Chapter 4. Series resonant converter

Chapter 4
THE SERIES RESONANT CONVERTER
T
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.
Vg
Q3
D3
D1
Q4
D4
D2
+
-
Q1
iS
+
Q2 vS
+
vT
L
C
Fig. 4.1.
Series resonant converter schematic.
D5
I
D6
+
vR D7 CF
V
-
-
+
D8
R
Principles of Resonant Power Conversion
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
(a)
The applied tank voltage is therefore constant and
equal to Vg–V. In normalized form, one obtains
MT = VT = 1 – M
Vg
(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
+
vT
-
+ vC Vg +-
iL > 0
L
direction of
current flow
C
CF
+
V
R
-
vT = Vg - V
(b)
jL > 0
.
would be changed, and an arc centered at a
MT = 1 - M
1 mC
different location would be obtained. A subinterval
utilizing the switch conduction state described Fig. 4.2 Q1 conduction subinterval, in
which Q1 and Q4 conduct, and iL > 0 so
above and in Fig. 4.2 is referred to in shorthand
that D5 and D8 conduct: a) circuit;
form as subinterval Q1.
b) normalized state plane trajectory.
2
Chapter 4.
(a)
Vg +-
+
iL < 0
vT
-
L
C
direction of CF
current flow
+
(a)
+ vC -
V
iL < 0
Vg +-
-
.
L
+
C
CF
V
-
vT = -Vg + V
jL
(b)
jL < 0
1
-
direction of
current flow
vT = Vg + V
(b)
jL
vT
+ vC -
+
R
The Series Resonant Converter
-1
MT = 1 - M
.
MT = -1 + M
mC
mC
jL < 0
Fig. 4.3
D1 conduction subinterval, in
which iL < 0 such that diodes D1, D4,
D6, and D7 conduct: a) circuit; b)
normalized state plane trajectory.
+
vT
(a)
+ vC -
Vg +-
iL > 0 L
C
CF
direction of
current flow
+
V
-
VT = -Vg - V
(b)
Fig. 4.4 Q2 conduction subinterval, in
which Q2 and Q3 conduct, and iL < 0 so
that D6 and D7 conduct: a) circuit;
b) normalized state plane trajectory.
Many other switch conduction states can
occur. Subinterval D1 is similar to subinterval Q 1,
except that the tank current iL(t) is negative. The
conducting devices are antiparallel diodes D1 and
D 4, and output rectifier diodes D6 and D 7. The
applied tank voltage is therefore VT = Vg+V, or in
normalized form,
.
jL > 0
MT = 1 + M
(4-4)
The circuit and state plane trajectory for this
mC subinterval are summarized in Fig. 4.3. Note that
MT = -1 - M -1
this switch conduction state can only occur in the
negative half-plane (jL < 0).
Fig. 4.5 D2 conduction subinterval, in
Symmetrical switch conduction states Q2
which iL > 0, such that diodes D2, D3,
(Fig. 4.4) and D2 (Fig. 4.5) can also occur, in
D5 and D8 conduct: a) circuit; b)
normalized state plane trajectory.
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).
3
Principles of Resonant Power Conversion
(a)
+
vT
+
(a)
-
iL > 0 L
C
open
circuit
-
o
oo
Vg +-
vT
iL > 0
+
Vg +-
V
L
C
+
V
direction of
current flow
-
-
oo
vT = -V
(b)
(b)
jL
.
jL = 0
jL
jL > 0
.
mC
mC does not change
mC
MT = -M
Fig. 4.6
Subinterval X, in which all four
rectifier diodes D5, D6, D7 and D8 are
reverse-biased. The inductor current
remains at zero, and the tank capacitor
voltage does not change: a) one possible
circuit topology; b) normalized state
plane trajectory.
Fig. 4.7
Subinterval P1, in which D2 and
Q4 (or Q1 and D3) conduct, and iL > 0 so
that D5 and D8 also conduct: a) circuit;
b) normalized state plane trajectory.
+
vT
(a)
Under certain conditions, it is possible
for all four uncontrolled rectifier diodes (D5, D 6,
+
D 7, D8) to become simultaneously reverse- Vg 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 (b)
initial value. This switch conduction state is
denoted “X”.
When phase control is used, two other
subintervals can occur: P 1 , which occurs for
i L > 0, is summarized in Fig. 4.7, and P 2 ,
which occurs for iL < 0, is summarized in Fig.
-
+ vC -
o
iL < 0
L
+
C
direction of
current flow
CF
V
-
vT = V
jL
.
MT = M
mC
jL < 0
4.8.
An operating mode is defined by a Fig. 4.8 Subinterval P2, in which D1 and Q3
(or Q2 and D4) conduct, and iL < 0 so that
sequence of subintervals which combine to form
D6 and D7 conduct: a) circuit; b)
normalized state plane trajectory.
a complete switching period. Discontinuous
4
Chapter 4.
The Series Resonant Converter
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.
5
Principles of Resonant Power Conversion
4 . 2 . State Plane and Charge Arguments for the k=1 Continuous Conduction
Mode
i L(t)
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 Q1
and Q4 off, and Q2 and Q3 are switched on.
Subinterval Q2 begins, and as shown in Fig.
ω0 t
v C (t)
+VC1
-VC1
v T(t)
Vg + V
Vg - V
-Vg + V
-Vg - V
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
6
ωT
γ= 0 S
2
γ
γ
β
subinterval : Q1
α
D1
β
Q2
α
D2
Fig. 4.9 Typical tank inductor current iL,
tank capacitor voltage, vC, and applied
tank voltage vT waveforms, for the
k = 1 continuous conduction mode.
Chapter 4.
The Series Resonant Converter
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.
jL
(a)
ω0 t = 0
radius MA1
jL
Q1 subinterval
β
.
(c)
.
ω0 t = β
1 - M + MA1 mC
1-M
.
.
.
ω0 t = γ + β -1 + M
mC
radius MA1
β
ω0 t = γ
Q2 subinterval
(b)
jL
jL
(d)
D2 subinterval
. .
1+M
radius MA2
ω0 t = α + β = γ
α
D1 subinterval
α
ω0 t = β
mC
. .
-1 - M
ω0 t = γ + α + β = 2γ
radius MA2
.
mC
ω0 t = γ + β
Fig. 4.10 Construction of the state plane trajectory for one complete switching period, in the
k=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
7
Principles of Resonant Power Conversion
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.
area = q
Hence, we have:
i L(t)
q = C (2VC1)
(4-5)
area = -q
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 | >
=
=
1
1
T
2 s
ω0 t
v C (t)
1T
s
2
+VC1
| iL(τ) | dτ
(4-6)
0
2q
Ts
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 = ITs
4C
-VC1
area = q
|i L(t)|
I=<|iL|>
(4-7)
or, in normalized form:
MC1 =
Jγ
2
(4-8)
where MC1 = VC1 / Vg.
Fig. 4.11 Use of charge arguments to relate
the peak capacitor voltage VC1, load
current I, and charge quantity q in the
series resonant converter for k=1 CCM.
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.
8
Chapter 4.
The Series Resonant Converter
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) =
Jγ
–1+M
2
(4-9)
Jγ
–1–M
2
(4-10)
and the radius MA2 of the D1 subinterval is
MA2 = (MC1) – (1+M) =
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).
jL
D2 subinterval
ω0 t = 2γ
.
-MC1
ω0 t = γ + β
α
ω0 t = 0
. .
. .
-1 + M
1+M
-1 - M
β
Q1 subinterval
1-M
radius MA1
.
MC1 =
ω0 t = β
Jγ
2
mC
radius MA2
ω0 t = γ
D1 subinterval
Q2 subinterval
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 ω 0 = 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
9
Principles of Resonant Power Conversion
jL
π − (π − α) − (π − β) = γ − π
MA2 = _J_γ
2 -1-M
Α
M
1
A
J_γ_ = 2
1
+
M
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
steadystate characteristics.
The lengths
of the two radii are
already known, and
the length of the
triangle base is the
distance between the
and
D2
Q1
C
α
.
B
π−α
π−β
−1 − Μ
.
β
1−Μ
mC
2
Fig. 4.13
Magnification of Fig. 4.12, for the solution of steady-state
conditions.
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
Θ
b
c
a2 = b2 + c2 - 2bc cosΘ
a
Fig. 4.14 The Law of Cosines
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
Jγ
Jγ
– 2( – 1 – M)( – 1 + M) cos(γ–π)
2
2
10
(4-12)
Chapter 4.
The Series Resonant Converter
Jγ
Jγ
– 1)2 + 2M2 + 2[( – 1)2 – M2] cos(γ)
2
2
(4-13)
Simplification yields:
4 = 2(
which can be rearranged to obtain:
1 =
(Jγ2 – 1)2
1 + cos γ
1 – cos γ
+ M2
2
2
(4-14)
Trigonometric identities can now be used to obtain:
M2 sin2
γ
Jγ
γ
+ ( – 1)2 cos2
= 1
2
2
2
(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.
no solutions
The uncontrolled output
J
for M ≥ 1
rectifier diodes do not allow the
(DCM instead)
load current to be negative, so
we must have J > 0. Also, with
valid
passive load requires
a passive load, M must be
solutions
I and V to have same
positive when J is positive.
polarity
1
Hence, the portions of the
M
ellipse that lie in the second,
bridge rectifier does not allow I < 0
third, and fourth quadrants are
not valid physical solutions.
Also, it is shown in section 4.4 Fig. 4.15 Elliptical output characteristic M vs. J (Eq. 4-15),
that the solution is not valid for
for a given γ in the k = 1 CCM.
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.
11
Principles of Resonant Power Conversion
5
F = 0.95
F = 1.0
F = 0.925
4
F = 0.9
J
3
2
F = 0.8
F = 0.7
1
2
π
F = 0.6
F = 0.5
0
0
0.2
0.4
0.6
0.8
1
M
Fig. 4.16
At
Output characteristics of the series resonant converter, operating in the k = 1
continuous conduction mode. Solutions occur over the range 0 ≤ M <1, 2/π ≤ J < ∞;
solutions not shown here for J > 6.
F = 0.5
(half resonance):
Then fs = 0.5 f0, and γ = π / 0.5 = 2π. The output characteristic, Eq. (4-15), becomes
M 2 ⋅ 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 π – 1) 2 ⋅ 0 = 1
2
(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.
12
Chapter 4.
The Series Resonant Converter
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. (415) yields
γ
1 – sin2
2
2 = 1
(J2π – 1) =
(γ ≠ π)
(4-20)
γ
cos2
2
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 J S C
(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.
10
8
6
Jsc
When M = 0, Eq. (4-15) becomes
(4-22)
(JSC2 γ – 1)2 cos2 2γ = 1
Solve for JSC:
γ
JSC = 2 (1 + | sec |) = 2F (1 + | sec π |)
2
2F
γ
π
4
2
0
0.5
0.6
0.7
0.8
0.9
1
F
The characteristics of a given load can be Fig. 4.17 Normalized short-circuit output
current JSC vs. switching frequency, in
superimposed
on
the
converter
output
the k=1 CCM.
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 = MQ
(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.
13
Principles of Resonant Power Conversion
F = 1.0
5
F = 0.95
4
F = 0.925
J
F = 0.9
3
ine
l
load
2
F = 0.8
F = 0.7
F = 0.6
1
2
π
Q=
0
0
R0
R
0.2
F = 0.5
0.4
0.6
0.8
1
M
Fig. 4.18
Resistive load line superimposed over the converter output characteristics.
F = 1.0
5
F = 0.95
4
F = 0.925
J
F = 0.9
3
e
2
ad
F = 0.8
lin
lo
F = 0.7
1
2
π
F = 0.6
F = 0.5
0
0
0.2
0.4
0.6
0.8
1
M
Fig. 4.19
Nonlinear load characteristic superimposed over the converter output characteristics.
14
Chapter 4.
The Series Resonant Converter
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
γ
MQγ
γ
+(
– 1)2 cos2
= 1
2
2
2
(4-26)
Now solve for M:
M2 [sin2
γ
Qγ 2 2 γ
+
cos
2
2
2
] – M Qγ cos2 2γ
γ
+ (cos2 – 1) = 0
2
(4-27)
Use of the quadratic formula yields:
Qγ
2
M =
γ
Qγ
tan2
+
2
2
2
1+ 2
Qγ
1±
2
(tan2 2γ ) (tan2 2γ
+
Qγ
2
2
)
(4-28)
1
Q=1
0.8
Q=2
0.6
M
Q=5
0.4
Q = 20
0.2
0
0.5
0.6
0.7
0.8
0.9
1
F
Fig. 4.20
Steady-state control characteristics M vs. F for various values of Q = R0/R in the k=1
continuous conduction mode, Eq. (4-28).
15
Principles of Resonant Power Conversion
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
Jγ
–1+M
2
2
= 22 +
Jγ
–1–M
2
2
– 2 (2)
Jγ
– 1 – M cos(π–α)
2
(4-29)
This equation can be solved for γ:
(1 + M) (1 – cos( α))
γ = 2
J
(M – cos(α))
(4-30)
This describes how the switching frequency varies for a given range of α and load. Equations (430) and (4-15) can now be used to eliminate γ, and to determine the α control characteristics. The
result is
J =
where
(1 + M) (1 – cos( α))
sin(α)
(M – cos(α)) π – tan–1
M – cos(α)
(4-31)
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
16
Chapter 4.
The Series Resonant Converter
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.
6
5
4
α = .3π
α = .2π
α=0
α = .4π
J
3
α = .5π
2
α = .6π
α = .7π
2/π
0
α=π
0
0.25
0.5
0.75
1
M
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 β
both 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
17
Principles of Resonant Power Conversion
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:
2q
Ts
where q, shown in Fig. 4.22, is:
I = <| iL |> =
(4-32)
Ts/2
q =
iL (t) dt
(4-33)
0
The average input current is:
Ts/2
< ig > = 1 1
2T s
iL (t) dt
(4-34)
0
since ig = iL when Q1 conducts. Substitution of Eqs. (4-32) and (4-33) into Eq. (4-34) yields:
< ig > =
2q
= I
Ts
(4-35)
18
Chapter 4.
Q1
X
Q2
The Series Resonant Converter
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:
X
iL
q
Pin = Vg < ig > = Pout = V I
ω 0t
γ−π
π
Use of Eq. (4-35) yields:
-q
Vg = V
vC
or:
VC1
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
-VC1
vT :
(4-36)
Vg -V
-Vg +V
q
q
|iL|
<|iL|>
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)
ω 0t
Elimination of q using Eq. (4-32) yields:
I Ts
VC1 =
4C
Fig. 4.23 Tank inductor current and
capacitor voltage waveforms for the k = 1
DCM.
or, in normalized form:
MC1 =
(4-38)
Jγ
2
(4-39)
(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:
19
Principles of Resonant Power Conversion
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 diL = vS – vC – vR
dt
(4-43)
During subinterval Q 1, vS = V g, and vR = V. We have already shown that V = Vg in this mode
(Eq. (4-37)), and so vL becomes
vL = L diL = – vC
dt
(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 diL = Vg + V – vC = 2Vg – vC
dt
(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.
20
Chapter 4.
k = 1 DCM
iL
The Series Resonant Converter
k = 1 CCM
iL
ω 0t
Q1
vC
X
Q2
ω 0t
X
vC
Q1
D1
Q2
D2
VC1
-VC1
vL
vL
Vg -V-vC(t)
Vg+V-VC1 > 0
Vg -V-vC(t)
Vg +V-vC(t)
2Vg
2V 2Vg
2V
-Vg +V-vC(t)
Fig. 4.23
-Vg +V-vC(t)
Comparison of tank waveforms of the k=1 continuous and discontinuous conduction
modes.
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
In normalized form, this can be written:
21
(4-46)
Principles of Resonant Power Conversion
1 + M – MC1 > 0
Substitution of Eqs. (4-37) and (4-40) into this expression yields:
(4-47)
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 >
jL
(a)
Q1
D1 A ?
. ..
Q1
D1
(1-M)
(1+M) mC
mC = MC1
at ω0 t = π
D1 B ?
(b)
jL
Q1
π, since such a subinterval would involve a
trajectory centered at mC = (1+M), and given either
X
. ..
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
(1-M)
(c)
jL
MC1 (1+M) mC
Q1
. ..
Q1
D1 mC
Fig. 4.24b.
(1-M) (1+M) MC1
In contrast, a CCM trajectory is shown in
D1
Fig. 4.24c. In this case, MC1 > (1+M), so that for
negative jL the trajectory is able to evolve in the Fig.4.24 Hypothetical state plane
trajectories: (a) for MC1 < (1+M),
clockwise direction about the center mC = (1+M).
trajectories A and B are impossible; (b)
Thus, the geometry of the state plane
actual k=1 DCM trajectory, with MC1 <
(1+M); (c) actual k=1 CCM trajectory,
trajectory gives a simple interpretation of the
with MC1 > (1+M).
boundary condition between the continuous and
discontinuous conduction modes.
22
Chapter 4.
The Series Resonant Converter
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 ≤
1
2
(4-49)
This is a useful mode, because the output is easily controllable: the converter output behaves as a
Q1
D1
π
π
X
Q2
D2
π
π
X
iL
γ
vL=LdiL
dt
γ
Vg -V-vC
Vg +V-vC
2V
2V Vg -V-vC(2π)
2Vg
vC
VC2
VC1
-VC1
Fig. 4.25
Tank waveforms for the k=2 discontinuous conduction mode.
23
ω0 t
Principles of Resonant Power Conversion
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.
iL
total charge = -q
ω0 t
Fig. 4.26
Tank inductor current waveform, with emphasis on total charge flowing
through the tank capacitor.
Hence:
2 VC2 =
q
C
(4-50)
The average output current I is:
I = <| iL |> =
2q
Ts
(4-51)
Elimination of q from Eqs. (4-50) and (4-51) yields:
24
Chapter 4.
The Series Resonant Converter
I = 4CVC2
Ts
(4-52)
Normalization of Eq. (4-52) and solution for the normalized peak capacitor voltage MC2 yields:
MC2 =
Jγ
2
(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.
jL
Q1
)
M
1-
D2
M
. . . .
-MC2
-(
. . . .
MC2 =
MC1
-MC1
(-1+M)
(-1-M)
C2
(1-M)
X
X
Jγ
2
(1+M)
mC
D1
MC2 - (1+M)
Q2
Fig. 4.27
State plane diagram for the k=2 discontinuous conduction mode.
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)
25
(4-54)
Principles of Resonant Power Conversion
Simplification yields
MC2 = 2 =
jL
Jγ
2
(4-55)
radius = MC2 -1-M
.
Now solve for the normalized load current J:
J = 4 = 4F
γ
π
(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 = 4 F
π
(-1-M)
radius = MC2 -1+M
.
-MC1
(1-M)
2
Fig. 4.28 Illustration of relations between
MC1, MC2 and M.
k=1 M=1
DCM
J
k=1 CCM
(4-57)
F = 1/2
2
π
with Q = R/R0. Solution for M then yields the
control plane characteristic:
M
1
This result is used in the next section, for
derivation of the mode boundaries.
1
3
1/π
k=1 DCM; M=1
M
(4-59)
k=2 DCM boundaries
1
Fig. 4.29 Output plane characteristics,
emphasizing k=2 DCM.
π
Q=
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:
1
3
.5/π
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
MC1 = MC2 – 2(MC2 – 1 – M) = 2M
F = 1/4
k=2 DCM
J = 4F
π
F=0
(4-58)
Q=0
M = 4 F
πQ
mC
k=2 DCM
M=4 F
πQ
Q
=
2/
3/π
=
Q
4/π
Q=
k=1
CCM
other modes
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
26
1
4
1
2
Fig. 4.30 Control plane characteristics,
emphasizing k=2 DCM.
F
Chapter 4.
The Series Resonant Converter
the load current J:
J = 4F ≤ 2
π
π
(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 >
1
3
(4-63)
To summarize, the k=2 DCM boundaries are:
1 > M >
1
3
2 > J > 0
π
1
2
(4-64)
> 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
f 0 < f < f0
S
k+1
k
(4-65)
27
Principles of Resonant Power Conversion
CCM WAVEFORMS
type k CCM; k odd
iL
γ
Q1
Q1
π
∫ ∫
π
D1
Q1
π
D2
π
D1
ω0 t
symmetrical
Q2
(k-1) complete half-cycles
type k CCM; k even
γ
iL
Q1
D1
π
π
D1
Q1
π
∫ ∫
Q1
π
D1
D2
Q2
symmetrical
k complete half-cycles
Fig. 4.31
Tank inductor current waveforms for the general type k continuous conduction
mode.
The output plane characteristics are elliptical, and are described by the equation
γ
Jγ
γ
+ 1 ( + (-1)k)2 cos2( ) = 1
)
2 2
2
2
ξ
where ξ is the subharmonic index,
2
M2 ξ sin2(
1 + (-1)k
2
The voltage conversion ratio M is restricted to the range
(4-66)
ξ = k+
(4-67)
0 ≤ M ≤ 1
ξ
(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
28
Chapter 4.
The Series Resonant Converter
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:
Qγ
2
M =
γ
Qγ
4
ξ tan2
+
2
2
2
(-1)k+1 +
1+
(ξ2 – cos2 2γ ) (ξ4tan2 2γ
Qγ 2
γ
cos2
2
2
+
Qγ
2
2
)
(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
f s < f0
k
(4-70)
In these modes, the output voltage is independent of both load current and switching frequency,
and is described by
(4-71)
M = 1
k
This mode occurs for the range of load currents
2(k + 1)
2(k – 1)
> J >
γ
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.
29
Principles of Resonant Power Conversion
(a)
iL
γ
Q1
π
π
X
Q1
∫ ∫
D2
π
D1
X
ω0 t
Q2
symmetrical
k complete half-cycles
(b)
4
π
J
1.0
8
3π
12
5π
k=1
DCM
k=3
k=5 DCM
DCM
etc.
1
5
Fig. 4.32
1
3
1
General type k discontinuous conduction mode, k odd: a) tank inductor current
waveform; (b) output characteristics.
Ig
1:1
k
I
0
0
Fig. 4.33
M
0
+
+
Vg
V
-
-
0
Steady-state equivalent circuit model for an odd discontinuous conduction mode: an
effective dc transformer.
30
Chapter 4.
The Series Resonant Converter
Type k DCM, k even
The type k discontinuous conduction modes, for k even, occur over the frequency range
f
fs < 0
k
(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 = 2k
γ
(4-74)
Operation in this mode occurs for:
(4-75)
1 > M > 1
k–1
k+1
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 Vg
into its dual, an output current source of value g V g [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].
31
Principles of Resonant Power Conversion
(a)
iL
γ
Q1
Q1
π
π
∫∫
π
D1
X
π
D2
D1
Q2
ω0 t
symmetrical
k complete half-cycles
(b)
J
1.0
2
π
F = .2
k=4
DCM
F = 0.4
etc.
k=2
F = 0.25
DCM
F = 0.1
F=.05
1
5
Fig. 4.34
1
3
M
1
General type k discontinuous conduction mode, k even: (a) tank inductor current
waveform; (b) output characteristics.
Ig = gV
I = gVg
0
0
+
g
Vg
0
Fig. 4.35
-
+
V
g = 2k
γR0
-
0
Steady-state equivalent circuit model for an even discontinuous conduction mode: an
effective gyrator. The converter exhibits current source characteristics.
32
Chapter 4.
1.0
The Series Resonant Converter
Q=0.2
0.9
Q=0.2
0.8
Q=0.35
0.7
Q=0.5
0.6 Q=0.35
Q=0.75
M
0.5
0.3
Q=1
Q=0.5
0.4
Q=0.75
Q=1
Q=1.5
Q=1.5
Q=2
Q=2
Q=3.5
Q=5
0.2
0.1
Q=3.5
Q=5
Q=10
Q=20
Q=10
Q=20
0.0
0.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
F
Fig. 4.36. Complete control plane characteristics of the series resonant converter, for the range
0.2 ≤ F ≤ 2.
M
k=1 DCM
1
k=2 DCM
1
5
k=3 DCM
k=4 DCM
k=5 DCM
etc.
1
5
1
4
k=2 CCM
1
3
k=0 CCM
k=1 CCM
k=3 CCM
1
2
1
3
1
2
1
Fig. 4.37. Complete control plane characteristics for continuous and discontinuous
conduction mode boundaries.
33
F
Principles of Resonant Power Conversion
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 zerovoltage 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 1
F
(4-76)
where INT(x) denotes the integer part of x. Next, the quantity k1 is determined:
k1 = INT 1 +
2
1 + Qπ
4 2F
(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.
34
Chapter 4.
The Series Resonant Converter
6
F = 1.05
F = 1.07
5
F = 1.10
4
J
F = 1.01
3
F = 1.15
2
F = 1.30
1
0
0
0.2
0.4
0.6
0.8
1
M
Fig. 4.38
Output characteristics in the k=0 continuous conduction mode (above resonance).
F = 1.0
J
F = .93
3
F = .96
F = .90
2.5
F = .85
2
k=1 CCM
1.5
F = .75
4
π
k=1 DCM
1
F = .5
2
π
F = .25
F = .1
k=2 DCM
0
0
0.2
0.4
0.6
0.8
1
M
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.
35
Principles of Resonant Power Conversion
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 Vg
device forward voltage drops are defined
as follows:
Vg
VT
transistor on-state voltage
VDin antiparallel diode forward voltage
VDout output diode forward voltage
VDin
+
VT
-
VDout
+
VT
+
V
-
The applied tank voltage waveform vT is
shown in Fig. 4.41 for the k=1 CCM.
Since the applied tank voltage is (by Fig. 4.40 Half-bridge circuit with nonideal
semiconductor devices.
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 =
1
2
VD in –
1
2
VT + actual Vg
(4-79)
VT
Vg +VDin +V+2VDout
Vg -VT -V-2VDout
Q1
D1
Q2
D2
-Vg +VT +V+2VDout
-Vg -VDin -V-2VDout
Fig. 4.41
Actual tank voltage waveforms for the half-bridge circuit of Fig. 4.40.
36
ω0 t
Chapter 4.
1 VDin
2
1 VT
2
actual
Vg
+
1 VDin
2
Fig. 4.42
ideal
switches
-
-
+
ideal diodes
(no forward
drop)
+
actual
Vg
1 VT
2
The Series Resonant Converter
VT
1 VDout +
2
1 VDin +
2
+
1 VT 2
-
+
+
-
+
actual
V
-
Half-bridge circuit with explicit voltage drops due to non-ideal
diodes and transistors.
Veff = 2 VD out +
1
2
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 = Veff
Vg eff
(4-81)
Jeff = I R0
Vg eff
(4-82)
2
M2eff ξ sin2(
γ
J γ
γ
+ 1 ( eff + (-1)k)2 cos2( ) = 1
)
2
2
2
2
ξ
(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
37
Principles of Resonant Power Conversion
same conclusions can be obtained more simply using the approximate sinusoidal methods of
chapter 2.
RS
RS
Re
L
C
esr
Vg
RS
RDout
RDout
RDout
RDout
+
RS
V
_
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
38
Chapter 4.
(a)
L
iL
I = 1n <| iL |>
C
+ vC -
I = 1n < | iL | >
(4-84)
To the extent that the waveshape of iL(t) does not
1:n
1 |iL|
n
I
Cf
+
V
–
(b)
The Series Resonant Converter
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:
Ts
I = 1
n Ts
0
ILP | sin ωt | dt = 1n 2ILP
π
(4-85)
Hence:
|iL(t)|
ILP = π I
2
ILP ≅ nπ I
2
ILP
I = <|iL|>
(c)
J
(4-86)
Or, in normalized form, one obtains:
JLP ≅ π J
2
(4-87)
Thus, we expect the peak tank current ILP to be
directly proportional to the load current I, and
3
essentially independent of the load voltage. We
JLP = 4
can plot contours of constant peak tank current in
J
=
3
LP
2
the output plane, to see how component stresses
JLP = 2
vary with operating point; Eq. (4-87) predicts that
such contours should appear as in Fig. 4.44(c).
1
The approximation used above coincides
with the sinusoidal approximation used in chapter
1
M
2. It is accurate for operation in the continuous
Fig. 4.44 Peak stress approximation where conduction mode, near resonance. However, in
(a) series resonant converter has a 1:n
the discontinuous conduction mode, or in the
transformer turns ratio; (b) dc load
continuous conduction mode near the DCM
current equals average rectified tank
inductor current; (c) result shows that
boundaries, the tank current is highly
peak tank current is directly proportional
nonsinusoidal, and Eqs. (4-84) - (4-87) become
to load current.
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
39
Principles of Resonant Power Conversion
sinusoidal tank waveforms, the peak tank capacitor voltage VCP is related to ILP through the
characteristic impedance R0:
VCP ≅ ILP R0 ≅ nπ I R0
2
(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 =
Jγ
–1+M
2
(4-89)
The peak normalized tank capacitor voltage MCP, shown in Fig. 4.45, was previously found. It is:
MCP = MC1 =
Jγ
2
(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):
40
Chapter 4.
jL
The Series Resonant Converter
peak value of jL:
Jγ
jLP = - 1 + M
2
(circle radius during Q1 conduction interval)
jLP
Q1
Jγ
2 -1
D2
+M
. .
.
. .
-1 + M
.
Jγ
2
mC
MC1 =
1+M
-1 - M
1-M
D1
peak value of mC:
Jγ
mCP = MC1 =
2
Q2
Fig. 4.45
State plane diagram for k=1 CCM.
γ
J +1–M
= LP
(4-91)
2
J
γ
= MCP
(4-92)
2
J
These expressions are then inserted into the converter output characteristic, Eq. (4-15), to eliminate
γ. Solving for J, one then obtains
JLP + 1 – M
J =
(4-93)
2–1
(J
–
M)
LP
π – tan–1
1 – M2
J =
π – tan–1
MCP
(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.
41
Principles of Resonant Power Conversion
5
F=0.975
JLP =7
M CP=7
4
F=0.95
JLP =5
3
J
M CP=5
F=0.9
JLP =3
2
F=0.8
M CP=3
F=0.7
1
F=0.6
JLP =1.5
MCP=2
F=0.5
0
0
0.25
0.5
0.75
1
M
Fig. 4.46
Superposition of peak tank current and voltage stress curves on the normalized
output characteristics, for the k=1 continuous conduction mode (below
resonance). Solid lines: curves of constant switching frequency; dashed lines:
contours of constant peak tank capacitor voltage; shaded lines: contours of
constant peak tank inductor current.
Similar analysis can be used to derive the component stresses for above resonance
operation. The results for the k=0 CCM are
MCP
(MCP + 1)2 – 1
1 – M2
J =
tan–1
JLP – 1 + M
J =
tan–1
–1 +
J =
tan–1
(JLP + M)2 – 1
1 – M2
J2
1 + LP
1 – M2
JLP
1 – M2
(4-95)
,
1 – M2
< JLP
M
(4-96)
,
42
1 – M2
> JLP
M
Chapter 4.
The Series Resonant Converter
5
M CP =7
JLP =7
F=1.05
4
F=1.07
M CP =5
JLP =5
3
J
F=1.1
M CP =3
F=1.15
2
JLP =3
F=1.3
M CP =1
1
F=1.5
JLP =1
0
0
0.25
0.5
0.75
1
M
Fig. 4.47. Superposition of peak stress curves on the normalized output plane, above
resonance (k=0 CCM). Solid curves are contours of constant switching
frequency, dashed curves are contours of constant peak tank capacitor voltage,
and shaded curves are contours of constant peak tank inductor current.
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)
43
Principles of Resonant Power Conversion
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 =
V
n 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 I
Vg
(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 =
V
n Vgmax
Mmax =
V
n Vgmin
(4-101)
The output power can be written:
P = IV =
(4-102)
J Vg V
2
= J V
n R0
M R0
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:
2
J = M n R0 P
V2
(4-103)
44
Chapter 4.
The Series Resonant Converter
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:
n Pmax R0
Jmax =
Vgmin V
(4-104)
while the minimum J occurs at minimum power and maximum input voltage:
(4-105)
n Pmin R0
Jmin =
Vgmax V
J
Jmax
P
Jmin
=
Pm
Vg = Vgmin
ax
Vg = Vgmax
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 ,
P min , Vgmax, and V gmin, the values of Mmin and
Jmin can be evaluated, and the operating region is
in
P = Pm
constructed. It can be seen that the maximum
Mmax
M
Mmin
switching frequency will occur at the point (Mmin,
J min ), i.e., at maximum input voltage Vgmax and Fig. 4.48 Voltage regulator operating
region, plotted in the output plane.
minimum power P min . 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 dcdc 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
45
Principles of Resonant Power Conversion
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.
A
Table 4.1. Comparison
Mmax Jmax Mmin Jmin fsmin
kHz
0.9
5.0 0.62 0.34 280
of designs,
L,
f0,
MHz µH
1.04 75
below resonance
C,
ILP, VCP, mode
nF
A
V
0.3 4.2 2080 k=1 CCM,
B
0.9
1.5
0.62
0.10
95
1.2
20
0.9
4.7
710
C
0.9
0.5
0.62
0.068 68
2.5
3.0
1.2
9.9
510
D
0.45
1.5
0.31
0.068 104
1.3
9.0
1.6
8.7
780
point
k=2 DCM
k=1 CCM,
k=2 DCM
k=2 DCM
k=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 f 0 , 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.
46
Chapter 4.
The Series Resonant Converter
Table 4.2. Comparison of designs, above resonance
L,
C,
ILP, VCP, mode
point Mmax Jmax Mmin Jmin fsmin
f0,
µH
nF
kHz
A
V
kHz
A
0.9
5.0
0.62 0.34 511
495
157
0.6
4.0
1940 k=0
B
0.9
1.5
0.62
0.10
220
200
117
5.4
3.9
545
C
0.9
1.0
0.62
0.068 157
137
113
12
3.8
350
D
0.9
0.2
0.62
0.014 45
28
111
290
3.9
50
E
0.45
1.0
0.31
0.068 133
95
81
34
8.8
287
CCM
k=0
CCM
k=0
CCM
k=0
CCM
k=0
CCM
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 f 0 . 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.
47
Principles of Resonant Power Conversion
REFERENCES
[1]
F.C. Schwarz, “An Improved Method of Resonant Current Pulse Modulation for Power Converters,” IEEE
Power 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,” IEEE
Transactions 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 Power
Electronics Specialists Conference, 1982 Record, pp. 85-100, June 1982.
[4]
R. King and T.A. Stuart, “Inherent Overload Protection for the Series Resonant Converter,” IEEE
Transactions 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 Transactions
on 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 Transactions
on 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 Series
Resonant Converter,” IEEE Power Electronics Specialists Conference, 1987 Record, pp. 545-556, (IEEE
Publication 87CH2459-6).
[8]
R.L. Steigerwald, “High Frequency Resonant Transistor Dc-Dc Converters,” IEEE Transactions on
Industrial Electronics, vol. IE-31, no. 2, pp. 181-191, May 1984.
[9]
S. Singer, “Loss-Free Gyrator Realization,” IEEE Transactions on Circuits and Systems,
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 25kW
X-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,” IEEE
Power 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 for
Low Switching Loss,” IEEE Power Electronics Specialists Conference, 1987 Record, pp. 527-536, June
1987.
48
vol. CAS-28,
Chapter 4.
The Series Resonant Converter
P ROBLEMS
1.
Using the phase plane, derive the k=0 continuous conduction mode characteristics. Sketch your results in the
output plane (M vs. J) and control plane (M vs F for resistive load).
2.
Analyze the k=3 discontinuous mode:
3.
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 capacitor
voltage boundary values;
d)
Solve for the output characteristics;
e)
Determine the complete set of conditions on normalized switching period and load current which
guarantee operation in this mode.
It is desired to obtain a converter with current source characteristics. Hence, a series resonant converter is
designed 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 mode
boundaries, evaluate the short-circuit current, and give analytical expressions for the output
characteristics.
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
I
v in
CF
V o ut
Design a half-bridge series resonant converter, as shown above, to meet the following specifications:
Input voltage Vin:
Output voltage Vout:
Output current I:
Maximum switching frequency:
Output voltage ripple:
134–176 volts
48 volts
1–10 amperes
750 kHz
no greater than 1 volt peak-to-peak
49
Principles of Resonant Power Conversion
Design the “best” converter that you can, which combines high efficiency with small volume. Use your
engineering judgement to select the operating mode, tank elements L and C, and transformer turns ratio to attain
what you consider to be the best combination of small transformer size, small C F , 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 mm 3 . Capacitor volume
scales as the product of capacitance and voltage rating, so that the volume of a 100V 5µF capacitor of the same
dielectric material is 500 mm 3 . 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 least
25% 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:
2 2
Kg =
ρ λ 1 i1 108
Pcu Ku B2max
cm5
where ρ is the resistivity of copper (1.724·10-6Ω·cm), λ1 is the applied primary volt-seconds, and i 1 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 constructed
in this manner are as follows:
Core type
vol., mm3
Kg , cm5
EE12
675
0.73·10-3
EE16
1125
2.0·10-3
EE19
1550
4.1·10-3
EE22
EE30
EE40
2425
8575
14700
8.3·10-3
86·10-3
0.21
EE50
EE60
31500
42250
0.91
1.4
Hence, attempt to minimize the total volume V tot 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 wire
size); (4) the value of CF required; and (5) the total volume Vtot as defined above.
50