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Switched-mode and Resonant dc Power Supplies491

Figure 15.2. Non-isolated forward converter (buck converter) where v 0≤ E 1:

(a) circuit diagram; (b) waveforms for continuous output current; and (c) waveforms

for discontinuous output current.

15.1.1 Continuous inductor current

The inductor current is analysed first when the switch is on, then when the switch is

off. When transistor T is turned on for period t T , the difference between the supply

voltage E i and the output voltage v0 is impressed across L. From V = Ldi/dt=L ∆i/ ∆t , the

current change through the inductor will be

i L = i o

Powe

L Li i∆ =

When T is switched off for the remaindiode D conducts and -v0 is impressed a

Li∆

Equating equations (15.1) and (15.2) gi

( - i o

E v

This expression shows that the inductor

o i

oi

v t I

E I τ= =

This equation shows that for a given in

the transistor conduction duty cycle

voltage. This confirms and validates t

voltage transfer function is independen

The inductor rms ripple current (and ca

r

1

2 3 2 3

L

L

vii

L

∆= =

while the inductor total rms current is

2 2 2

rms r L L L Li I i I

= + = +

The switch and diode average and rms c

( )1

T i o

D o i

I I I

I I I

δ

δ

= =

= − = −

If the average inductor current, henceminimum inductor current levels are giv

½ L Li I∧

= + ∆

and

½ L Li I= − ∆∨

respectively, where ∆i L is given by e




i i = i L

15.3.1 Continuous inductor current

The circuit voltage and current waveforms for continuous inductor conduction are

shown in figure 15.5b. The inductor current excursion, which is the input current

excursion, during the switch on-time t T and switch off-time τ - t T , is given by

( - )( )o i i

L T T

v E E i t t

L Lτ ∆ = − = (15.43)

Figure 15.5. Non-isolated, step-up, flyback converter (boost converter) where v 0≥E 1:(a) circuit diagram; (b) waveforms for continuous input current; and

(c) waveforms for discontinuous input current.

Powe

that is, after rearranging, the voltage tra

o

i

v

E

where δ = t T / τ and t T is the transistor on

the maximum input current, Li∧

, using e

(

½ ½

½ 1-1

L L L i

o o

i I i I

I v

Lδ

δ

∧

= + ∆ = +

= +

− while the minimum inductor current, Li

∨

( )

½ ½

½ 1-1

L L L i

o o

i I i I

I v

Lδ

δ

== − ∆ = −

= − −

∨

For continuous conduction 0 Li ≥ , that

½ L

E I

≥

The inductor rms ripple current (and inp

r

2 3

L

L

ii

∆=

The harmonic components in the input

( 2 2

2 si

2 1

i

in

E I

n

τ

π =

−

while the inductor total rms current is2 2 2


= + = +


( )1

T i o i

D i o

I I I I

I I I

δ δ

δ

= − = =

= − =

Switch utilisation ratio

The switch utilisation ratio, SUR, is a m

handling capabilities are utilised in any




15.4 The buck-boost converter

The basic buck-boost flyback converter circuit is shown in figure l5.5a. When transistor

T is on, energy is transferred to the inductor. When the transistor turns off, inductor

current is forced through the diode. Energy stored in L is transferred to C and the load

R. This transfer action results in an output voltage of opposite polarity to that of the

input. Neither the input nor the output current is continuous, although the inductor

current may be continuous or discontinuous.

Figure 15.6. Non-isolated, step up/down flyback converter (buck-boost converter)where v o ≤ 0: (a) circuit diagram; (b) waveforms for continuous inductor current; and

(c) discontinuous inductor current waveforms.

Powe

15.4.1 Continuous choke current

Various circuit voltage and current w

operating in a continuous inductor cond

Assuming a constant input and output

by

L

Ei

L∆ =

thus

o

i

v E =

where δ = t T / τ . For δ<½ the output ma

while for δ > ½ the output is greater in

The maximum and minimum inductor c

( ½ 1-1

L

o o I v

i Lδ

∧

= + −

and

( ½ 1-1

L

o o I v

i Lδ

∨

= − −

The inductor rms ripple current (and inp

r

2 3

L

L

ii

∆=

while the inductor total rms current is

2 2 2


= + = +


( )1

T i L

D L o

I I I

I I I

δ

δ

= =

= − =

Switch utilisation ratio

The switch utilisation ratio, SUR, is a m

handling capabilities are utilised in any

SUR

where p is the number of power sw

converter. The switch maximum in




(max)

(min)

2 ( 1)

D o

o

o

i

t v L

v I

E

=+

(15.106)

The minimum capacitance is specified by the maximum allowable ripple voltage, that

is

(min)2

i D

o o

Q i t C

e e

∧

∆= =

∆ ∆

that is

(max)

(min)

(min)

( 1)

o D

o

o

i

I t C v

e E

=∆ +

(15.107)

The ripple voltage is dropped across the capacitor equivalent series resistance, which is

given by

(max)

i

oe

ESRi∧

∆= (15.108)

The frequency varies as a function of load current. Equation (15.104) gives

(max)

(max)2

i oo T I I i t

f f

∧

= =

therefore

(max)

(max)

o

o

I f f

I = × (15.109)

and

(min)

(min) (max)

(max)

o

o

I f f

I = × (15.110)

Example 15.4: Buck-boost flyback converter

The 10kHz flyback converter in figure 15.6 is to operate from a 50V input and

produces an inverted non-isolated 75V output. The inductor is 300µH and the resistive

load is 2.5Ω.

i. Calculate the duty cycle, hence transistor off-time, assuming continuous

inductor current.

ii. Calculate the mean input and output current.

iii. Draw the inductor current, showing the minimum and maximum values.

iv. Calculate the capacitor rms ripple current and output p-p ripple voltage if C =

10,000µF.

Powe

v. Determine

• the critical load resistance.

• the minimum inductance for

vi. At what load resistance does th

output current?

vii What is the output voltage if t

critical resistance?

Solution

i. From equation (15.87), which assum

1

o

i

v

E

δ

δ = −

−

that is

75V

50V 1

δ

δ =

−That is, τ = 1/ f s = 100 µs with a 60µs s

ii. The mean output currento

I is give

/o o

I v R=

From power transfer considerations

/i L o o i

I I v I E= =

10 A

I T

60µs

I o=30A

i C (A )

30

0

40

50

0 50

0 50

i L (A )

80

75

70




var

var var

1

1or

crit

T T crit T o

o

T T o

Rt t R t I

v R

t t I R

α α

= × = ×

(15.126)

That is, once discontinuous inductor current commences, if the switching frequency f s

remains constant, regulation of the output voltage vo can be maintained if the switch

on-state period t T is varied

• proportionally with the square root of the load current,o

I

• inversely with the square root of the load resistance, √ Rcrit

• inversely with the square root of the output voltage, √vo.

Example 15.5: Reversible forward converter

The step-down reversible converter in figure 15.7a operates at a switching frequency of

10 kHz. The output voltage is to be fixed at 48 V dc across a 1 Ω resistive load. If theinput voltage E i =192 V and the choke L = 200µH:

i. calculate the switch T on-time duty cycle δ and switch on-time t T

ii. calculate the average load currento

I , hence average input currenti

I

iii. draw accurate waveforms for

• the voltage across, and the current through L; v L and i L

• the capacitor current, ic

• the switch and diode voltage and current; vT , v D, iT , i D

iv. calculate

• the maximum load resistance Rcrit before discontinuous inductor

current with L=200µH and

• the value to which the inductance L can be reduced before dis-continuous inductor current, if the maximum load resistance is 1Ω.

Solution

i. The switch on-state duty cycle δ can be calculate from equation (15.113), that is

58

48V2 1 = ¼

192V

o

i

v

E δ δ − = = ⇒ =

Also, from equation (15.113), for a 10kHz switching frequency, the switching period τ is 100µs and the transistor on-time t T is given by

58=

100µs

T T t t

δ τ

= =

whence the transistor on-time is 62½µs and the diode conducts for 37½µs.

Powe

ii. The average load current is o

o

v I

R=

From power-in equals power-out, the a

/i o o i

I v I E=

iii. The average output current is the a

is given by equation (15.113), that is

192V - 4=

200µ

L L L

i i i∆ = − =

∧ ∨

iv. Critical load resistance is given by

o

crit

o

v R

I ≤

(V )

(A )

V T r a n

V D i o d e

I ca p

0 2 5 5 0

0 2 5 5 0

0 2 5 5 0

V L (V )

1 4 4

- 2 4 0

I L (A )

4 8

0

I av e

Figure: Example




example 15.4 and the essential results from example 15.4 are summarised in Table 15.3and transferred to the secondary where appropriate. The waveform answers to part iii

are shown in figure 15.12.

Table 15.3 Transformer coupled flyback converter analysis

parameter value

for primary analysis

transfer factor nT = 3→

valuefor

secondary analysis

E i V 50 3 150

v o V 75 3 225

R L Ω 2½ 32

22½

C o µF 10,000 3-2

1100

I o(ave) A 30 ⅓ 10

P o W 2250 invariant 2250

I i(ave) A 45A ⅓ 15A

δ p.u. ⅗ invariant ⅗

τ µs 100 invariant 100

t on µs 60 invariant 60

t off µs 40 invariant 40

f s kHz 20 invariant 20

∆i L A 5 ⅓ 1⅔

L I ∧

A 80 ⅓ 80/3

L I ∨

A 70 ⅓ 70/3

i Crms A rms 21.3 ⅓ 7.1

R crit Ω 37½ 32 337½

Lcrit µH 20 32

180

V Dr V 125 3 375

∆v o mV 180 3 540

∆v o / v o p.u. 0.24% invariant 0.24%

Powe

Figure 15.12. Currents for the

Example 15.8: Transformer coupled

The 10kHz forward converter in figur1:3:2 (1:n f/b:n sec) step-up transformer

magnetising inductance is 1.2mH, reinductance is 800µH.

i. Calculate the maximum switch d

continuous inductor current.

At the maximum duty cycle:

ii. Calculate the mean input and outpu

iii. Draw the transformer currents, sho

iv. Determine

• the critical load resistance• the minimum inductance for c

Solution

The maximum duty cycle is determinethe primary and the feedback windin(15.154)

δ ∧

The maximum conduction time is 2

secondary output voltage is therefore

I primary

I secondary

70A

80/

70/3A

0µs

I transformer

I o =

10A




Voltage and impedance matching, for example voltage step-up, can be obtained byusing a transformer coupled circuit as shown in figure 15.20b. When a transformer is

used as in figure 15.20c, a centre tapped secondary can reduce the number of highfrequency rectifying diodes from four to two, but diode reverse voltage rating isdoubled from V s to 2V s. Secondary copper winding utilisation is halved.

A further modification to any series converter is to used the resonant capacitor to forma split dc rail as shown in figure 15.20c, where each capacitance is ½C R. In ac terms the

resonant capacitors are in parallel, with one charging while the discharges, and visaversa, such that their voltage sum is V s.

Figure 15.20. Series load resonant converters variations:

(a) half bridge, split dc supply rail; (b) transformer couple full bridge; and (c) split resonant capacitor, with a centre tapped output rectifier stage.

V D-B V H-B

LR

½C R

½C R

V s C o

R

R c

V D-B

V H-B

C R

LR V s C o

R

R c

V D-B

V H-B

LR

C R

C large

C large

½V s

V s ½V s

C o R

R c

(a)

(b)

(c)

+

+

+

Powe

(a)

(b)

C R

I o

V s ++

LR

LR

V s

C R

V H-B

15.9.2 Parallel loaded resonant dc-to

The basic parallel load resonant dc-toequivalent circuit is shown in figure 1circuit produces a near constant current

be greater than the input voltage V s, thaThe capacitor voltage and inductor curr15.21b, for a constant current load I o, a

( ) (

sin

L o Lo

s

o

o

i t I i I

V I Z

= + −

= +

( ) (

( 1 cos

c s s Co

s

v t V V v

V ω

= − −

= −

Figure 15.21. Paralle(a) circuit (b) equivalent ac circuit; (c

and (d) series-par

The relationship between the output v

determined from the equivalent circu

resistance has been replaced by its equ

fundament frequency magnitude, 4 s

V π

assumed to be sinusoidal.




100µH/1 14.6

0.47µF

o

c

Z Q

R= = Ω =

For this high Q, the circuit resonant frequency and damped frequency will be almost

the same, that is

1/

1 / 100µH 0.47µF 146 krad/s

2

146krad/s /2 = 23.25 kHz

or = 43 s

o R R L C

f

f

T

ω ω

π

π

µ

≈ =

= × =

=

=

ii. For zero current switching, the load current must be greater than the resonantcurrent, that is

/ 340V/14.6Ω = 23.3Ao s o

I V Z > =

iii. The commutation period comprises the four intervals, I to IV , shown in figure

15.24b.Interval I

The switch turns on and the inductor current rises from 0A to 23.3A in a time given by

/

=100µH×23.3A/340V = 6.85µs

I R st L I V = ∆

Intervals II and III These two interval take just over half a resonant cycle, which takes 43µs/2 = 21.5µs to

complete. Assuming action is purely sinusoidal resonance with a 23.3A offset, from 0Ato a maximum of 23.3A and down to -23.3A then from

/ sin for

23.3 23.3A sin gives

= ¾×43µs

= 32.25µs

o s o I V Z t t

A t

t

ω ω π

ω

= >

= − ×

The capacitor voltage at the end of this period is given by

32

(1 cos )

340V (1 cos )

340V

cIV sV V t ω

π

= −

= × −

=

Interval IV The capacitor voltage must discharge from 340V dc to zero volts, providing the 23.3Aload current. That is

/

= 340V×0.47µF/23.3A = 6.86µs

IV cIV R ot V C I = ×

Powe

The minimum commutation cycle timemaximum operating frequency is 21.7k

iv. The output voltage vo is the averagis in parallel with the resonant capacito

in figure 15.24b gives equation (15.190

( )

(

[

4

1

5

32.25µs

11 cos

1

340V 146µs

1 3340V +1

46µs 2

113292Vµs + 11

46µs

t

o st

o

v V t dtt

ω

π

= − +

= × × −

= × ×

= ×

∫

∫

The maximum output voltage is 314

energy based equation (15.189):

(

(

½

340V½ 6.85µ

46µs

s

o I II III IV

V v t t t

τ += + +

= × ×

Example 15.11: Zero-voltage, resona

The zero voltage resonant switch c

following conditions:V s = 192V I o =

L R = 10µH C RDeterminei. the switching frequency f s for

ii. switch average current andiii. the peak switch/diode/capacito

iv. the minimum output current

Solution i.

1 1

10µH 0.1µo

R R L C

ω = = ×




Table 15.5. Transfer functions with constant input voltage, E i , with respect to I o

converter E i

constant step-down step-up step-up/down

referenceequation

(15.4) (15.44) (15.74)

currentconduction

o

i

v

E δ =

1

1

o

i

v

E δ =

−

1

o

i

v

E

δ

δ

−=

−

referenceequation

(15.21) (15.59) (15.91)

discontinuouscurrent

2

1

21

o

i o

i

v

E LI

E δ τ

=

+

2

12

o i T

i o

v E t

E LI

δ = +

2

2

o i

i o

v E

E LI

δ τ = −

normalised

8

i E

I L

τ ∧

=o

@ δ = ½

2

1

11

4

o

i o

v

E I

I δ ∧

=

+ ×

o

21 4 /o o

i

v I

E I

δ ∧

= +

o

24 /o o

i

v I

E I

δ ∧

= −

o

change of variable

24 1 o

io

oo

i

v

E I

v I

E

δ

∧

−

=

24

1

o

oo

i

I

v I

E

δ ∧

=

−

24o

oo

i

I

v I

E

δ ∧

−=

boundary 4 1o oo

i i

o

v v I

E E

I

∧

= −

2

4 1o

io

oo

i

v

E I

v I E

∧

−

=

2

4

1

o

io

oo

i

v

E I

v I E

∧

−

=

+

duty cycleall with boundary

( )4 1o I

I

δ δ ∧ = −

o

½

1

oo

i

o

i

v I

E I v

E

δ

∧ ×

=−

o ½ 1oo

i

v I

E I

δ ∧

= × −

o

½ oo

i

v I

E I

δ ∧

= ×

o

Powe

Table 15.6. Transfer functions with

v o

constant step-dow

reference equation (15.4)

current conductiono

i

v

E δ=

reference equation (15.20

equation2

1o

i

v

E δ = −

normalised4

127

o

i

v

E δ= −

I ∧

i max @ δ = δ = ⅔

I ∧

i

4

27 I

∧

= ×i

change of variable227

4i

I

I

δ ∧

=

i

boundary27

4 1iv I

E I ∧

= −

i

duty cycle4

27

1

δ =

boundary227

4i

I

I

δ ∧ =

i




Table 15.7. Transfer functions with constant output voltage, v o, with respect to I o

converter v o

constant step-down step-up step-up/down

reference equation (15.4) (15.44) (15.74)

current conductiono

i

v

E δ =

1

1

o

i

v

E δ =

−

1

o

i

v

E

δ

δ

−=

−

reference equation (15.20) (15.60) (15.91)

equation2

21o i

i i

v LI

E E δ τ = − 2

1

12

o

i T i

i

v

E t E

LI

δ =

−

2

2

o o

i i

v v

E LI

τδ =

normalised

2

2

11

4

o oo

i i

v v I

E E I

δ ∧

= − × ×

o

2

2

1

1 4 /

o

ioo

i

v

E v I

E I

δ ∧

= − × o

2

/o oo

i i

v v I

E E I

δ ∧

= × o

I ∧

omax @ δ = δ = 0 δ = ⅓ δ = 0

I ∧

o

2

ov

I L

τ ∧

=o

4

27 2

ov

I L

τ ∧

= ×o

2

ov

I L

τ ∧

=o

change of variable

2

2

1 o

io

oo

i

v

E I

v I E

δ

∧

−

=

2274

1

o

o oo

i i

I

v v I

E E

δ ∧

=

−

2

2o

oo

i

I

v I E

δ ∧

−=

boundary 1 oo

io

v I

E I

∧= −

3

274 1o

io

oo

i

v

E I

v I E

∧

− =

2

1

1

o

oo

i

I

v I E

∧

−=

+

duty cycle

1

o

o

oi

i

I

v I v E

E

δ

∧

=

−

o 4

27 1o oo

i i

v v I

E E I

δ ∧

= −

×

o

o o

i

v I

E I

δ ∧

=

o

boundary 1 o I

I

δ ∧= −

o

( )2

274 1o

I

I

δ δ ∧

= −

o

1 o I

I

δ ∧

= −

o



Power Electronics 588

discontinuous

o o I / I i i I / I

o

i

v

E

o

i

v-

E

1

d = ¾

d = ¼

d = ½

d = ¼

d = ½

d = ¼

d = ½

d = ½

d = ¼

d = ¼

d = ½

d = ¾

0

d = 0

d = 1

1

d = 0

d = 1

d = ½

d = ¼

d = 0d = 0

2

-2

-1

1

½

-½

24

o

i

v

E

δ−

( )2

4

1

o

i

o

i

v

E

v

E

−

+

4 1v vo o

E E i i

−

24 1

v

o E ivo

E i

δ

−

24

1o

i

v E

δ

−

( )

( )2

4 1o

i

o

i

v E

v

E

−

1

1o

i

v

E δ=

−

o

i

v

E

δ=

1

o

i

v

E

δ

δ

−=

−

2

1

o

i

o

i

v

E

v

E

δ

−

1o

i

o

i

v

E

v

E

−

1

1o

i

v

E δ=

−

1

o

i

v

E

δ

δ

−=

−

o

i

v

E δ= ( )

2274 1 o

i

v E

δ −

δ= 2

i i I / I

?

2

1

o

i

o

i

v

E

v

E

−+

2

274 1 o o

i i

v v E E

−

E i constant E i constant0

0d 2

dis continuous

discontinuous

continuous

Figure 15.26. Characteristics for three dc-dc converters,

when the input voltage E i is held constant.




i i I / I o o

I / I

o

i

v

E

o

i

v-

E

d = ¾

d = ¼

d = ½

d = ¼

d = ½

d = ¼

d = ½

d = ½

d = ¼

d = ¼

d = ½

d = ¾

0

d = 0

d = 1

1

d = 0

d = 1

d = ½

d = ¼

d = 0d = 0

2

-2

-1

1

½

-½

1

24

o

i

v

E

δ−

( )2

4

1

o

i

o

i

v

E

v

E

−

+

4 1v vo o

E E i i

−

24 1 vo E i

vo

E i

δ −

2

4

1o

i

v E

δ

−

( )( )

2

4 1o

i

o

i

v E

v

E

−

1

1o

i

v

E δ=

−

o

i

v

E δ=

1

o

i

v

E

δ

δ

−=

−

( )

2

2

o

i

v

E

δ−

21

2

vo E

i

vo E

i

δ

−

( )

2274

1o o

i i

v v

E E

δ

−

( )3

274

1o

i

o

i

v

E

v E

−

1 o

i

v

E −

( )2

1

1 o

i

v

E

−

+

1

o

i

v

E

δ

δ

−=

−

vo E i

δ=

1

1

o

i

v

E δ=

−

v o constant v o constant

0 0

00

discontinuous

discontinuous

continuous

Figure 15.27. Characteristics for three dc-dc converters,

when the output voltage vo is held constant.



Power Electronics 590

Reading list

Fisher, M. J., Power Electronics,

PWS-Kent Publishing, 1991.

Hart, D.W., Introduction to Power Electronics,

Prentice-Hall, inc, 19974.

Hnatek, E. R., Design of Switch Mode Power Supplies,

Van Nostrand Reinhold, 1981.

Mohan, N., Power Electronics, 3rd

Edition,

Wiley International, 2003.

Thorborg, K., Power Electronics – in theory and practice,Chartwell-Bratt, 1993.

http://www.ipes.ethz.ch/

Problems

15.1. An smps is used to provide a 5V rail at 2.5A. If 100 mV p-p output ripple is

allowed and the input voltage is 12V with 25 per cent tolerance, design a flyback buck-

boost converter which has a maximum switching frequency of 50 kHz.

15.2. Derive the following design equations for a flyback boost converter, which

operates in the discontinuous mode.

(max)

(min)(max)

(min)

(min)

(min) (max)

(min)(max)

(min)

(min) (max)

12 constant

1

2

i

i

i

i

o

o D

oi

i

o i o o i

T

o io

T o

o o

vi I t

v E f

E

v E I v E

L t f f v E I i

i t eQC ESR

e e i

τ

∧

∧

∧

∧

= × × = =

− −

= = = × −

∆∆= = =

∆ ∆

15.3. Derive design equations for the forward non-isolated converter, operating in

the continuous conduction mode.

15.4. Prove that the output rms ripple current for the forward converter in figure

15.2 is given by / 2 3oi∆ .

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