T SERIES RESONANT CONVERTERecee.colorado.edu/~ecen5817/notes/ch4.pdf · he objective of this chapter is to describe the operation of the series resonant converter in ... turns ratio

Chapter 4

THE SERIES RESONANT CONVERTER

he objective of this chapter is to describe the operation of the series resonant converter in

detail. The concepts developed in chapter 3 are used to derive closed-form solutions for

the output characteristics and steady-state control characteristics, to determine operating mode

boundaries, and to find peak component stresses. General results are presented, for every

continuous and discontinuous conduction mode using frequency control. The origin of the

discontinuous conduction modes is explained.

These results are used to consider three design problems. First, the variation of peak

component stresses with the choice of worst-case operating point is investigated, and some

guidelines regarding the choice of transformer turns ratio and tank characteristic impedance are

discussed. Second, the effects of variations in input line voltage and output load current are

examined using the converter output characteristics. Finally, switching frequency variations are

considered, and the tradeoff between transformer size and tank capacitor voltage is exposed.

+-D5

D6

D7

D8

D1

D2

D3

D4Q4

Q3

Q2

Q1

Vg

I

R+

-V

+

-vS

iS

L C +

-vR

+ vT -

CF

Fig. 4.1. Series resonant converter schematic.

T

Principles of Resonant Power Conversion

2

4 . 1 . Subintervals and Modes

The series resonant converter, Fig. 2.1, is reproduced in Fig. 4.1. It can be seen that the

instantaneous voltage vT(t) applied across the tank circuit is equal to the difference between the

switch voltage vS(t) and the rectifier voltage vR(t):

vT(t) = vS(t) – vR(t) (4-1)

These voltages, in turn, depend on the conducting state of the controlled switch network and

uncontrolled rectifier network.

A subinterval is defined as a length of time for which the conducting states of all of the

semiconductor switches in the converter remain fixed; during each subinterval, vS(t), vR(t), and

vT(t) are constant. For example, consider the case where transistors Q1 and Q4 conduct, and iL(t)

is positive so that diodes D5 and D8 also conduct, as in Fig. 4.2a. In this case, we have

vs = +Vg

vR = +V (4-2)

vT = Vg – V

The applied tank voltage is therefore constant and

equal to Vg–V. In normalized form, one obtains

MT = VTVg

= 1 – M(4-3)

Hence, according to section 3.3, the normalized

state plane trajectory for this subinterval is a

circular arc centered at MT = 1-M, as shown in Fig.

4.2b. The radius depends on the initial conditions.

Note that, since we have assumed that iL(t) is

positive and diodes D5 and D8 conduct, this

particular switch conducting state can occur only in

the upper half-plane (jL > 0). For negative jL,

diodes D6 and D7 would conduct instead, MT

would be changed, and an arc centered at a

different location would be obtained. A subinterval

utilizing the switch conduction state described

above and in Fig. 4.2 is referred to in shorthand

form as subinterval Q1.

.MT = 1 - M 1 mC

jL > 0

(b)

+-VgR

+

-V

L C CF

+ vC -

vT = Vg - V

iL > 0

+ vT -(a)

direction of current flow

Fig. 4.2 Q1 conduction subinterval, inwhich Q1 and Q4 conduct, and iL > 0 sothat D5 and D8 conduct: a) circuit;b) normalized state plane trajectory.

Chapter 4. The Series Resonant Converter

3

Many other switch conduction states can

occur. Subinterval D1 is similar to subinterval Q1,

except that the tank current iL(t) is negative. The

conducting devices are antiparallel diodes D1 and

D4, and output rectifier diodes D6 and D7. The

applied tank voltage is therefore VT = Vg+V, or in

normalized form,

MT = 1 + M (4-4)

The circuit and state plane trajectory for this

subinterval are summarized in Fig. 4.3. Note that

this switch conduction state can only occur in the

negative half-plane (jL < 0).

Symmetrical switch conduction states Q2

(Fig. 4.4) and D2 (Fig. 4.5) can also occur, in

which iL, vS, vR, and vT have the opposite polarity

from states Q1 and D1 respectively. These correspond to MT = -1+M (Q2) and MT = -1-M (D2).

.MT = 1 - M1

mC

(b)

jL < 0jL

+-VgR

+

-V

L C CF

+ vC -+ vT -(a)

iL < 0

vT = Vg + V


Fig. 4.3 D1 conduction subinterval, inwhich iL < 0 such that diodes D1, D4,D6, and D7 conduct: a) circuit; b)normalized state plane trajectory.

.-1

mC

(b)

jL < 0

jL

MT = -1 + M

+-Vg

+

-V

L C

+ vC -+ vT -(a)

iL < 0


vT = -Vg + V

CF

Fig. 4.4 Q2 conduction subinterval, inwhich Q2 and Q3 conduct, and iL < 0 sothat D6 and D7 conduct: a) circuit;b) normalized state plane trajectory.


+-Vg

+

-V

L C CF

+ vC -

iL > 0

+ vT -(a)

VT = -Vg - V

.-1 mC

jL > 0

(b)

MT = -1 - M

Fig. 4.5 D2 conduction subinterval, inwhich iL > 0, such that diodes D2, D3,D5 and D8 conduct: a) circuit; b)normalized state plane trajectory.


4

Under certain conditions, it is possible

for all four uncontrolled rectifier diodes (D5, D6,

D7, D8) to become simultaneously reverse-

biased. When this occurs, the circuit topology is

as given in Fig. 4.6. The tank inductor is then

zero, and the tank capacitor voltage remains at its

initial value. This switch conduction state is

denoted “X”.

When phase control is used, two other

subintervals can occur: P1, which occurs for

i L > 0, is summarized in Fig. 4.7, and P2,

which occurs for iL < 0, is summarized in Fig.

4.8.

An operating mode is defined by a

sequence of subintervals which combine to form

a complete switching period. Discontinuous

+-VgL CiL > 0

+ vT -(a)

o

o

+

-V

o

o

opencircuit

.jL

jL = 0 mC

mC does not change

(b)

Fig. 4.6 Subinterval X, in which all fourrectifier diodes D5, D6, D7 and D8 arereverse-biased. The inductor currentremains at zero, and the tank capacitorvoltage does not change: a) one possiblecircuit topology; b) normalized stateplane trajectory.

+-Vg

L CiL > 0

+ vT -(a)o

+

-Vdirection of

current flow

vT = -V

.

jL

mC

(b)

jL > 0

MT = -M

Fig. 4.7 Subinterval P1, in which D2 andQ4 (or Q1 and D3) conduct, and iL > 0 sothat D5 and D8 also conduct: a) circuit;b) normalized state plane trajectory.

.mC

(b)

jL < 0

jL

MT = M

+-Vg

+

-V

L CCF

+ vC -+ vT -

(a)

i L < 0


o

vT = V

Fig. 4.8 Subinterval P2, in which D1 and Q3(or Q2 and D4) conduct, and iL < 0 so thatD6 and D7 conduct: a) circuit; b)normalized state plane trajectory.


5

conduction modes contain at least one X subinterval, while continuous conduction modes contain

no X subintervals. As seen later in this chapter, the different modes cause the series resonant

converter to exhibit widely varying terminal characteristics.


6

4 . 2 . State Plane and Charge Arguments for the k=1 Continuous Conduction

Mode

State plane trajectory

Typical inductor current iL(t), capacitor

voltage vC(t), and applied tank voltage vT(t)

waveforms are diagrammed in Fig. 4.9 for the k=1

continuous conduction mode. This mode is

defined by the subinterval sequence Q1-D1-Q2-D2.

In this mode, the switching period begins when the

control circuit switches transistors Q1 and Q4 on,

with the inductor current iL(t) positive. The state

plane trajectory for this subinterval is given in Fig.

4.10a; it begins at ω0t = 0 with some initial values

of tank inductor current and capacitor voltage. The

tank rings with a circular state plane trajectory

centered at MT = 1–M until, at ω0t = β, the

inductor current rings negative, and subinterval D1

begins. The normalized state plane trajectory then

follows a circular arc centered at MT = 1+M.

At time ω0t = β+α ≡ γ (one half switching

period), the control circuit switches transistors Q1and Q4 off, and Q2 and Q3 are switched on.

Subinterval Q2 begins, and as shown in Fig.

4.10c, the trajectory continues along a circular arc

centered at MT = -1+M until the inductor current

again reaches zero. The output bridge rectifiers

then switch, and subinterval D2 begins. The

trajectory follows an arc centered at MT = -1-M for

the remainder of the switching period. The

switching period ends when the control circuit

switches Q2 and Q3 off, and Q1 and Q4 on.

If the converter operates in equilibrium,

then the trajectory begins and ends at the same

point in the state plane, and the tank waveforms are

Vg - V

Vg + V

-Vg + V

-Vg - V

subinterval :

β βα α

Q1 D1 Q2 D2

vT(t)

+VC1

-VC1

vC(t)

i L(t)

γ γ

ω0t

γ = ω0TS2

Fig. 4.9 Typical tank inductor current iL,tank capacitor voltage, vC, and appliedtank voltage vT waveforms, for thek = 1 continuous conduction mode.


7

periodic. Otherwise, a transient occurs in which the trajectory for each switching period begins at

a different point, and follows a different path in the state plane. If the circuit is stable, then the

trajectory eventually converges to a single closed path, and the waveforms become periodic. To

find the converter steady-state characteristics, we need to solve the geometry of this closed path,

and to relate it to the load current using charge arguments.

.1 - M

β

1 - M + MA1 mC

.

jL

ω0t = βradius MA1

(a)

ω0t = 0

Q1 subinterval

.mC

.

jL

ω0t = β1 + M

radius MA2 αω0t = α + β = γ

(b)

D1 subinterval

mC

.

jL

. .ω0t = γ + β

αω0t = γ + α + β = 2γ

-1 - M

(d)

radius MA2

D2 subinterval

mC

.

jL

-1 + M.β radius MA1

.ω0t = γ

ω0t = γ + β

(c)

Q2 subinterval

Fig. 4.10Construction of the state plane trajectory for one complete switching period, in thek=1 CCM: a) subinterval Q1; b) subinterval D1; c) subinterval Q2; d) subinterval D2.

Capacitor charge arguments

The inductor current waveform of Fig. 4.9 contains only one positive-going and one

negative-going zero crossing per switching period; this is true throughout the k=1 continuous

conduction mode. In consequence, the discussion of section 3.1, regarding tank capacitor charge

variation, applies directly to this case and leads to a result nearly identical to Eq. (3-6). The tank


8

inductor current coincides with the tank capacitor current in the series tank, and hence the tank

capacitor voltage vC(t) increases when the inductor current iL(t) is positive. During the half

switching period where iL(t) is positive, the capacitor voltage increases from its negative peak -VC1

to its positive peak +VC1. The total change in vC is the peak-to-peak value 2VC1. This

corresponds to a total increase in charge q on the capacitor, given by the integral of the positive

portion of the iL as shown in Fig. 4.11.

Hence, we have:

q = C (2VC1) (4-5)

This q is also directly related to the dc load

current I. The load current is the dc component, or

average value, of the rectified tank inductor current

| iL |:

I = < | iL | >

= 112 Ts

| iL(τ) | dτ0

12 Ts

(4-6)

=

2qTs

since the integral in Eq. (4-6) is equal to the charge

q. We can now eliminate q from Eqs. (4-5) and

(4-6), and solve for VC1:

VC1 = ITs4C (4-7)

or, in normalized form:

MC1 = Jγ2

(4-8)

where MC1 = VC1 / Vg.

This is a useful result, because it allows us to relate the load current to the peak capacitor

voltage. In the normalized state plane, we can now find the circle radii directly in terms of the

normalized load voltage and current M and J.

area = -q

area = q

area = q

+VC1

-VC1

vC(t)

i L(t)

|i L(t)|

ω0t

I=<|iL|>

Fig. 4.11Use of charge arguments to relatethe peak capacitor voltage VC1, loadcurrent I, and charge quantity q in theseries resonant converter for k=1 CCM.


9

4 . 3 . Solution of the k=1 Continuous Conduction Mode Characteristics

The radii MA1 and MA2 of the state plane trajectory, re-drawn in Fig. 4.12, can now be

found. At time ω0t = β, it can be seen that the radius MA1 of the Q1 subinterval is

MA1 = (MC1) – (1–M) = Jγ2

– 1 + M (4-9)

and the radius MA2 of the D1 subinterval is

MA2 = (MC1) – (1+M) = Jγ2

– 1 – M (4-10)

So we know the radii and centers of the circular arcs in terms of the normalized output voltage M,

current J, and control input γ (or switching frequency fs = π f0 / γ).

..

Q1 subinterval

1 + M

D1 subinterval

D2 subinterval

.mC

jL

-1 + M

β

ω0t = γ

Q2 subinterval

1 - M. .α

ω0t = 2γ ω0t = 0

radius MA1

-1 - Mω0t = γ + β

radius MA2

-MC1

ω0t = β

.MC1 = Jγ2

Fig. 4.12 Complete normalized state-plane trajectory, for steady-state operation in the k=1 CCM.

Finally, it is desired to find a closed-form expression that relates the steady-state output

voltage, output current, and the control input; i.e., we want to directly relate M, J, and γ. In

steady-state, the endpoint of the state plane plot after one switching interval ( at time ω0 = 2γ =ω0Ts) coincides with the initial point (at time ω0t = 0), and the trajectory is closed. For a given M

and J in the k=1 CCM, there is a unique set of values of γ, α, β, MA1, and MA2 which cause this


10

to happen, and

which can be found

by solving the

geometry of the state

plane. With these

values, the triangle

of Fig. 4.13 is

formed, whose

solution yields the

converter steady-

state characteristics.

The lengths

of the two radii are

already known, and

the length of the

triangle base is the

distance between the

Q1 and D2

subinterval centers,

or 2. Hence, the

lengths of the three sides of the triangle are known,

and are functions of only M, J, and γ.The included angles can be found using

simple geometry. The two angles adjacent to the

base are (π – α) and (π – β), which are functions

of the unknowns α and β. The remaining angle

can be found knowing that the three angles of the

triangle must sum to π, and is given by

π – (π – α) – (π – β) = γ – π (4-11)

since γ = α + β. Note that this is a function of the control input γ alone, and does not depend

separately on α or β.

The Law of Cosines, Fig. 4.14, can now be used to relate the top angle and the three sides,

and hence to find how M and J depend on γ. One obtains:

(2)2 = (Jγ2

– 1 – M)2 + (Jγ2

– 1 + M)2

– 2(Jγ2

– 1 – M)(Jγ2

– 1 + M) cos(γ–π)(4-12)

.α

π − (π − α) − (π − β) = γ − π

jL

−1 − Μ

2

Α

C

. βπ − βπ − α

mC1 − Μ

B

M = - 1 + M

A1 Jγ__2M

=

- 1

- M

A2

Jγ __ 2

Fig. 4.13 Magnification of Fig. 4.12, for the solution of steady-stateconditions.

a

b cΘ

a2 = b2 + c2 - 2bc cosΘ

Fig. 4.14 The Law of Cosines


11

Simplification yields:

4 = 2(Jγ2

– 1)2 + 2M2 + 2[(Jγ2

– 1)2 – M2] cos(γ) (4-13)

which can be rearranged to obtain:

1 = (Jγ2

– 1)2 1 + cos γ

2 + M2

1 – cos γ2

(4-14)

Trigonometric identities can now be used to obtain:

M2 sin2 γ2

+ (Jγ2

– 1)2 cos2γ2

= 1 (4-15)

This is the desired closed-form solution for the series resonant converter operating in the k=1

continuous conduction mode, with frequency control.

Output characteristics

At a given switching frequency fs, corresponding to a given γ = π f0 / fs, Eq. (4-15) shows

that the relation between M and J is an ellipse, centered at M = 0 and J = 2/γ. A typical ellipse is

plotted in Fig. 4.15.

The uncontrolled output

rectifier diodes do not allow the

load current to be negative, so

we must have J > 0. Also, with

a passive load, M must be

positive when J is positive.

Hence, the portions of the

ellipse that lie in the second,

third, and fourth quadrants are

not valid physical solutions.

Also, it is shown in section 4.4

that the solution is not valid for

M≥1; instead, the k=1

discontinuous conduction mode occurs for M=1. Hence, the solution is valid only for 0 ≤ M < 1

and J > 0.

Equation (4-15) is plotted in Fig. 4.16. It can be seen that as the load current (or J) is

increased, the output voltage (or M) decreases. Hence, the output impedance of the open-loop

converter is substantial. It is instructive to examine some limiting cases.

M

J

passive load requiresI and V to have same

polarity

bridge rectifier does not allow I < 0

no solutions for M ≥ 1(DCM instead)

1

validsolutions

Fig. 4.15 Elliptical output characteristic M vs. J (Eq. 4-15),for a given γ in the k = 1 CCM.


12

F = 1.0F = 0.95

F = 0.8

F = 0.9

F = 0.7

F = 0.5

F = 0.62π

M

J

0

2

3

4

5

0.2 0.4 0.6 0.8 1

F = 0.925

0

1

Fig. 4.16 Output characteristics of the series resonant converter, operating in the k = 1continuous conduction mode. Solutions occur over the range 0 ≤ M <1, 2/π ≤ J < ∞;solutions not shown here for J > 6.

At F = 0.5 (half resonance):

Then fs = 0.5 f0, and γ = π / 0.5 = 2π. The output characteristic, Eq. (4-15), becomes

M2 ⋅ 0 + (π J – 1)2 ⋅ 1 = 1 (4-16)

or,

J = 2 / π (4-17)

which is independent of M. The ellipse collapses to a horizontal line, and the converter operates as

a current source.

At F = 1.0 (resonance):

Then fs = f0, and γ = π / 1 = π. The output characteristic, Eq. (4-15), becomes

M 2 ⋅ 1 + (J π2

– 1)2 ⋅ 0 = 1(4-18)

or,

M = 1 (4-19)

which is independent of J. The ellipse collapses to a vertical line, and the converter operates as a

voltage source. For 0.5 < F < 1, the converter operates as neither a voltage source nor a current

source.


13

Value of J when M = 1

The critical minimum value of J occurs when M = 1; for J less than this value, the

converter does not operate in k = 1 CCM, and Eq. (4-15) is not valid. Plugging M = 1 into Eq. (4-

15) yields

(J π2

– 1)2 = 1 – sin2

γ2

cos2γ2

= 1 (γ ≠ π) (4-20)

or,

J = 4 / γ (4−21)which varies between 2 / π and 4 / π for F between 0.5 and 1.

Output short circuit current JS C

When M = 0, Eq. (4-15) becomes

(JSC γ2

– 1)2 cos2γ2

= 1 (4-22)

Solve for JSC:

JSC = 2γ (1 + | sec

γ2

|) = 2Fπ

(1 + | secπ2F

|)

(4-23)

Equation (4-23) is plotted in Fig. 4.17. It can be

seen that the converter short-circuit current is

inherently limited, except at resonance.

The characteristics of a given load can be

superimposed on the converter output

characteristics, allowing graphical determination of

output voltage vs. switching frequency. For example, with a linear resistive load,

I = V / R (4-24)

or, in normalized form,

J = M Q (4-25)

with Q = R0 / R. Equation (4-25) describes a line with slope Q, as shown in Fig. 4.18. The

intersection of this load line with the converter elliptical output characteristic is the steady-state

operating point for a given switching frequency. As shown in the example of Fig. 4.19, nonlinear

load characteristics can also be superimposed, and the operating point determined graphically.

F

Js

c

0

2

4

6

8

10

0.5 0.6 0.7 0.8 0.9 1

Fig. 4.17 Normalized short-circuit outputcurrent JSC vs. switching frequency, inthe k=1 CCM.


14

load line

F = 1.0

F = 0.95

F = 0.8

F = 0.9

F = 0.7

F = 0.5

F = 0.62π Q =

R0

R

M

J

0

2

3

4

5

0.2 0.4 0.6 0.8 1

F = 0.925

0

1

Fig. 4.18 Resistive load line superimposed over the converter output characteristics.

load

line

F = 1.0

F = 0.95

F = 0.8

F = 0.9

F = 0.7

F = 0.5

F = 0.62π

M

J

0

2

3

4

5

0.2 0.4 0.6 0.8 1

F = 0.925

0

1

Fig. 4.19 Nonlinear load characteristic superimposed over the converter output characteristics.


15

Control plane characteristics

It is also instructive to plot the voltage conversion ratio M vs. normalized switching

frequency F. Doing so requires knowledge of the load characteristics, so that J can be eliminated

from Eq. (4-15). In the case of a resistive load satisfying V = I R, Eq. (4-25) can be substituted

into Eq. (4-15), yielding:

M2 sin2 γ2

+ (MQγ2

– 1)2 cos2γ2

= 1 (4-26)

Now solve for M:

M2 [sin2 γ2

+ Qγ2

2cos2

γ2

] – M Qγ cos2γ2

+ (cos2γ2

– 1) = 0(4-27)

Use of the quadratic formula yields:

M =

Qγ2

tan2 γ2

+ Qγ2

2 1 ± 1 + 2

Qγ2(tan2 γ

2) (tan2 γ

2 +

Qγ2

2) (4-28)

Q = 1

Q = 2

Q = 5

Q = 20

F

M

0

0.2

0.4

0.6

0.8

1

0.5 0.6 0.7 0.8 0.9 1

Fig. 4.20 Steady-state control characteristics M vs. F for various values of Q = R0/R in the k=1continuous conduction mode, Eq. (4-28).


16

To obtain the correct solution, in which M > 0, the plus sign should be used. Equation

(4-28), together with the identity γ = π / F = π f0 / fs, is a closed-form representation of the control

characteristics M vs. F for the k=1 CCM. It is plotted in Fig. 4.20 for various values of Q, or load

resistance R. As R is decreased, corresponding to heavy loading, the Q is increased. Loading the

converter causes the output voltage to decrease, and results in a peaked characteristic near

resonance.

Control of diode conduction angle α

Another popular scheme for controlling the series resonant converter when it operates

below resonance is known as diode conduction angle control, or “α control”. Rather than using a

voltage controlled oscillator to cause the switching frequency and γ to be directly dependent on a

control signal (known as frequency control, or “γ control”), the α controller causes the diode

conduction time and angle α to be directly dependent on the control signal. The switching

frequency varies indirectly, and depends on both α and the load current.

This control scheme requires a current monitor circuit which senses the zero crossings of

the tank current waveform. A timing circuit then causes the transistors to switch on after a delay

which is proportional to a control voltage. The transistor off time, which coincides with the diode

conduction time, is therefore proportional to the control voltage.

To understand the converter characteristics under α control, we need to eliminate γ from the

k=1 CCM solution, Eq. (4-15), in favor of α . This can be done by again referring to Fig. 4.13.

Application of the Law of Cosines, using the included angle at vertex C, yields

Jγ2

– 1 + M2 = 22 +

Jγ2

– 1 – M2 – 2 (2)

Jγ2

– 1 – M cos(π–α) (4-29)

This equation can be solved for γ:

γ = 2J

(1 + M) (1 – cos(α))

(M – cos(α)) (4-30)

This describes how the switching frequency varies for a given range of α and load. Equations (4-

30) and (4-15) can now be used to eliminate γ, and to determine the α control characteristics. The

result is

J = (1 + M) (1 – cos(α))

(M – cos(α)) π – tan–1 sin(α)M – cos(α)

(4-31)

where 0 ≤ tan–1(•) ≤ π/2, and M > cos(α).

Equation (4-31) describes the output characteristics under α control, and is plotted in Fig.

4.21. It can be seen that the output characteristics resemble hyperbolae, with vertical asymptotes


17

M = cos(α). Also, comparison with Fig. 4.16 reveals that the switching frequency approaches

resonance (fs →f0) as α→0, and fs→0.5f0 as α→π. Decreasing α causes M and/or J to increase.

It can also be seen that, for α < π/2, the converter short-circuit current is not inherently limited.

α = π

α = .7π

α = .6π

α = .5π

α = .4πα = .3π

α = .2π

2/π

M

J

0

2

3

4

5

6

0 0.25 0.5 0.75 1

α = 0

Fig. 4.21 k=1 CCM output plane characteristics, diode conduction angle control.

Mode boundaries, k=1 CCM

So far, we have studied only the k=1 continuous conduction mode, characterized by the

subinterval sequence Q1–D1–Q2–D2. The state plane diagram of Fig. 4.12 and the succeeding

analysis are both based on the assumption that the transistors and diodes conduct in the order given

by this sequence.

As stated previously, the diode conduction angle α (i.e., the lengths of the D1 and D2

subintervals) vanishes as the switching frequency approaches resonance. Increasing the switching

frequency beyond resonance must therefore cause a different subinterval sequence to occur.

Likewise, as fs approaches 0.5f0, the diode conduction angle α and transistor conduction angle βboth approach π, or a complete resonant half-period. For the series resonant converter, no ringing

subinterval can extend through an angle of more than π radians, because the state plane centers and

output diode switching boundary both lie on the jL = 0 axis. Therefore, decreasing the switching


18

frequency below fs = 0.5f0 must cause new subintervals to occur. Hence, the k=1 CCM and Eq.

(4-15) are restricted in validity to the range 0.5f0 ≤ fs ≤ f0.

By examination of Fig. 4.16, it can be seen that the k=1 CCM solutions do not extend

beyond the range J ≥ 2/π. If it is desired to operate the converter at light loads corresponding to J

< 2/π, then a different mode must be used, most likely with a different range of switching

frequencies. In addition, as shown in the next section, the k=1 CCM is restricted to the range 0 ≤M < 1.

4 . 4 . Discontinuous Conduction Modes

At light loads, all four bridge rectifier diodes can become reverse-biased during part of the

switching period, causing the converter to operate in a discontinuous conduction mode. More that

one discontinuous conduction mode is possible, depending on the load current. Each of these is

characterized by a sequence of subintervals ending in subinterval X, Fig. 4.6.

The k=1 discontinuous conduction mode

This mode is defined by the subinterval sequence Q1-X-Q2-X. As shown in Fig. 4.22,

transistor Q1 conducts for a complete tank half-period. The four bridge rectifier diodes then

become reverse-biased, and subinterval X occurs for the remainder of the half switching period.

As in the k=1 continuous conduction mode, the dc load current I is given by:

I = <| iL |> = 2qTs

(4-32)

where q, shown in Fig. 4.22, is:

q = iL(t) dt0

Ts/2

(4-33)

The average input current is:

< ig > = 112Ts

iL(t) dt0

Ts/2

(4-34)

since ig = iL when Q1 conducts. Substitution of Eqs. (4-32) and (4-33) into Eq. (4-34) yields:

< ig > = 2qTs

= I(4-35)


19

So the converter dc input and output currents are

equal. If the converter is lossless and operates in

equilibrium, then the input and output powers must

be equal; this implies that the voltages are also

equal:

Pin = Vg < ig > = Pout = V I (4-36)

Use of Eq. (4-35) yields:

Vg = V

or:

M = 1 (4-37)

Hence, in the k=1 DCM, the converter dc

conversion ratio M is unity, and is independent of

the values of load current and switching frequency.

The usual tank capacitor charge arguments

can be used to complete the solution and compute

the peak tank capacitor voltage VC1. During the Q1

conduction subinterval, the charge on the tank

capacitor changes by an amount q, corresponding

to an increase in voltage of 2VC1 (see Fig. 4.23).

Hence:

q = C · (2VC1) (4-38)

Elimination of q using Eq. (4-32) yields:

VC1 = I Ts4 C (4-39)

or, in normalized form:

MC1 = J γ2 (4-40)

Equation (4-40) happens to be identical to the result for the k=1 CCM, Eq. (4-8). Beware, this

does not occur for all other operating modes.

This mode cannot occur above resonance. The switching period must be long enough that

the tank can ring through one complete Q1 subinterval of length π during each half switching

period of length γ. Hence, a necessary condition for the occurrence of the k=1 DCM is:

iL

X X

π γ − π

q

-q

|iL| <|iL|>

ω0t

q q

ω0t

Q1 Q2

VC1

vC

-VC1

Vg-V -Vg+VvT :

Fig. 4.23 Tank inductor current andcapacitor voltage waveforms for the k = 1DCM.


20

γ > π (4-41)

or, in terms of F:

F > 1 (4-42)

An additional necessary condition for occurrence of the k=1 DCM is given in the next subsection.

Reason for occurrence of the k=1 DCM

Why does the tank stop ringing at the end of the Q1 subinterval? As suggested previously,

the reason is that all four bridge rectifier diodes become reverse-biased at this instant. Physical

arguments are used in this subsection to prove this assertion, and to derive the conditions on load

current and frequency which lead to operation in this mode. These arguments also have a very

simple state-plane interpretation.

As seen in Fig. 4.1, the voltage applied to the tank inductor vL is:

vL = L diLdt

= vS – vC – vR (4-43)

During subinterval Q1, vS = Vg, and vR = V. We have already shown that V = Vg in this mode

(Eq. (4-37)), and so vL becomes

vL = L diLdt

= – vC (4-44)

for subinterval Q1. vL(t) is plotted in Fig. 4.23. At ω0t = 0, this is a positive quantity since, as

shown in Fig. 4.22, vC(0) = -VC1. So initially, diL/dt is positive and iL increases. At ω0t = π/2,

vL(t) passes through zero, and iL begins to decrease. At ω0t = π, iL reaches zero.

Can the inductor current iL continue to decrease for ω0t ≥ π? This is possible only if the

applied inductor voltage vL continues to be negative. Note that, if iL rings negative, then the bridge

rectifier will switch from vR = +V to vR = –V, and subinterval D1 will occur. The applied tank

inductor voltage, Eq. (4-41), would then become:

vL = L diLdt

= Vg + V – vC = 2Vg – vC (4-45)

for subinterval D1, with vC(π) = VC1. Note that it is possible for this voltage to be either positive

or negative at ω0t = π, depending on whether or not VC1 is greater than 2Vg. In the k=1

discontinuous conduction mode, 2Vg–VC1 is a positive quantity. As a result, diodes D6 and D7

cannot turn on at ω0t = π: doing so would require that iL become negative, which cannot occur if

vL is positive (since iL(π)=0). Instead, all four bridge rectifier diodes become reverse-biased.


21

X X

iL

ω0t

Vg+V-VC1 > 0

2V 2Vg

vL

vC

VC1

-VC1

Q1 Q2

k = 1 DCM k = 1 CCMiL

ω0t

vC

Q1 Q2 D2D1

Vg-V-vC(t)

-Vg+V-vC(t) 2V

vLVg-V-vC(t)

-Vg+V-vC(t)

Vg+V-vC(t)

2Vg

Fig. 4.23 Comparison of tank waveforms of the k=1 continuous and discontinuous conductionmodes.

The inductor voltage and current remain at zero for the rest of the half-switching-period. Inductor

voltage and current waveforms for the k=1 DCM and k=1 CCM are compared in Fig. 4.23.

Hence, the requirement for the k=1 DCM to occur is:

Vg + V – vC(π) = 2Vg – vC(π) > 0 (4-46)

In normalized form, this can be written:


22

1 + M – MC1 > 0 (4-47)

Substitution of Eqs. (4-37) and (4-40) into this expression yields:

J < 4γ (4-48)

This is the basic condition for operation in this mode. It can be seen that the k=1 DCM occurs at

light load.

A simple state plane interpretation

The above arguments can be given a simple

geometrical interpretation in the state plane. As

shown in chapter 3, for the series tank circuit the

state plane trajectories evolve in the clockwise

direction about the applied tank voltage. Consider

the hypothetical state plane trajectories of Fig.

4.24a. At ω0t = π, jL reaches zero and mC = MC1.

The figure is drawn for the case MC1 < (1+M).

Note that a D1 subinterval cannot occur for ω0t >

π, since such a subinterval would involve a

trajectory centered at mC = (1+M), and given either

by hypothetical trajectory A or B. Trajectory A is

impossible, because subinterval D1 cannot occur

except for negative jL. Trajectory B is also

impossible, because it does not travel clockwise

about the center mC = (1+M). Hence, there can be

no D1 subinterval, and instead an X subinterval

occurs (as described in Fig. 4.6) in which mC

remains constant and equal to MC1, as shown in

Fig. 4.24b.

In contrast, a CCM trajectory is shown in

Fig. 4.24c. In this case, MC1 > (1+M), so that for

negative jL the trajectory is able to evolve in the

clockwise direction about the center mC = (1+M).

Thus, the geometry of the state plane

trajectory gives a simple interpretation of the

boundary condition between the continuous and

discontinuous conduction modes.

Q1

Q1

D1

D1

D1

...(1+M)(1-M)

jL

A

B

?

?

mC

mC = MC1at ω0t = π

Q1

...(1+M)(1-M)

jL

mCMC1

X

Q1

. ..(1+M)(1-M)

jL

mCMC1

D1

Q1

D1

(a)

(b)

(c)

Fig.4.24 Hypothetical state planetrajectories: (a) for MC1 < (1+M),trajectories A and B are impossible; (b)actual k=1 DCM trajectory, with MC1 <(1+M); (c) actual k=1 CCM trajectory,with MC1 > (1+M).


23

The k=2 discontinuous conduction mode

In the k=2 discontinuous conduction mode, the tank rings for two complete half-cycles

during each half-period of length Ts/2. After the two complete half-cycles, the output bridge

rectifier diodes become reverse-biased. Waveforms for this mode are given in Fig. 4.25.

For this mode to occur, the switching period must be at least as long as twice the tank

natural period T0. Hence,

γ ≥ 2π, or F ≤ 12

(4-49)

This is a useful mode, because the output is easily controllable: the converter output behaves as a

π π π π

γ γ

ω0t

D1 D2Q2Q1 X XiL

vL=LdiLdt Vg-V-vC

Vg+V-vC 2V

2V2VgVg-V-vC(2π)

vC

VC1

-VC1

VC2

Fig. 4.25 Tank waveforms for the k=2 discontinuous conduction mode.


24

current source, of value controllable by the switching frequency. Also, the switch turn-on and

turn-off transitions both occur at zero current, so switching losses are low. A disadvantage of this

mode is its higher peak transistor currents than in the continuous conduction mode, and hence

higher conduction losses.

As is shown in this subsection, the load current and switching frequency are directly related

for the k=2 DCM, and hence a wide load current specification implies that the switching frequency

must vary over a wide range. This is less of a disadvantage than one might at first think, because

the transformer can be sized to the maximum switching frequency (0.5 f0). Operating point

variations that cause the switching frequency to vary below this value do not necessitate use of a

larger, lower frequency transformer.

Analysis

First, capacitor charge arguments are used to relate the peak tank capacitor voltage to the dc

load current. The tank inductor current iL(t), which coincides with the tank capacitor current, is re

drawn in Fig. 4.26. The total charge contained in the negative portion of the iL(t) waveform is

defined as –q. The peak-to-peak tank capacitor voltage is 2VC2, which represents a change in

capacitor charge of q.

ω0t

iLtotal charge = -q

Fig. 4.26 Tank inductor current waveform, with emphasis on total charge flowingthrough the tank capacitor.

Hence:

2 VC2 = qC

(4-50)

The average output current I is:

I = <| iL |> = 2qTs

(4-51)

Elimination of q from Eqs. (4-50) and (4-51) yields:


25

I = 4CVC2Ts

(4-52)

Normalization of Eq. (4-52) and solution for the normalized peak capacitor voltage MC2 yields:

MC2 = Jγ2

(4-53)

This result is again similar to the k=1 DCM and k=1 CCM cases. It states that the peak tank

capacitor voltage is proportional to the load current.

The state plane diagram for the k=2 DCM is given in Fig. 4.27. The switching period

begins with (mC, jL) = (–MC1,0). Subintervals Q1 and D1 are of length π, and are given by

semicircles in the state plane. Their radii are MC2–(1-M) and MC2–(1+M), respectively, as

indicated on the diagram. Subinterval X ends the half switching period, with (mC, jL) = (+MC1,0).

The converter output characteristics for this mode can be found in a manner similar to that

used for the k=1 CCM. If the converter operates in steady state, then the state plane diagram is

closed, as indicated in Fig. 4.27. The ending point of the D2 subinterval must therefore coincide

with the beginning of the Q1 subinterval. The portion of the state plane in the vicinity of this point

is magnified in Fig. 4.28. It can be seen that, in steady state, the sum of the radii of the D2 and Q1

subintervals is equal to the distance between their centers, or

2 = (MC2 – 1 – M) + (MC2 – 1 + M) (4-54)

...mC

..

jL

.-MC2

(-1-M)

-MC1

X

MC1. .(-1+M) (1-M) (1+M)

MC2 = Jγ2

M

- (1

-M)

C2

M - (1+M)C2

D1

D2

Q2

Q1

X

Fig. 4.27 State plane diagram for the k=2 discontinuous conduction mode.


26

Simplification yields

MC2 = 2 = Jγ2

(4-55)

Now solve for the normalized load current J:

J = 4γ = 4

π F (4-56)

Hence, the load current depends on switching

frequency but not on output voltage, and the

converter behaves as a current source in this mode.

As sketched in Fig. 4.29, the k=2 DCM output

characteristics are straight horizontal lines.

With a resistive load, we have

J = MQ = 4π

F (4-57)

with Q = R/R0. Solution for M then yields the

control plane characteristic:

M = 4π

FQ (4-58)

Thus, the control plane characteristics (M vs. F,

for a linear resistive load) vary linearly with F.

The solution can be completed by solving

for the normalized voltage MC1. By inspection of

the state plane diagram of Fig. 4.27, it can be seen

that MC1 is equal to MC2, minus twice the radius of

the D1 subinterval:

MC1 = MC2 – 2(MC2 – 1 – M) = 2M (4-59)

This result is used in the next section, for

derivation of the mode boundaries.

k=2 DCM boundaries

As noted previously, this mode is restricted

to the frequency range γ ≥ 2π, or F ≤ 0.5. Since,

according to Eq. (4-56), J and F are directly

related, this restriction also places an upper limit on

jL

mC

.(1-M)(-1-M)

.-MC1

2

radius = MC2-1-M radius = MC2-1+M

Fig. 4.28Illustration of relations betweenMC1, MC2 and M.

k=1 M=1 DCM

13

1 M

k=1 CCM

k=2 DCMJ = 4

πF

F = 1/4

F = 1/2

F = 0

2π

J

Fig. 4.29Output plane characteristics,emphasizing k=2 DCM.

13

1

M

F12

14

Q =

0.5

/πQ

= 1

/π

Q = 2

/π

Q = 3/π

Q = 4/π

other modes

k=2 DCMM = 4

π FQ

k=1 DCM; M=1

k=1CCM

Fig. 4.30Control plane characteristics,emphasizing k=2 DCM.


27

the load current J:

J = 4π

F ≤ 2π

(4-60)

Hence, the k=2 DCM is restricted to the portion of the output plane below J = 2/π.

In addition, for the k=2 DCM to occur, the output bridge rectifier (1) must continue to

conduct at ω0t = π (between the Q1 and D1) subintervals, and (2) must become reverse-biased at

ω0t = 2π (after the D1) subinterval. By use of the state plane, it can be seen that, requirement (2)

leads to the constraint:

MC1 > 1 – M (4-61)

and requirement (1) leads to:

MC2 > 1 + M (4-62)

By substitution of Eqs. (4-55) and (4-59), and after a small amount of algebra, these two

constraints become:

1 > M > 13

(4-63)

To summarize, the k=2 DCM boundaries are:

1 > M > 13

2π

> J > 0(4-64)

12 > F > 0

The k=2 DCM output plane boundaries are given in Fig. 4.29, and the control plane characteristics

in Fig. 4.30. As M→1, the converter enters the k=1 DCM, while for M→1/3, the converter enters

a higher-order (k≥2) continuous or discontinuous conduction mode.

4 . 5 . A General Closed-Form Solution

The steady-state solutions for all frequency-controlled continuous and discontinuous

conduction modes are stated here.

Type k CCM

The type k continuous conduction mode occurs over the frequency range

f0k + 1

< fS < f0k (4-65)


28

The output plane characteristics are elliptical, and are described by the equation

M2 ξ2 sin2(γ

2) + 1

ξ2 (Jγ

2 + (-1)k)2 cos2(γ

2) = 1 (4-66)

where ξ is the subharmonic index,

ξ = k + 1 + (-1)k

2(4-67)

The voltage conversion ratio M is restricted to the range

0 ≤ M ≤ 1ξ

(4-68)

At light load, the converter may enter a discontinuous conduction mode.

Typical tank current waveforms are shown in Fig. 4.31. For k even, diode D1 conducts

first, for a fraction of a half resonant cycle. If k is odd, then transistor Q1 conducts first, for a time

CCM WAVEFORMS

type k CCM; k odd

ω0t

Q1

D1 D1

D2

Q2

symmetrical

γ

πππ π

∫ ∫Q1 Q1

(k-1) complete half-cycles

type k CCM; k even

iL

∫ ∫Q1 D2

γ

Q1 Q1

D1 D1 Q2

symmetrical

k complete half-cycles

ππ

ππ

D1

iL

Fig. 4.31 Tank inductor current waveforms for the general type k continuous conductionmode.


29

less than one complete half-cycle. In either case, this is followed by (ξ - 1) complete half-cycles of

ringing. The half-switching-period is then concluded by a subinterval shorter than one complete

resonant half-cycle, in which the device that did not initially conduct is on. The next half switching

period then begins, and is symmetrical.

The steady-state control plane characteristic can be found for a resistive load R by

substituting J = M Q into Eq. (4-66), where Q = R0 / R. Use of the quadratic formula and some

algebraic manipulations yields:

M =

Qγ2

ξ4 tan2

γ2

+ Qγ2

2 (-1)k+1 + 1 +

(ξ2 – cos2

γ2) (ξ4

tan2 γ2

+ Qγ2

2)

Qγ2

2 cos2

γ2

(4-69)

This is the closed-form relationship between the switching frequency and the voltage conversion

ratio, for a resistive load. It is valid for any continuous conduction mode k.

Type k DCM, k odd

The type k discontinuous conduction modes, for k odd, occur over the frequency range

fs < f0k (4-70)

In these modes, the output voltage is independent of both load current and switching frequency,

and is described by

M = 1k

(4-71)

This mode occurs for the range of load currents

2(k + 1)γ

> J > 2(k – 1)

γ(4-72)

In the odd discontinuous conduction modes, the tank current rings for k complete resonant

half-cycles. All four output bridge rectifier diodes then become reverse-biased, and the tank

current remains at zero until the next switching half-period begins, as illustrated in Fig. 4.32.

A dc equivalent circuit for the SRC operating in an odd discontinuous conduction mode is

given in Fig. 4.33; it is a “dc transformer” with effective turns ratio 1:1/k. Converters are not

usually designed to operate in these modes, because the output voltage cannot be controlled by

variation of the switching frequency. Nonetheless, all converters designed to perform below

resonance can operate in one or more odd discontinuous conduction modes if the load current is

sufficiently small, and hence the converter designer should be aware of their existence.


30

0 0

0 0

+

-

+

-

VVg

1 : 1k

IIg

Fig. 4.33 Steady-state equivalent circuit model for an odd discontinuous conduction mode: aneffective dc transformer.

ω0t

Q1

D1

D2

Q2

symmetrical

γ

ππ

∫ ∫Q1


iL

π

X

(a)

X

(b)

1.0

J

M13

15

1

4π

83π12

5π

etc.

k=1DCM

k=5DCM

k=3DCM

Fig. 4.32 General type k discontinuous conduction mode, k odd: a) tank inductor currentwaveform; (b) output characteristics.


31

Type k DCM, k even

The type k discontinuous conduction modes, for k even, occur over the frequency range

fs < f0k (4-73)

These modes have current source characteristics, in which the load current is a function of

switching frequency and input voltage, but not of the load voltage. The output relation is:

J = 2 kγ

(4-74)

Operation in this mode occurs for:

1k – 1

> M > 1k + 1

(4-75)

In the even discontinuous conduction modes, the tank current rings for k complete resonant

half-cycles. All four output bridge rectifier diodes then become reverse-biased, and the tank

current remains at zero until the next switching half-period begins, as illustrated in Fig. 4.34a.

The series resonant converter possesses some unusual properties when operated in an even

discontinuous conduction mode. A dc equivalent circuit is given in Fig. 4.35; it is a gyrator with

gyration conductance g = 2k/γR0. The gyrator has the property of transforming circuits into their

dual networks; in this case, the gyrator characteristic effectively turns the input voltage source Vginto its dual, an output current source of value g Vg [9]. Some very large converters have been

designed to operate purposely in the k = 2 DCM at light load [ref Hughes] and even at full load

[10].


32

ω0t

Q1

D1 D1

D2

Q2

symmetrical

γ

πππ

∫ ∫Q1


iL

π

X

(a)

(b)

1.0

J

2π

M

F = 0.4

F = 0.25

F = 0.1

k=2

DCM

F = .2

F=.05

k=

4 D

CM

13

15

etc.

1

Fig. 4.34 General type k discontinuous conduction mode, k even: (a) tank inductor currentwaveform; (b) output characteristics.

0 0

0 0

+

-

+

-

VVg

Ig = gV I = gVg

g

g = 2kγR0

Fig. 4.35 Steady-state equivalent circuit model for an even discontinuous conduction mode: aneffective gyrator. The converter exhibits current source characteristics.


33

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

Q=0.2

Q=0.35

Q=0.5

Q=0.75

Q=1

Q=1.5

Q=2

Q=3.5Q=5Q=10Q=20

F

MQ=0.2

Q=0.35

Q=0.5

Q=0.75Q=1

Q=1.5

Q=2Q=3.5

Q=5

Q=10Q=20

Fig. 4.36. Complete control plane characteristics of the series resonant converter, for the range0.2 ≤ F ≤ 2.

M

F

k=0 CCM k=1 CCM

k=2 DCM

k=1 DCM

k=3 DCM

k=5 DCM

k=4 DCM

k=3

CC

M

k=2

CC

M

1

1

12

13

14

15

etc.

12

13

15

Fig. 4.37. Complete control plane characteristics for continuous and discontinuousconduction mode boundaries.


34

Composite characteristics

The complete control plane characteristics can now be plotted using Eqs. (4-65) - (4-75).

The result is shown in Fig. 4.36, and the mode boundaries are explicitly diagrammed in Fig. 4.37.

It can be seen that, for operation above resonance, the only possible operating mode is the k=0

CCM, and that the output voltage decreases monotonically with increasing switching frequency.

Reduction in load current (or increase in load resistance, which decreases the Q) causes the output

voltage to increase. A number of successful designs that operate above resonance and utilize zero-

voltage switching have been documented in the literature.

Operation below resonance is complicated by the presence of subharmonic and

discontinuous conduction modes. The k=1 CCM and k=2 DCM are well behaved, in that the

output voltage increases monotonically with increasing switching frequency. Increase in load

current again causes the output voltage to decrease. Successful designs which operate in these

modes and employ zero-current switching are numerous. However, operation in the higher-order

modes (k=2 CCM, k=4 DCM, etc.) is normally avoided.

Given F and Q, the operating mode can be evaluated directly, using the following

algorithm. First, the continuous conduction mode k corresponding to operation at frequency F

with heavy loading is found:

k = INT 1F (4-76)

where INT(x) denotes the integer part of x. Next, the quantity k1 is determined:

k1 = INT 12

+ 14

+ Qπ2F

(4-77)

The converter operates in type k CCM provided that:

k1 > k (4-78)

Otherwise, the converter operates in type k1 DCM. A computer listing is given in Appendix 1, in

which the conversion ratio M is computed for a given F and Q. First, the above algorithm is used

to determine the operating mode. Then, the appropriate equation (4-69), (4-71), or (4-74) is

evaluated to find M. The function works correctly for every frequency-controlled mode.

Output plane characteristics for the k=0 CCM, plotted using Eq. (4-66), are shown in Fig.

4.38. The constant-frequency curves are elliptical, and all pass through the point M=1, J=0.

Output plane characteristics which combine the k=1 CCM, k=1 DCM, and k=2 DCM are shown in

Fig. 4.39. These were plotted using Eqs. (4-66), (4-71), and (4-74). It can be seen that the

constant-frequency curves are elliptical in the continuous conduction mode, vertical (voltage source

characteristic) in the k=1 DCM, and horizontal (current source characteristic) in the k=2 DCM.


35

M

J

0

1

2

3

4

5

6

0 0.2 0.4 0.6 0.8 1

F = 1.30

F = 1.15

F = 1.10

F = 1.07

F = 1.05 F = 1.01

Fig. 4.38 Output characteristics in the k=0 continuous conduction mode (above resonance).

M

J

0

1

1.5

2

2.5

3

0 0.2 0.4 0.6 0.8 1

2π

F = .5

F = .75

F = .85

F = .90

F = .93F = .96

F = 1.0

F = .25F = .1

k=2 DCM

k=1 CCM

k=1 D

CM

4π

Fig. 4.39 Composite output characteristics for the k=1 CCM, k=1 DCM and k=2 DCM modes.Mode boundaries are indicated by heavy lines.


36

4 . 6 . Converter Losses

If the transistors and diodes can be modeled as constant voltage drops while they conduct,

then there is a relatively easy way to adapt the ideal analysis of the previous sections to include

these losses. The transistor and diode forward voltage drops cause a different voltage to be applied

to the tank circuit during each subinterval, which can be accounted for by changing the effective Vg

and V seen by the tank circuit.

Let us consider the half-bridge

example of Fig. 4.40. The semiconductor

device forward voltage drops are defined

as follows:VT transistor on-state voltage

VDin antiparallel diode forward voltage

VDout output diode forward voltage

The applied tank voltage waveform vT is

shown in Fig. 4.41 for the k=1 CCM.

Since the applied tank voltage is (by

assumption) constant during each

subinterval, the normalized state plane trajectories are again a series of circular arcs, and the

preceding analytical method can be used to solve this converter. In fact, it is not necessary to

rederive the entire analysis, because the converter circuit of Fig. 4.42 (in which the input and

output voltages are modified but the converter is otherwise ideal) exhibits exactly the same tank

voltage waveform, vT of Fig. 4.41. In consequence, the ideal k=1 CCM solution previously

derived, Eqs. (4-15) and (4-28), also apply to this converter, but with the input and output

voltages replaced by the effective values:

Vg eff = 12 VD in – 1

2 VT + actual Vg (4-79)

+V-

VDout

VDin

VT+ -

Vg

Vg

VT

+

-

Fig. 4.40Half-bridge circuit with nonidealsemiconductor devices.

Vg -VT -V-2VDout

Vg +VDin +V+2VDout

-Vg +VT +V+2VDout

-Vg -VDin -V-2VDout

Q1

Q2 D2

D1

VT

ω0t

Fig. 4.41 Actual tank voltage waveforms for the half-bridge circuit of Fig. 4.40.


37

Veff = 2 VD out + 12 VD in + 1

2 VT + actual V

(4-80)

Note that these effective values must also be substituted into the normalization base quantities. The

result is given below:

Meff = VeffVg eff (4-81)

Jeff = I R0Vg eff (4-82)

Meff2 ξ2

sin2(γ2) + 1

ξ2 (Jeff γ

2 + (-1)k)2 cos2(γ

2) = 1

(4-83)

The effect of the semiconductor forward voltage drops is primarily to shift the output plane

characteristics somewhat to the left, and hence to lower the actual output voltage.

It is also possible, but much more complicated, to account for resistive losses such as tank

inductor and capacitor equivalent series resistance (esr), shown in Fig. 4.43. These losses

effectively damp the tank circuit, and cause the normalized state plane trajectories to become spirals

of decreasing radius. Closed-form solutions are not known; instead, computer iteration can be

used to evaluate the highly transcendental equations that are obtained. The results show that the

converter is very sensitive to small resistive loss elements when operated at large values of

normalized current J. This coincides with operation near resonance with a large Q = J/M. The

+

V-

VT+ -

+-

+-

+-

+- +

-+-+-

actual

idealswitches

ideal diodes(no forward drop)

actual Vg

actual Vg

12

VDin

12

VDin

12

VT

12

VT

12

VT

12

VDin

12

VDout

Fig. 4.42 Half-bridge circuit with explicit voltage drops due to non-idealdiodes and transistors.


38

same conclusions can be obtained more simply using the approximate sinusoidal methods of

chapter 2.

Vg

RS RS

RSRS

+V_

RDoutRDout

RDout

RDoutesrL CRe

Fig. 4.43 Half-bridge circuit with explicit resistances due to non-ideal diodes and switches.

4 . 7 . Design Considerations

It may not be initially apparent how to best design a series resonant converter to meet given

design objectives, yet a sub-optimal design may exhibit significantly higher tank current and

voltage stresses than necessary, and/or may cause the switching frequency to vary over an

unacceptably wide range. In this section, component stresses are analyzed, and the results are

overlaid on the output plane plots derived in the previous sections. Also, specifications on the

input voltage and output power ranges are translated into a region in the same output plane, so that

it can be seen how component stresses and switching frequency vary with operating point. One

can then choose the transformer turns ratio and tank characteristic impedance, as well as tank

inductance and capacitance, to obtain a good converter design. A 600W full bridge example is

given.

Approximate peak component stresses

In the continuous conduction mode, the tank inductor current and tank capacitor voltage

peak magnitudes are both approximately proportional to the load current, and nearly independent of

the converter input and output voltages. This is true because the dc load current I is equal to the

average rectified tank inductor current <| iL |>. With a 1:n transformer turns ratio, we have


39

I = 1n <| iL |> (4-84)

To the extent that the waveshape of iL(t) does not

vary with converter operating point, there is a

direct relationship between the peak value of iL and

its average rectified value <|iL|> = nI. Indeed this

is the case near resonance, where iL is nearly

sinusoidal. Equation (4-84) then becomes:

I = 1n Ts

ILP | sin ωt | dt0

Ts

= 1n 2ILPπ

(4-85)

Hence:

ILP ≅ nπ2

I(4-86)

Or, in normalized form, one obtains:

JLP ≅ π2

J(4-87)

Thus, we expect the peak tank current ILP to be

directly proportional to the load current I, and

essentially independent of the load voltage. We

can plot contours of constant peak tank current in

the output plane, to see how component stresses

vary with operating point; Eq. (4-87) predicts that

such contours should appear as in Fig. 4.44(c).

The approximation used above coincides

with the sinusoidal approximation used in chapter

2. It is accurate for operation in the continuous

conduction mode, near resonance. However, in

the discontinuous conduction mode, or in the

continuous conduction mode near the DCM

boundaries, the tank current is highly

nonsinusoidal, and Eqs. (4-84) - (4-87) become

poor approximations.

The tank capacitor voltage waveform is directly related to the tank inductor current, since

these components are connected in series. Hence, the peak tank capacitor voltage is also directly

proportional to the load current, and is essentially independent of the load voltage. In the case of

|iL(t)|

ILP

I = <|iL|>

ILP = π2

I

(b)

JLP = 2

JLP = 3

JLP = 4

J

M1

3

2

1

(c)

1:n

L C

+ vC -

Cf V

+

–

1n |iL| I

I = 1n < | iL | >

(a)

i L

Fig. 4.44 Peak stress approximation where(a) series resonant converter has a 1:ntransformer turns ratio; (b) dc loadcurrent equals average rectified tankinductor current; (c) result shows thatpeak tank current is directly proportionalto load current.


40

sinusoidal tank waveforms, the peak tank capacitor voltage VCP is related to ILP through the

characteristic impedance R0:

VCP ≅ ILP R0 ≅ nπ2

I R0 (4-88)

Discussion

Suppose that we have constructed a converter which is capable of producing the given rated

output power Pmax at the normalized operating point M = 0.5, J = 5. This converter could also

produce twice the rated power, or 2Pmax, by operating at the point M = 1.0, J = 5; i.e., by

doubling the output voltage without changing the load current. This is true because the peak

component stresses are independent of M and depend only on J. The component currents will be

nearly the same in both cases, and hence the losses will be almost the same also.

Furthermore, a converter with lower peak currents could be constructed by halving the

transformer turns ratio n, so that the converter operates at rated output voltage and power with M =

1. Equation (4-84) predicts that this would cause the peak tank current to be reduced by a factor of

two. Component stresses and losses would be reduced accordingly.

The conclusion is that the converter should be designed to operate with M as close to unity

as other considerations will allow. This implies that the transformer turns ratio should be

minimized, and it leads to low peak tank and transistor currents. Converters designed to operate

with lower-than-necessary values of M do not fully utilize the power components.

Exact peak component stresses

Exact expressions for the peak tank current and voltage can be derived using the state plane

diagram. For example, the state plane diagram for the k=1 CCM is reproduced in Fig. 4.45, with

the peak values of jL and mC identified. It can be seen that the peak normalized tank inductor

current is given by the circle radius during the Q1 conduction subinterval:

JLP = J γ2

– 1 + M (4-89)

The peak normalized tank capacitor voltage MCP, shown in Fig. 4.45, was previously found. It is:

MCP = MC1 = J γ2

(4-90)

We wish to plot these component stresses in the output plane to determine how stresses

vary with operating point. To do so, we need to express JLP and MCP as functions of J and M,

with γ eliminated. The easiest way to do this is to solve for γ/2 in Eqs. (4-89) and (4-90):


41

γ2

= JLP + 1 – M

J(4-91)

γ2

= MCPJ

(4-92)

These expressions are then inserted into the converter output characteristic, Eq. (4-15), to eliminate

γ. Solving for J, one then obtains

J = JLP + 1 – M

π – tan–1 (JLP – M)2 – 1

1 – M2

(4-93)

J = MCP

π – tan–1 (MCP – 1)2 – 1

1 – M2

(4-94)

Here, it is necessary to use care to select the correct branch of the arc tangent function. When the

denominator is written as shown, then the correct answer will be obtained when the arc tangent

function is defined to lie in the domain – π/2 ≤ tan–1(•) ≤ + π/2.

Equations (4-93) and (4-94) can now be used to overlay contours of constant peak tank

stress on the output plane characteristics. The result is given in Fig. 4.46. It can be seen that the

contours are nearly horizontal lines, and do not differ much from Fig. 4.44.

.. 1 + M.mC

jL

-1 + M

1 - M. .

-1 - M.MC1 =

Jγ2

peak value of mC:

mCP = MC1 = Jγ2

D2

Q2

Q1

D1

Jγ2 - 1 + M

jLP

peak value of jL:

jLP = Jγ2

- 1 + M

(circle radius during Q1 conduction interval)

Fig. 4.45 State plane diagram for k=1 CCM.


42

Similar analysis can be used to derive the component stresses for above resonance

operation. The results for the k=0 CCM are

J = MCP

tan–1 (MCP + 1)2 – 1

1 – M2

(4-95)

J = JLP – 1 + M

tan–1 (JLP + M)2 – 1

1 – M2

, 1 – M2

M < JLP

J =

–1 + 1 + JLP2

1 – M2

tan–1 JLP

1 – M2

, 1 – M2

M > JLP

(4-96)

M

J

0

1

2

3

4

5

0 0.25 0.5 0.75 1

F=0.975

F=0.95

F=0.9

F=0.8

F=0.7MCP=3

MCP=5

MCP=7

JLP=7

JLP=5

JLP=3

F=0.6

F=0.5

JLP=1.5

MCP=2

Fig. 4.46 Superposition of peak tank current and voltage stress curves on the normalizedoutput characteristics, for the k=1 continuous conduction mode (belowresonance). Solid lines: curves of constant switching frequency; dashed lines:contours of constant peak tank capacitor voltage; shaded lines: contours ofconstant peak tank inductor current.


43

Two cases occur in Eq. (4-96), depending on whether the peak occurs at the beginning of the

switching period or at some time in the middle of the period. Equations (4-95) and (4-96) can now

be used to generate the plot of Fig. 4.47, where peak tank stresses are overlaid on the converter

output characteristics for the k=0 CCM.

Use of the output plane

In a typical voltage regulator design, the output voltage is regulated to a given constant

value V. The input voltage Vg and output power P, as well as the output current I = P/V, vary

over some specified range:

Vgmax ≥ Vg ≥ Vgmin (4-97)

Pmax ≥ P ≥ Pmin (4-98)

M

J

0

1

2

3

4

5

0 0.25 0.5 0.75 1

F=1.05

F=1.07

F=1.1

F=1.15

F=1.3

F=1.5

MCP =1

MCP =3

MCP =5

MCP =7

JLP=7

JLP=5

JLP=3

JLP=1

Fig. 4.47. Superposition of peak stress curves on the normalized output plane, aboveresonance (k=0 CCM). Solid curves are contours of constant switchingfrequency, dashed curves are contours of constant peak tank capacitor voltage,and shaded curves are contours of constant peak tank inductor current.


44

where Pmax = V Imax and Pmin = V Imin

To regulate the output voltage, the controller will vary the switching frequency fs over some range,

and consequently the operating point will vary. It is desired to choose the transformer turns ratio

and the values of tank inductance and capacitance such that a good design is obtained, in which the

tank capacitor voltage and inductor current are low, the range of switching frequency variations is

small, and the transformer and tank elements are small in size.

Note that the specifications (4-97) and (4-98) do not, by themselves, determine the

operating region of the normalized operating plane. By changing the tank characteristic impedance

R0 and transformer turns ratio n, the operating region can be moved to any arbitrary range of M

and J. This follows from the definitions of M and J:

M = Vn Vg (4-99)

Here, V and the range of Vg are specified, but the transformer turns ratio n is not. So we can

choose n, and hence also scale the range of M, arbitrarily, subject to M ≤ 1. Also:

J = n R0 IVg

(4-100)

The range of I and Vg are specified, and n is given by the choice of M from Eq. (4-99) above. But

we can still choose the tank characteristic impedance R0 arbitrarily, and hence we can also scale the

range of J arbitrarily. So selection of an operating region, or M and J, is equivalent to choosing n

and R0.

Let us, therefore, map Eqs. (4-97) and (4-98) into the normalized output plane, for some

given values of n and R0. Equations (4-97) and (4-99) imply that, as the input voltage varies from

its maximum to minimum values, the voltage conversion ratio will also vary from Mmin to Mmax,

where:

Mmin = Vn Vgmax

Mmax = Vn Vgmin

(4-101)

The output power can be written:

P = I V = J Vg Vn R0

= JM

V2

R0

(4-102)

The expression on the right of Eq. (4-102) is the most useful for design because Vg has been

eliminated in favor of the constant specified value of V. Solution for J yields:

J = M n2 R0 P

V2(4-103)


45

For a given value of power P, J is proportional to M: if the input voltage varies (causing M to vary

as given by Eq. (4-99) then J will also vary as given by Eq. (4-103). The peak J is given at

maximum power and minimum input voltage:

Jmax = n Pmax R0Vgmin V

(4-104)

while the minimum J occurs at minimum power and maximum input voltage:

Jmin = n Pmin R0Vgmax V

(4-105)

The region defined by the specifications can be

plotted using Eqs. (4-101) and (4-103) - (4-105),

as in Fig. 4.48. This region can be overlaid on the

converter output characteristics, to see how

switching frequency and component stresses must

vary as the operating point changes. For the k=0

CCM example shown, the design (Mmax, Jmax) =

(0.8, 4) has been selected. Given values for Pmax,

Pmin, Vgmax, and Vgmin, the values of Mmin and

Jmin can be evaluated, and the operating region is

constructed. It can be seen that the maximum

switching frequency will occur at the point (Mmin,

Jmin), i.e., at maximum input voltage Vgmax and

minimum power Pmin. The component stresses are

highest at maximum power Pmax.

600W design example

The preceding arguments are next applied to the example of a 500W off-line full-bridge dc-

dc converter. The converter specifications are

input voltage range 255 ≤ Vg ≤ 373

output power range 60 ≤ Pout ≤ 600

maximum switching frequency fsmax = 1 MHz

regulated output voltage V = 24

Designs operating below resonance, for various choices of Mmax and Jmax, are compared in Table

4.1. Given the specifications and the choice of Mmax and Jmax, values of Mmin and Jmin can be

computed from Eqs. (4-101), (4-104), and (4-105). The converter operating region can then be

superimposed on the converter characteristics, as in Fig. 4.48, and the normalized peak stresses

Jmin

J

M

Jmax

Mmin Mmax

P = Pm

ax

P = PminVg

= V

gmax

Vg

= V

gmin

Fig. 4.48Voltage regulator operatingregion, plotted in the output plane.


46

and switching frequency variations can be determined graphically. Alternatively, the exact

equations can be evaluated to find these quantities. The tank resonant frequency f0 is then chosen

such that the maximum switching frequency, which occurs at the point (Mmax, Jmax) for below

resonance operation, coincides with 1 MHz. The required transformer turns ratio n is found by

evaluation of Eq. (4-101), and the required tank characteristic impedance R0 can then be found

using Eq. (4-104). Values of L, C, ILP, and VCP can then be evaluated.

Table 4.1. Comparison of designs, below resonancepoint Mmax Jmax Mmin Jmin fsmin

kHzf0,MHz

L,µH

C,nF

ILP,A

VCP,V

mode

A 0.9 5.0 0.62 0.34 280 1.04 75 0.3 4.22080 k=1 CCM,k=2 DCM

B 0.9 1.5 0.62 0.10 95 1.2 20 0.9 4.7 710k=1 CCM,k=2 DCM

C 0.9 0.5 0.62 0.068 68 2.5 3.0 1.2 9.9 510k=2 DCM

D 0.45 1.5 0.31 0.068 104 1.3 9.0 1.6 8.7 780k=1 CCM,k=2 DCM

All of the designs of Table 4.1 operate in the k=2 discontinuous conduction mode at light

load; design C operates in DCM even at full load. It is assumed that the transistors are allowed to

conduct only once per half-switching-period, and hence the converter operates in k=2 DCM for all

switching frequencies below 0.5 f0, and for all values of J below 2/π. To ensure that the rated

output voltage can be obtained when semiconductor forward voltage drops and other losses are

accounted for, the maximum allowable value of M is taken to be 0.9 rather than 1.0. These

designs all operate with zero current switching.

Point A, (Mmax, Jmax) = (0.9, 5.0), exhibits the lowest peak current, 4.2A referred to the

transformer primary. However, the peak tank capacitor voltage is very high, 2080V referred to the

primary side. This voltage can be reduced by reducing the choice of Jmax, for example, to 1.5 as

in point B. This has little effect on the peak current, but the peak capacitor voltage is reduced to

710V. Further reductions in Jmax further reduce the peak capacitor voltage to some extent;

however, the peak tank current is increased as the discontinuous conduction mode boundary is

approached. At point C, where the converter operates in DCM for all load currents, the peak tank

current is approximately twice that of points A and B. Point D illustrates the effect of a suboptimal

choice of Mmax, 0.45 rather than 0.9; this also approximately doubles the peak current.

Several designs for the same specifications are illustrated in Table 4.2 for the above

resonance case. No discontinuous conduction mode occurs in this case. The design procedure is

similar to the below resonance case, except that the maximum switching frequency now occurs at

minimum output power and maximum input voltage, i.e., at the point (Mmin, Jmin). These designs

operate with zero voltage switching.


47

Table 4.2. Comparison of designs, above resonancepoint Mmax Jmax Mmin Jmin fsmin

kHzf0,

kHzL,µH

C,nF

ILP,A

VCP,V

mode

A 0.9 5.0 0.62 0.34 511 495 157 0.6 4.0 1940 k=0CCM

B 0.9 1.5 0.62 0.10 220 200 117 5.4 3.9 545k=0CCM

C 0.9 1.0 0.62 0.068 157 137 113 12 3.8 350k=0CCM

D 0.9 0.2 0.62 0.014 45 28 111 290 3.9 50 k=0CCM

E 0.45 1.0 0.31 0.068 133 95 81 34 8.8 287k=0CCM

Point A is again at Mmax = 0.9, Jmax = 5.0. The design is not significantly different from

the corresponding design below resonance. The peak current is again approximately 4A, and the

peak tank capacitor voltage is again very high (approximately 2kV). Reducing Jmax has the

favorable effect of reducing this voltage. In fact, since there is no discontinuous conduction mode,

VCP can be made arbitrarily small by reducing Jmax a sufficient amount. The peak tank inductor

current is slightly smaller than in the below-resonance case. The problem with doing so is the

large range of switching frequency variations. For example, point D illustrates a design with a

peak capacitor voltage of 50 volts and peak current of 3.9A; the problem with this design is the

minimum switching frequency of 45kHz. This design will require quite large magnetics. A

compromise is point C, for which the peak capacitor voltage is 350V, with low peak inductor

current of 3.9A and a minimum switching frequency of 157kHz. Point E illustrates again the

effect of a suboptimal choice of Mmax; this leads to increased peak tank current.

It is apparent that, for operation above resonance, large switching frequency variations can

occur. This is undesirable because it requires that the transformer and filter components be sized to

the minimum switching frequency, which can be quite low. The benefits of high frequency

operation are then lost. Several control schemes have been described in the literature which

circumvent this problem. Constant frequency operation can be obtained by duty cycle [7] or phase

control [13-15] of the transistor bridge.

Large switching frequency variations can also occur below resonance, but these do not lead

to a large transformer. The reason for this is the occurence of the k=2 DCM at all switching

frequencies below 0.5 f0. In this mode, no voltage is applied to the transformer during the

discontinuous (X) subintervals. The transformer can therefore be designed as if its minimum

frequency is 0.5 f0, and the low values of fsmin in Table 4.1 are not a problem. Filter size is also

not adversely affected by the wide range of switching frequency variations in this case, because the

low frequency operating points coincide with the low output current points, where less filtering is

needed. The output capacitor value is therefore determined by the maximum power point, which

occurs at high frequency.


48

REFERENCES

[1] F.C. Schwarz, “An Improved Method of Resonant Current Pulse Modulation for Power Converters,” IEEEPower Electronics Specialists Conference, 1975 Record, pp. 194-204, June 1975.

[2] R. King and T. Stuart, “A Normalized Model for the Half Bridge Series Resonant Converter,” IEEETransactions on Aerospace and Electronic Systems, March 1981, pp. 180-193.

[3] V. Vorperian and S. Cuk, “A Complete DC Analysis of the Series Resonant Converter,” IEEE PowerElectronics Specialists Conference, 1982 Record, pp. 85-100, June 1982.

[4] R. King and T.A. Stuart, “Inherent Overload Protection for the Series Resonant Converter,” IEEETransactions on Aerospace and Electronic Systems, vol. AES-19, no. 6, pp. 820-830, Nov. 1983.

[5] R. Oruganti and F.C. Lee, “Resonant Power Processors, Part 1: State Plane Analysis,” IEEE Transactionson Industry Application, vol. IA-21, Nov/Dec 1985, pp. 1453-1460.

[6] A. Witulski and R. Erickson, “Steady-State Analysis of the Series Resonant Converter,” IEEE Transactionson Aerospace and Electronic Systems, vol. AES-21, no. 6, pp. 791-799, Nov. 1985.

[7] Steven G. Trabert and Robert W. Erickson, "Steady-State Analysis of the Duty Cycle Controlled SeriesResonant Converter,” IEEE Power Electronics Specialists Conference, 1987 Record, pp. 545-556, (IEEEPublication 87CH2459-6).

[8] R.L. Steigerwald, “High Frequency Resonant Transistor Dc-Dc Converters,” IEEE Transactions onIndustrial Electronics, vol. IE-31, no. 2, pp. 181-191, May 1984.

[9] S. Singer, “Loss-Free Gyrator Realization,” IEEE Transactions on Circuits and Systems, vol. CAS-28,no. 2, pp. 26-34, January 1988.

[10] Chambers, 15kHz 35kW high voltage converter using $15 SCR’s.

[11] Y. Cheron, H. Foch, and J. Salesses, “Study of a Resonant Converter Using Power Transistors in a 25kWX-ray Tube Power Supply,” IEEE Power Electronics Specialists Conference, Proceedings ESA Sessions,pp. 295-306, June 1985.

[12] A. Witulski and R. Erickson, “Design of the Series Resonant Converter for Minimum Component Stress,”IEEE Transactions on Aerospace and Electronic Systems, July 1986.

[13] SRC phase control reference Vandelac PESC 87

[14] F.S. Tsai, P. Materu, and F.C. Lee, “Constant Frequency, Clamped Mode Resonant Converters,” IEEEPower Electronics Specialists Conference, 1987 Record, pp. 557-566, June 1987.

[15] SRC phase control ref GE scheme (high voltage converter) PESC 90

[16] K.D.T. Ngo, “Analysis of a Series Resonant Converter Pulsewidth-Modulated of Current-Controlled forLow Switching Loss,” IEEE Power Electronics Specialists Conference, 1987 Record, pp. 527-536, June1987.


49

PROBLEMS

1 . Using the phase plane, derive the k=0 continuous conduction mode characteristics. Sketch your results in theoutput plane (M vs. J) and control plane (M vs F for resistive load).

2 . Analyze the k=3 discontinuous mode:

a) Draw the phase plane diagram;

b) Draw waveforms for the inductor current, capacitor voltage, and inductor voltage;

c) Use tank capacitor charge arguments to relate the normalized load current to normalized capacitorvoltage boundary values;

d) Solve for the output characteristics;

e) Determine the complete set of conditions on normalized switching period and load current whichguarantee operation in this mode.

3 . It is desired to obtain a converter with current source characteristics. Hence, a series resonant converter isdesigned for operation in the k=2 discontinuous mode. The switching frequency is chosen to be fs = 0.225

fo, where fo is the tank resonant frequency (consider only open-loop operation). The load R is a linear

resistance whose value can change to any positive value.

a) Plot the output characteristics (M vs J), for all values of R in the range [0,∞]. Label modeboundaries, evaluate the short-circuit current, and give analytical expressions for the outputcharacteristics.

b) Over what range of R (referred to the tank characteristic impedance Ro) does the converter operate as

intended, in the k=2 discontinuous mode?

4 . Derive the equations for the peak tank stresses in the k=0 continuous conduction mode, Eqs. (4-95) and (4-96).

5 . Design of a high density series resonant converter

L C

CF

vinI

V o ut

Design a half-bridge series resonant converter, as shown above, to meet the following specifications:

Input voltage Vin: 134–176 volts

Output voltage Vout: 48 volts

Output current I: 1–10 amperesMaximum switching frequency: 750 kHzOutput voltage ripple: no greater than 1 volt peak-to-peak


50

Design the “best” converter that you can, which combines high efficiency with small volume. Use yourengineering judgement to select the operating mode, tank elements L and C, and transformer turns ratio to attainwhat you consider to be the best combination of small transformer size, small CF, high minimum switching

frequency, low peak tank capacitor voltage, and low peak transistor current.The volume of a 50V, 1µF, X7R ceramic chip capacitor is approximately 50 mm3. Capacitor volume

scales as the product of capacitance and voltage rating, so that the volume of a 100V 5µF capacitor of the samedielectric material is 500 mm3. Capacitors are available with voltage ratings of 50V, 75V, 100V, 200V, 300V,400V, and 500V, and essentially any capacitance value. You must choose a capacitor with a voltage rating at least25% greater than the actual peak voltage applied by your design.

Use a ferrite EE core for the transformer. Estimate the transformer size using the Kg method, allowing a

fill factor Ku of 0.5, total copper loss Pcu no greater than 0.5W, and peak flux density Bmax of 0.1T. The core

geometrical constant Kg of the center-tapped transformer is defined as follows:

Kg = ρ λ1

2 i12 108

Pcu Ku Bmax2

cm5

where ρ is the resistivity of copper (1.724·10-6Ω·cm), λ1 is the applied primary volt-seconds, and i1 is the applied

primary rms current.For this problem, you may neglect core loss. This is not valid in general, especially for 750kHz

transformers, but is a simplifying assumption for this problem. The estimated volumes of transformers constructedin this manner are as follows:

Core type vol., mm 3 K g , cm 5

EE12 675 0.73·10-3

EE16 1125 2.0·10-3

EE19 1550 4.1·10-3

EE22 2425 8.3·10-3

EE30 8575 86·10-3

EE40 14700 0.21

EE50 31500 0.91EE60 42250 1.4

Hence, attempt to minimize the total volume Vtot of the transformer, output filter capacitor, and tank capacitor,

while keeping the peak currents reasonably low. You may neglect the size of all other components.Specify: (1) your choices for L, C, and transformer turns ratio; (2) the range of M, J, and fs over which

your design will operate; (3) the transformer core size required (don’t bother to compute number of turns or wiresize); (4) the value of CF required; and (5) the total volume Vtot as defined above.

Documents

T SERIES RESONANT CONVERTERecee.colorado.edu/~ecen5817/notes/ch4.pdf · he objective of this chapter is to describe the operation of the series resonant converter in ... turns ratio