Single-phase converter systems containing ideal rectifiers

8/18/2019 Single-phase converter systems containing ideal rectifiers

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Fundamentals of Power Electronics 51 Chapter 18: PWM Rectifiers

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M+*)/A 8,.)0/* *4*1*+,

Instead of a capacitor, and inductor or higher-order LC network could

store 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 less

expensive

So a single big capacitor is the best solution


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>+)(8= 3())*+,

A problem caused by the large energy storage capacitor: the large

inrush current observed during system startup, necessary to charge

the 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 converter

diode to the capacitor, as long as v(t ) < vg(t ).

Additional circuitry is needed to limit the magnitude of this inrush

current.

Converters having buck-boost characteristics are capable of

controlling the inrush current. Unfortunately, these converters exhibit

higher transistor stresses.


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H.?K@)*C(*+3A 1.2*4 .@ 23K23 3.+9*),*)

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


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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 storage

capacitor

vC (t)

+

–

Dc-dc

converter

+ –

Pload V

$ pac(t)%T s

A complete low-frequency system model:

• Difference between rectifier output power and dc-dc converter

input 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 the

rectifier average power is equal to the load power


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5*),()70, i 2(t )

V g,rms >> vg,rms(t )

V control >> vcontrol(t )

Let with

The averaged model predicts that the rectifier output current is

i2(t ) T 2 L=

p(t )T 2 L

v(t )T 2 L

=vg,rms

2 (t )

Re(vcontrol(t )) v(t ) T 2 L

= f vg,rms(t ), v(t ) T 2 L, vcontrol(t ))


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Table 18.1 Small-signal model parameters for several types of rectifier control schemes

Controller type g2 j 2 r2

Average current control with

feedforward, Fig. 18.140

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

PavVV control

V 2

Pav

2PavVV g,rms

PavVV control

V 2

Pav

2PavVV g,rms

PavVV control

V 22Pav

2PavVV g,rms

PavVV control

V 2

Pav

2PavVV g,rms

2PavVD

V 2

Pav


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'.+8,0+, -.?*) 4.02

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)%T s

Rectifier and dc-dc converter operate with same average power

Incremental resistance R of constant power load is negative, and is

R = – V Pav

which is equal in magnitude and opposite in polarity to rectifier

incremental output resistance r2 for all controllers except NLC


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!"#$ &'()*+,- *'..). /,( )00+1+),12+, 33& 4+-4567/*+82 9)18+0+)9.

Objective: extend procedure of Chapter 3, to predict the output

voltage, duty cycle variations, and efficiency, of PWM CCM low

harmonic rectifiers.

Approach: Use the models developed in Chapter 3. Integrate over

one ac line cycle to determine steady-state waveforms and average

power.

Boost example

+ –

Q1

L

C R

+

v(t)

–

D1

vg(t)

ig(t) R L i(t)

+ –

R

+

v(t)

–

vg(t)

ig(t) R L

i(t) DRon

+ –

D' : 1V F

Dc-dc boost converter circuit Averaged dc model


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&'()*+,- 84) /15(1 :''.8 9)18+0+)9

Rvac(t)

iac(t) +

vg(t)

–

ig(t)

+

v(t)

–

id (t)

Q1

L

C

D1

controller

i(t)

R L

+ –

R

+

v(t) = V

–

vg(t)

ig(t) R L

i(t) = I d(t) Ron

+ –

d'(t) : 1V F

id (t)

C (large)

Boost

rectifier

circuit

Averaged

model


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!"#$#! ?@A9)..+', 0'9 1',89'**)9 (782 121*) !"#$

+ –

R

+

v(t) = V

–

vg(t)

ig(t) i(t) = I

d(t) Ron

d'(t) : 1

id (t)

C (large)

Solve input side of

model:

ig(t )d (t ) Ron = vg(t ) – d '(t )v

with ig(t ) =vg(t )

Re

eliminate ig(t ):

vg t

Red (t ) Ron = vg(t ) – d '(t )v

vg(t ) = V M sin !t

solve for d (t ):

d (t ) =v – vg(t )

v – vg(t ) Ron Re

Again, these expressions neglect converter dynamics, and assume

that the converter always operates in CCM.


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!"#$#G ?@A9)..+', 0'9 84) (1 *'/( 1799),8

+ –

R

+

v(t) = V

–

vg(t)

ig(t) i(t) = I

d(t) Ron

d'(t) : 1

id (t)

C (large)

Solve output side of

model, using chargebalance on capacitor C :

I = id T ac

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 ) =v g

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 .


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H1 *'/( 1799),8 %

I = id T ac= 2

T ac

V M 2

Re

1 – Ron Re

sin2 !t

v – V M Ron

Resin !t

dt

0

T ac/2

Now substitute vg (t ) = V M sin &t , and integrate to find "id (t )#T ac:

This can be written in the normalized form

I = 2T ac

V M 2

VRe1 –

Ron Re

sin2 !t

1 – a sin !t dt

0

T ac/2

with a = V M

V

Ron Re


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I,8)-9/8+',

By waveform symmetry, we need only integrate from 0 to T ac/4. Also,make the substitution % = &t :

I = V M

2

VRe1 –

Ron Re

2*

sin2 -

1 – a sin -d -

0

*/2

This integral is obtained not only in the boost rectifier, but also in the

buck-boost and other rectifier topologies. The solution is

4*

sin2 -

1 – a sin -d -

0

*/2

= F (a) = 2a 2*

– 2a – * +4 sin

– 1a + 2 cos – 1 a

1 – a 2

• Result is in closed form

• a is a measure of the loss

resistance relative to Re

• a is typically much smaller than

unity


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F4) +,8)-9/* &J'K

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

4*

sin2 -

1 – a sin -d -

0

*/2

= F (a) = 2a 2*

– 2a – * +4 sin

– 1a + 2 cos – 1 a

1 – a 2

F (a) & 1 + 0.862a + 0.78a 2

Approximation via

polynomial:

For | a | ' 0.15, thisapproximate expression iswithin 0.1% of the exact

value. If the a2 term is

omitted, then the accuracydrops to ±2% for | a | ' 0.15.The accuracy of F(a)

coincides with the accuracy

of the rectifier efficiency (.


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!"#$#L D'*78+', 0'9 1',=)98)9 )00+1+),12 (

Converter average input power is

Pin = pin(t ) T ac=

V M 2 Re

Average load power is

Pout = VI = V V M

2

V Re1 –

Ron Re

F (a)

2with a =

V M V

Ron Re

So the efficiency is

/ = Pout

Pin= 1 –

Ron Re

F (a)

Polynomial approximation:

/ & 1 – Ron Re

1 + 0.862 V M

V

Ron Re

+ 0.78 V M

V

Ron Re

2


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/ = Pout

Pin= 1 –

Ron Re

F (a)

0.0 0.2 0.4 0.6 0.8 1.0

0.75

0.8

0.85

0.9

0.95

1

V M /V

/

Ron/ Re

= 0. 05

R o n/ Re

= 0. 1

R o n / R e

= 0. 1 5

R o n / R e =

0. 2

• To obtain highefficiency, choose V

slightly larger than V M

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

• Losses other than

MOSFET on-resistance

are not included here


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Let us design for a given efficiency. Consider the following

specifications:Output voltage 390 V

Output power 500 W

rms input voltage 120 V

Efficiency 95%

Assume that losses other than the MOSFET conduction loss are

negligible.

Average input power is

Pin = out

/ = 500 W

0.95 = 526 W

Then the emulated resistance is

Re =V g, rms

2

Pin=

(120 V)2

526 W = 27.4 0


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H).+-, )@/>A*)

Also, M V

= 120 2 V390 V

= 0.435

0.0 0.2 0.4 0.6 0.8 1.0

0.75

0.8

0.85

0.9

0.95

1

V M /V

/

Ron/ Re

= 0. 05

R o n/ Re

= 0. 1

R o n / R e = 0

. 1 5

R o n / R e

= 0. 2

95% efficiency with

V M /V = 0.435 occurs

with Ron/ Re ) 0.075.

So we require a

MOSFET with on

resistance of

Ron ) (0.075) Re

= (0.075) (27.4 0) = 2 0

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Single-phase converter systems containing ideal rectifiers