)
–
ig(t)
ig(t)vg(t)
+
v(t)
–
i(t)
Q1
L
C
D1
d(t)
• DC output voltage ≥ peak AC input voltage
• Controller varies duty cycle as necessary to make ig(t) proportionalto vg(t)
Fundamentals of Power Electronics 12 Chapter 18: PWM Rectifiers
Variation of duty cycle in boost rectifier
M(d(t)) =v(t)vg(t)
= VVM sin (ωt)
Since M ≥ 1 in the boost converter, it is required that V ≥ VM
If the converter operates in CCM, then
M(d(t)) = 11 – d(t)
The duty ratio should therefore follow
d(t) = 1 –vg(t)
Vin CCM
Fundamentals of Power Electronics 13 Chapter 18: PWM Rectifiers
CCM/DCM boundary, boost rectifier
Inductor current ripple is
∆ig(t) =vg(t)d(t)Ts
2L
ig(t) Ts=
vg(t)Re
Low-frequency (average) component of inductor current waveform is
The converter operates in CCM when
ig(t) Ts> ∆ig(t) ⇒ d(t) < 2L
ReTs
Substitute CCM expression for d(t):
Re < 2L
Ts 1 –vg(t)V
for CCM
Fundamentals of Power Electronics 14 Chapter 18: PWM Rectifiers
CCM/DCM boundary
Re < 2L
Ts 1 –vg(t)V
for CCM
Note that vg(t) varies with time, between 0 and VM. Hence, thisequation may be satisfied at some points on the ac line cycle, and notat others. The converter always operates in CCM provided that
Re < 2LTs
The converter always operates in DCM provided that
Re > 2L
Ts 1 –VM
V
For Re between these limits, the converter operates in DCM when vg(t)is near zero, and in CCM when vg(t) approaches VM.
Fundamentals of Power Electronics 15 Chapter 18: PWM Rectifiers
Static input characteristicsof the boost converter
A plot of input current ig(t) vs input voltage vg(t), for various duty cyclesd(t). In CCM, the boost converter equilibrium equation is
vg(t)V
= 1 – d(t)
The input characteristic in DCM is found by solution of the averagedDCM model (Fig. 11.12(b)):
Beware! This DCM Re(d) fromChapter 11 is not the same as therectifier emulated resistance Re = vg/ig
Solve for input current:
p(t)+–
+
V
–
vg(t)+–
2Ld 2Ts
p(t)V – vg(t)
ig(t)
ig(t) =vg(t)
2Ld 2Ts
+p(t)
V – vg(t)
p(t) =vg
2(t)
2Ld 2Ts
with
Fundamentals of Power Electronics 16 Chapter 18: PWM Rectifiers
Static input characteristicsof the boost converter
Now simplify DCM current expression, to obtain
2LVTs
ig(t) 1 –vg(t)V
= d 2(t)vg(t)V
CCM/DCM mode boundary, in terms of vg(t) and ig(t):
2LVTs
ig(t) >vg(t)V
1 –vg(t)V
Fundamentals of Power Electronics 17 Chapter 18: PWM Rectifiers
Boost input characteristicswith superimposed resistive characteristic
0 0.25 0.5 0.75 1
0
0.25
0.5
0.75
1
d =
0
d =
0.2
d =
0.4
d =
0.6
d =
0.8
d =
1
CCM
DCM
j g(t)
=2L VT
si g
(t)
mg(t) =vg(t)
V
i g(t)
= v g(t)/
R e
2LVTs
ig(t) 1 –vg(t)V
= d 2(t)vg(t)V
vg(t)V = 1 – d(t)
2LVTs
ig(t) >vg(t)V
1 –vg(t)V
CCM:
DCM:
CCM when
Fundamentals of Power Electronics 18 Chapter 18: PWM Rectifiers
Open-loop DCM approach
We found in Chapter 11 that the buck-boost, SEPIC, and Cukconverters, when operated open-loop in DCM, inherently behave asloss-free resistors. This suggests that they could also be used asnear-ideal rectifiers, without need for a multiplying controller.
Advantage: simple control
Disadvantages: higher peak currents, larger input current EMI
Like other DCM applications, this approach is usually restricted to lowpower (< 200W).
The boost converter can also be operated in DCM as a low harmonicrectifier. Input characteristic is
ig(t) Ts=
vg(t)Re
+vg
2(t)
Re v(t) – vg(t)
Input current contains harmonics. If v is sufficiently greater than vg,then harmonics are small.
Fundamentals of Power Electronics 19 Chapter 18: PWM Rectifiers
Other similar approaches
• Use of other converters (in CCM) that are capable of increasing thevoltage:
SEPIC, Cuk, buck-boost
Flyback, isolated versions of boost, SEPIC, Cuk, etc.
• Boundary or critical conduction mode: operation of boost or otherconverter at the boundary between CCM and DCM
• Buck converter: distortion occurs but stresses are low
• Resonant converter such as parallel resonant converter or somequasi-resonant converters
• Converters that combine the functions of rectification, energystorage, and dc-dc conversion
Fundamentals of Power Electronics 20 Chapter 18: PWM Rectifiers
18.2.2 DCM flyback converter
Flyback converter
D
Rvac(t)
iac(t)+
vg(t)
–
ig(t)
+
v(t)
–
i(t)
CL
Q1
D1
n : 1EMI filter
Operation in DCM: we found in Chapter 11 that the converter inputport obeys Ohm’s law with effective resistance Re = 2n2L/D2Ts.
Hence, simply connect input port to AC line.
Fundamentals of Power Electronics 21 Chapter 18: PWM Rectifiers
Averaged large-signal model
Averaged model
Rvac(t)
iac(t)+
vg(t)
–
⟨ ig(t) ⟩T
+
⟨v(t)⟩T
–
C
EMI filter
2n2LD2Ts
⟨ p(t) ⟩T
D
⟨ i(t) ⟩T
• Under steady-state conditions, operate with constant D
• Adjust D to control average power drawn from AC line
Fundamentals of Power Electronics 22 Chapter 18: PWM Rectifiers
Converter design
Select L small enough that DCMoperation occurs throughout AC linecycle. DCM occurs provided thatd3 > 0, or
d1Ts
Ts
td2Ts d3Ts
i1(t)ipkArea q1
i1(t) Tsd2(t) < 1 – D
d2(t) = Dvg(t)nV
But
Substitute and solve for D:
D < 1
1 +vg(t)nV
Converter operates in DCM inevery switching period whereabove inequality is satisfied.
To obtain DCM at all points oninput AC sinusoid: worst case is atmaximum vg(t) = VM :
D < 1
1 +VM
nV
Fundamentals of Power Electronics 23 Chapter 18: PWM Rectifiers
Choice of Lto obtain DCM everywhere along AC sinusoid
We have: D < 1
1 +VM
nV
with Vrms
Vac,rms= R
Re
Substitute expression for Re to obtain
D = 2nVVM
LRTs
Solve for L:
L < Lcrit =RT s
4 1 + nVVM
2
Worst-case design
For variations in loadresistance and ac inputvoltage, the worst caseoccurs at maximum loadpower and minimum ac inputvoltage. The inductanceshould be chosen as follows:
L < Lcrit-min =RminTs
4 1 + nVVM-min
2
Fundamentals of Power Electronics 24 Chapter 18: PWM Rectifiers
18.3 Control of the Current Waveform
18.3.1 Average current control
Feedforward
18.3.2 Current programmed control
18.3.3 Critical conduction mode and hysteretic control
18.3.4 Nonlinear carrier control
Fundamentals of Power Electronics 25 Chapter 18: PWM Rectifiers
18.3.1 Average current control
+–
+
v(t)
–
vg(t)
ig(t)
Gatedriver
Pulse widthmodulator
CompensatorGc(s)
+–
+–
Currentreference
vr(t)
va(t)≈ Rs ⟨ ig(t)⟩Ts
LBoost example
Low frequency(average) componentof input current iscontrolled to followinput voltage
Fundamentals of Power Electronics 26 Chapter 18: PWM Rectifiers
Block diagram
Boost converter
Controller
Rvac(t)
iac(t)+
vg(t)
–
ig(t)
ig(t)vg(t)
+
v(t)
–
i(t)
Q1
L
C
D1
vcontrol(t)
Multiplier X
+–vr (t)
= kx vg(t) vcontrol(t)
Rsva(t)
Gc(s)
PWM
Compensator
verr(t)
• Currentreferencederivedfrom inputvoltagewaveform
• Multiplier allowscontrol of emulatedresistance value
• Compensation ofcurrent loop
Fundamentals of Power Electronics 27 Chapter 18: PWM Rectifiers
The emulated resistance
Boost converter
Controller
Rvac(t)
iac(t)+
vg(t)
–
ig(t)
ig(t)vg(t)
+
v(t)
–
i(t)
Q1
L
C
D1
vcontrol(t)
Multiplier X
+–vr (t)
= kx vg(t) vcontrol(t)
Rsva(t)
Gc(s)
PWM
Compensator
verr(t)
• Current sensor hasgain Rs :
va(t) = Rs ig(t) Ts
va(t) ≈ vr(t)
• If loop is welldesigned, then:
• Multiplier:vr(t) = k xvg(t) vcontrol(t)
• Hence the emulated resistance is:
Re =vg(t)
ig(t)=
vr(t)k xvcontrol(t)
va(t)Rs
Re vcontrol(t) =Rs
k xvcontrol(t)
which can be simplified to
Fundamentals of Power Electronics 28 Chapter 18: PWM Rectifiers
System model using LFRAverage current control
Rvac(t)
iac(t)
Re
+
–
vcontrol(t)
⟨v(t)⟩Ts
+
–
Ideal rectifier (LFR)
C
Re(t)Re(t) =
Rs
k x vcontrol(t)
⟨i(t)⟩Ts⟨ig(t)⟩Ts
⟨ p(t)⟩Ts
⟨vg(t)⟩Ts
Fundamentals of Power Electronics 29 Chapter 18: PWM Rectifiers
Use of multiplier to control average power
+–
+
v(t)
–
vg(t)
ig(t)
Gatedriver
Pulse widthmodulator
CompensatorGc(s)
+–
+–vref1(t)kx xy
xy
Multiplier
vg(t)
vcontrol(t) Gcv(s)+
–
Voltage reference
C
vref2(t)
v(t)
verr(t)
va(t)
Pav =V g,rms
2
Re= Pload
As discussed in Chapter17, an output voltagefeedback loop adjusts theemulated resistance Re
such that the rectifierpower equals the dc loadpower:
An analog multiplierintroduces thedependence of Re
on v(t).
Fundamentals of Power Electronics 30 Chapter 18: PWM Rectifiers
Feedforward
+–
+
v(t)
–
vg(t)
ig(t)
Gatedriver
Pulse widthmodulator
CompensatorGc(s)
+–
+–vref1(t)
x
y
multiplier
vg(t)
vcontrol(t) Gcv(s)+
–
Voltage reference
k vxyz2z
Peakdetector VM
vref2(t)
va(t)
Feedforward is sometimesused to cancel outdisturbances in the inputvoltage vg(t).
To maintain a given powerthroughput Pav, the referencevoltage vref1(t) should be
vref 1(t) =Pavvg(t)Rs
V g,rms2
Fundamentals of Power Electronics 31 Chapter 18: PWM Rectifiers
Feedforward, continued
+–
+
v(t)
–
vg(t)
ig(t)
Gatedriver
Pulse widthmodulator
CompensatorGc(s)
+–
+–vref1(t)
x
y
multiplier
vg(t)
vcontrol(t) Gcv(s)+
–
Voltage reference
k vxyz2z
Peakdetector VM
vref2(t)
va(t)
vref 1(t) =k vvcontrol(t)vg(t)
V M2
Pav =k vvcontrol(t)
2Rs
Controller with feedforwardproduces the following reference:
The average power is thengiven by
Fundamentals of Power Electronics 32 Chapter 18: PWM Rectifiers
Modeling the inner wide-bandwidthaverage current controller
+–
+–
L
C R
+
⟨v(t)⟩Ts
–
⟨vg(t)⟩Ts⟨v1(t)⟩Ts
⟨i2(t)⟩Ts
⟨i(t)⟩Ts
+
⟨v2(t)⟩Ts
–
⟨i1(t)⟩Ts
Averaged switch network
Averaged (but not linearized) boost converter model:
vg(t) Ts= Vg + vg(t)
d(t) = D + d(t) ⇒ d'(t) = D' – d(t)
i(t)Ts
= i1(t) Ts= I + i(t)
v(t)Ts
= v2(t) Ts= V + v(t)
v1(t) Ts= V1 + v1(t)
i2(t) Ts= I2 + i2(t)
In Chapter 7,we perturbedand linearizedusing theassumptions
Problem: variations in vg,i1 , and d are not small.
So we are faced with thedesign of a controlsystem that exhibitssignificant nonlineartime-varying behavior.
Fundamentals of Power Electronics 33 Chapter 18: PWM Rectifiers
Linearizing the equations of the boost rectifier
When the rectifier operates near steady-state, it is true that
v(t)Ts
= V + v(t)
with
v(t) << V
In the special case of the boost rectifier, this is sufficient to linearizethe equations of the average current controller.
The boost converter average inductor voltage is
Ld ig(t) Ts
dt= vg(t) Ts
– d'(t)V – d'(t)v(t)
substitute:
Ld ig(t) Ts
dt= vg(t) Ts
– d'(t)V – d'(t)v(t)
Fundamentals of Power Electronics 34 Chapter 18: PWM Rectifiers
Linearized boost rectifier model
Ld ig(t) Ts
dt= vg(t) Ts
– d'(t)V – d'(t)v(t)
The nonlinear term is much smaller than the linear ac term. Hence, itcan be discarded to obtain
Ld ig(t) Ts
dt= vg(t) Ts
– d'(t)V
Equivalent circuit:
+–
L
+– d'(t)Vvg(t) Ts
ig(t) Ts
ig(s)d(s)
= VsL
Fundamentals of Power Electronics 35 Chapter 18: PWM Rectifiers
The quasi-static approximation
The above approach is not sufficient to linearize the equations needed todesign the rectifier averaged current controllers of buck-boost, Cuk,SEPIC, and other converter topologies. These are truly nonlinear time-varying systems.
An approximate approach that is sometimes used in these cases: thequasi-static approximation
Assume that the ac line variations are much slower than the converterdynamics, so that the rectifier always operates near equilibrium. Thequiescent operating point changes slowly along the input sinusoid, and wecan find the slowly-varying “equilibrium” duty ratio as in Section 18.2.1.
The converter small-signal transfer functions derived in Chapters 7 and 8are evaluated, using the time-varying operating point. The poles, zeroes,and gains vary slowly as the operating point varies. An average currentcontroller is designed, that has a positive phase margin at each operatingpoint.
Fundamentals of Power Electronics 36 Chapter 18: PWM Rectifiers
Quasi-static approximation: discussion
• In the literature, several authors have reported success using thismethod
• Should be valid provided that the converter dynamics are suffieientlyfast, such that the converter always operates near the assumedoperating points
• No good condition on system parameters, which can justify theapproximation, is presently known for the basic converter topologies
• It is well-understood in the field of control systems that, when theconverter dynamics are not sufficiently fast, then the quasi-staticapproximation yields neither necessary nor sufficient conditions forstability. Such behavior can be observed in rectifier systems. Worst-case analysis to prove stability should employ simulations.
Fundamentals of Power Electronics 37 Chapter 18: PWM Rectifiers
18.3.2 Current programmed control
Boost converter
Current-programmed controller
Rvg(t)
ig(t)
is(t)vg(t)
+
v(t)
–
i2(t)
Q1
L
C
D1
vcontrol(t)
Multiplier Xic(t)
= kx vg(t) vcontrol(t)
+–
+ +
+
–
Comparator Latch
ia(t)Ts0
S
R
Q
ma
Clock
Current programmedcontrol is a naturalapproach to obtain inputresistor emulation:
Peak transistor current isprogrammed to followinput voltage.
Peak transistor currentdiffers from averageinductor current,because of inductorcurrent ripple andartificial ramp. This leadsto significant inputcurrent waveformdistortion.
Fundamentals of Power Electronics 38 Chapter 18: PWM Rectifiers
CPM boost converter: Static input characteristics
Mode boundary: CCM occurs when
ig(t) Ts>
TsV2L
vg(t)V 1 –
vg(t)V
or, ic(t) >TsVL
maLV +
vg(t)V 1 –
vg(t)V
It is desired that ic(t) =vg(t)Re
0
0.2
0.4
0.6
0.8
1
0.0 0.2 0.4 0.6 0.8 1.0vg(t)
V
j g(t)
=i g
(t)
Ts
Rba
se
VCCM
DCM
R e =
Rba
se
R e =
4R ba
se
R e =
0.3
3Rba
seR e
= 0
.5R ba
se
R e =
2R ba
se
R e =
0.2
R base
Re
= 0
.1R
base
Re= 10
R base
ma = V2L
Rbase = 2LTs
Static input characteristics ofCPM boost, with minimumslope compensation:
Minimum slope compensation:ma = V
2L
ig(t) Ts=
vg(t)Lic
2(t) fsV – vg(t) vg(t) + maL
in DCM
ic(t) – 1 –vg(t)
V ma+vg(t)
L Ts in CCM
Fundamentals of Power Electronics 39 Chapter 18: PWM Rectifiers
Input current waveformswith current mode control
0.0
0.2
0.4
0.6
0.8
1.0
ωt
ig(t)Peak ig
Sin
usoi
d
ma = V2L
Rbase = 2LTs
R e =
0.1
R base R e =
0.3
3Rba
se
R e =
2R ba
se
• Substantialdistortion canoccur
• Can meetharmonic limitsif the range ofoperating pointsis not too large
• Difficult to meetharmonic limitsin a universalinput supply
Fundamentals of Power Electronics 40 Chapter 18: PWM Rectifiers
18.3.3 Critical conduction modeand hysteretic control
ωt
ig(t)
ωt
ig(t)
ton
Variable switchingfrequency schemes
• Hysteretic control
• Critical conductionmode (boundarybetween CCM andDCM)
Hystereticcontrol
Criticalconductionmode
Fundamentals of Power Electronics 41 Chapter 18: PWM Rectifiers
An implementation of critical conduction control
Boost converter
Controller
Rvac(t)
iac(t)+
vg(t)
–
ig(t)
vg(t)
+
v(t)
–
i(t)
Q1
L
C
D1
vcontrol(t)
Multiplier X
vr (t)= kx vg(t) vcontrol(t)
Rs
va(t)
Comparator
+
–
S
R
QZero currentig detector
EMI filter
Latch
Fundamentals of Power Electronics 42 Chapter 18: PWM Rectifiers
Pros and cons of critical conduction control
• Simple, low-cost controller ICs
• Low-frequency harmonics are very small, with constant transistoron-time (for boost converter)
• Small inductor
• Increased peak current
• Increased conduction loss, reduced switching loss
• Requires larger input filter
• Variable switching frequency smears out the current EMI spectrum
• Cannot synchronize converter switching frequencies
Fundamentals of Power Electronics 43 Chapter 18: PWM Rectifiers
Analysis
ωt
ig(t)
ton
Transistor is on for fixed time ton
Transistor off-time ends wheninductor current reaches zero
Ratio of vg(t) to ⟨ ig(t)⟩ is
Re = 2Lton
On time, as a function of loadpower and line voltage:
ton = 4LPV M
2
Inductor volt-second balance:
vgton + vg – V toff = 0
Solve for toff:
toff = ton
vg
V – vg
Fundamentals of Power Electronics 44 Chapter 18: PWM Rectifiers
Switching frequency variations
Solve for how the controller varies the switching frequency over the acline period:
Ts = toff + ton Ts = 4LPV M
21
1 –vg(t)
V
For sinusoidal line voltage variations, the switching frequency willtherefore vary as follows:
fs = 1Ts
=V M
2
4LP1 –
VM
V sin (ωt)
Minimum and maximum limits on switching frequency:
max fs =V M
2
4LPmin fs =
V M2
4LP1 –
VM
V
These equations can be used to select the value of the inductance L.
Fundamentals of Power Electronics 45 Chapter 18: PWM Rectifiers
18.3.4 Nonlinear carrier control
• Can attain simple control of input current waveform without sensingthe ac input voltage, and with operation in continuous conductionmode
• The integral of the sensed switch current (charge) is compared to anonlinear carrier waveform (i.e., a nonlinear ramp), on a cycle-by-cycle basis
• Carrier waveform depends on converter topology
• Very low harmonics in CCM. Waveform distortion occurs in DCM.
• Peak current mode control is also possible, with a different carrier
Fundamentals of Power Electronics 46 Chapter 18: PWM Rectifiers
Controller block diagramNonlinear carrier charge control of boost converter
S
R Q
Boost converter
Nonlinear-carrier charge controller
Rvg(t)
ig(t)
is(t)
+
v(t)
–
Q1
L
C
D1
vcontrol(t)
+–
+
–
Comparator Latch
Ts0
Clock
n : 1
Ci
– vi (t) +
is/n vi (t)
Nonlinear carriergenerator
vc(t) Q
TsdTs0
is(t)
vc(t)
vi (t)
Fundamentals of Power Electronics 47 Chapter 18: PWM Rectifiers
Derivation of NLC approach
The average switch current is
is(t) Ts= 1
Tsis(τ)dτ
t
t + Ts
We could make the controller regulate the average switch current by
• Integrating the monitored switch current
• Resetting the integrator to zero at the beginning of eachswitching period
• Turning off the transistor when the integrator reaches areference value
In the controller diagram, the integrator follows this equation:
vi(t) = 1Ci
is(τ)n dτ
0
dTsfor 0 < t < dTs
vi(dTs) =is Ts
nCi fsfor interval 0 < t < Ts
i.e.,
Fundamentals of Power Electronics 48 Chapter 18: PWM Rectifiers
How to control the average switch current
Input resistor emulation:
ig(t) Ts=
vg(t) Ts
Re(vcontrol)
Relate average switch current to input current (assuming CCM):
is(t) Ts= d(t) ig(t) Ts
Relate input voltage to output voltage (assuming CCM):
vg(t) Ts= d′(t) v(t)
Ts
Substitute above equations to find how average switch current shouldbe controlled:
is(t) Ts= d(t) 1 – d(t)
v(t)Ts
Re(vcontrol)
Fundamentals of Power Electronics 49 Chapter 18: PWM Rectifiers
Implementation using nonlinear carrier
is(t) Ts= d(t) 1 – d(t)
v(t)Ts
Re(vcontrol)Desired control, from previous slide:
Generate carrier waveform as follows (replace d by t/Ts ):
vc(t) = vcontrolt
Ts1 – t
Tsfor 0 ≤ t ≤ Ts
vc(t + Ts) = vc(t)
The controller switches the transistor off when the integrator voltageequals the carrier waveform. This leads to:
vi(dTs) = vc(dTs) = vcontrol(t) d(t) 1 – d(t)
is(t) Ts
nCi f s= vcontrol(t) d(t) 1 – d(t)
Re(vcontrol) = d(t) 1 – d(t)v(t)
Ts
is(t) Ts
=v(t)
Ts
nCi fsvcontrol(t)
Fundamentals of Power Electronics 50 Chapter 18: PWM Rectifiers
Generating the parabolic carrier
+–vcontrol(t)
Integrator with reset Integrator with reset
vc(t)
Removal of dccomponent
Clock
(one approach, suitable for discrete circuitry)
Note that no separate multiplier circuit is needed
Fundamentals of Power Electronics 51 Chapter 18: PWM Rectifiers
18.4 Single-phase converter systemscontaining ideal rectifiers
• It is usually desired that the output voltage v(t) be regulated withhigh accuracy, using a wide-bandwidth feedback loop
• For a given constant load characteristic, the instantaneous loadcurrent and power are then also constant:
pload(t) = v(t)i(t) = VI
• The instantaneous input power of a single-phase ideal rectifier isnot constant:
withvg(t) = VM sin (ωt) ig(t) =
vg(t)Re
so pac(t) =V M
2
Resin2 ωt =
V M2
2Re1 – cos 2ωt
pac(t) = vg(t)ig(t)
Fundamentals of Power Electronics 52 Chapter 18: PWM Rectifiers
Power flow in single-phase ideal rectifier system
• Ideal rectifier is lossless, and contains no internal energy storage.
• Hence instantaneous input and output powers must be equal
• An energy storage element must be added
• Capacitor energy storage: instantaneous power flowing intocapacitor is equal to difference between input and output powers:
pC(t) =dEC(t)
dt=
d 12 CvC
2 (t)
dt= pac(t) – pload(t)
Energy storage capacitor voltage must be allowed to vary, inaccordance with this equation
Fundamentals of Power Electronics 53 Chapter 18: PWM Rectifiers
Capacitor energy storage in 1¿ system
Pload
pac(t)
t
vc(t)
=d 1
2 CvC2 (t)
dt= pac(t) – pload(t)
Fundamentals of Power Electronics 54 Chapter 18: PWM Rectifiers
Single-phase system with internal energy storage
vac(t)
iac(t)
Re
+
–
Ideal rectifier (LFR)
C
i2(t)ig(t)
⟨ pac(t)⟩Ts
vg(t)
i(t)
load
+
v(t)
–
pload(t) = VI = Pload
Energy storagecapacitor
vC(t)
+
–
Dc–dcconverter
Energy storage capacitorvoltage vC(t) must beindependent of input andoutput voltage waveforms, sothat it can vary according to
=d 1
2 CvC2 (t)
dt= pac(t) – pload(t)
This system is capable of
• Wide-bandwidth control ofoutput voltage
• Wide-bandwidth control ofinput current waveform
• Internal independent energystorage
Fundamentals of Power Electronics 55 Chapter 18: PWM Rectifiers
Hold up time
Internal energy storage allows the system to function in othersituations where the instantaneous input and output powers differ.
A common example: continue to supply load power in spite of failureof ac line for short periods of time.
Hold up time: the duration which the dc output voltage v(t) remainsregulated after vac(t) has become zero
A typical hold-up time requirement: supply load for one completemissing ac line cycle, or 20 msec in a 50 Hz system
During the hold-up time, the load power is supplied entirely by theenergy storage capacitor
Fundamentals of Power Electronics 56 Chapter 18: PWM Rectifiers
Energy storage element
Instead of a capacitor, and inductor or higher-order LC network couldstore the necessary energy.
But, inductors are not good energy-storage elements
Example
100 V 100 µF capacitor
100 A 100 µH inductor
each store 1 Joule of energy
But the capacitor is considerably smaller, lighter, and lessexpensive
So a single big capacitor is the best solution
Fundamentals of Power Electronics 57 Chapter 18: PWM Rectifiers
Inrush current
A problem caused by the large energy storage capacitor: the largeinrush current observed during system startup, necessary to chargethe capacitor to its equilibrium value.
Boost converter is not capable of controlling this inrush current.
Even with d = 0, a large current flows through the boost converterdiode to the capacitor, as long as v(t) < vg(t).
Additional circuitry is needed to limit the magnitude of this inrushcurrent.
Converters having buck-boost characteristics are capable ofcontrolling the inrush current. Unfortunately, these converters exhibithigher transistor stresses.
Fundamentals of Power Electronics 58 Chapter 18: PWM Rectifiers
Universal input
The capability to operate from the ac line voltages and frequenciesfound everywhere in the world:
50Hz and 60Hz
Nominal rms line voltages of 100V to 260V:
100V, 110V, 115V, 120V, 132V, 200V, 220V, 230V, 240V, 260V
Regardless of the input voltage and frequency, the near-ideal rectifierproduces a constant nominal dc output voltage. With a boostconverter, this voltage is 380 or 400V.
Fundamentals of Power Electronics 59 Chapter 18: PWM Rectifiers
Low-frequency model of dc-dc converter
C
i2(t)
i(t)
load
+
v(t)
–
pload(t) = VI = Pload
Energy storagecapacitor
vC(t)
+
–
Dc-dcconverter
+–Pload V
Dc-dc converter produces well-regulated dc load voltage V.
Load therefore draws constant current I.
Load power is therefore the constant value Pload = VI.
To the extent that dc-dc converter losses can be neglected, then dc-dcconverter input power is Pload , regardless of capacitor voltage vc(t).
Dc-dc converter input port behaves as a power sink. A low frequencyconverter model is
Fundamentals of Power Electronics 60 Chapter 18: PWM Rectifiers
Low-frequency energy storage process, 1¿ system
vac(t)
iac(t)
Re
+
–
Ideal rectifier (LFR)
C
i2(t)ig(t)
vg(t)
i(t)
load
+
v(t)
–
pload(t) = VI = Pload
Energy storagecapacitor
vC(t)
+
–
Dc-dcconverter
+–Pload V
⟨ pac(t)⟩Ts
A complete low-frequency system model:
• Difference between rectifier output power and dc-dc converterinput power flows into capacitor
• In equilibrium, average rectifier and load powers must be equal
• But the system contains no mechanism to accomplish this
• An additional feeback loop is necessary, to adjust Re such that therectifier average power is equal to the load power
Fundamentals of Power Electronics 61 Chapter 18: PWM Rectifiers
Obtaining average power balance
vac(t)
iac(t)
Re
+
–
Ideal rectifier (LFR)
C
i2(t)ig(t)
vg(t)
i(t)
load
+
v(t)
–
pload(t) = VI = Pload
Energy storagecapacitor
vC(t)
+
–
Dc-dcconverter
+–Pload V
⟨ pac(t)⟩Ts
If the load power exceeds the average rectifier power, then there is anet discharge in capacitor energy and voltage over one ac line cycle.
There is a net increase in capacitor charge when the reverse is true.
This suggests that rectifier and load powers can be balanced byregulating the energy storage capacitor voltage.
Fundamentals of Power Electronics 62 Chapter 18: PWM Rectifiers
A complete 1¿ systemcontaining three feedback loops
Boost converter
Wide-bandwidth input current controller
vac(t)
iac(t)+
vg(t)
–
ig(t)
ig(t)vg(t)
+
vC(t)
–
i2(t)
Q1
L
C
D1
vcontrol(t)
Multiplier X
+–vref1(t)
= kxvg(t)vcontrol(t)
Rsva(t)
Gc(s)
PWM
Compensator
verr(t)
DC–DCConverter Load
+
v(t)
–
i(t)
d(t)
+–Compensatorand modulator
vref3
Wide-bandwidth output voltage controller
+–Compensatorvref2
Low-bandwidth energy-storage capacitor voltage controller
vC(t)
v(t)
Fundamentals of Power Electronics 63 Chapter 18: PWM Rectifiers
Bandwidth of capacitor voltage loop
• The energy-storage-capacitor voltage feedback loop causes thedc component of vc(t) to be equal to some reference value
• Average rectifier power is controlled by variation of Re.
• Re must not vary too quickly; otherwise, ac line current harmonicsare generated
• Extreme limit: loop has infinite bandwidth, and vc(t) is perfectlyregulated to be equal to a constant reference value
• Energy storage capacitor voltage then does not change, andthis capacitor does not store or release energy
• Instantaneous load and ac line powers are then equal
• Input current becomes
iac(t) =pac(t)vac(t)
=pload(t)vac(t)
=Pload
VM sin ωt
Fundamentals of Power Electronics 64 Chapter 18: PWM Rectifiers
Input current waveform, extreme limit
vac(t)
iac(t)
t
iac(t) =pac(t)vac(t)
=pload(t)vac(t)
=Pload
VM sin ωt THD → ∞
Power factor → 0
So bandwidth ofcapacitor voltageloop must belimited, and THDincreases rapidlywith increasingbandwidth
Fundamentals of Power Electronics 65 Chapter 18: PWM Rectifiers
18.4.2 Modeling the outer low-bandwidthcontrol system
This loop maintains power balance, stabilizing the rectifier outputvoltage against variations in load power, ac line voltage, andcomponent values
The loop must be slow, to avoid introducing variations in Re at theharmonics of the ac line frequency
Objective of our modeling efforts: low-frequency small-signal modelthat predicts transfer functions at frequencies below the ac linefrequency
Fundamentals of Power Electronics 66 Chapter 18: PWM Rectifiers
Large signal modelaveraged over switching period Ts
Re(vcontrol)⟨ vg(t)⟩Ts
vcontrol
+
–
Ideal rectifier (LFR)
acinput
dcoutput
+–
⟨ ig(t)⟩Ts
⟨ p(t)⟩Ts
⟨ i2(t)⟩Ts
⟨ v(t)⟩TsC Load
Ideal rectifier model, assuming that inner wide-bandwidth loopoperates ideally
High-frequency switching harmonics are removed via averaging
Ac line-frequency harmonics are included in model
Nonlinear and time-varying
Fundamentals of Power Electronics 67 Chapter 18: PWM Rectifiers
Predictions of large-signal model
Re(vcontrol)⟨ vg(t)⟩Ts
vcontrol
+
–
Ideal rectifier (LFR)
acinput
dcoutput
+–
⟨ ig(t)⟩Ts
⟨ p(t)⟩Ts
⟨ i2(t)⟩Ts
⟨ v(t)⟩TsC Load
vg(t) = 2 vg,rms sin ωt
If the input voltage is
Then theinstantaneous poweris:
p(t)Ts
=vg(t) Ts
2
Re(vcontrol(t))=
vg,rms2
Re(vcontrol(t))1 – cos 2ωt
which contains a constant term plus a second-harmonic term
Fundamentals of Power Electronics 68 Chapter 18: PWM Rectifiers
Separation of power source into its constant andtime-varying components
+
–
⟨ i2(t)⟩Ts
⟨ v(t)⟩TsC Load
V g,rms2
Re–
V g,rms2
Recos2 2ωt
Rectifier output port
The second-harmonic variation in power leads to second-harmonicvariations in the output voltage and current
Fundamentals of Power Electronics 69 Chapter 18: PWM Rectifiers
Removal of even harmonics via averaging
t
v(t)
⟨ v(t)⟩T2L
⟨ v(t)⟩Ts
T2L = 12
2πω = π
ω
Fundamentals of Power Electronics 70 Chapter 18: PWM Rectifiers
Resulting averaged model
+
–
⟨ i2(t)⟩T2L
⟨ v(t)⟩T2LC Load
V g,rms2
Re
Rectifier output port
Time invariant model
Power source is nonlinear
Fundamentals of Power Electronics 71 Chapter 18: PWM Rectifiers
Perturbation and linearization
v(t)T2L
= V + v(t)
i2(t) T2L= I2 + i2(t)
vg,rms = Vg,rms + vg,rms(t)
vcontrol(t) = Vcontrol + vcontrol(t)
V >> v(t)
I2 >> i2(t)
Vg,rms >> vg,rms(t)
Vcontrol >> vcontrol(t)
Let with
The averaged model predicts that the rectifier output current is
i2(t) T2L=
p(t)T2L
v(t)T2L
=vg,rms
2 (t)
Re(vcontrol(t)) v(t)T2L
= f vg,rms(t), v(t)T2L
, vcontrol(t))
Fundamentals of Power Electronics 72 Chapter 18: PWM Rectifiers
Linearized result
I2 + i2(t) = g2vg,rms(t) + j2v(t) –vcontrol(t)
r2
g2 =df vg,rms, V, Vcontrol)
dvg,rmsvg,rms = Vg,rms
= 2Re(Vcontrol)
Vg,rms
V
where
– 1r2
=df Vg,rms, v
T2L, Vcontrol)
d vT2L
v T2L= V
= –I2
V
j2 =df Vg,rms, V, vcontrol)
dvcontrolvcontrol = Vcontrol
= –V g,rms
2
VRe2(Vcontrol)
dRe(vcontrol)
dvcontrolvcontrol = Vcontrol
Fundamentals of Power Electronics 73 Chapter 18: PWM Rectifiers
Small-signal equivalent circuit
C
Rectifier output port
r2g2 vg,rms j2 vcontrol R
i2
+
–
v
v(s)vcontrol(s)
= j2 R||r21
1 + sC R||r2
v(s)vg,rms(s)
= g2 R||r21
1 + sC R||r2
Predicted transfer functions
Control-to-output
Line-to-output
Fundamentals of Power Electronics 74 Chapter 18: PWM Rectifiers
Model parameters
Table 18.1 Small-signal model parameters for several types of rectifier control schemes
Controller type g2 j2 r2
Average current control withfeedforward, Fig. 18.14
0
Current-programmed control,Fig. 18.16
Nonlinear-carrier charge controlof boost rectifier, Fig. 18.21
Boost with critical conduction mode control, Fig. 18.20
DCM buck-boost, flyback, SEPIC,or Cuk converters
Pav
VVcontrol
V 2
Pav
2Pav
VVg,rms
Pav
VVcontrol
V 2
Pav
2Pav
VVg,rms
Pav
VVcontrol
V 2
2Pav
2Pav
VVg,rms
Pav
VVcontrol
V 2
Pav
2Pav
VVg,rms
2Pav
VDV 2
Pav
Fundamentals of Power Electronics 75 Chapter 18: PWM Rectifiers
Constant power load
vac(t)
iac(t)
Re
+
–
Ideal rectifier (LFR)
C
i2(t)ig(t)
vg(t)
i(t)
load
+
v(t)
–
pload(t) = VI = Pload
Energy storagecapacitor
vC(t)
+
–
Dc-dcconverter
+–Pload V
⟨ pac(t)⟩Ts
Rectifier and dc-dc converter operate with same average power
Incremental resistance R of constant power load is negative, and is
R = – V 2
Pav
which is equal in magnitude and opposite in polarity to rectifierincremental output resistance r2 for all controllers except NLC
Fundamentals of Power Electronics 76 Chapter 18: PWM Rectifiers
Transfer functions with constant power load
v(s)vcontrol(s)
=j2
sC
v(s)vg,rms(s)
=g2
sC
When r2 = – R, the parallel combination r2 || R becomes equal to zero.The small-signal transfer functions then reduce to
Fundamentals of Power Electronics 77 Chapter 18: PWM Rectifiers
18.5 RMS values of rectifier waveforms
t
iQ(t)
Doubly-modulated transistor current waveform, boost rectifier:
Computation of rms value of this waveform is complex and tedious
Approximate here using double integral
Generate tables of component rms and average currents for variousrectifier converter topologies, and compare
Fundamentals of Power Electronics 78 Chapter 18: PWM Rectifiers
RMS transistor current
t
iQ(t)RMS transistor current is
IQrms = 1Tac
iQ2 (t)dt
0
Tac
Express as sum of integrals overall switching periods containedin one ac line period:
IQrms = 1Tac
Ts1Ts
iQ2 (t)dt
(n-1)Ts
nTs
∑n = 1
Tac/Ts
Quantity in parentheses is the value of iQ2, averaged over the nth
switching period.
Fundamentals of Power Electronics 79 Chapter 18: PWM Rectifiers
Approximation of RMS expression
IQrms = 1Tac
Ts1Ts
iQ2 (t)dt
(n-1)Ts
nTs
∑n = 1
Tac/Ts
When Ts << Tac, then the summation can be approximated by anintegral, which leads to the double-average:
IQrms ≈ 1Tac
limTs→0
Ts1Ts
iQ2 (τ)dτ
(n-1)Ts
nTs
∑n=1
Tac/Ts
= 1Tac
1Ts
iQ2 (τ)dτ
t
t+Ts
0
Tac
dt
= iQ2 (t)
Ts Tac
Fundamentals of Power Electronics 80 Chapter 18: PWM Rectifiers
18.5.1 Boost rectifier example
For the boost converter, the transistor current iQ(t) is equal to the inputcurrent when the transistor conducts, and is zero when the transistoris off. The average over one switching period of iQ2(t) is therefore
iQ2
Ts= 1
TsiQ
2 (t)dtt
t+Ts
= d(t)iac2 (t)
If the input voltage isvac(t) = VM sin ωt
then the input current will be given by
iac(t) =VM
Resin ωt
and the duty cycle will ideally be
Vvac(t)
= 11 – d(t)
(this neglectsconverter dynamics)
Fundamentals of Power Electronics 81 Chapter 18: PWM Rectifiers
Boost rectifier example
Duty cycle is therefore
d(t) = 1 –VM
Vsin ωt
Evaluate the first integral:
iQ2
Ts=
V M2
Re2 1 –
VM
V sin ωt sin2 ωt
Now plug this into the RMS formula:
IQrms = 1Tac
iQ2
Ts0
Tac
dt
= 1Tac
V M2
Re2 1 –
VM
V sin ωt sin2 ωt0
Tac
dt
IQrms = 2Tac
V M2
Re2 sin2 ωt –
VM
V sin3 ωt0
Tac/2
dt
Fundamentals of Power Electronics 82 Chapter 18: PWM Rectifiers
Integration of powers of sin θ over complete half-cycle
1π sinn(θ)dθ
0
π
=
2π
2⋅4⋅6 (n – 1)1⋅3⋅5 n
if n is odd
1⋅3⋅5 (n – 1)2⋅4⋅6 n
if n is even
n 1π sinn (θ)dθ
0
π
1 2π
2 12
3 43π
4 38
5 1615π
6 1548
Fundamentals of Power Electronics 83 Chapter 18: PWM Rectifiers
Boost example: transistor RMS current
IQrms =VM
2Re
1 – 83π
VM
V = Iac rms 1 – 83π
VM
V
Transistor RMS current is minimized by choosing V as small aspossible: V = VM. This leads to
IQrms = 0.39Iac rms
When the dc output voltage is not too much greater than the peak acinput voltage, the boost rectifier exhibits very low transistor current.Efficiency of the boost rectifier is then quite high, and 95% is typical ina 1kW application.
Fundamentals of Power Electronics 84 Chapter 18: PWM Rectifiers
Table of rectifier current stresses for various topologies
Tabl e 18.3 Summary of rectifier current stresses for several converter topologies
rms Average Peak
CCM boost
Transistor Iac rms 1 – 83π
VM
V Iac rms
2 2π 1 – π
8VM
V Iac rms 2
Diode Idc163π
VVM
Idc 2 IdcV
VM
Inductor Iac rms Iac rms2 2
π Iac rms 2
CCM flyback, with n:1 isolation transformer and input fi lter
Transistor,xfmr primary
Iac rms 1 + 83π
VM
nV Iac rms
2 2π Iac rms 2 1 +
Vn
L1 Iac rms Iac rms2 2
π Iac rms 2
C1 Iac rms8
3πVM
nV0 Iac rms 2 max 1,
VM
nV
Diode,xfmr secondary
Idc32
+ 163π
nVVM
Idc 2Idc 1 + nVVM
Fundamentals of Power Electronics 85 Chapter 18: PWM Rectifiers
Table of rectifier current stressescontinued
CCM SEPIC, nonisolated
Transistor Iac rms 1 + 83π
VM
V Iac rms
2 2π Iac rms 2 1 +
VM
V
L1 Iac rms Iac rms2 2
π Iac rms 2
C1 Iac rms8
3πVM
V0 Iac rms max 1,
VM
V
L2 Iac rmsVM
V32
Iac rms
2VM
V Iac rms
VM
V2
Diode Idc32
+ 163π
VVM
Idc 2Idc 1 + VVM
CCM SEPIC, with n:1 isolation transformer
transistor Iac rms 1 + 83π
VM
nV Iac rms
2 2π Iac rms 2 1 +
VM
nV
L1 Iac rms Iac rms2 2
π Iac rms 2
C1,
xfmr primary Iac rms
83π
VM
nV0 Iac rms 2 max 1,
n
Diode,xfmr secondary
Idc32
+ 163π
nVVM
Idc 2Idc 1 + nVVM
with, in all cases, Iac rms
Idc= 2 V
VM
, ac input voltage = VM sin(ωt)
dc output voltage = V
Fundamentals of Power Electronics 86 Chapter 18: PWM Rectifiers
Comparison of rectifier topologies
Boost converter
• Lowest transistor rms current, highest efficiency
• Isolated topologies are possible, with higher transistor stress
• No limiting of inrush current
• Output voltage must be greater than peak input voltage
Buck-boost, SEPIC, and Cuk converters
• Higher transistor rms current, lower efficiency
• Isolated topologies are possible, without increased transistorstress
• Inrush current limiting is possible
• Output voltage can be greater than or less than peak inputvoltage
Fundamentals of Power Electronics 87 Chapter 18: PWM Rectifiers
Comparison of rectifier topologies
1kW, 240Vrms example. Output voltage: 380Vdc. Input current: 4.2Arms
Converter Transistor rmscurrent
Transistorvoltage
Diode rmscurrent
Transistor rmscurrent, 120V
Diode rmscurrent, 120V
Boost 2 A 380 V 3.6 A 6.6 A 5.1 A
NonisolatedSEPIC
5.5 A 719 V 4.85 A 9.8 A 6.1 A
IsolatedSEPIC
5.5 A 719 V 36.4 A 11.4 A 42.5 A
Isolated SEPIC example has 4:1 turns ratio, with 42V 23.8A dc load
Fundamentals of Power Electronics 88 Chapter 18: PWM Rectifiers
18.6 Modeling losses and efficiencyin CCM high-quality rectifiers
Objective: extend procedure of Chapter 3, to predict the outputvoltage, duty cycle variations, and efficiency, of PWM CCM lowharmonic rectifiers.
Approach: Use the models developed in Chapter 3. Integrate overone ac line cycle to determine steady-state waveforms and averagepower.
Boost example
+– Q1
L
C R
+
v(t)
–
D1
vg(t)
ig(t)RL i(t)
+– R
+
v(t)
–
vg(t)
ig(t)RL
i(t)DRon
+ –
D' : 1VF
Dc-dc boost converter circuit Averaged dc model
Fundamentals of Power Electronics 89 Chapter 18: PWM Rectifiers
Modeling the ac-dc boost rectifier
Rvac(t)
iac(t)+
vg(t)
–
ig(t)
+
v(t)
–
id(t)
Q1
L
C
D1
controller
i(t)
RL
+– R
+
v(t) = V
–
vg(t)
ig(t)RL
i(t) = Id(t) Ron
+ –
d'(t) : 1VF
id(t)
C(large)
Boostrectifiercircuit
Averagedmodel
Fundamentals of Power Electronics 90 Chapter 18: PWM Rectifiers
Boost rectifier waveforms
0
2
4
6
8
10
0
100
200
300
vg(t)
vg(t)
ig(t)
ig(t)
0° 30° 60° 90° 120° 150° 180°
d(t)
0
0.2
0.4
0.6
0.8
1
0° 30° 60° 90° 120° 150° 180°
0
1
2
3
4
5
6
id(t)
i(t) = I
ωt0° 30° 60° 90° 120° 150° 180°
Typical waveforms
(low frequency components)
ig(t) =vg(t)Re
Fundamentals of Power Electronics 91 Chapter 18: PWM Rectifiers
Example: boost rectifierwith MOSFET on-resistance
+– R
+
v(t) = V
–
vg(t)
ig(t) i(t) = I
d(t) Ron
d'(t) : 1
id(t)
C(large)
Averaged model
Inductor dynamics are neglected, a good approximation when the acline variations are slow compared to the converter natural frequencies
Fundamentals of Power Electronics 92 Chapter 18: PWM Rectifiers
18.6.1 Expression for controller duty cycle d(t)
+– R
+
v(t) = V
–
vg(t)
ig(t) i(t) = I
d(t) Ron
d'(t) : 1
id(t)
C(large)
Solve input side ofmodel:
ig(t)d(t)Ron = vg(t) – d'(t)v
with ig(t) =vg(t)Re
eliminate ig(t):
vg(t)Re
d(t)Ron = vg(t) – d'(t)v
vg(t) = VM sin ωt
solve for d(t):
d(t) =v – vg(t)
v – vg(t)Ron
Re
Again, these expressions neglect converter dynamics, and assumethat the converter always operates in CCM.
Fundamentals of Power Electronics 93 Chapter 18: PWM Rectifiers
18.6.2 Expression for the dc load current
+– R
+
v(t) = V
–
vg(t)
ig(t) i(t) = I
d(t) Ron
d'(t) : 1
id(t)
C(large)
Solve output side ofmodel, using chargebalance on capacitor C:
I = id Tac
id(t) = d'(t)ig(t) = d'(t)vg(t)
Re
Butd’(t) is:
d'(t) =vg(t) 1 –
Ron
Re
v – vg(t)Ron
Re
hence id(t) can be expressed as
id(t) =vg
2(t)Re
1 –Ron
Re
v – vg(t)Ron
Re
Next, average id(t) over an ac line period, to find the dc load current I.
Fundamentals of Power Electronics 94 Chapter 18: PWM Rectifiers
Dc load current I
I = id Tac= 2
Tac
V M2
Re
1 –Ron
Resin2 ωt
v –VM Ron
Resin ωt
dt
0
Tac/2
Now substitute vg (t) = VM sin ωt, and integrate to find ⟨ id(t)⟩Tac:
This can be written in the normalized form
I = 2Tac
V M2
VRe1 –
Ron
Re
sin2 ωt
1 – a sin ωtdt
0
Tac/2
with a =VM
VRon
Re
Fundamentals of Power Electronics 95 Chapter 18: PWM Rectifiers
Integration
By waveform symmetry, we need only integrate from 0 to Tac/4. Also,make the substitution θ = ωt:
I =V M
2
VRe1 –
Ron
Re
2π
sin2 θ1 – a sin θ
dθ0
π/2
This integral is obtained not only in the boost rectifier, but also in thebuck-boost and other rectifier topologies. The solution is
4π
sin2 θ1 – a sin θ
dθ0
π/2
= F(a) = 2a2π – 2a – π +
4 sin– 1 a + 2 cos– 1 a
1 – a2
• Result is in closed form
• a is a measure of the lossresistance relative to Re
• a is typically much smaller thanunity
Fundamentals of Power Electronics 96 Chapter 18: PWM Rectifiers
The integral F(a)
F(a)
a
–0.15 –0.10 –0.05 0.00 0.05 0.10 0.150.85
0.9
0.95
1
1.05
1.1
1.15
4π
sin2 θ1 – a sin θ
dθ0
π/2
= F(a) = 2a2π – 2a – π +
4 sin– 1 a + 2 cos– 1 a
1 – a2
F(a) ≈ 1 + 0.862a + 0.78a2
Approximation viapolynomial:
For | a | ≤ 0.15, thisapproximate expression iswithin 0.1% of the exactvalue. If the a2 term isomitted, then the accuracydrops to ±2% for | a | ≤ 0.15.The accuracy of F(a)coincides with the accuracyof the rectifier efficiency η.
Fundamentals of Power Electronics 97 Chapter 18: PWM Rectifiers
18.6.3 Solution for converter efficiency η
Converter average input power is
Pin = pin(t) Tac=
V M2
2Re
Average load power is
Pout = VI = VV M
2
VRe1 –
Ron
Re
F(a)2 with a =
VM
VRon
Re
So the efficiency is
η =Pout
Pin= 1 –
Ron
ReF(a)
Polynomial approximation:
η ≈ 1 –Ron
Re1 + 0.862
VM
VRon
Re+ 0.78
VM
VRon
Re
2
Fundamentals of Power Electronics 98 Chapter 18: PWM Rectifiers
Boost rectifier efficiency
η =Pout
Pin= 1 –
Ron
ReF(a)
0.0 0.2 0.4 0.6 0.8 1.00.75
0.8
0.85
0.9
0.95
1
VM /V
η
Ron/Re
= 0.05
Ron/Re
= 0.1
Ron/Re
= 0.15
Ron/Re
= 0.2
• To obtain highefficiency, choose Vslightly larger than VM
• Efficiencies in the range90% to 95% can then beobtained, even with Ron
as high as 0.2Re
• Losses other thanMOSFET on-resistanceare not included here
Fundamentals of Power Electronics 99 Chapter 18: PWM Rectifiers
18.6.4 Design example
Let us design for a given efficiency. Consider the followingspecifications:
Output voltage 390 VOutput power 500 Wrms input voltage 120 VEfficiency 95%
Assume that losses other than the MOSFET conduction loss arenegligible.
Average input power is
Pin =Poutη = 500 W
0.95= 526 W
Then the emulated resistance is
Re =V g, rms
2
Pin=
(120 V)2
526 W= 27.4 Ω
Fundamentals of Power Electronics 100 Chapter 18: PWM Rectifiers
Design example
Also, VM
V = 120 2 V390 V
= 0.435
0.0 0.2 0.4 0.6 0.8 1.00.75
0.8
0.85
0.9
0.95
1
VM /V
η
Ron/Re
= 0.05
Ron/Re
= 0.1
Ron/Re
= 0.15
Ron/Re
= 0.2
95% efficiency withVM/V = 0.435 occurswith Ron/Re ≈ 0.075.
So we require aMOSFET with onresistance of
Ron ≤ (0.075) Re
= (0.075) (27.4 Ω) = 2 Ω