AN1679/D
How to deal with Leakage
Elements in Flyback
Converters
Prepared by: Christophe Basso
ON Semiconductor
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APPLICATION NOTE
Introduction
Several ways exist to depict how the magnetic variables
are organized within the transformer. The reluctance method
keeps the analogy with our electric circuits and therefore
appears easier to us non−magnetic experts. Remember that
the electrical resistance of a metallic piece tends to brake the
current flow. That is to say, we impose a current variation dI
(Amps) through the resistance R (W) of the metallic part by
applying a voltage variation dV (Volts) across it, or
It is difficult to find a good tradeoff between cost and
performance when designing offline Flyback power
supplies. The basic problem is the presence of the various
leakage inductances of the transformer. Those leakage
inductances cause voltage overshoot (which, in turn, calls
for power switches with increased voltage rating), divert a
portion of the primary current to a clamp, and degrade the
voltage regulation in power supplies with primary
regulation or with multiple outputs.
To evaluate the effect of the leakage inductances on the
circuit performance in the design phase is a challenging task.
Here a simulation program can be quite helpful. This paper
reviews the magnetic and electric models of the
two−winding and three−winding transformers, discusses
how to extract the inductance values of the models by
measurements and calculations, and how to feed a SPICE
model with. Furthermore, the paper analyzes the influence
of the primary leakage inductance on the switch voltage
waveform, provides guidelines on how to protect the switch
from the voltage overshoot, and considers the effect of the
secondary leakage inductances on the cross regulation of
multiple−output Flyback power supplies.
dl + dV
R
With magnetic materials, we could compare the current with
the flux f (Weber) which is constrained to circulate in the
magnetic material by an external force F, the magnetomotive
force (mmf) expressed in Amps−turns. This force is
elaborated by a winding of N turns in which a current I flows.
The resulting force F is equal to N x I. The obstacle which
brakes the circulation of the flux is the reluctance ℜ of the
magnetic medium and can be explained by:
+ dF + N @ di
df
dq
That is to say, how much of magnetomotive force F shall I
apply to the magnetic material in order to obtain a change in
the flux f. The analogy between electricity and magnetism
is depicted by Figure 1.
The Reluctance Concept
Since the transformer, or coupled inductances for a
Flyback, is the key player in our setbacks, it is important to
spend some time to understand the way the inductive
leakage elements are spread to finally measure them and
feed our simulator model with.
I
f
3
1
+
+
R
V
N.I
2
ℜ
4
Figure 1. Analogy Between Electricity and Magnetism
© Semiconductor Components Industries, LLC, 2005
September, 2005 − Rev. 3
1
Publication Order Number:
AN1679/D
AN1679/D
To physically define ℜ, let’s start again with the definition
of the resistance R of a metallic bar:
We already defined the term:
df
di
l
R+ m
s @ Ae
when we talked about reluctance:
with: lm length of the material, s the material conductivity and
Ae the material area (also noted S for section). Applying the
same analogy as before and replacing the electrical
conductivity s by m, the magnetic permeability, we define the
reluctance by:
R+
df
+N
di
By introducing this last equation in the definition of the
self−inductance L, we come to:
L+N
2
lm
with m + mO @ mRȀ
m @ Ae
or by defining the permeance P as the reciprocal of
reluctance
the remaining symbols being the same as above. As one can
see, the reluctance depends on the magnetic medium by its
relative permeability mR. In the air, like in a gap or if flux lines
were closing externally to the magnetic material (e.g., a leaky
choke), mR drops to 1 and the reluctance simply becomes:
air
(P + 1 ),
the
Figure 2 shows a transformer wound across a square
magnetic material. The primary side made of Np turns
creates the necessary force F which gives birth to two
components: fm who links both windings, but also fl1
which does not couple to the secondary and corresponds to
a leakage path through the air. This leakage path could be
responsible for ElectroMagnetic Interferences (EMI). Due
to fm, a current Io circulates in the secondary, but this
current also gives birth to another leakage flux fl2 whose
polarity is opposite of that of fm. It is important to note that
fm produces Io while fl2 is a consequence of it.
Faraday’s law of electromagnetic induction states that the
voltage v appearing across an inductor L in which a current
I circulates can be calculated by:
v + L @ di
dt
This law can be expressed in a different way when applied
to a single coil made of N turns of wire:
df
dt
we
define
the
df
di
Ip
Io
fl1
Np
fl2
Ns
Vexcitation
to:
The Two−winding Transformer Magnetic Model
Inductance and Permeance
L+N@
simplifies
We will see later on how powerful this concept is to easily
make inductance appear.
with mo the air magnetic permeability, 4 ⋅ p ⋅ 10−7henry.meter
in the International System (SI).
By equating both equations,
self−inductance of a single coil by:
calculation
L + N 2 @ P,
lm
,
mo @ Ae
v+N@
inductance
Vo
fm
Magnetic material with a
linear reluctance ℜm
Figure 2. A Simple Two−winding Transformer with its Associated Fluxes
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Applying the technique depicted in Figure 1, we are now
able to make a reluctance model immediately highlighting
the way the various fluxes combine, as Figure 3 portrays.
Ns @ Io
Np
fm − fl2
fm + fl1
ℜmaterial
fm
+
fl1
Np.Ip
is the output current reflected to the primary. Figure 5 details
this next step.
2
+
fl2
Ns.Io
fl1
fm
Rload
Ns
Io.Ns/Np
4
6
Io
Np 2
Ns 2
fl2
Figure 6 unveils the final drawing.
Ns.Io
2
Np
Each permeance in the network is once again scaled by the
reference winding Np to come up with a full inductive
network (remember L = N2 ⋅ P). According to Faraday’s law,
scaling (fm+fl1) by Np has the dimension of a voltage
(Vp), since V.dt = N.df.
As we previously stated, fl2 is the consequence of Io, the
secondary current. Since this leakage inductance Ll2
“belongs” to the secondary, we keep it at the primary due to
the reflection ratio
Pl3
Np.Ip
5
Figure 5. First Stage Toward the Final Inductance
Model
The fluxes definitions are given beneath the drawing. You
calculate the fluxes (f) the same way as if you were
evaluation a current by solving for a voltage (N.I) across a
resistance (ℜ). Now let’s apply duality rules by which a
series reluctance becomes a parallel permeance (and vice
versa) and a voltage source turns into a current source. The
result appears in Figure 4 where the current source is
reversed to account for the flux circulation (fm−fl2).
3
Np.Pm
Ip
−
Figure 3. The Reluctance Model of Transformer in
Figure 2
Pl1
4
Vo
Np @ Ip
Np @ Ip * Ns @ Io
fl1 +
, fl2 + Ns @ Io , fm +
1
2
material
air
air
Np.Pl2
2
ℜair2
ℜair1
1
Np.Pl1
+
fm + fl1
1
3
which is equivalent to the previous equation. It is now
pertinent to make the turn ratio appear, since
Ll2.(Np/Ns)2
Ll1
Pmaterial
1
2
Lm
Np
6
+
Vp
Figure 4. Permeance Version of the Two−winding
Transformer
Ip
The voltage induced across the left arm is fm+fl1, while
the (output) voltage generated at the right arm is fm−fl2.
The goal of the next stages is to unveil the inductances. To
do so, we will scale everything according to Np, our winding
reference. If a current source N.I naturally forces a current
flow into a resistance R, the voltage across R is obviously
(N.I)xR. If we scale the current source by 1/N, we must
multiply the resistance by N: V = Ix(N.R). As a
consequence, the input source becomes Ip and every
permeance is multiplied by Np. But we still need to keep the
terms including Io (fm and fl2) constant to their previous
values once the Np multiplication is done: fl2 = Ns.Io.Pl2
or, after multiplying Pl2 by
3
Rload
Ns
5
Io
4
Figure 6. The Final Electrical Model of the
Two−winding Transformer
Ll1
6
Ll2
1
+
Vp
Ip
7
3
Lm
Np
5
Rload Vo
Ns
4
Io
Figure 7. A More Practical Electrical Model
N : phil2 + Ns @ Io @ (Pl2 @ Np)
Np
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Vo
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This model, called the p−model or the T−model, was
introduced many years ago. Reference [1] and [2] offer a
comprehensive description of the different methods used to
model the magnetics in general.
Figure 6 configuration is not easy to implement in a
simulator. Reflecting the secondary leakage back to its place
facilitates the handling of this model, as Figure 7 depicts.
Finally, your simulation could be implemented using the
schematic in Figure 8. It uses a perfect transformer, as
described in [4], where the leakage and resistive elements
have been added.
3
1
You often read: “short the secondary and measure the
leakage inductance on the primary.” This statement depends
on the model you have adopted for simulation [3]. For
instance, in the model shown in Figure 7, shorting the
secondary does not give you the final primary leakage value
… Watch for the steps:
1. Inject a sinusoidal voltage Vp on the primary and
measure the open−circuit voltage on the secondary
Vs.
Compute N +
Lm
Ǹ
1*
5
6
Ll1
2
Figure 8. The Complete Two−winding Transformer
SPICE Model
Np
Vp
+
Ns
Vs
2. Measure the primary inductance, the secondary
being open. This gives you Lpsopen
3. Repeat step 2 but the secondary is now shorted by
a wire. You obtain Lpsshort.
4. Compute the coupling coefficient k with:
k+
4
X1
XFMR
Rp
Determining the Physical Values of the Two−winding p
Model
Ll2
Rs
Np:Ns
The Three−winding Transformer
A three−winding transformer is often implemented in
either dual−output SMPS or primary regulated supplies. As
you imagine, the numerous fluxes coming into play greatly
complicate the determination of the model. Figure 9 offers
for reference a model for a mesh transformer and highlights
the position of the fluxes. Similar exercise can be undertaken
for a junction transformer, more commonly used in Flyback
transformers. More information can also be found in [8].
Figure 9 portrays the three leakage fluxes, fl1, fl2 and
fl3 but also the linking fluxes between the three windings:
f12, f13 and f23. As you can see in Figure 10 the
reluctance model differs in complexity from the Figure 2
counterpart. For the sake of clarity, we will not go through
the transformation steps as we did before but rather directly
jump to the results given in [1].
Lps
short
Lps open
5. Compute Ll1 with:
Ll1 + (1 * k) @ Lps open
6. Compute Ll2 with:
Ll2 + (1 * k) @ Lps open @ 12
N
7. Compute Lm with: Lm = k . Lpsopen
8. Measure with an ohm−meter the primary and
secondary DC resistances, respectively Rp and Rs.
f12
f1 + f2
I1
f13
N1
N2
I3
V2
V3
V1
f23
fl1
N3
fl3
fl
I2
fl2
Linear Reluctance ℜm
f2
Figure 9. The Three−winding Transformer Complicates the Flux Graph
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f12
R12
f1
Rm
f2
2
fl3
+
N1l1
3
Rm
f1 + f2
Rm
R13
R23
4
fl3
f3
fl1
N3l3
Rl1
8
1
f23
Rl3
+
+
N2l2
fl2
Rl2
Figure 10. The Reluctance Model (Figure 9) Becomes Difficult to Manipulate
Determining the Physical Values of the Three−winding
p−model
These results are gathered upon Figure 11 drawing where
the three leakage elements appear in series with each
winding. The surprise comes from the primary leakage
element which depends upon the leakage air path P1 but also
from the permeance P23 between both secondary windings.
As a matter of fact, if you improve the coupling between
both secondaries (e.g., by twisting the wires) you increase
the primary leakage inductance. Reference [1] paper also
demonstrated how the leakage elements stay practically
independent of the air−gap length: the coupling coefficient
diminishes as the air−gap increases (the magnetizing
inductance becomes smaller) but leakage elements stay
constant.
4
15
2
5
Rs2
10
A
Ll1
Rs1
Ll2
These measurements require a RLC bridge or a network
analyzer operating at a frequency high enough to neglect the
winding resistance (if you measure an impedance) but low
enough to minimize the interwinding capacitances. In our
numerical application, we selected 100 kHz which is the
operating frequency of our converter. The network analyzer
gives you a complete impedance−phase plot and ensures that
you stay on the inductive portion during the measurement.
With a standard LRC−meter, if changes in the operating
frequency do not lead to big variations of the values you
read, then you are well on the inductive portion.
The 4 W transformer used in our experiment was made by
Thomson Television Components (Gray, France) on an
small E core with a special insulation material [5]. As you
will discover through the following steps, the results express
how you combine the various leakage elements from the
measurement data.
1. Inject a sinusoidal voltage Vp on the primary and
measure the open−circuit voltages on the
secondaries Vspower and Vsauxiliary.
1
11
Mo
1
6
16
Ll3
Rs3
7
12
B
Compute A +
Vs power
13
Vp
and B +
Vs power
Vp
Measures gave A = 0.0817 and B = 0.156
2. Measure the inductance L1 seen from the Primary,
the secondaries open: L1 = Ll1 + Mo = 3.62 mH
3. Measure the inductance L2 seen from Primary
with the Power Winding open, the Auxiliary
shorted:
Figure 11. Final Equivalent Model Showing Every
Leakage Elements
Leakage Elements Definitions:
L11 [ n12.(P1 ) P23) Primary Winding
L12 [ n12.(P1 ) P23) Primary Winding
L2 + Ll1 )
Mo @ Ll32
B
Mo ) Ll32
B
L13 [ n12.(P1 ) P13) Power Winding
Mo [ n2.(Pm)
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Measure the inductance L3 seen from Primary with the
Power Winding shorted, the Auxiliary open:
L3 + Ll1 )
Mo @ Ll22
A
Mo ) Ll22
L4 + Ll2 ) A 2 @
Mo @ Ll32
B
Mo ) Ll32
B
ƫ
+ 1.405 mH
Please note that L1−L4 could also be impedances Z1−Z4.
You should then divide each value by 2 ⋅ p ⋅ Fmeasure to obtain
the inductance value. We now have a system of four
equations with four unknowns. Feeding a math processor
with these equations give the solutions in a snap−shot:
+ 127 mH
A
Measure the inductance L4 seen from the Power Winding,
the Auxiliary shorted and the Primary open:
LI1 + L1 *
ƪ
ǸL3.L2 * L3.L1 * L1.L2 ) L12 ) L4.L1 A*2 L4.L3 + 58.5 mH
Ll2 +
A 2 @ (Ll1 * L1) @ (L3 * Ll1)
+ 466 nH
L3 * L1
Ll3 +
B 2 @ (Ll1 * L1) @ (L2 * Ll1)
+ 3.558 mH
L2 * L1
Mo + L1 * Ll1 + 3.56 mH
Series resistances are measured with a 4−wire ohm−meter and included in the SPICE model.
Some Precisions on the p−models
however convenient to pick−up the real ratio to make the
model close to its physical counterpart.
At the opposite, the three−winding model is unique. It
does not need any arbitrary parameters to reveal the values
of the elements its made of. The p−model cannot be used for
transformers with more than three windings.
If you go through the equations that lead to the first
two−winding transformer, you come−up with a set of three
equations for four elements in the model. We say the model
is underdetermined and we obviously need to give an
arbitrary value to the primary/secondary turn ratios. It is
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Three−winding SPICE Model
The SPICE of the above transformer mixes a pair of two−winding transformers as reference [4] details. For clarity reasons,
we keep the leakage elements external to this ideal structure.
.SUBCKT XFMR−AUX 1 2 3 4 10 11 {RATIO_POW = ??? RATIO_AUX=???}
*
* RATIO_POW = A
* RATIO_AUX = B
*
1
1
3
4
10
2
2
11
3
4
10
11
* 1:A
* 1:B
RP 1 2 1MEG
E1 5 4 1 2 {RATIO_POW}
F1 1 2 VM1 {RATIO_POW}
RS1 6 51 1U
VM1 5 6
E2 20 11 2 1 {RATIO_AUX}
F2 2 1 VM2 {RATIO_AUX}
RS2 21 61 1U
VM2 20 21
.ENDS
pitfall, simply add a 100 Meg resistor between nodes 4 and
11 and the ground. INTUSOFT’s IsSPICE4 (San−Pedro,
CA) elegantly implements .OPTIONS RSHUNT=100Meg
which places a shunt element between each node and
ground, thus easing the DC convergence.
Positive ratios for A and B describe pin 3 and 10 as the
positive outputs (e.g., for a FORWARD) by respect to 1. By
entering negative values for A and B, you simulate separate
or both windings in Flyback mode.
Floating nodes, as encountered in isolated supplies, often
generate matrix errors in SPICE simulators. To avoid this
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The Switching Effect of the Leakage Inductance in
Flyback Structures
whose amplitude depends on the various parasitic elements
the transformer is made of. If we stick to Figure 12
references, the peak voltage can be expressed by:
As we can see in the sketch in Figure 12, the primary
leakage element LI1 does not participate to the
primary/secondary energy transfer. When the main switch
SW1 closes, the primary current builds−up until the
adequate threshold is reached and resets the internal latch
(current mode supply). The current flows through Mo but
also through LI1. As a matter of fact, when SW1 opens, both
magnetic fields collapse and the voltage across the
inductances reverses in an attempt to keep the amps−turn
constant: Mo energy is thus coupled to the secondary and
gives birth to the output current charge. Since LI1 cannot
find a circulating path, you observe a large positive spike
Rs1
+
Cprimary
1 nF
2
Ll1
)V
)V
DC
FLYBACK
) C oss
primary
VO
IC = 5
CO
Rload
Ip = 4.58 A
Y1 = 2 A/div
LEAKAGE CURRENT
4
LO
V3
330
C
the square root representing the characteristic impedance of
the parasitic LC network. This network is also guilty of the
oscillations which ring after the spike.
With the numerical values from Figure 12, we come−up
to a theoretical drain voltage of 1514 V (VfDOUT=800 mV)
at the opening, confirmed by an IsSPICE4 simulation.
1
X1
XFMR
3
Ǹ
Dout
1:0.05
5
Vpeak + Ip @
x = 2 ms/div
Vp = 1524 V
Ll1
80 mH
+
8
VPWM
Y2 = 500 V/div
Vdrain
Vdrain
6
SW1
Fosc = 443 kHz
Coss
470 pF
Figure 12. Simulating the Effects of the Primary Leakage Inductance Ll1
If we have a lateral MOSFET in place of SW1, (e.g., with
an MC33363 or MC33370 recently introduced by ON
Semiconductor (Phoenix, AZ)), the internal body−diode is
weak and cannot accept any avalanche energy: the transistor
is destroyed and a mean has to be found to protect it against
this trouble.
Now, let’s assume we implement a switching component
which withstands an accidental avalanche due to its internal
body−diode. It could be an MTD1N60E or even the new
MMG05N60D IGBT, specifically tailored for low−power
offline supplies. In that case, the leakage spike grows−up
until the avalanche is reached, thus clipping the voltage
(Figure 13).
Vclamp + Vin (passive network)
BVDSS (MOSFET alone)
Vout.N + Vin
Ip
0
Dt
Figure 13. Turn−off Waveforms with a MOSFET
Clipping the Peak
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drain voltage to rise above Vin + Vclamp, the clipping voltage
we want to impose (Figure 14). Let’s us first consider that the
voltage across the RC network is constant (ripple is low
compared to average voltage) and equals Vclamp. The power
dissipated at turn−off in the clamp is:
In lack of any external clipping network, we can calculate
the amount of energy ET dissipated in the transistor every
time it opens. As we said, the leakage inductance tries to
keep the current circulating at its level (Ip, when the
transistor opens) during Dt and pushes the drain voltage up
to BVDSS. Ip(t) can be expressed by:
P
+1@V
@ Ip @ Dt @ F
clamp
2 clamp
.
Ip(t) + Ip @ Dt * t
Dt
Dt has already been defined, but this time, BVDSS is replaced
by Vclamp. Once introduced in the previous equation, we
obtain the power dissipated in the clipping network:
You calculate the energy by integrating over time the
cross−over area between current and voltage:
Dt
E +
T
ŕ Id(t) @ VDS(t) @ dt + 12 @ Ip @ BVDSS @ Dt
0
V
clamp
P
+ 1 @ Ip 2 @ L
@F@
clamp
leak
2
V
* Vout @ N
clamp
.
.
The time Dt is given by the reset−voltage applied across the
leakage inductance: BVDSS − Vout ⋅ N, neglecting the output
diode forward voltage. It is then easy to define
Since this power will mainly be dissipated by the resistor
Rclamp, we can write the following equality:
V
V
clamp2
clamp
+ 1 @ Ip 2 @ L
@F@
leak
2
V
* Vout @ N
R
clamp
clamp
.
L
@ Ip
leak
Dt +
BV
* Vout @ N
DSS
.
By solving for Rclamp we calculate its value for a given level
of clamping voltage:
By introducing this equation into the previous one, we get:
BV
DSS
P + 1 @ Ip 2 @ L
@F@
T
leak
2
BV
* Vout @ N
DSS
.
R
+
clamp
It is important to minimize the ripple level Vripple
superimposed on Vclamp. A capacitor Cclamp will fulfill this
function. If we agree that the amount of charges Q will
equally split between Rclamp and Cclamp at turn−off, we can
write: Vripple ⋅ Cclamp = IRclamp ⋅ T. Knowing that IRclamp =
Vclamp / Rclamp and then solving for Cclamp it comes:
This results depicts the average power the transistor would
be the seat of if no mean were implemented to re−route the
energy spike elsewhere. We shall then implement a clipping
network to prevent the MOSFET from entering the
avalanche.
Preventing the avalanche
V
clamp
C
+
clamp
V
@F@R
ripple
clamp
By wiring a RC network from drain to VCC via an
ultra−fast diode (e.g., an MUR1100E), we will prevent the
3
4
R1
+
Rclamp
Cclamp
D1
5
Cout
1
Lprim
3 mH
Vin
2@V
@ (V
* Vout @ N)
clamp
clamp
2
L
@ Ip @ F
leak
2
Vout
6
9
Dclamp
Lleak
80 mH
7
From PWM
MTD1N60E
8
Figure 14. An RC Network Clamps the Drain Voltage at Turn−off
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Using a Zener Diode
VDCRAIL + Vz
A zener diode can be used to replace the RC network when
the clamping level needs to be more precisely defined.
However, the cost associated with this active solution is
higher than the passive one. The electrical schematic is as
follows:
DCrail
Vdrain−source
D1
N:1
Vout
1
Cout
Lp
Dclamp
4
8
VDCRAIL + VO.N
5
Ip
2
Izener diode
R1
Dt
Lleak
Figure 16. Clipping Waveforms with a Zener Diode
7
When the Leakage Inductance Brakes the Energy
Transfer...
Drv
When the transistor is closed, as shown in Figure 15, input
energy from the bulk capacitor is stored into the core’s gap.
The total involved energy is equal to
Figure 15. Clipping with a Zener Defines a More
Precise Protection Level
E
V
) Vz
CC
P + 1 @ Ip 2 @ L
@F@
T
leak
2
Vz * Vout @ N
The average power in the zener diode can easily be found
from the simplified drawing in Figure 16. After integration,
it is found that
E zener + 1 Vz @ Ip @ Dt
2
where Dt is imposed by the reset−voltage across Lleak:
L
@ Ip
leak
Vz * Vout @ N
If we introduce this last equation into the previous one, the
power dissipated in the zener diode is:
Pz +
+ 1 @ Ip 2 @ Lp
2
When the transistor opens, the voltage across the primary
inductance reverses and the primary/secondary energy
transfer should immediately take place. However, the
primary current also circulated in the leakage inductance
and causes an additional reversal voltage which sums−up to
the previous one: the clipping diode is forward biased and
routes some current to the clamp network, transforming the
leakage energy into heat. Unfortunately, this leakage path
causes some of the energy stored in the gap to be also
transformed into heat: any current flowing in the leakage
inductance forces the same current to flow through the
mutual inductance. As a matter of fact, the leakage
inductance delays the primary to secondary energy transfer
(by Dt) until its current has dropped to zero. This parasitic
inductance also diverts a substantial amount of the stored
energy thus hurting the open−loop gain. To account for it, the
primary current will be higher, forcing you to design a larger
core transformer. You can diminish the above Dt either by
decreasing Lleak or by increasing the reset voltage upon it,
(Vclamp − Vout.N). Figure 17 shows the behavior of the
various currents at the opening and details the timing
involved with a 1:1 transformer.
As previously written, the power dissipated at turn−off in
the transistor can be expressed by:
Dt +
stored
Vz @ Ip 2 @ L
@F
leak
2 @ (Vz * Vo @ N)
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Iprimary
Ip
Ipx
Vin/Lp
Vo @ N
Lp
ILleak
Iprimary
ILleak
Isecondary
V
* Vo @ N
clamp
L
ton
leak
Dt
Figure 17. The Primary Leakage Effect Delays the Energy Transfer by t
To calculate the percentage of Ip who flows into the
clamp, you can calculate the point at which Isecondary
catches−up with Iprimary. This point is called Ipx on
Figure 17. As we previously said,
Dt +
This last equation gives you the effective percentage of
primary current stolen by the leakage inductance.
Reference [6] details an interesting study of the primary
leakage effects in a dual−transistor Flyback converter.
L
@ Ip
leak
V
* Vout @ N
clamp
The Secondary Inductance Degrades the
Cross−regulation
When you design primary regulated power supplies, a
good cross−regulation is needed to avoid the output to run
out of specs. Unfortunately, the leakage elements will, once
again, play in your disfavor. Figure 18 explains how the
secondary leakage inductance, lumped in series with the
power winding, generates a parasitic voltage which
subtracts from the power voltage. This voltage simply
equals:
The slope of the decreasing primary current is simply
N @ Vout
Lp
but this equation can also be written as:
N @ Vout
Ip * Ipx
+
Dt
Lp
Replacing Dt and solving for Ipx gives:
Ipx
+1*
Ip
Lp @
ǒ
L
L
leak
V
clamp
*1
Vo @ N
lumped
Ǔ
Llumped
1
+
secondary
Dt
Vleak
10
POWER
N
POWER
15
6
8
4
1
dI
5
2
N
@
1
Aux
TO FEEDBACK CIRCUIT
+
Veq
Aux
9
TO FB
Figure 18. The Parasitic Voltage Subtracts from the Power...
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11
Figure 19. ...but Adds to the Auxiliary Level
AN1679/D
To calculate how this parasitic voltage Vleak reflects to the
auxiliary winding, we can state that
In primary regulated supplies, the primary current at the
opening is almost entirely diverted into the main secondary
output because the auxiliary circuit only requires a
negligible portion of it. However, in real multi−output
Flyback converters, the split depends on the current imposed
by the converter loads. If the sum of all these output currents
leads to the well known triangular waveform, individual
shapes can be quite strange in discontinuous mode when a
load changes its nominal value. These phenomena are
described in details by Lloyd DIXON in reference [7].
Vout + Vaux @ N * V
leak
should equal
(Vaux ) Veq ) @ N
When solved, it comes that
Veq + * 1 @ V
leak
N
Putting the Model to Work
Figure 19 properly represents how this source now reflected
to the secondary adds to the auxiliary winding and creates
the well known cross−regulation error: the feedback senses
a higher auxiliary winding, due to Veq, and decreases the
duty−cycle forcing Vout to diminish.
Several methods exist to counteract the effects of the
secondary leakage inductance. The simplest one consists in
adding a resistance in series with the auxiliary winding. This
resistance will be the seat of a voltage whose polarity fights
Vleak and cancels the previous effect. However, the results are
load dependent and does never give an optimum regulation.
Due to ideal components, a SPICE simulator can add the
parasitic elements and shows how they interact with each
other. Figure 20 presents a complete offline AC/DC wall
adapter implemented around the recent MC33363. The
corresponding INTUSOFT’s IsSPICE4 and MICROSIM’s
PSPICE models are available for download at
http://motserv.indirect.com/home2/models/bin/smps_circu
its.html.
VleakSec
RATIO_POW = −0.0817
RATIO_AUX = −0.156
R13
100 k
Vclamp
21
29
C9
1 nF
R1
3.5
D2
1N4937
Lp
3.56 M
7
X1
MC33363
V1
320
1
Aux
5
14
R7
15 k
13
9
MC33363
Osc
13
18
R2
600 M
IRipple1
9
C1
100 mF
28
D4
1N4148
22
R9
8.2 k
2.2 nF
C8
100 mF
15
Aux
Vaux
R5
110 k
11
7
10
8
9
R8
4.7 k
Verr
23
C2
10 mF
C4
26
R6
24 k
27
R4
100 pF C3
230 k
19
R10
921
1.2 nF
Figure 20. IsSPICE4 Simulations Help to Understand How the Phenomena Take Place
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12
24
20
R3
9.8
IsecAux
C7
Idrain
R12
600 M
IRipple2
IleakP
10
R11
400 M
12
6
C6
C5
820 p 100 nF
17
R13
536
VsecA
Vdrain
3
4
4
1
26
L3
10 mH
6
LI3
3.55 m
LI1
58 m
16
4
IsecP
Isnubb
+
D1
1N5821
VsecP
2
12
LI2
466 nH
R14
278 M
Vout
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Conclusion
In this schematic, the aforementioned parasitic inductive
elements have been incorporated and simulated. Figures 21
and 22 respectively detail the auxiliary/power voltages
observed on the real PCB and the simulated ones.
The good agreement between the figures, corroborated by
the output voltage measurements, confirms the validity of
the model and the measurement method. It also highlights
the good secondary coupling coefficient of the transformer
we used. However, a refined model should include the
numerous capacitive leakage elements but also the core
material model.
When designing Switch−Mode Power Supplies, the
understanding of how the various leakage elements interact
within the system is the key to designing reliable and rugged
systems. Due to easy measurement methods, the evaluation
of these parasitic components becomes simple. With the
help of a SPICE simulator and adequate models, the effect
can then immediately be visualized and properly
compensated.
References
1. Transformer modeling and design for leakage
control, Shi−ping HSU, R.D. MIDDLEBROOK,
Slobodan CUK, Advances in Switched−Mode
Power conversion, Volumes I & II, TESLAco
2. Modern DC−to−DC switchmode power converter
circuits, Rudolf. P. SEVERNS,
Gordon E. BLOOM
http:\\www.ejbloom.com
3. Alimentation à découpage, convertisseurs à
résonance, J.−P. FERRIEUX, F. FOREST,
MASSON FRANCE
4. Improved SPICE model simulates transformers
physical processes, Larry MEARES, Charles
HYMOWITZ, EDN August 19 1993
5. Thomson Television Components France, B.P. 24,
70101 Gray Cedex France, Fax.: 33 3 84 65 18 45,
e−mail: [email protected]
6. 150 W Flyback regulator, R. PATEL, D. REILLY
and R. ADAIR, UNITRODE Power Supply
Design Seminar SEM−500
7. The effects of leakage inductance on multi−output
Flyback circuits, Lloyd DIXON, UNITRODE
Power Supply Design Seminar SEM−500
8. SMPS Simulations with SPICE3, Steven M.
Sandler, MacGraw−Hill, ISBN 0−07−0913227−8
Figure 21. Measured Auxiliary and Power
Secondary Voltages
Vsec Aux
Vsec Power
1 ms/div
Figure 22. Simulated Secondary Waveforms
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