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Transformer & Magnetic Design
Transformer Core Materials & Geometries Core Losses Peak Flux Density Selection Maximum Transformer Core Output Power Flyback Transformer (Magnetics) Powdered Molypermalloy (MPP) Cores Transformer Copper Losses
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Transformer & Magnetic Design
Why do we need a transformer?
Isolation for off-line operation Obtain the output voltage for a given duty
cycle Multiple outputs
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Selection of Topologies
Consideration of the voltage stress of the power switch at high
line the current stress of power switch at maximum
output the operating frequency and transformer core
size
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Transformer Core Materials
Ferrites Mixture of iron, manganese or zinc oxides Ceramic ferromagnetic material (brittle!) High resistance, no eddy current loss Core loss is mainly due to hysteresis loss Manufacturers: Ferroxcube-Philips, Magnetics
Inc., Ceramic Magnetics Inc., Fairite, Ferrite International, TDK, Thomson-CSF, etc.
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Transformer Core Geometries
Pot cores (up to 125W) Almost entirely enclosed by
ferrite material therefore excellent EMI performance
Narrow notch for coil leads Restrict to low voltage and
current applications
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Transformer Core Geometries
EE cores (5W to 10kW) Most widely used core No narrow notch restricting
coil leads Poor EMI performance Good thermal performance
due to unimpeded airflow
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Transformer Core Geometries
EC & ETD cores Improved EE core Round centre leg Reduce the mean length of a
turn for a given core area by 11%
Lower Copper loss
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Transformer Core Geometries
RM or “Square” cores Compromise between a pot
and an EE core Good EMI performance No narrow notch Unimpeded airflow
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Transformer Core Geometries
PQ cores Optimum ratio of volume to
radiating surface and coil winding area
Minimizing temperature rise for a given power
Minimizing volume for a given power
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Transformer Core Geometries
LP cores Designed for low profile
applications Minimizing leakage
inductance with long centre legs
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Transformer Core Geometries
UU and UI cores Mainly for high-voltage
or ultra-high power applications
Large window area permits thick wire size
Large leakage inductance
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MKS Units
Flux Φ in weber = B × Area in m2 Flux Density B in tesla Magnetic Field Strength H in A/m H = NI/l where magnetic path length l is in m And B = µ0H in free space where µ0 is 4π ×10-7 H/m
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CGS Units
Flux Φ in maxwell = B × Area in cm2 Flux Density B in gauss Magnetic Field Strength H in oersted H = (0.4 π NI)/l where magnetic length l is in cm Why CGS? Because B = H numerically in free
space
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Conversion Between Units
1 weber = 108 maxwell 1 tesla = 104 gauss 1 A/m = 0.0126 oersted Faraday’s Law of induction V = N (dΦ/dt) in MKS V = N (dΦ/dt) × 10-8 in CGS, i.e., if Φ is
calculated from gauss × cm2
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At 100oC, complete saturation occurs at a flux density of 3200 G
Hysteresis Curve
At 100oC, residual flux density is about 900 G
3F3
1mT = 10G
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Peak flux density
Core Losses
Mainly due to hysteresis loss; expressed in mW/cm3
3F3
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Core Losses
Core loss ∝ (Bmax) 2.7
Core loss ∝ f 1.7
Value of core losses are usually given for bipolar magnetic circuits (first and third-quadrant operation) by various manufacturers
For unipolar circuits, e.g., forward, divide by 2
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Table of Core Losses
Page A.1
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Peak Flux Density Selection
Although BH loop is still linear up to 2000G, peak flux density should be chosen at 1600G such that power switch will not be destroyed due to saturation which may occur with fast load change if peak flux density is too high
3F3
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Peak Flux Density Selection
How to detect the occurrence of saturation in practice?
Current Waveform Saturation starts
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Maxi. Transformer Core Output Power
Forward Topology Po = (0.5 Bmax f Ae Ab)/Dcma Watts Peak flux density Bmax in Gauss Switching frequency f in kHz Core area Ae in cm2
Bobbin winding area Ab in cm2
Dcma in circular mils per rms ampere (Inverse of current density)
1mT = 10G
1A/m = 0.0126 Oersted
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What is Dcma in circular mils per rms ampere? Area per unit ampere 500 cma means that a wire with a diameter of √ 500 mils (1 mil = 1/1000 of an inch) carries 1A rms
Maxi. Transformer Core Output Power
√500 = 22.3 mil = 0.566 mm 1A
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Maxi. Transformer Core Output Power
How did we come up with this equation? Po = (0.5 Bmax f Ae Ab)/Dcma Watts
Voltage Current
Assume that Efficiency is 0.8 Space factor SF is 0.4 Duty cycle is 0.4
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Maxi. Transformer Core Output Power
What is space factor ? The fraction of bobbin winding area occupied
by the winding of primary and secondaries
4mm
4mm
Why SF is 0.4?
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Maxi. Transformer Core Output Power
We can increase the output power by Higher Efficiency (Mission Impossible!) Higher SF (Difficult!) Higher duty cycle (Yes, using different reset
arrangement, but higher voltage stress on power switch)
Higher switching frequency (Yes, but higher switching and core losses)
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Primary & Secondary Windings
Forward Topology Np = (40000 Vdc )/( Ae ΔB f) Turns Core area Ae in cm2
ΔB = Bmax ( Use 1600 Gauss) Switching frequency in kHz DC input voltage Vdc in volts Assume that duty cycle is 0.4
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Forward Topology Vo ≅ 0.4 Vdc (Ns/Np) Ns can be easily found Wire size of primary winding is (988 Po)/Vdc
in circular mils Wire size of secondary winding is 316 Io in
circular mils Assume that duty cycle is 0.4 and efficiency is 0.8
Primary & Secondary Windings
500×(1/0.8)×(1/0.4)×√0.4
500×√0.4
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Maxi. Transformer Core Output Power
Push-Pull Topology Po = (Bmax f Ae Ab)/Dcma Watts Peak flux density Bmax in Gauss Switching frequency f in kHz Core area Ae in cm2
Bobbin winding area Ab in cm2
Dcma in circular mils per rms ampere (Inverse of current density)
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Maxi. Transformer Core Output Power
Push-Pull Topology Po = (Bmax f Ae Ab)/Dcma Watts Output is two times compared to that of
Forward Topology Why? Because the flux change is doubled Core is warmer due to double in core losses
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Primary & Secondary Windings
Push-Pull Topology Np = (40000 Vdc )/( Ae ΔB f) Turns Core area Ae in cm2
ΔB = 2 Bmax ( Use 2 ×1600 Gauss) Switching frequency in kHz DC input voltage Vdc in volts Assume that duty cycle is 0.4
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Primary & Secondary Windings
Push-Pull Topology Vo ≅ 0.8 Vdc (Ns/Np) Ns can be easily found Wire size of primary winding is (494 Po)/Vdc
in circular mils Wire size of secondary winding is 316 Io in
circular mils Assume that duty cycle is 0.4 and efficiency is 0.8
500×(1/0.8)×(1/0.4)×√0.4 1/2 ×
500×√0.4
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Maxi. Transformer Core Output Power
Half- or Full-Bridge Topologies Po = (1.4 Bmax f Ae Ab)/Dcma Watts Peak flux density Bmax in Gauss Switching frequency f in kHz Core area Ae in cm2
Bobbin winding area Ab in cm2
Dcma in circular mils per rms ampere (Inverse of current density)
1mT = 10G
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Maxi. Transformer Core Output Power
Half- or Full-Bridge Topologies I thought a Full-Bridge delivers twice the
output power of a Half-Bridge!! (NOT TRUE in transformer)
If the same core size is used both for Half-Bridge and Full-Bridge, no. of primary turns must be doubled in Full-Bridge which leads to a reduction in wire size, current, by half. (Equal power)
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Primary & Secondary Windings
Half-Bridge Topology Np = (20000 Vdc )/( Ae ΔB f) Turns Core area Ae in cm2
ΔB = 2 Bmax ( Use 2 ×1600 Gauss) Switching frequency in kHz DC input voltage Vdc in volts Assume that duty cycle is 0.4
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Primary & Secondary Windings
Half-Bridge Topology Vo ≅ 0.4 Vdc (Ns/Np) Ns can be easily found Wire size of primary winding is (1397 Po)/Vdc
in circular mils Wire size of secondary winding is 316 Io in
circular mils Assume that duty cycle is 0.4 and efficiency is 0.8
500×(1/0.8)×(1/0.8)×√0.8 2 ×
Two current pulses per period
500×√0.4
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Primary & Secondary Windings
Full-Bridge Topology Np = (40000 Vdc )/( Ae ΔB f) Turns Core area Ae in cm2
ΔB = 2 Bmax ( Use 2 ×1600 Gauss) Switching frequency in kHz DC input voltage Vdc in volts Assume that duty cycle is 0.4
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Primary & Secondary Windings
Full-Bridge Topology Vo ≅ 0.8 Vdc (Ns/Np) Ns can be easily found Wire size of primary winding is (699 Po)/Vdc
in circular mils Wire size of secondary winding is 316 Io in
circular mils Assume that duty cycle is 0.4 and efficiency is 0.8
500×√0.4
500×(1/0.8)×(1/0.8)×√0.8
Two current pulses per period
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Table of Maxi. Output Power
Page A2-5
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Transformers (Forward-Type Converters)
In Forward-type converters, when the power switch is ON, primary current flows into a dot and secondary current flows out of a dot
Primary and secondary mmfs in transformer are balanced.
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Transformers (Forward-Type Converters)
Mmfs are balanced except magnetization inductance in the primary winding of a Forward converter and that’s why we need a reset winding
No core saturation if the peak flux density is chosen properly
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Flyback Transformer (Magnetics)
However in Flyback, no mmf balance during ON time therefore cores can reach saturation easily if they are ungapped
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Flyback Transformer (Magnetics)
Recall Buck-Boost topology, the unisolated version of Flyback. There is a coil for storing energy
Flyback transformer is actually a set of magnetic coupled coils
Energy is stored in primary winding during the ON time, believe or not, most of the energy is stored in the air gap!
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Flyback Transformer (Magnetics)
Turn ratio (Np/Ns) voltage stress on power switch Vstress = 1.3 Vdcmax + (Np/Ns)(Vo + 1)
Choose Vstress with 30% margin Vstress = 0.7 Vbreakdown
Voltage spike due to leakage inductance
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Find TONmax from
For DCM, dead time of 0.2T TONmax= (Np/Ns)(Vo+ 1)(0.8 T)/(Vdc+(Np/Ns)(Vo+ 1)) For CCM, TONmax= (Np/Ns)(Vo+ 1)T/ Vdc
Flyback Transformer (Magnetics)
Vdc/L (Np/Ns)(Vo+ 1)/L
TON TOFF
0.2T
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Flyback Transformer (Magnetics)
Peak current DCM Ip = Vdc TONmax/ Lp CCM Ip = 1.25 Pomax /(Vdc (TONmax/ T)) is half of that
of DCM Assume that efficiency is 0.8
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Flyback Transformer (Magnetics)
Primary inductance DCM Lp = (Vdc TONmax)2/(2.5 T Pomax) CCM Lp = (Vdc TONmax)2/(2.5 T Pomin)
Assume that efficiency is 0.8
Pomax = (0.8 × 0.5LpIp2)/T
Inductance at mode boundary
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Flyback Topology DCM Wire size of primary winding is 289 Ip √ TONmax/ T in circular mils Wire size of secondary winding is 289 (Np/Ns)( Ip √ (T-TONmax)/ T ) in circular mils
Primary & Secondary Windings
500/√3
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Flyback Topology CCM Wire size of primary winding is 500 Ip √ TONmax/ T in circular mils Wire size of secondary winding is 500 (Np/Ns)( Ip √ (T-TONmax)/ T ) in circular mils
Primary & Secondary Windings
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Flyback Topology Np = 1000 √Lp/ Alg Turns Alg, the inductance per 1000 turns with air gap Avoiding saturation, Np Ip must be less than N Isat given by the
manufacturer at the chosen air gap Bigger airgap; smaller Alg; higher N Isat
Primary & Secondary Windings
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What if Alg is not given? Well, Al , the inductance per 1000 turns of
ungapped core is always given Alg can be estimated as Al(l/µ)/(lair + l/µ) l is the magnetic path length lair is the air gap length µ is the effective core permeability
Primary & Secondary Windings
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What if N Isat is not given? First find the Bsat in Gauss for the chosen
material NIsat can be estimated as (Bsat (lair + l/µ))/(0.4 π) l is the magnetic path length in cm lair is the air gap length in cm µ is the effective core permeability
Primary & Secondary Windings
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Primary & Secondary Windings
Alg = Al(l/µ)/(lair + l/µ)
NIsat = (Bsat (lair + l/µ))/(0.4 π)
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Powdered Molypermalloy (MPP) Cores
Powder is Square Permalloy 80, an alloy of nickel, iron, and molybdenum
Powdered particles are mixed with plastic resin and cast in the shape of toroids
The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again.
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Powdered Molypermalloy (MPP) Cores
Each particle is encapsulated in a resin envelope that provides the core with distributed air gap
Behaves like gapped core with less EMI For building inductors in power supplies,
cores of permeability greater than 125 is rarely chosen because of the saturation caused by DC bias current
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Powdered Molypermalloy (MPP) Cores
10% drop in inductance
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Powdered Molypermalloy (MPP) Cores
At 10% drop in inductance, the values of H in Oersteds for materials having µ of 14, 26, 60, 125 are 170, 85, 36, and 19
Maximum ampere-turn can be calculated as NI = H lm/(0.4 π) lm is the length of magnetic path in cm Nmax and Lmax are tabulated in given tables
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Powdered Molypermalloy (MPP) Cores
Page A6-8 Nmax Lmax
NI = H lm/(0.4 π)
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Transformer Copper Losses
At low frequency Copper loss = (Irms)2 Rdc
Rdc is length of wire times resistive per unit length
At high frequency, there are losses due to Skin effect Proximity effect
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Transformer Copper Losses
Both skin and proximity effects arise from eddy currents which are induced by varying, high-frequency, magnetic field.
Skin effect -- the magnetic field is generated from the current carried by the wire itself
Proximity effect -- the magnetic field is generated from the current in adjacent wires
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Skin Effect
Skin effect causes current in a wire flow only on the surface of the wire
Reduction in the effective cross-sectional area Increase the resistance at high frequency At high frequency Copper loss = (Irms)2 Rac
Rac can be found from the ratio of Rac/ Rdc
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Skin Depth
The thickness at which the current density is decreased to 36% of the value at surface
Skin depth d = √1/(πfµσ) in m f is frequency in Hz µ is permeability in H/m σ is the conductivity in S/m For copper, d = 2.1mm at 1kHz and 0.067 mm
at 1MHz
1A/mm2
0.36A/mm2
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Skin Depth
Iron has a conductivity that is only about 6 times less than that of copper, not too bad indeed!
Why not use cheap iron for transmission of power?
Because iron has high µ value, the skin depth is very small even at low frequency, 0.7 mm at 50Hz
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Table of Rac/ Rdc (Sine-Wave Currents)
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Table of Rac/ Rdc (Square-Wave Currents)
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Skin Effect
Skin effect is more profound when wire diameter is large but thick wire still yields less loss! Getting worse when frequency is high
Use a number of small-diameter wires instead of a single large-diameter wire because the total area in annular skin of the small-diameter wires is bigger
Litz wire; take good care in soldering
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Skin Effect
Skin effect is more profound for square-wave currents because they have a lot of high-frequency harmonics
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Proximity Effect
Proximity effect is caused by high-frequency magnetic fields arising from currents in adjacent wires or winding layers
Effective area for current flow
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Proximity Effect
Proximity effect is more profound when the no. of winding layers is large
1A 2A 3A
0A –1A –2A A 3-layer secondary
winding in a EE core
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Proximity Effect
Primary/secondary without interweaving
Pri-- Layer 1
Pri-- Layer 2
Sec-- Layer 2
Sec-- Layer 1
Core centre leg mmf
Higher Rac/ Rdc
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Proximity Effect
Primary/secondary with interweaving
Pri-- Layer 1
Pri-- Layer 2
Sec-- Layer 2
Sec-- Layer 1
Core centre leg mmf
Lower Rac/ Rdc
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