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Calculations for Flybacks
By Kirby Creel ,Senior Design Enginee, Datatronics,
Romoland, Calif.
When charging deibrillator capacitors, a novel
approach can be used with conidence to come
up with a more exacting design, eliminat-
ing the cumbersome aspects of many other
approaches.
voltage is a common approach. he voltage
can charge a capacitor for a high-energy
pulse. Such an approach is used in deibril-
lator capacitors, photolash capacitors, strobe
capacitors and ignition circuits to name a few. Using a new
step-by-step procedure, it’s possible to quickly realize an ini-
tial lyback transformer design for charging a capacitor in
a stated amount of time.
Following this procedure eliminates “cut and try” and
over-design approaches. It also allows designers to select
critical values with conidence, and can be used to provide
insight as to the efects of holding an element constant
and varying other elements. he variables included are
frequency, voltage, pulse width, peak current, load capaci-
tance and eiciency.
Before we delve into this novel approach, we need to
understand the pros and cons of lyback topologies. Ad-
vantages include circuit simplicity, a high-voltage output
that’s not dependent on a large transformer turns ratio, a
self-limiting circuit that can be short circuited without any
damage, and an output that can be regulated over a large
range. A lyback also can provide voltage isolation, allows
for multiple isolated outputs where one output can be used
for a low-ratio feedback voltage, and does not require a
smoothing choke.
he disadvantages of lyback topologies include the need
for a large and oten bulky transformer, a fast-switching
output that can generate problematic EMI signals, leakage
inductance that must be kept low for good eiciency and a
circuit that can be damaged once a load is removed without
using a feedback loop.
Fig. 1 shows a simpliied circuit and Fig. 2 shows ideal-
ized waveforms for a frequency of 50 kHz and a pulse on
time (T ON ) of 45%.
he lyback operates by storing energy on the “charge”
portion of the cycle and delivers the stored energy to the
load on the discharge cycle. In the case of a lyback, the
transformer is oten described as a coupled inductor. Due
T1
D1
R1
+
C1
R2
0RIMARYCURRENT
RAMPSCHARGE
Vdc
Q1
M S
U1
M S
Pulse
control
3ECONDARYCURRENT
DISCHARGE
$EADTIME
Fig. 1. A lyback loop for pulse control keeps a lyback transformer
from being damaged once a load is removed.
Fig. 2. Dead time for a lyback transformer increases as the
storage capacitor approaches a full charge.
Power Electronics Technology January 2008
26
www.powerelectronics.com
Expedite Transformer
U sing a lyback topology to generate a high
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Parameter Description
Capacitor 100 μF
Charge voltage 2000 Vdc
Charge time 10 sec
Circuit Discontinuous-mode lyback
Frequency 50 kHz
Maximum duty cycle 45%
Maximum on time 9 μs ( t)
Input voltage 12 Vdc
Eficiency See the fourth step and Fig. 3
Primary inductance To be decided
Peak current To be decided
Table. Values for a typical charging circuit.
voltage, (t) is the time duration from start to inish of
the applied pulse and (i) is the change in current over
the same interval. If i starts at zero, i is equal to the peak
current (I PEAK ):
I
=
Vt
L
() .
∆
(Eq. 2)
Eq. 2 is Eq. 1 solved for I PEAK . For example, if V A = 12 V,
t is 0 μs to 15 μs, L = 60 μH and I PEAK = 3 A.
he energy stored in the inductance is:
2
2 (Eq. 3)
where U is energy measured in Joules, L is in henries
and I PEAK is in amperes. In the previous example, the energy
stored in each pulse is 270 μJ.
It is during the discharge of this stored energy that the
greatest advantage of the lyback is realized. he output
voltage will rise to whatever level is needed to cause current
to low, thus dissipating the stored energy. he voltage of
the output has limits to be sure, but within the insulation
structure, transistor breakdown, and circuit design, and
taking losses into account, the voltage can rise to very
high levels.
hough ancient technology now, the most common
example was the lyback transformer in color TVs with a
cathode ray tube. hese transformers could generate volt-
ages greater than 35,000 V, voltages so high that, when the
= (
LI PEAK
)
,
to the diode polarity, current only lows in the secondary
side during discharge. During the charge cycle, energy is
stored in the primary inductance by a current ramp. he
dead time shown in Fig. 2 ensures that the lyback is dis-
continuous. As the capacitor approaches full charge, the
dead time increases.
he primary current ramp (charge) follows the induc-
tance formula:
L
=
∆
∆ (Eq. 1)
where L is the inductance in henries, V A is the applied
Vt
i
() ,
T
O
www.powerelectronics.com
27
Power Electronics Technology January 2008
A
PEAK
U
A
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flyback transformer
-AXIMUMEFFICIENCY
forms. In a deibrillator, pulse control will be a voltage feed-
back loop that ixes the number of Joules to be delivered to
the patient. During successive resuscitation attempts, the
level will increase. For photolashes, the charge level is ixed.
he capacitor will be charged and additional pulses will only
be applied as a refresh.
In photolash applications, the dead time may be limited
to speed up the charge time. he low dead time and vari-
able discharge produces the characteristic of an increasing
high-pitch sound. Variations in the pulse-control element
are almost endless.
%NERGYSTORED*OULES
Design Example
With the background provided, we can now tackle the
problem of charging a capacitor to a given voltage in a
stated amount of time. Designs begin with a list of known
values.
Let’s illustrate the process by considering the following
example of a typical design problem, for which the values
are listed in the table . he application is for charging a de-
ibrillator capacitor. (Caution: he charged capacitor used
in this example can provide a lethal shock.)
he irst step is to determine the number of Joules re-
quired to charge the capacitor:
Fig. 3. Energy storage eiciency in a lyback circuit is fairly linear up to
about 100 J, before it decreases asymptotically.
circuits malfunctioned, the TV could generate damaging
X-rays. he analysis is somewhat simpliied because the
TV lyback transformer performs more functions than just
generating high voltage. he design of the lyback circuit
and transformer for power transformation is well illustrated
in Abraham I. Pressman’s book Switching Power Supply
Design . [1]
he block in Fig. 1 labeled “pulse control” can take many
CV CAP
)
2
100 10 2000
2
−
6
(
)
2
U
= =
=
200 J,
(Eq. 4)
2
Power Electronics Technology January 2008
28
www.powerelectronics.com
(
×
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flyback transformer
) 0%!+ !
4
6DC
igure calculated in the previous equation can be solved with
an almost ininite number of solutions. he remainder of the
design requirements and the 500-µJ calculation limits the
answer to one solution (as shown next).
Eq. 1 is an inductance formula, while Eq. 3 is solved for
the inductance. he right-hand side of both equations are
equal to each other. he resulting expression, with only
one unknown, can be solved for the peak current. Having
found the peak current, the value is entered into the origi-
nal equation and solved for the inductance. For a “sanity
check,” both formulas are solved for the inductance. he
equal results provide conidence that the calculations were
performed correctly.
L Et
i
note that iI
$
2
, M (
M &
#
2
6DC
1
5
0ULSE
CONTROL
=
∆
∆
U
=
LI
PEAK
2
.
2
(
∆
=
PEAK
)
Solve for L:
Fig. 4. This completed design for a lyback transformer circuit with
component values is based on the calculations in this article.
12 910
(
× −
6
) .
2U
L
=
L=
I
I
2
PEAK
PEAK
2 500 10
× −
6
) .
where U is energy in Joules, V CAP is the capacitor voltage
and C is the capacitance in Farads.
he next step is to calculate the number of charging
pulses (N) in the stated time:
N =× =
L
=
I
2
PEAK
12 910
(
×
−
6
)
2 500 10
(
×
−
6
)
=
I I
PEAK PEAK
10810110
2
s 50,000 pulses/s , . (Eq. 5)
In the third step, calculate the energy required per charg-
ing pulse (U P ):
U
10
500 000
. ×
−
4
= ×
−
3
I
I
2
PEAK
PEAK
Joules
N
200
500 000
J/pulse.µ (Eq. 6)
he fourth step is to make an estimate of the circuit
eiciency and include the factor in calculating the energy
that must be supplied. All calculations up to this point are
based on the assumption that there were no losses in the
switching transistor, diode and transformer (winding or
core). Switching transistor, diode, and transformer losses
are shown to a irst order. Second-order losses are those for
winding capacitance and leakage inductance.
Fig. 3 provides an estimation of a typical loss factor ex-
pressed as an eiciency igure. he losses must be included
in the power supplied from the dc source, as shown in the
next calculation. Note that Fig. 3 is an estimate and results
will vary. Use the result from Eq. 4 (in this case, 200 J) to
ind the eiciency value. A number of variables in circuit
design, layout, transformer design, components and others
will afect the result. (For very high-voltage designs, even
leakage current paths across the surface of the pc board and
leakage within the capacitor must be considered.)
U
Efficiency
P = = =
400
I
I
I
2
110
10810
9 259
×
−
3
,
PEAK
=
.
. A.
Calculate the inductance by substituting the value of I PEAK
in both Eq. 1 and Eq. 3 as a check:
×
−
4
PEAK
PEAK
=
−
12 910
9 259
11 66
(
×
6
)
2 500 10
9 259
11 66
(
×
−
6
)
L
=
L
=
.
. H.
(. )
. H.
2
µ
Fig. 4 is the completed design with the calculated values
included.
A similar application to charging a deibrillator capaci-
tor is charging a photolash capacitor. Linear Technology’s
LT3468 IC performs most functions in a small footprint.
he IC is designed around the speciic application with some
other applications mentioned in the data sheet. One limita-
tion of this device is the breakdown voltage of the switching
transistor, which at 70 V at 25°C makes it better suited for
lower-voltage applications.
he LT3468 data sheet [2] provides design information,
circuits and typical waveforms, and gives a designer good
insight into the process, possibly sparking new avenues of
application. (Note: he data sheet provides a warning related
to working with high voltage. Many of the circuits discussed
can provide lethal shocks when working properly.)
For further illustration of the design process, the follow-
ing low-power example was built and tested. Test data, notes
on circuit operation and waveform are shown. he design
L
=
µ
L
=
P
=
Joules
from dc source
U
= =
J
0.8
µ
500
µ e.
P
In the ith and inal step, solve for the unknowns. here
are two unknowns and two equations previously presented
that will provide the answer. he unknowns are the trans-
former primary inductance and the peak current. he 500-µJ
www.powerelectronics.com
29
Power Electronics Technology January 2008
(
400
J/puls
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flyback transformer
T1 (see text)
12 V
600 V DC
12 V
+
1N6625
D ischarge
100 M F
+
10 M F
1
4
100 M 7
A
6 M F
Output
monitor
(see note)
1 M 7
7
24 7
3
5
B
6
8
1N
5817
Q1
1 k 7
Notes:
1. U1 = UC2845A, U2 = CD 4013G
U3 = TLV 3702I, Q1 = IRFD220
2. Capacitor charge may be terminated
manually or automatically (see below).
3. Configure to record total count.
Charge time = total count/pulses per second.
0.010
M F
U1
3
1 k 7
10 7
17.4 k 7
0.001
M F
7.5 k 7
4
1
0.001 M F
5
2
100 k 7
3.6 k 7
INPUT
Counter
(note 3)
A
Manual
“stop”
required
54.54 V DC
= Full charge
600 V DC
DVM
input =
10 M 7
+12 V +12 V
10 k 7
10 k 7
12 V
+12 V
B
8 14
START
STOP
A
Automatic
“stop” A
600-Vdc
D
5
9.774
M 7
Q
24 k 7
(2 PL)
1
0.01 M F
(2 PL)
RESET
4
+12 V
U2
+12 V
SET
226 k 7
No
connection
Q
6
8
2
5
3
B
CLK
B
24 k 7
3
U3
10
7
11
1
2
4
Z(v) = 1.235 V
(LM385-1.2)
Fig. 5. This circuit can be used to test the transformer for the lyback topology shown in Fig. 4.
starts with this list of given values:
C = 6 µF
V = 600 Vdc
Charge time = 10 sec
Circuit = lyback
Frequency = 50 kHz
Maximum duty cycle = 45%
Maximum on time = 9 µs
Input = 12 Vdc
Using the technique shown previously, the inductance
and peak charging current are calculated here:
Eq. 4 = 1.08 J
Eq. 5 = 500,000 pulses
Eq. 6 = 2.16 µJ/pulse
Eq. 7 = 4.32 µJ/pulse (eiciency is estimated to be 50%).
12 910
(
×
−
6
)
2 43210
(.
×
−
6
)
=
I
I
2
PEAK
PEAK
1081064 10
.
×
−
4
.
×
−
6
=
I
I
2
PEAK
PEAK
I
I
2
86410
10810
.
.
×
×
−
6
PEAK
=
=
80
mA.
−
4
PEAK
Calculate the inductance by substituting the value of
I PEAK in both Eqs. 1 and 3.
−
12 910
008
135
(
×
6
)
2 500 10
008
(
×
−
6
)
L
=
L
=
(. )
. mH.
Knowing the nominal primary inductance and the peak
current, the design of the transformer can proceed. he
transformer design for a lyback circuit does not follow
normal transformer design procedures. he lyback trans-
former can be viewed as two inductors sharing a common
core. (Reference 3 given at the end of this article illustrates
that selection of a transistor or the turns ratio is the irst step.
Chapter 7 of the same reference provides general design
guidelines. Reference 4 also provides detailed transformer
design information.)
.
. mH.
2
L
=
L
=
135
Et
i
LI
2
L
=
U
=
PEAK
2
(note that
∆
iI
=
PEAK
)
Solve for L:
12 910
(
× −
6
) .
2
U
L
=
L
=
I
2
I
PEAK
PEAK
2432 10
(.
× −
6
) .
L
=
I
2
PEAK
Power Electronics Technology January 2008
30
www.powerelectronics.com
∆
∆
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