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MICHAEL BAIRANZADE - profil-zyak-2012
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SEMICONDUCTOR APPLICATION NOTE by AN1543/D
Prepared by: Michaël Bairanzade
Power Semiconductor
Applications Engineer
Motorola SPS Toulouse
Fluorescent Lamp OperationABSTRACT
When the lamp is off, no current flows and the apparent
With a continuous growth rate of 20% per year, electronic
impedance is nearly infinite. When the voltage across the
lamp ballasts are widely spread over the world. Even though
electrodes reaches the V value, the gas mixture is highlytrigthe light out of a fluorescent tube has a discontinuous
ionized and an arc is generated across the two terminals of the
spectrum, the higher efficiency brought by the electronic
lamp. This behavior is depicted by the typical operating curve
control of these lamps make them the best choice to save the
shown in Figure 1.
energy absorbed by the lighting systems.
A few years ago, the lack of reliable and efficient power
Itransistors made the design of such circuits difficult! Today,
thanks to the technology improvements carried out by
MOTOROLA, design engineers can handle all of the problems
linked with the power semiconductors without sacrificing the
global efficiency of their circuits.
IThis Application Note reviews basic electronic lamp ballast nom
concepts and gives the design rules to build industrial circuits.
SUMMARY V
1. MAIN PURPOSE Von
– Fluorescent tube basic operation
– Standard electromagnetic ballast
Vstrike– Electronic circuits
2. HALF BRIDGE CIRCUIT DESIGN
3. DIMMABLE CIRCUIT
4. NEW POWER SEMICONDUCTORS
5. CONCLUSIONS
6. APPENDIX
Figure 1. Typical Low Pressure Fluorescent
Tube I/V CharacteristicELECTRONIC LAMP BALLAST
Main Purpose
The value of V is a function of several parameters:strikeTo generate the light out of a low pressure fluorescent lamp,
– gas filling mixturethe electronic circuit must perform four main functions:
– gas pressure and temperature
a – provide a start–up voltage across the end electrodes – tube length
of the lamp. – tube diameter
b – maintain a constant current when the lamp is – kind of electrodes: cold or hot
operating in the steady state.
Typical values of V range from 500 V to 1200 V.strikec – assure that the circuit will remain stable, even under
Once the tube is on, the voltage across it drops to the
fault conditions.
on–state voltage (V ), its magnitude being dependent uponond – comply with the applicable domestic and international
the characteristics of the tube. Typical V ranges from 40 Vonregulations (PFC, THD, RFI, and safety).
to 110 V.
Obviously, a high end electronic lamp ballast will certainly The value of V will vary during the operation of the lampon
include other features like dimming capability, lamp weareout but, in order to simplify the analysis, we will assume, in a first
monitoring, and remote control, but these are optional and will approximation, that the on–state voltage is constant when the
be analyzed separately. tube is running in steady state.
MOTOROLA 1 Motorola, Inc. 1995
LConsequently, the equivalent steady state circuit can be
described by two back to back zener diodes as shown in
Figure 2, the start–up network being far more complex,
particularly during the gas ionization. This is a consequence
of the negative impedance exhibited by the lamp when the
S
TUBE Tvoltage across its electrodes collapses from V to V .strike on C BI–METALLIC
MAINS TRIGGER
220 V– 50 Hz
BALLAST
Figure 3. Standard Ballast Circuit for
Fluorescent Tube
The operation of a fluorescent tube requires severalVac Von
components around the tube, as shown in Figure 3. The gas
mixture enclosed in the tube is ionized by means of a high
voltage pulse applied between the two electrodes.
To make this start–up easy, the electrodes are actually
made of filaments which are heated during the tube ionization
start–up (i.e.: increasing the electron emission), their
deconnection being automatic when the tube goes into theFigure 2. Typical Fluorescent Tube Equivalent
steady state mode. At this time, the tube impedanceCircuit in Steady State
decreases toward its minimum value (depending upon the
tube internal characteristics), the current in the circuit being
Up to now, there is no model available to describe the start limited by the inductance L in series with the power line.
up sequence of these lamps. However, since most of the The starting element, commonly named starter”, is an
phenomena are dependent upon the steady state essential part to ignite the fluorescent tube. It is made of a
characteristics of the lamp, one can simplify the analysis by bi–metallic contact, enclosed in a glass envelope filled with a
assuming that the passive networks control the electrical neon based gas mixture, and is normally in the OPEN state.
behavior of the circuit. When the line voltage is applied to the circuit, the
Obviously, this assumption is wrong during the time elapsed fluorescent tube exhibits a high impedance, allowing the
from V to V , but since this time interval is very short, thestrike on voltage across the starter” to be high enough to ionize the
results given by the proposed simple model are accurate neon mixture. The bi–metallic contact gets hot, turning ON the
enough to design the converter. contacts which, in turn, will immediately de–ionize the
When a fluorescent tube is aging, its electrical starter”. Therefore, the current can flow in the circuit, heating
characteristics degrade from the original values, yielding less up the two filaments. When the bi–metallic contact cools
light for the same input power, and different V and Vstrike on down, the electrical circuit is rapidly opened, giving a current
voltages. variation in the inductance L which, in turn, generates an
A simple, low cost electronic lamp ballast cannot optimize overvoltage according to Lenz’s law.
the overall efficiency along the lifetime of the tube, but the Since there is no synchronization with the line frequency
circuit must be designed to guarantee the operation of the (the switch operates on a random basis), the circuit opens at
lamp even under the worst case end of life” conditions. a current level anywhere between maximum and zero.
As a consequence, the converter will be slightly oversized If the voltage pulse is too low, the tube doesn’t turn ON and
to make sure that, after 8000 hours of operation, the system the start–up sequence is automatically repeated until the
will still drive the fluorescent tube. fluorescent tube ionizes. At that time, the tube impedance falls
to its minimum value, yielding a low voltage drop across its end
Controlling the Fluorescent Lamp electrodes and, hence, across the switch. Since the starter
As already stated, both the voltage and the current must be can no longer be ionized, the electrical network of the
accurately controlled to make sure that a given fluorescent filaments remains open until the next turn–on of the circuit.
lamp operates within its specifications. We must point out that the fluorescent tube turns off when
The most commonly used network is built around a large the current is zero: this is the source of the 50 Hz flickering in
inductor, connected in series with the lamp, and associated a standard circuit. It’s an important problem which can lead to
with a bi–metallic switch generally named the starter”. visual problems due to the stroboscopic effect on any rotating
Figure 3 gives the typical electrical schematic diagram for the machines or computer terminals.
standard, line operated, fluorescent tube control.
2 MOTOROLA
To take care of this phenomena, the fluorescent tubes, at
%least those used in industrial plants, are always set on a dual
basis in a single light spreader, and are fed from two different
phases (real or virtual via a capacitor) in order to eliminate the
flickering.
The value of the inductor L is a function of the input line
frequency (50 Hz or 60 Hz), together with the characteristics
of the lamp.
The impedance of L is given by Equation 1:
Z = L* (1)L
with: = 2* *F
F = in Herz L = in Henry Z = in Ohm
F
50 Hz 10 kHz 1 MHz
Computing the value of L is straightforward. Assuming a
European line (230 V/50 Hz) and a 55 W tube (V = 100 V,on Figure 4. Typical Fluorescent Lamp Efficiency as a
V = 800 V), then:trig Function of the Operating Frequency
Ptube
I (2)RMS The electronic circuit one can use to build a fluorescentVon
lamp controller can be divided into two main groups:
I 55 100 0.55 ARMS A – Single switch topology, with unipolar AC current,
To limit the steady state current, the impedance must be (unless the circuit operates in the parallel resonant
equal to: mode).
B – Dual switch circuit, with a bipolar AC output current.
Line–Von (3)Z The manufacturers of the fluorescent lamps highlyIRMS
recommend operating the tubes with a bipolar AC current.
Z (230–100) 0.55 238 This avoids the constant bias of the electrodes as an
Anode–Cathode pair which, in turn, decreases the expectedTherefore, the inductor must have a value of (assuming the
lifetime of the lamp.pure Ohmic resistance of the total circuit being negligible):
In fact, when a unipolar AC current flows into the tube, the
Z electrodes behave like a diode and the material of the cathodeL
2* *F side is absorbed by the electron flow, yielding a rapid wear out
of the filaments.
L 238 (2 * * 50) 0.75 H As a consequence, all of the line operated electronic lamp
ballasts are designed with either a dual switch circuit (the onlyIn order to minimize the losses generated into the inductor
one used in Europe), or a single switch, parallel resonantby Joule’s effect, the DC resistance must be kept as low as
configuration (mainly used in countries with 110 V lines),possible: this is achieved by selecting a current density of 4
2 providing an AC current to the tubes.A/mm maximum for the copper.
A few low power, battery operated fluorescent tubes areHowever, the end value of the wire diameter used to
driven with a single switch flyback topology. But, the outputmanufacture the inductor will be limited by the cost, the size
transformer is coupled to the tube by a capacitive network andand the weight expected for a given inductor.
The trigger switch S is a standard device. the current through the lamp is alternating. However, the
The electromechanical ballast has two main drawbacks: filaments (if any) cannot be automatically turned off by this
simple configuration and the global efficiency is downgraded
a – Ignition of the lamp is not controlled.
accordingly.
b – Light out of the lamp flickers at the same frequency as
The dual switch circuits are divided into two main
the AC line voltage.
topologies:
But, on the other hand, the magnetic ballast provides a very
A1 – Half bridge, series resonant.
low cost solution for driving a low pressure fluorescent tube.
A2 – Current fed push–pull converter.
To overcome the flickering phenomenon and the poor
The half bridge is, by far, the most widely used in Europestart–up behavior, the engineers have endeavored to design
(100% of the so–called Energy Saving Lamps and Industrialelectronic circuits to control the lamp operation at a much
applications are based on this topology), while the push–pullhigher frequency. The efficiency (Pin/Lux) of the fluorescent
is the preferred solution in the USA with around 80% of thelamp increases significantly as shown in Figure 4, as soon as
electronic lamp ballasts using this scheme today (see typicalthe current through the lamp runs above a few kiloHertz.
schematic diagram Figure 5).
MOTOROLA 3
pwwhT1
Q1
C2
LDRIVER pC3
V INETWORK Q2 on tube
LINE
FILTER
C1
LINE
220 V
FLUORESCENT
TUBE
Figure 5. Typical Current Fed, Push–Pull Converter
Both of these topologies have their advantages and
T1drawbacks, the consequence for the associated power
Q1
transistors being not at all negligible as shown by Table 1.
Np
VCCTable 1. Main Characteristics of the
Dual Switches Topologies
Np
Parameters Half Bridge Push–Pull LOAD
Q2
V 700 V* 1100 V to 1600 V*(BR)CER
Inrush Current 3 to 4 times I ** 2 to 3 times I **nom nom
Nbt window 2.60 s–3.60 s 1.90 s–2.30 ssi
Drive High & Low side Low side only
Nb
Intrinsic Galvanic no yes
Isolation
Figure 6. Basic Single Transformer CircuitNotes: *numbers are typical for operation on a 230 V line.**I : current into the transistors in steady state.nom
SINGLE TRANSFORMER BASIC OPERATION
The circuit uses the same core to drive the transistors and
PUSH PULL TOPOLOGY to supply the power to the load. Operation is based on the
The main advantage of the current fed push–pull converter, saturation of the core when the magnetic flux exceeds the
besides the common grounded Emitter structure, is the maximum value the core can sustain. Although this is a very
ruggedness of this topology since it can sustain a short circuit low cost solution, it is not commonly used for power above a
of the load without any damage to the semiconductors few tens of watts, because the global efficiency is downgraded
(assuming they were sized to cope with the level of current and by the dual mode operation of the output transformer (i.e.,
voltage generated during such a fault condition). This is a saturable and linear). Figure 6 gives the typical schematic
direct benefit of the current mode brought by the inductor in diagram.
series with the V line. However, the imbalance in both theCC
power transistors and the magnetic circuit leads to high TWO TRANSFORMER OPERATION
voltage spikes that make this topology difficult to use for line At high load currents and high frequency, the transformer
voltage above 120 V. Additionally, it is not practical to dim the requirements for the dual role of frequency control and
fluorescent tubes when they are driven from a push–pull efficient transformation of output voltage becomes a difficult
circuit, the half bridge, series resonant topology being a far problem in the single transformer design. For this reason, the
better solution. two transformer design, depicted in Figure 7, is more
The push–pull converter can be designed with either one advantageous. The operation of this circuit is similar to the one
single transformer, as shown in Figure 6, or by using a transformer case, except that only the small core T2 need be
separate core to build the oscillator (see Figure 7). saturated. Since the magnetization current of T2 is small, high
current levels due to transformer saturation magnetic flux are
reduced significantly when compared to the one transformer
design. Of course, the stresses applied to the power
semiconductors are reduced in the same ratio. Another major
4 MOTOROLA
mmmmLadvantage of the two transformer inverter design is that the
operating frequency is determined by V , a voltage easilyFB
regulated to provide a constant frequency to drive the power
T1transformer. Q1
C1
T1Q1 VCC
T2
Np C2
R1 LOADNV bCC
Q2
Nb
Np R2 NLOAD b
C3 NR1 Q2 b
VFB
Figure 9. Typical Current Fed,
Sinusoidal Output Converter
Figure 7. Basic Two Transformer Circuit
STARTING CIRCUIT
In general, the basic circuits depicted in Figures 6 and 7 will
not oscillate readily, unless some means is provided to begin
oscillation. This is especially true at full load and low
temperature. A simple, commonly used starting circuit is
shown in Figure 8. In this design, resistors R1 and R2 form a
simple voltage divider to bias the transistors to conduction
before the oscillation starts.
I , FULL LOADC
R1
Q1 T1
T2
NC1 p
NV bCC
Nb R2
Np
LOAD I , UNLOADED, TWO TRANSFORMER CONVERTERC
R3 Q2
V = p *VCCVCE
Figure 8. Basic Starting Circuit
SINUSOIDAL OUTPUT INVERTER
The basic inverters discussed above have an output
frequency and voltage directly proportional to the supply
voltage, the output being a square wave. To get a sinusoidal
ICoutput, or a tightly controlled frequency together with an easily
regulated output voltage, the inverter must be modified from
the basic circuit. A simple but efficient way, is to use a current
fed topology, with an inductor connected between the primary
CURRENT FED, RESONANT CIRCUITof the output transformer and the supply line as shown in
Figure 9. When the circuit is tuned with the capacitors C1 and
Figure 10. Typical Push–Pull Waveforms
C2, then the voltage across the switches is sinusoidal, yielding
minimum switching losses into the silicon. Typical waveforms
are given in Figure 10.
MOTOROLA 5
C2
Q1 100 nF
I 400 VtubeR2R1
330 k
D2
1N4937C6C5 INPUT
T1
D1RECTIFIER
V C VLINE onL p trigpA
C3FILTER Q2
C1 100 nF
R3C4 22 nF 400 V
D3
1N4937
LINE
220 V
D4 R4
FLUORESCENT TUBEDIAC 32 V 10
Figure 11. Typical Half Bridge Topology
Transistor selection criteria: By arranging the windings as depicted in Figure 11, the
voltage V is negative for the upper switch and positive forBB• Select the Collector current capability to sustain the peak
the lower one. This forward biases Q2 and the Collectorvalue during either the unloaded or short circuit conditions.
current of this transistor keeps rising until the core of T1
• Select the V to avoid avalanche under the worst(BR)CES saturates .
case conditions (i.e., high line, unloaded operation).
From electromagnetic circuit theory, the magnitude of the
• Define the storage time window to make sure the devices current in the secondaries of T1 is given by Equation 4:
will be tightly matched, thus minimizing the magnetic
Npimbalance into the output transformer. (4)I I *B C Ns• Make sure the load line never goes outside either the
FBSOA or RBSOA maximum ratings of the selected Of course, the value of I must be large enough to fullyB
transistors. saturate the transistor, even under worst case conditions:
ICHALF BRIDGE TOPOLOGY ANALYSIS (5)IB
Basic Circuit
with = intrinsic current gain of the transistorThe basic schematic diagram of the half bridge, self
oscillant topology is given in Figure 11. The two transistors Q1 On the other hand, the V voltage developed across theBB
& Q2 are the active side of the bridge, capacitors C2 & C3 secondaries must be limited to a value lower than the
being the passive arm. V of the transistors, otherwise the Base–Emitter(BR)EBO
junction goes in avalanche and the global efficiency can be
Operation Description downgraded.
The oscillations are generated by means of the saturable Moreover, one must point out that, even if the transistor can
transformer T1. Since the two transistors are biased in the off sustain a Base–Emitter avalanche (assuming that the
state via the low Base–Emitter impedance provided by the associated energy E is within the V maximum rating),j (BR)EBO
secondaries of transformer T1, this circuit cannot start by such a continuous mode of operation may make the transient
itself, unless there is an imbalance between the high side and and long term behavior of the converter more difficult to
the low side of the converter. But, such an imbalance will predict.
severely downgrade the operation once the converter begins. However, there is no problem if the Base–Emitter junction
Therefore, it is preferable to have a pair of matched transistors is forced into the avalanche mode during start–up because,
and to start the converter with the network built around the under these conditions, the energy dissipated into the junction
Diac D4, capacitor C1 and resistor R1. is very low and can be absorbed by the silicon.
When the line voltage is applied, capacitor C1 charges The V voltage is given by another electromagneticBB
exponentially through resistor R1. When the voltage across equation:
C1 reaches the trig value of D4, the diac turns on, discharging
NsC1 into the Base–Emitter network of Q2. This transistor turns (6)V V *BB B Npon and the change in collector current (dI/dt) through the
primary of T1, generates a voltage across each of the As a safety rule of thumb, in steady state V < V .BB (BR)EBO
secondaries of T1. The load being highly inductive,the Collector current will rise
with a slope given by Equation 7:
dI Vc CC (7)
dt L
6 MOTOROLA
WbWStart–up Sequence
The value of Q is dictated by the needs of the application,The start–up voltage (V ) is generated by the seriesstrike
and the associated components must be sized accordingly.resonant network built with the inductor L and the capacitor C,
Since the inductor L is a direct function of the output power andthe behavior of this network being predictable with Equations
operating conditions, the designer has no other choice than to8 to 15 given below.
adjust the values of capacitor C and resistor R to set up theThe resonant frequency is :
Quality factor, keeping in mind the DC resistance of the
1 filaments.(8)fo
2* *(L*C)
HALF BRIDGE DESIGN
The Quality Factor Q is given by :
Note: the design proposed herein assumes a 220 V–50 Hz
L* input line voltage together with a single four foot 55 W tube.(9)Q
R The V voltage is 100 V, the V being 800 V. Nominalon trig
operating frequency: 35 kHz.with R = sum of the DC resistance in the circuit.
Designing a converter for the lamp ballast application is not
This factor can also be expressed by Equation 10: very difficult, but there are many steps and iterations that must
be performed first. Unfortunately, there is no accurate and1 L
(10)Q * simple model available, at the time of this publication, toR C
simulate an electronic lamp ballast. However, the simple
Out of resonance, the impedance of the RLC series circuit equivalent circuit given in Figure 13 is helpful to perform the
is given by Equation 11: first calculations when designing this kind of circuit.
2 RL1 p I2 tube(11)Z R L –
C
At resonance, the L term equals the 1/C and cancel each
other:
HALF FLUORESCENT1 V VPP onL (12) BRIDGE TUBEC
Therefore, the impedance is minimum and equals the DC
resistance:
P = V *Iout on tube
Z = R (13)
At resonance, the current in the circuit is maximum and
Figure 13. Basic Equivalent Circuit in Steady–State
follows Ohm’s law:
V The first step is to define the chopper frequency, since mostCC
I (14) of the critical parameters are dependent upon this criteria.R
The topology being a self oscillant, Half Bridge will permit
At the same time, the voltage across the capacitor is
the design to make the manufacturing of the electronic circuit
maximum as stated by Equation 15:
as simple as possible.
V = V *Q (15) The selected core used to build the converter must meet theC CC
following specifications:
The behavior of an R/L/C resonant circuit is depicted by
Figure 12. Depending upon the L/R ratio, the curve is more or The core must:
less flattened. This is described as the selectivity of the R/L/C a – be saturable
network. b – exhibit a BH curve as square as possible
c – be available at the lowest possible cost
L
lowZ By re–arranging Ampere’s equation, we can compute the( ) R
operating frequency for a self oscillant converter based on a
saturable core:
4V 10P*
(16)F
4*N *B *AeP SZ = R
With: V = voltage across the Primary windingP
N = number of turns of the PrimaryP
L B = core saturation flux in TeslaShigh
R 2Ae = Core cross section in cm
F = frequency in Herz
Z′ = R′
F
fo
Figure 12. Typical R/L/C Series Network Behavior
MOTOROLA 7
SSSwwWBase current) reverses, and the transistor will rapidly switchCare must be taken, not to try to cut cost in the base drive
network as the dynamic parameters of the power transistors off the Collector current.
will likely to not be optimized. In fact, the storage time will The oscillograms given in Figures 15 and 16 show the
probably be greater than the computed operating chopper typical Base bias for a standard converter using this
frequency. technique. Based on these oscillograms, it’s clear that the
The graph given in Figure 14 gives the typical storage time turn–off mechanism, with typical timing values around 4 s, is
variation, as a function of the bias conditions, for a bipolar not negligible and must be taken into account during the
transistor. More detailed information is available from the design.
designer’s data sheet provided by MOTOROLA.
IB15
STORAGE TIME
I = I = 160 mAB1 B2
L = 200 Hc
4 V = 300 Vclamp
tsi
3
ZERO
2
H = 5 s/DIV
V = 100 mA/DIV
1 IB2
0
0 1.0 2.0
Figure 15. Typical Base Current Waveform
I (A)C
Figure 14. Typical Storage Time Variation as a
COLLECTORFunction of the Collector Current H = 5 s/DIV
CURRENT
V = 500 mA/DIV
The turn–off mechanism of the transistor is two fold:
a – When the current increases in the Primary winding N ,P
the magnetic flux increases accordingly, and the operating
point of the core moves toward the B .sat
At this point, the core goes into the saturation area and its
relative permeability – r– collapses from its nominal value
down to unity.
With a typical r of 6000, this large variation makes the
Primary/Secondary coupling nearly negligible and,
consequently, the V voltage starts to drop, yielding lessBB
forward bias to the Base network.
Figure 16. Typical I WaveformC
b – As the Collector current increases, the operating point
of the transistor moves along its H = f(I ) curve.FE C Therefore, the practical operating frequency will be
In the meantime, the Base current is limited to the N /N dependent upon the core used to build the saturableS P
transformer T1, and the absolute value of the Collector currentratio, as stated by Equation 4.
storage time (t ). This is shown by Equation 17:siOne must remember that the V voltage is a function ofBB
the dI /dt, the absolute magnitude of the Collector current 1 1C (17)F
I being irrelevant. T 2*tC si
When the V voltage drops, the available Base driveBB with: T = period depending upon core T1
decreases and the transistor will rapidly leave the
t = storage timesisaturation region. Consequently, the Collector current
The factor 2 stands for the half bridge topology used.decreases and the dI /dt reverses from a positive goingC
The design of a saturable transformer is bounded by severalslope to a negative going slope.
parameters:
If the transistor is driven from a current transformer, then
a – Magnetic material availabilitythe same mechanism applies for the available Base
b –Core shapes available. (Toroids are preferredcurrent, as stated by Equation 4.
because they have the highest r and square BH
These two points are cumulative and, as soon as the characteristic.)
primary current decreases, the core starts to recover from the
c – Manufacturing costs
flux saturation, the V voltage (or the magnetically inducedBB
8 MOTOROLA
mmmmmmmm
tsi ( s)From the toroid data book provided by LCC, let us try theThe typical B/H curves given in Figure 17 are provided by
the manufacturers of cores for the different material they may FT6.3 toroid (external diameter = 6.30 mm) with I = half IP C
propose in their portfolio of products. Most of the time, the data peak:
sheets show the upper side of the curves, the characteristic
I HE* Sbeing absolutely symmetrical on the X axis. NP IPOn the other hand, the shape of the curve, i.e. the B value,sat
can be controlled by using an air gap to increase the
1.60 0.40*reluctance of the core. Of course, this is not possible for the NP 0.35
toroidal cores.
N 1.82 turnP
Using the next available toroid, a FT10 (external diameter
B = 10 mm)
2.50 0.40*NP 0.35
NON SATURABLE MATERIAL:
N 2.85 turnsPMAINLY USED TO BUILD
OUTPUT TRANSFORMERS. The selection of a core from the ‘off the shelf’ standard
H products (see the preferred models given in Table 2), depends
upon the expected frequency, the cost, and the availability.
As an example, let us select the T10 toroid with the A4 ferrite
SATURABLE MATERIAL, WITH A material, the r being higher than 6000. To simplify the
SQUARE B/H CHARACTERISTIC, manufacture of this transformer we will make a first iteration
USEFUL TO BUILD OSCILLATORS. with N = 3 turns, assuming V = 1 V across the primary.P P
The characteristic curves of this core show that the
NOTE: Drawing is not to scale.
saturation flux is 510 mT at room temperature (T = +25°C), the
2cross sectional area being 0.08 cm .
Figure 17. Typical B/H Curves
These parameters yield a theoretical operating frequency
of:
To build the transformer, one can either use the data
41 10*provided by the manufacturers of the cores, using the B=f(H) F
4*3*0.51*0.08curves, or pre–define the type of core that would best fit the
application.
F 20424 Hz
This can be derived from Equation 18, which gives the
Using, as a first analysis, a storage time of 5.5 s for theminimum electromagnetic field needed to saturate a given
power transistors, as given by the data sheet, the practical ONcore:
time (t ) per switch will be:on
N IP* P
H (18)S 1 –6I t 5.5 * 10E on 2 * 20424
H = saturation field in A/cmwith: S t = 29.98 son
N = number of turns on the primaryP
I = current into NP P yielding a typical operating frequency of:
I = effective core perimeter in cmE
F = 1/(2*t )onSince H must be higher than H (the intrinsic fieldS O
sustainable by the material as defined by the data sheet), then –6F = 1/(2*29.98*10 ) = 16677 Hzone can compute the perimeter of the core, assuming a given
number of turns, by rearranging Equation 8: This is below the expected operating frequency as specified
above.
N IP* P Performing the same analysis with the FT6.3 toroid yieldsI (18a)E HS a typical operating frequency of 53 kHz, with N = 2 turns, aP
value well within the expected range. This toroid will be theOf course, in order to end up with lower cost, it’s preferable
final choice for this design.to use a standard core and to run several iterations of the
above equation, using N as a variable.P
Since the current keeps increasing during the storage time Table 2. Popular Available Toroids
of the transistor, one cannot use the calculated I peak valueC Ext. Dia. I AE
to saturate the core because, in this case, the current will be 2Toroid mm cm cm
much higher than the expected one. On the other hand, it’s not
FT6.3 6.30 1.60 0.032very easy to anticipate all of the tolerance at this point of the
FT10 10.00 2.50 0.08design; therefore, as a rule of thumb, the first pass can be
FT16 16 4.00 0.20
made by using I /2 to compute the oscillator.P
MOTOROLA 9
mmm
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