R.W.Ericksonecee.colorado.edu/~ecen5807/course_material/Lecture40.pdfElectronics 78 Rectifiers RMS transistor current t is i Q (t) I Qrms = 1 T ac i Q 2 (t) dt 0 T ac ver ained period:

R. W. Erickson Department of Electrical, Computer, and Energy Engineering

University of Colorado, Boulder

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


RMS transistor current

t

iQ(t)RMS transistor current is

IQrms = 1TaciQ2 (t)dt

0

Tac

Express as sum of integrals overall switching periods containedin one ac line period:

IQrms = 1TacTs 1Ts

iQ2 (t)dt(n-1)Ts

nTs

∑n = 1

Tac/Ts

Quantity in parentheses is the value of iQ2, averaged over the nthswitching period.


Approximation of RMS expression

IQrms = 1TacTs 1Ts

iQ2 (t)dt(n-1)Ts

nTs

∑n = 1

Tac/Ts

When Ts


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 =1Ts

iQ2 (t)dtt

t+Ts

= d(t)iac2 (t)If the input voltage is

vac(t) = VM sin ωt

then the input current will be given by

iac(t) =VMRe

sin ωt

and the duty cycle will ideally beV

vac(t)= 11 – d(t)

(this neglectsconverter dynamics)


Boost rectifier example

Duty cycle is therefored(t) = 1 – VMV sin ωt

Evaluate the first integral:

iQ2 Ts =V M2

Re21 – VMV sin ωt sin

2 ωt

Now plug this into the RMS formula:

IQrms = 1TaciQ2 Ts0

Tac

dt

= 1TacV M2

Re21 – VMV sin ωt sin

2 ωt0

Tac

dt

IQrms = 2TacV M2

Re2sin2 ωt – VMV sin

3 ωt0

Tac/2

dt


Integration of powers of sin θ over complete half-cycle

1π sin

n(θ)dθ0

π

=2π2⋅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)dθ0π

1 2π

2 12

3 43π

4 38

5 1615π

6 1548


Boost example: transistor RMS current

IQrms =VM2Re

1 – 83πVMV = Iac rms 1 –

83π

VMV

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.


Table of rectifier current stresses for various topologies

Table 18. 3 Summary of rectifier current stresses for several converter topologies

rms Average Peak

CCM boostTransistor Iac rms 1 – 83π

VMV

Iac rms 2 2π 1 –π8

VMV

Iac rms 2

Diode Idc 163πV

VM Idc 2 Idc VVM

Inductor Iac rms Iac rms 2 2π Iac rms 2

CCM flyback, with n:1 isolation transformer and input filterTransistor,xfmr primary

Iac rms 1 + 83πVMnV

Iac rms 2 2π Iac rms 2 1 +Vn

L1 Iac rms Iac rms 2 2π Iac rms 2

C1 Iac rms 83πVMnV

0 Iac rms 2 max 1,VMnV

Diode,xfmr secondary

Idc 32 +163π

nVVM

Idc 2Idc 1 + nVVM


Table of rectifier current stressescontinued

CCM SEPIC, nonisolatedTransistor Iac rms 1 + 83π

VMV

Iac rms 2 2π Iac rms 2 1 +VMV


C1 Iac rms 83πVMV

0 Iac rms max 1,VMV

L2 Iac rmsVMV

32

Iac rms2

VMV

Iac rmsVMV 2

Diode Idc 32 +163π

VVM

Idc 2Idc 1 + VVM

CCM SEPIC, with n:1 isolation transformertransistor Iac rms 1 + 83π

VMnV

Iac rms 2 2π Iac rms 2 1 +VMnV


C1,xfmr primary

Iac rms 83πVMnV

0 Iac rms 2 max 1, n

Diode,xfmr secondary

Idc 32 +163π

nVVM

Idc 2Idc 1 + nVVM

with, in all cases, Iac rmsIdc= 2 VVM

, ac input voltage = VM sin(ω t)

dc output voltage = V


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


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


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


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) = I

d(t) Ron

+ –

d'(t) : 1VF id(t)

C(large)

Boostrectifiercircuit

Averagedmodel


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


Example: boost rectifierwith MOSFET on-resistance

+– R

+

v(t) = V

–

vg(t)

ig(t) i(t) = I

d(t) Ron

d'(t) : 1id(t)

C(large)

Averaged model

Inductor dynamics are neglected, a good approximation when the acline variations are slow compared to the converter natural frequencies


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) : 1id(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)RonRe

Again, these expressions neglect converter dynamics, and assumethat the converter always operates in CCM.


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) : 1id(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 –

RonRe

v – vg(t)RonRe

hence id(t) can be expressed as

id(t) =vg2(t)Re

1 – RonRe

v – vg(t)RonRe

Next, average id(t) over an ac line period, to find the dc load current I.


Dc load current I

I = id Tac =2

TacV M2Re

1 – RonResin2 ωt

v – VM RonResin 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 = 2TacV M2VRe

1 – RonResin2 ωt

1 – a sin ωtdt

0

Tac/2

with a = VMVRonRe


Integration

By waveform symmetry, we need only integrate from 0 to Tac/4. Also,make the substitution θ = ωt:

I = V M2

VRe1 – RonRe

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


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


The integral F(a)

F(a)

a–0.15 –0.10 –0.05 0.00 0.05 0.10 0.15

0.85

0.9

0.95

1

1.05

1.1

1.15

4π


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.


18.6.3 Solution for converter efficiency η

Converter average input power is

Pin = pin(t) Tac =V M22Re

Average load power is

Pout = VI = VV M2VRe

1 – RonReF(a)2 with a =

VMV

RonRe

So the efficiency is

η = PoutPin= 1 – RonRe

F(a)

Polynomial approximation:

η ≈ 1 – RonRe1 + 0.862 VMV

RonRe+ 0.78 VMV

RonRe

2


Boost rectifier efficiency

η = PoutPin= 1 – RonRe

F(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

R on/R e = 0.1

5

R on/R e =

0.2

• To obtain highefficiency, choose Vslightly larger than VM

• Efficiencies in the range90% to 95% can then beobtained, even with Ronas high as 0.2Re

• Losses other thanMOSFET on-resistanceare not included here


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 isPin =

Poutη =

500 W0.95 = 526 W

Then the emulated resistance is

Re =V g, rms2

Pin= (120 V)

2

526 W = 27.4Ω


Design example

Also, VMV =

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

R on/R e = 0.1

5

R on/R e =

0.2

95% efficiency withVM/V = 0.435 occurswith Ron/Re ≈ 0.075.

So we require aMOSFET with onresistance ofRon ≤ (0.075) Re

= (0.075) (27.4Ω) = 2 Ω

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R.W.Ericksonecee.colorado.edu/~ecen5807/course_material/Lecture40.pdfElectronics 78 Rectifiers RMS transistor current t is i Q (t) I Qrms = 1 T ac i Q 2 (t) dt 0 T ac ver ained period: