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Flyback Converters for Dummies
A simple flyback converter high voltage power supply for NIXIE
tubes.
Ronald Dekker
Special thanks to Frans Schoofs, who really understands how
flyback converters work
introductionWhat you need to know about inductorsThe boost
converterA simple boost converter high voltage supply for NIXIEsAn
inductor test benchWhat you need to know about transformersThe
flyback converterA flyback converter high voltage supply for
NIXIEsback to homepage
If you are interested in FlybackConverters you might want to
keep track of my present project: theTracer:
a miniature radio-tube curve-tracer
Click here to read about my low-noise 6 to 90 V converter
project
which replaces the anode battery inbattery tube receivers.
introduction
In the NIXIE clocks that I have built, I did not want to have
the big and ugly mains transformer in theactual clock itself.
Instead I use an AC adapter that fits into the mains wall plug.
This means that Ihave to use some sort of an up-converter to
generate the 180V anode supply for the NIXIEs.
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This page describes a simple boost converter and a more
efficient flyback converter both of which canbe used as a high
voltage power supply for a 6 NIXIE tube display. Frans Schoofs
beautifullyexplained to me the working of the flyback converter and
much of what he explained to me you findreflected on this page. I
additionally explain the essentials of inductors and transformers
that you needto know. This is just a practical guide to get you
going, it is not a scientific treatise on the topic.
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What you need to know about inductors
Consider the simple circuit consisting of a battery connected to
an inductor with inductance L andresistance R (Fig. 1). When the
battery is connected to the inductor, the current does not
immediatelychange from zero to its maximum value V/R. The law of
electromagnetic induction, Faraday's lawprevents this. What happens
instead is the following. As the current increases with time, the
magneticflux through this loop proportional to this current
increases. The increasing flux induces an e.m.f. in thecircuit that
opposes the change in magnetic flux. By Lenz's law, the induced
electric field in the loopmust therefore be opposite to the
direction of the current. As the magnitude of the current
increases,the rate of the increase lessens and hence the induced
e.m.f. decreases. This opposing e.m.f. results ina linear increase
in current at a rate I=(V/L)*t. The increase in current will
finally stop when itbecomes limited through the series resistance
of the inductor. At that moment the amount of magneticenergy stored
in the inductor amounts to E=0.5*L*I*I.
Figure 1
In words: the inductor does not allow for any abrupt changes in
the current. When a change in appliedvoltage occurs, the inductor
will always generate an e.m.f. that counteracts this change. When
thecircuit is interrupted for instance, the inductor will still try
to maintain the current flowing by generatinga very high voltage
over its terminals. Usually this will result in a spark in which
the magnetic energystored in the inductor is released. This
particular behavior of inductors is used in boost converters
toboost the voltage to levels above the battery voltage.
Materials like ferrites can be used to increase the magnetic
flux in an inductor. When a magnetic field isapplied to a ferrite
the small magnetic domains in the ferrite will align with this
field and increase its
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magnitude. In this way inductors can be made smaller and with
lesser turns and thus with smallerseries resistances (smaller
losses). Note that the flipping of these domains costs some energy,
but ingood ferrites this can be very small.
With increasing magnetic flux more and more magnetic domains
point into the direction of the field. Ata certain point all the
magnetic domains point into the direction of the field and at that
point we saythat the ferrite saturates. Any further increase in
currentwill only result in a small increase of flux, basically as
ifthe ferrite was not present. Since most ferrites have avery high
permeability, already small currents can resultin a high magnetic
flux. As a result the ferrite willsaturate at a current which is
not practical for powerconversion applications
Ferrite cores for inductors and transformers for
powerapplications therefore have an air gap. An air gapreduces the
effective permeability and thus themagnetic flux. The larger the
air gap, the stronger thereduction in flux an the higher the
maximum current the inductor can handle. We say that the
magneticenergy is stored in the air gap. The photograph shows
several inductors for DC/DC converterssalvaged from old PCBs from
PCs, Laptops etc. If you consider playing with DC/DC converters it
isbest to buy at least one decent inductor with a known inductance,
series resistance and maximumcurrent. The inductor in front of the
picture is the 100uH "reference" inductor I use.
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The boost converter
The boost converter is perhaps the simplest of all switched mode
converters. It uses a single inductorwithout the need for
"difficult" transformers. It's working can best be explained with
the simplifiedcircuit diagram given in Fig. 2. Here the transistor
is represented by an ideal switch and the controlcircuitry has been
omitted. The dissipation by the NIXIE tubes is represented by the
load resistorRload. A high voltage capacitor C is used to buffer
the output voltage. In a typical configuration theinput voltage
would be something like Vbat=12V and the output voltage
Vout=180V.
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Figure 2 Simplified circuit diagram of a boost converter.
At t=0 the switch closes (Fig. 2A). As a result the current
through the inductor will start to increaselinearly according to
I=(Vbat/L)*t. At a certain moment the switch is opened by the
control circuit (Fig.2B). The current at that monent has reached a
certain value Ipeak. We have seen in the previoussection that the
inductor wants to keep the current flowing through it's windings
constant, whatever ittakes. The switch is open, so the only way the
inductor can achieve this is to forward bias diode D sothat the
current (and thus the energy) can be dumped in the buffer capacitor
C. Now remember thatthe capacitor was charged to 180V! So in order
to forward bias the diode, the inductor has to generatean e.m.f.
(or induction voltage) of something like 180-12=168V., something
like a "controlled spark.The current now quickly drops according to
I=Ipeak-(Vout/L)*t. For Vbat=12V and Vout=180V thismeans that it
will take only a fifteenth (180/12) of the time it took to reach
Ipeak when the switch wasclosed, to drop again from Ipeak to 0 now
the switch is open. After a certain time the whole processrepeats
at a rate of f times per second.
So far so good. However, the boost converter has a serious
disadvantage. To understand this we firsthave to consider the
switch that we have been using. In a real circuit most likely a
power MOStransistor will be used as the switching element. In the
boost converter this transistor will have tohandle both a high
current when the switch is closed and a high blocking voltage when
the switch isopen! For the transistor this is a difficult
combination. In order to make the transistor withstand highblocking
voltages, the manufacturer of the transistor has to include regions
in the transistor that willaccommodate these voltages so that the
intrinsic transistor will not breakdown. However, when theswitch is
closed (transistor conducting), these regions will result in
additional parasitic series resistancesand thus in an increased
Ron. This is the reason why transistors with a high breakdown
voltage alwayshave a higher Ron than transistors with a lower
breakdown voltage. Since the currents can be quitehigh, this
inevitably means losses in the form of dissipation in the
transistor. As we will see in one ofthe next sections this problem
is solved in the fly-back converter by the use of a
transformer.
By balancing the amount of power stored in the inductor to the
amount of power dissipated in the loadit is possible the calculate
the output voltage of the boost converter.Every second the amount
of power dissipated by the load is:
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If T is the total cycle time, and x the fraction of T that the
switch is closed, then the maximum currentin the inductor is:
The energy per package delivered by the inductor is:
In one second f=1/T packages are delivered so the amount of
energy delivered per second is:
Since in steady-state the amount of energy delivered should
equal the amount of energy used [1]=[2]:
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A simple boost converter high voltage supply for NIXIEs
If you want to build a simple DC/DC converter to lighten up your
NIXIEs and you don't care to muchabout the conversion efficiency,
even if it means a (small) heatsink for the power transistor, then
theboost converter is the best choice. But even if you think of
building a real fly-back converter than it isa good idea to start
with a simple boost converter. The boost converter only requires an
"of the shelf"inductor and when you have it working it is easily
converted into a fly-back converter by a few
smallmodifications.
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Figure 3 Simple 12-180V boost converter using the 555 as
controller.
The circuit is very simple and closely follows the circuit
topology of Fig. 2. For the transistor I haveused a BUZ41A. This
transistor is rated at a maximum Vds=500V and an on resistance of
1.5ohm at4.5A. Equivalent or better types like the IRF730 will also
perform well. The diode should be a fastswitching type like the
BYW95C or better. An old (computer) power supply will yield you
most ofthese components. The inductor I picked from a catalogue and
is 100H with a few tenths of an ohmseries resistance capable of
handling several Amps of current.
The most interesting aspect of the circuit is how an ordinary
555 is used to regulate the output voltage.Now, there are hundreds
of switched mode controllers ICs on the market which are all better
suited forthis job than the 555. The problem with all these ICs is
that if you build a nice NIXIE clock usingthem, and at one moment
in the future the IC breaks down, it is more than likely that it is
alreadyobsolete and out of production. The 555 is (very) cheap,
performs well enough and most likely willremain in production
forever.
To understand how the controller works it is best to first
understand how the 555 functions. On theinternet you may find a
number of excellent 555 tutorials [1,2]. Without R3 and T1 the 555
isconfigured as a normal astable multivibrator running at a
frequency of:
Without any feedback, the output voltage at this frequency will
be well in excess of 200V. However,the voltage divider formed by
R4, R5 and R6 has been designed and adjusted in such a way that
whenthe output voltage reaches 180 V, T1 just starts to conduct.
This is at a base-emitter voltage of ca.
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0.8V. Now remember that the 555 works by charging and
discharging the capacitor between 1/3Vccand 2/3Vcc as defined by an
internal resistor ladder network. When T1 starts to conduct it will
pulldown the internal supply voltage of this network resulting in a
smaller voltage swing and hence a higherfrequency. From the last
equation in the previous section we learn that a higher frequency
(smaller T)will result in a lower output voltage. In this way the
output voltage will settle at a value determined byR5. For T1 I
have used a high voltage type. There is really no need for that and
any small signal npntransistor with a decent gain will work. A
drawback of such a simple controller is that the circuit has
noprotection at all against short circuits or overload situations.
An accidental short circuit of the outputwill therefore always
result in a defect power transistor (as I have experienced quite a
number oftimes).
Figure 4 Testing the boost converter using a dummy load (and one
NIXIE).
If you are in the testing phase, and do not want to connect the
power supply to the NIXIEs yet, it isbest to connect a dummy load
to the output since the circuit is not designed to work without a
load. Ialways first find out what the current is that I want to
operate my NIXIEs on. I usually choose a valuewell below the
operating condition specified in the datasheet. This will greatly
extend the lifetime of thetubes. Using a high voltage supply I
select the supply voltage and the load resistor in such a way
thatwith a minimum of current the brightness of the tube is still
good enough. Once the total current andthe voltage are known an
equivalent load resistor can be calculated from Rload=Vout/Itotal.
During thetesting phase this resistor connected to the output
replaces the NIXIE tubes.
A few words about safety. Although the 180 Volts are generated
starting with an innocent 12 Volt anaccidental contact with the
charged buffer capacitor will be a painful, possibly a lethal
experience.Always be very careful ! I always place a small neon
indicator lamp at the output of the converter(even in the final
clock) to clearly indicate that a dangerous voltage is present at
the output.Additionally during testing I permanently have a
20kohm/V multi-meter connected to the output so thatI always know
the output voltage. Finally, the advice of my father who was from
the radio tube area:always keep one hand in your pocket when
touching the circuit when it is switched on. In that way thecurrent
can never pass your heart.
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An inductor test bench
When you want to start experimenting with boost or fly-back
converters it is good idea to buy at leastone inductor with known
parameters that may act as a kind of reference device for the
inductors ortransformers that you make yourself. I use a 100H
inductor with about 0.2ohm series resistancecapable of handling
several Amps of current. It is especially designed for SMP
applications. The circuitdepicted in Fig. 5 allows you to compare
"an unknown" inductor (or transformer) with the
referenceinductor.
Figure 5 Circuit diagram of the inductor test bench.
The circuit is designed to test the inductor as closely as
possible under conditions that occur in theboost converter
presented in the last section or in the fly-back converter to be
presented in one of thenext sections. Basically, the circuit is
little more than the inductor which is connected to the 12V
powersupply by transistor T1. The current through the inductor is
measured by the small series resistor R4.A voltage drop of 100mV
over R4 corresponds to a current of approximately 1A. When the
transistoris opened, the inductor can dump its energy in diode D3.
Since the voltage drop over the diode is only0.6V, it will take
about 12/0.6=20 times as long for the current to drop to zero
(remember I=(V.t)/L).This is the reason why the gate of the
transistor is driven with a highly asymmetric signal generated
bythe oscillator around N1-N6. The transistor on-time is determined
by C1 and R1+R2. R2 is set so thatthe transistor on-time is equal
to the transistor on-time in the converter under normal load.
Thetransistor off-time is determined by C1 and R3 and about a
factor 20 longer than the on-time.
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Figure 6 The inductor test bench circuit (left) and a
measurement off the reference inductor (right).
In Fig. 6 (right) you find a measurement of the reference
inductor. We find that with a supply voltageof 12V the current
through the inductor reaches a value of I=V/R=0.361/0.11=3.28A in
27.1s. SinceI=(V/L)t we find for the inductance
L=(V/I)t=(12/3.28)27.1=97.6H. Not bad! At a little bit
highercurrent we observe a sharp increase in the current through
the inductor. This is the point where theferrite saturates. The
inductor should not be used beyond this point.
You may now want to try different inductors e.g. inductors
salvaged from old (computer) powersupplies. Switch S1 make it easy
to compare these inductors with the reference inductor.
Anotherimportant parameter to watch is the current consumption of
the test-bench. An increase in switchinglosses in the inductor core
is reflected by an increase in power consumption.
An alternative simple and quick way to measure the inductanceof
an unknown inductor can be found here.
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What you need to know about transformers
This section deals with a few basic things you need to know
about transformers in order to understandfly-back converters. In
Fig. 7 I have tried to sketch an elementary inductor and its
schematicequivalent. Note that both windings have a certain
direction and that equal directions are indicated by adot.
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Figure 7 Basic transformer with open secondary windings
In this example we assume that the primary side of the
transformer has a certain number of turns withinductance L1. The
secondary side of the transformer has ten times that number of
turns. As a resultthe secondary side will have an inductance
L2=10^2*L1=100*L1. First consider the case that thesecondary
windings are not connected. When a voltage source is connected to
the primary coil thecurrent through the primary winding will start
to increase linearly at a rate I=(V/L1)*t. Since with openterminals
at the secondary side no secondary current can flow, the
transformer will behave as a normalinductor with inductance L1. The
increasing primary current will generate a magnetic flux not
onlythrough the primary windings, but the same flux will also flow
through the secondary windings. It iseasy to see from reasons of
symmetry that if the secondary coil would be identical to the
primary coilthe voltage at the primary and secondary side would be
equal. In this case we have 10 times thenumber of turns at the
secondary side. This can be seen as a series connection of 10 coils
eachcarrying a voltage of 10V so that in total 100V is induced at
the primary side. The voltage of 100V atthe output remains as long
as the current continues to increase linearly. In practice this
means until thecurrent reaches its compliance or until the core
saturates.
Figure 8 Basic transformer with closed secondary windings
Next the secondary winding is connected to some load which will
allow for a current to flow (Fig. 8).If the primary winding is now
connected to some voltage source, a current through the
primarywinding will start to flow, resulting in magnetic flux in
the direction as indicated by the arrow. Thismagnetic flux will
obviously also flow through the secondary winding. We have seen
that an inductorresists a change in magnetic flux. To counteract
the increasing flux, a current flowing in oppositedirection through
the secondary winding will start to flow as indicated in Fig. 8
Resulting in a voltagedrop over the load as indicated.
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Figure 9 The transformer in flyback
Finally the voltage source at the primary side is suddenly
removed (Fig. 9). The only way thesecondary winding can prevent a
sudden collapse of flux is to reverse the direction of the
currentflowing through the secondary winding. As a result alsothe
voltage drop over the load will reverse.Note that the voltage over
the load will increase to any value that is needed in order to
maintain aconstant flux. The magnetic energy stored in the inductor
is dumped into the load and the secondarycurrent decreases at a
rate Vout/L2
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The flyback converter
Figure 10 depicts the basic elements from the flyback converter.
Again all control circuitry is omitted,and the switching MOSFET is
represented by an ideal switch.
Figure 10 Phase one, storing energy in the transformer.
For the moment we assume that at t=0 the buffer capacitor is
charged to the nominal output voltageVout and that the current
through the primary windings of the transformer is zero. At t=0 the
switchcloses and a current will start to flow through the primary
winding. This will induce a voltage over thesecondary winding with
a polarity as indicated (see previous section). Since the diode is
reverse biased
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no secondary current will flow, so basically the secondary
winding is "not connected". In other wordsat the primary side of
the transformer we "just see an inductor". As a result the primary
current willstart to increase lineary according to I=(12/L1)*t.
During the time the switch is closed the voltageinduced over the
secondary windings will be n*12V. This means that the diode
minimally has to blocka reverse voltage of n*12+Vout
Figure 11 Phase two, dumping the energy from the transformer
into the buffer capacitor.
At a certain moment the switch will open (Fig. 11). Lets call
the current that was flowing through theprimary winding at the
moment just before the switch was opened Ipeak. The energy then
stored atthe moment of opening is 0.5*L1*(Ipeak*Ipeak). The
transformer wants to sustain the magnetic flux.Since the circuit at
primary side is open the only way the inductor can do this is by
inducing a voltageat the secondary side high enough (>Vout) to
forward bias the diode. The initial value of the currentwill be
I2=Ipeak/n. During the time that the diode is forward biased, the
voltage over the secondarywinding will equal Vout+0.8V. The 0.8V is
the voltage drop over the diode and can for a high outputvoltage
like in a NIXIE converter be neglected. The transformer will
transform this voltage down toVout/n. So the total voltage that the
switch has to block in open position is 12+(Vout/n).
Actually this is the big advantage of a flyback converter over a
boost converter. In a boost converterthe switch (MOSFET) has to
carry a large current during the on phase and a high voltage during
theoff phase. In the flyback converter the voltage during the off
phase is transformed down to a valuedetermined by the ratio in
turns. This means that a MOSFET with a much lower Ron can be used
(seesection on the boostconverter). Similarly, in the boost
converter the diode has to carry both the high oncurrent and a high
reverse voltage. In the flyback converter the diode at the
secondary side only has toblock a high voltage while the current is
low (Ipeak/n). This makes it possible to select a diode withsmaller
capacitances and hence higher switching speed. All this results in
reduced losses and anincreased efficiency.
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Figure 12 Phase three, energy dump completed discharge of
drain-source capacitor
This continues until all energy stored in the transformer is
dumped in the buffer capacitor. At thatmoment I2 becomes zero (Fig.
12). At that moment the e.m.f induced at the primary side (Vout/n)
willvanish. However, the parasitic capacitance of the switch
(source-drain capacitance of the MOSFET)will be charged to
(Vout/n)+12 V. At the primary side now a series resonant tank is
formed with acharged capacitor (Fig. 12 right). This will cause a
dampened oscillation.
Figure 13 Voltage over the switch during all three phases
Figure 13 schematically shows the drain-source voltage (the
voltage over the switch) during all thephases of the converter just
described. During phase the switch is closed. What we see is the
voltagedrop over the switch caused by the non-zero on resistance.
During this phase the current will increaselinearly, so also the
voltage drop over Ron will increase linearly. At point b the switch
opens. Thesecondary current will start to flow and the output
voltage wil appear down transformed over theprimary winding. The
total blocking voltage over the switch will be 12+(Vout/n) (Fig
13c). At point dall the energy is dumped in the capacitor and the
secondary current drops to zero causing the inducede.m.f. at
primary side to vanish. The charged drain-source capacitor, now
suddenly connected in serieswith the inductance of the primary
winding will result in a dampened oscillation (Fig. 12e). At point
fthe switch closes again, and any remaining energy in the LC tank
will be dissipated in the transistor.
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Figure 14 Stray inductance.
This leaves just one small phenomenon to be explained. No
transformer is ideal. There will always bemagnetic field lines
generated by the primary windings which are not (fully) enclosed by
the secondarywindings. This will cause a stray inductance that can
be modeled as a small inductor in series with theprimary winding of
the transformer (Fig. 14). We have seen that all the energy that is
stored in thetransformer is dumped in the buffer capacitor. This
does not hold for the (small) amount of energystored in the stray
inductance. So the sudden opening of the switch will cause a sharp
voltage peak,just as with any inductor which is suddenly
disconnected from a DC current. The small stray inductorin series
with the source-drain capacitance will cause a dampened high
frequency oscillation (Fig. 15).
Figure 15 High frequency oscillations due to energy stored in
the stray inductance.
If needed the switching transistor can be protected from the
high voltage peak by an RC snubbernetwork or a zenerdiode which
limits the maximum source-drain voltage.
Finally you can check for yourself the equation derived for the
output voltage of the boost converteralso holds for the flyback
converter. This is not really surprising, like in the boost
converter the flybackconverter is based on the dumping of the
energy from an inductor or the primary winding of a
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tranformer in the load. The transformer just serves to lower the
voltage over the switch.
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A flyback converter high voltage supply for NIXIEs.
After all what has been said so far, the circuit diagram of the
flyback converter will hold no surprises(Fig.16). Literally the
only difference with the boost converter is that the inductor is
replaced by atransformer, and that the transistor has been replaced
for a BUZ21. The BUZ21 has a much lower onresistance (Ron=0.085
ohm) as compared to the BUZ41A (Ron=1.5 ohm) but also a lower
drain-source breakdown voltage (100V versus 500V).
Figure 16 Circuit diagram of the Flyback converter.
The difficult part of the circuit is the transformer. Well it is
not exactly difficult, but the problem is thatyou have to make it
yourself. What makes things worse is that finding a suitable
ferrite core can sometimes be difficult since component vendors
often only have a few types on stock. The E-shape ferritecore that
I use measures 20x20x5 mm (Fig. 16) I got them from Paul van de
Broek who always helpsme when I need something special.
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Figure 17 The ferrite core that I use (20x20x5 mm).
So what is the strategy for finding the number of turns you need
on the ferrite core that you have?Well first of all I always start
with my inductor test-bench so that I can compare what I have
madewith the reference 100 H inductor. If this is your first
fly-back converter it might be illustrative to firsttry the ferrite
core without an airgap. Mind everybody always says airgap, but what
they actually meanis a spacer, often made from plastic (cellotape).
So start with say 10 or 20 windings without an airgap.What you
probably will see in the test-bench is a too high inductance
(slower increase of current ascompared to the 100 H inductor). At
the same time you will find the ferrite saturating at a lowcurrent.
It is now time to include the spacer. Attach a peace of cello tape
and cut the excess amount oftape with a razor blade so that only
the touching surfaces of the ferrite are covered with tape. If
youtry the inductor now you will find a much lower inductance and a
higher saturation current. Probablyyou will need to add or remove
some turns to get an inductance of 100 H (same slope). For
theprimary winding I use 0.4 (or 0.5) mm diameter insulated copper
wire. When you have determined theproper number of primary turns,
the secondary winding consists of ten times that number of turns.
Forthe secondary windings I use something like 0.1-0.15 mm diameter
wire. I always include a layer oftape in between two layers of
secondary windings to prevent arcing. The transformers that I use
have22 primary turns and 220 secondary turns.
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Figure 18 Two examples of the Flyback converter built on a peace
of breadboard.
Figure 19 shows the drain-source voltage of power MOSFET
measured with a 1:10 reduction probe.The 1- on the left axis marks
the 0 V input level. The image is not very sharp due to some
trigger jittercaused by a 50Hz ripple on the power supply.
Nevertheless, several features from Fig. 15 can berecognized. The
repetition frequency is 32 kHz and the maximum blocking voltage of
the transistor isabout 31 V according to theory. The voltage over
the transistor almost swings for two full periods untilthe
transistor switches on again. The high frequency oscillations due
to the stray inductance are there,but difficult to see on the
photograph. The increasing voltage drop over Ron during the on
phase isclearly visible.
Figure 19 Drain-source voltage of power MOSFET measured with a
1:10 reduction probe.
The total converter can easily be built in an area of less than
4x4 cm. To increase the lifetime of my
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1/6/2015 Flyback Converters for Dummies
http://www.dos4ever.com/flyback/flyback.html 18/18
tubes I usually run them on as low as current as possible.
Typically 1-1.5 mA. This means that theconverter has to generate
for 6 digits about 6 to 7 watts. The efficiency is ca. 80%. This is
notspectacular but good enough for such a simple circuit. If you
decide to built one: have fun, be carefuland good luck!
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Web links
[1] http://www.williamson-labs.com/555-tutorial.htm [2]
http://www.uoguelph.ca/~antoon/gadgets/555/555.html
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