Before reading this section, please read the
the circuits in this tutorial can be simulated in
LTspice®. If you are new to
LTspice, please have a
look at my
Sometimes an isolated voltage needs to be produced.
For low power applications, a flyback converter is
another form of dc/dc converter to fulfil this need.
Flyback converters are very similar to boost
converters in their architecture and performance,
except with a flyback converter the inductor is
replaced with the primary of a transformer and the
output is taken from the secondary. However, in the
flyback architecture, the primary and secondary
windings can be treated as 2 separate inductors, so
the flyback converter’s operation is very similar to
that of a boost converter.
following tutorial describes how to design a
boundary mode flyback circuit where the circuit
operates on the boundary between continuous and
A typical flyback converter is shown in FIG 1
This circuit converts 12V to a pseudo-isolated 5V
and can support a load of 1A. This circuit is
adequate to demonstrate the operation of a flyback
converter, even though it does not offer full
isolation – the feedback resistors R1 and R2 violate
the isolation barrier.
Ignore components R4, D2, Q2 and C5. These just
provide a linear regulator function to allow the
LTC3873 to be used with high input voltages.
When the FET switches on, the current ramps up
as with a boost converter.
So in the case of FIG 1, the current ramps up at a
or 400,000A per second. Thus if the MOSFET switches
off after 1us, the current through the primary will
have ramped up by 400mA.
When the MOSFET switches off, the voltage on both
the primary and secondary windings increase in an
effort to maintain the current flow. The winding
that conducts first is the one in which the current
flows and in the circuit in FIG 1, diode D1 conducts
causing the energy to flow into the output
capacitors C2, C6, C7.
In a boost converter, the inductor current ramps up
linearly during charge and ramps down linearly
during discharge. In a flyback converter, when the
current ramps up in the primary, energy is built up
in the transformer core. When the FET switches off
this energy is dumped into the secondary allowing
current to flow in the secondary, so the primary
current collapses immediately to zero. This causes
an immediate rise in current in the secondary, after
which the current in the secondary ramps down
linearly as it discharges. For the sake of analysis,
it is often better to treat the primary and
secondary transformer windings as 2 separate
inductors where energy is built up in the primary
and dumped into the secondary. Although a flyback
converter can be considered as consisting of 2
separate inductors, classic transformer action still
happens. When the FET switches off, the primary
current collapses to zero causing an opposing rise
in current in the secondary. The rise in secondary
current is proportional to the collapse in primary
current and is determined by the turns ratio of the
The discharge current in the secondary winding
charges the output capacitor slightly after which
the MOSFET switches on and the process starts again.
The output voltage is monitored by the feedback
resistors R1 and R2 and when the junction of these
resistors reaches a certain point, the chip
terminates the drive to the MOSFET.
Like with the boost converter, during charging the
primary and secondary currents ramp up and down
according to the equation
During the charging of the primary, L is the
inductance of the primary. During discharge, L is
the inductance of the secondary.
The LTspice circuit of FIG 1 can be downloaded here
(right click over the link and save as a '.asc'
LTC3873 Flyback Converter
LTC3873 datasheet can be downloaded here:
We mentioned earlier that the primary current ramps
up at 400,000A per second. Our LTspice simulation
shows a current ramp of 408kA per second (see FIG
The secondary inductance is 3.3uH, so the current
ramps down at a rate of
or 1.66 million amps per second. In this case the
output voltage is 5V and the voltage across the
diode is 0.5V so the voltage across the secondary
during discharge is 5.5V.
Our LTspice simulation shows a current ramp of
1600kA per second.
FIG 2 below shows the gate waveform (green), the
primary current (blue) and secondary current (red).
The LTspice circuit in FIG 1 shows the transformer
modelled as 2 inductors. The ratio of inductance is
equal to the square of the turns ratio, so the turns
ratio of this transformer is 3:1. Since a
transformer is a purely passive device, when the
voltage is stepped down, the current is
stepped up by an equal amount and vice versa.
Thus when the current stops in the primary (in FIG 2
this is approx. 1.4A), a current is produced in the
secondary of 3x1.4A = 4.2A.
It is interesting to note that the value of di/dt
is determined ONLY by the inductance
value and the voltage across the inductor. The controller IC has nothing to do with
setting the inductor ramp current.
Like with a boost converter, it is useful to know
the duty cycle of the flyback converter.
We have already stated that the flyback converter is
similar to a boost converter, so it makes sense to
start with the duty cycle of a boost converter.
We determined that the duty cycle of a boost
Now, with a boost converter when the FET switches
off the voltage across the coil rises until the
rectifier diode conducts. So the output voltage is
equal to the sum of the coil voltage and the input
So, for a boost converter we can write
With a flyback converter, the output voltage (on the
secondary) is referenced to ground, not the input
voltage. Looking from the primary side, the voltage
across the primary is equal to the output voltage
divided by the turns ratio. So if the output voltage
is 5V and we have a 10:1 turns ratio then the
voltage across the primary is 50V.
So if we call Vout’ the output voltage as seen from
the primary side we can say
In FIG 1, the turns ratio (primary to secondary) is
3:1. The output voltage is 5V, so the voltage across
the secondary is 5.5V including the diode drop. When
this is reflected to the primary this results in a
voltage of 16.5V across the primary. With an input
voltage of 12V, this results in a duty cycle of
In LTspice, we can measure the ON time as 3.5us and
a period of 6.14us, giving a measured duty cycle of
With a boost converter, the current in the inductor
must be continuous for this equation to hold true. A
similar situation is true with the flyback, but a
current in either the primary or secondary
must be flowing for the duty cycle equation to hold
Similar to the boost converter, the duty cycle is
only dependent on the input voltage, output voltage
and turns ratio. It has nothing to do with the
controller IC or the inductor value. Hence if the
load current increases, after the circuit has
settled down, the duty cycle will remain unchanged,
but the primary and secondary ripple currents will
not ramp all the way down to zero. So instead of
having a triangular waveform starting from zero, the
primary current will have a front end vertical step
from zero to a certain value, then ramp from there.
Likewise, the secondary current will not ramp down
to zero, but have a vertical step down to zero when
the MOSFET switches on.
The Effect of Changing the Turns Ratio
Picking a primary to secondary turns ratio equal to
Vin:Vout ensures the converter operates with a duty
cycle of 50%, as seen above. If the input voltage
goes up, the duty cycle goes down to maintain
regulation and vice versa.
If the transformer turns ratio is changed, the duty
cycle changes accordingly, but changing the turns
ratio also has an effect on the surrounding
For a high primary to secondary turns ratio, from
any given output voltage will result in a higher
Vout’ resulting in a higher duty cycle, so the
designer has to be careful that the maximum duty
cycle specification of the controller is not
violated. A higher Vout’ also means a higher voltage
across the primary side FET.
Since the duty cycle is higher (the ON time of the
primary FET is longer), the time to discharge the
secondary is proportionately shorter, hence more
current needs to flow in the secondary circuit to
maintain the output current. This means a higher
current flows in the rectifier diode, possibly
resulting in increased heat dissipation.
However, when the primary FET switches ON, a high
turns ratio means less reverse voltage is applied
across the secondary diode.
Similarly if the designer chooses a low primary to
secondary turns ratio, the duty cycle is lower and
the voltage across the primary side FET (Vout’) is
lower. However, since the ON time of the FET is
shorter, more current needs to flow in the primary
FET resulting in higher switching and conduction
Since the ON time is shorter, the discharge time of
the secondary is longer so the current in the
secondary diode is lower. However, the reverse
voltage across the secondary diode is higher.
Having a turns ratio equal to something other than
Vin:Vout is not necessarily a bad thing. The turns
ratio is just another variable in the design process
that could be used to the designer’s advantage.
Flyback Converter Design
We are now going to use a circuit similar to FIG 1,
but this time to boost a voltage of 5V to 12V that
can support a load of 100mA. We are going to use the
LTC3873-5, a fixed frequency 200kHz controller.
Unlike with a boost converter where inductors are
available in many different values, the perfect
transformer turns ratio might be hard to come by.
Therefore it is wise to start the design by choosing
a transformer that is close to where we need to be
and adjusting the components in the design to
compensate for its shortcomings. Many power supply
books explain how to calculate the ideal transformer
primary inductance and turns ratio, but most
engineers do not have the luxury of a custom
transformer service available.
The turns ratio affects the peak primary current,
peak secondary current and the duty cycle. In our
example, if the turns ratio is too low (less turns
on the primary) the duty cycle decreases and the
primary current increases (this is logical because
if the MOSFET is on for a shorter duration, more
current needs to be ramped in the primary per
cycle). If the turns ratio is too high the output
voltage reflected back to the primary will be
greater meaning a higher voltage MOSFET will be
needed. In addition, the duty cycle will be longer
(the ON time of the MOSFET), so the OFF time (the
time when the secondary current is supplying the
output capacitor) will be shorter, so more secondary
current needs to flow to furnish the load.
A primary to secondary turns ratio of approximately
Vin:Vout is a good place to start. So with 5V input
and 12V output a turns ratio of 1:3 is a good
The circuit we are going to design will assume the
secondary current ramps down to zero after which the
primary MOSFET immediately switches back on again to
begin charging the primary. In other words, the part
is working on the boundary between continuous
conduction mode (where current is always flowing in
either the primary or the secondary) and
discontinuous mode (where there is a region of zero
current in both windings before the MOSFET starts
charging the primary again). We will design for
maximum load, therefore if the load current
decreases, there will be a delay between the
secondary current ramping down to zero and the
primary current charging again (discontinuous mode).
FIG 3 shows our outline architecture.
Referring to FIG 2 earlier in the text, we see that
the secondary current is triangular ramping down
from a peak value to (nearly) zero. The area under
this red waveform needs to have an average of 100mA
to support our load. Since the waveform is
triangular, the peak of the current needs to be at
least 2 x 100mA = 200mA. However, the secondary
current is only present for a certain amount of time
(dictated by the duty cycle), so the current needs
to be considerably greater than this.
A flyback converter has a duty cycle of
(i.e. Vout’ is the output voltage, but as seen by
the primary side)
Earlier we calculated the duty cycle ignoring the
diode drop. We can now include it without too much
hardship. If we include a diode drop of 0.3V in our
duty cycle equation, our value for Vout’ is 12.3 x
0.33 = 4.1V, so the duty cycle is
Obviously the lower the output voltage, the more
influential the diode drop is in our duty cycle
The primary current flows during the ON time of the
MOSFET and the secondary current flows during the
OFF time of the MOSFET, so the secondary current is
only present for (1-DC) during each cycle. Therefore
to get the desired peak secondary current, we need
to divide the peak 200mA current (calculated above),
by (1-0.45) to get the true secondary peak current
Thus the peak desired current in the secondary is
For a more
mathematical derivation of the above procedure,
A 364mA peak current in the secondary, with a turns
ratio of 1:3 means a peak primary current of 1.09A
(going from secondary to primary, the voltage goes
down so the current goes up in this configuration).
A 200kHz switching frequency has a period of 5us, so
with a duty cycle of 45%, this represents an ON time
0.45 x 5us = 2.25us
during the MOSFET ON time, the voltage across the
primary is 5V and we need a current ramp from zero
to 1.09A in 2.25us, so this implies a primary
inductance of 10.32uH
So we need to pick a transformer with a primary
inductance of approximately 10uH, with a saturation
current rating of at least 1.09A and a turns ratio
of 1:3. This is quite a specific requirement and the
reader will probably find it impossible to find a
transformer with such characteristics. All is not
lost however, as magnetics companies are now
introducing general purpose transformers with 6
windings on one bobbin that can be configured in any
way. Wurth Electronics provide such a series of
parts with the WE-FLEX Flexible Transformers.
Since we have more current on the primary than
secondary, it makes sense if our design has 3
windings in parallel for the primary and 3 windings
in series for the secondary. This ensures each of
our 3 windings shares the 1.09A current and gets us
our 1:3 turns ratio.
749196221 is a suitable
We must ensure that the transformer has a current
rating of at least 1.09A. If we exceed this current,
the ferrite that the transformer is wound on will
saturate and lose its magnetic properties. Thus the
inductance value of the primary will collapse and
from the equation
the current will increase rapidly, thus blowing up
the primary side MOSFET.
Now, parallel inductors wound on the same ferrite do
not follow the same laws as separate inductors
placed in parallel. They keep the same overall
inductance value, but share the current. This is
explained the following Wurth document:
in series and parallel
has an inductance of
11.6uH per winding and 3 in parallel will still give
a primary inductance of 11.6uH. As they share the
current, the saturation current of 0.84A allows us
to have (3 x 0.84A = 2.52A) peak primary inductor
If a suitable transformer cannot be found, select a
dc/dc converter with an adjustable switching
frequency (hence changing the parameter ‘dt’ in the
above equation) and repeat the steps above.
The current sense threshold on the LTC3873-5 is
95mV, so a current sense resistor of 63mOhms ensures
a peak current of 1.5A. The above calculations have
not accounted for the efficiency of the dc/dc
converter, so a current sense resistor of approx.
80% of this value is a more realistic figure. A
current sense resistor of 50m Ohms should suffice.
With an output voltage of 12V, the voltage on the
secondary winding is 12.3V (including a diode drop
of 0.3V). Therefore when the MOSFET switches off,
the primary winding develops a voltage of 12.3V/3 =
4.1V. Thus the MOSFET drain has to withstand a
voltage across its drain of 4.1V + Vin = 9.1V.
Now, it is worth choosing a MOSFET with a drain
source rating much higher than this since any
transformer will not perfectly couple the primary
energy into the secondary. This term is called
leakage inductance and can be modelled as an
inductance in series with the primary that is not
coupled to the secondary. Thus it stores energy
that is not dumped into the secondary during the
flyback cycle. This energy manifests itself as a
voltage spike on the drain of the MOSFET. This spike
increases with primary current and if it is high
enough, a snubber network may be required to reduce
In our design a MOSFET with a drain source voltage
of 30V should suffice.
The drain current of the MOSFET needs to be higher
than the peak current set by the current sense
resistor. Choosing a MOSFET with a drain current of
3A leaves us plenty of headroom.
The gate source turn on voltage of the MOSFET needs
to be much lower than the voltage of the gate pulses
coming out of the chip (5V).
The above parameters represent the bare minimum
characteristics of the MOSFET. However, to get a
good design, we must ensure that the losses in the
MOSFET are as low as possible. The MOSFET switch
presents 2 losses in the circuit: switching losses
and conduction losses.
The switching losses result from current flowing
through the MOSFET at the same time that a voltage
is across the MOSFET (so power is generated in the
MOSFET), during the turn on and turn off times of
the MOSFET. For a given gate drive coming out of the
controller IC, the lower the Gate-Source capacitance
of the MOSFET, the quicker the MOSFET will turn on.
Thus the Qg specification of the MOSFET is important
and should be as low as possible. The Qg of the
MOSFET will also have an impact on the heat
dissipation of the chip, especially if the input
voltage to the chip is high.
Charge is dictated by the equation:
Charge (Q) = Current (I) x Time (s)
Since Frequency is the inverse of Time, we can write
So we can calculate the current needed to flow into
the chip, just to charge the gate capacitance of the
FET. Since heat is the product of voltage and
current, if the gate charge is high and/or the
switching frequency is high, the heat dissipation in
the chip will be high if the input voltage is high.
Once the MOSFET has switched on, the MOSFET presents
a small dc resistance between its Drain and Source
terminals. This is the MOSFETs ‘Drain Source on
resistance’ or Rdson. Again, this needs to be as low
Now, MOSFET manufacturers reduce the ON resistance
of the MOSFET by constructing many parallel
conduction paths between the Drain and Source. Thus,
like connecting resistors in parallel, the ON
resistance comes down with more parallel paths.
However, in connecting Drain Source paths in
parallel, a negative effect is that the Gate Source
capacitance (Qg) is also connected in parallel, so a
low ON resistance (and hence low conduction loss)
sometimes implies a high gate source capacitance
(hence high switching loss). Thus the MOSFET that is
chosen should be a compromise between these two
characteristics. In addition, high current MOSFETs
tend to come in much larger packages, so meeting the
ideals of low ON resistance and low Qg might violate
a space requirement spec, so the selection process
has to start over. Engineering, as ever, is a
Indeed looking at the selection tables of the MOSFET
manufacturers, it is better to select a MOSFET with
a low ON resistance (less than 10mOhms), then filter
this selection to remove MOSFETs with a Qg of
greater than 10nC, then select a MOSFET from this
list, as long as the Gate turn on voltage, Vds and
Id can be met. Starting by selecting MOSFETs with a
Vds of between 20V and 30V might rule out some
higher voltage FETs that are better suited to lower
Failing that, download all the results to a
spreadsheet and sort from there. I have never had
much luck with the parametric searches on MOSFET
Alternatively, download all the MOSFET
characteristics into a spreadsheet, remove the ones
that don't meet the VDS and ID requirements, then
add a column called FOM (Figure of Merit). This
column should contain the value RDSON x QG. Then
sort by this column and pick the FET with the lowest
FOM. This part will be the best compromise between
RDSON and QG and ideal for the top MOSFET.
The Fairchild FDS6680 represents a good compromise
between low ON resistance and low gate charge, but
its SO8 package is large and therefore might be
unsuitable for compact designs.
Rectifier Diode Choice
When the MOSFET switches off, the secondary voltage
ramps rapidly in order to maintain current flow.
Many diodes are not fast enough to react to this
voltage change, resulting in a large spike on the
Drain of the MOSFET. This can (and does) destroy the
Therefore Schottky diodes should be used in all
dc/dc converter designs. Ultra fast diode have a
response time of 10’s of nanoseconds, standard
rectifier diodes have a response time of several
microseconds, whereas a Schottky has a response time
in the order of a few nanoseconds. Schottky diodes
also have a much lower forward voltage drop (0.3V)
compared with standard rectifiers (0.6V) so half the
power is wasted as a result of VxI losses.
When choosing a Schottky diode, the key parameters
are: forward voltage drop (should be as low as
possible), forward current (this should be greater
than the peak secondary current) and reverse voltage
rating. When the FET is charging the primary, there
will be a voltage across the primary equal to the
input voltage. With a flyback converter, the anode
of the secondary rectifier diode will see the same
voltage, but multiplied by the turns ratio. So in
our example, for a 5V input, the secondary winding
will develop 15V across it during the charging of
the primary. Its cathode will be held as the output
voltage (12V), so the diode needs to withstand a
reverse voltage of 27V.
In this design example, the MBRS340 is a good choice
with a reverse voltage rating of 40V and a forward
voltage of 0.53V at 3A peak current.
Output Capacitor Choice
Unlike the buck converter that has a continuous
current flowing from the inductor into the output
capacitor, the flyback converter output capacitor
has to keep the output voltage alive when the
primary is being charged.
The output ripple is made up of 2 components: the
ripple caused by the output capacitor discharging
when the primary is being charged and the ripple
caused by the inrush current from the secondary
winding into the ESR of the output capacitor. Low
ESR tantalum capacitors normally have an ESR of
<100m Ohms and ceramic capacitors are significantly
less. Putting 2 capacitors in parallel will double
the value and half the ESR.
The ripple caused by the discharge of the output
capacitor while the inductor is charging is dictated
where i is the load current in Amps, C
is the output capacitance in Farads and dv/dt
is the change in output voltage with time.
Earlier we calculated that the MOSFET switches on
for a period of 2.25us. If we require a discharge
ripple of 0.5% (60mV) with a load current of 100mA,
this implies we need a capacitance of
The ripple caused by the ESR is a product of the
peak secondary current and the ESR. In our example
the peak current is 364mA and the ESR is of a
typical tantalum capacitor is 70m Ohms, giving a
ripple of 25.5mV.
Therefore a capacitor of 4.7uF and an ESR of 70m
Ohms should enable us to comfortably meet our ripple
Other Points to Note
feedback resistor values can be calculated using the
Feedback Resistor Calculator:
Feedback Resistor Calculator
The feedback resistors were set to give a voltage of
12.04V. Keep the resistor values in the less than
500k. If they are too high, they will form a low
pass filter (and hence a phase shift) with the input
capacitance of the feedback pin. Higher resistor
values are also more susceptible to induced noise
from the inductor. If the values are too low, they
will drain current from the output unnecessarily.
Please refer to the datasheet for full information
on the operation of the LTC3873-5
This text has explained the basics of flyback
converter switched mode power supply design and is
applicable to most flyback converters. Refer to the
individual datasheets for a complete guide to
designing with that particular part.
The final LTspice circuit can be downloaded here
(right click over the link and save as a '.asc'
LTC3873-5 flyback converter
LTspice is a registered trademark of Linear