transformer & magnetic design - michael...

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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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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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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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RM or “Square” cores   Compromise between a pot

and an EE core   Good EMI performance   No narrow notch   Unimpeded airflow

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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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LP cores   Designed for low profile

applications   Minimizing leakage

inductance with long centre legs

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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


√500 = 22.3 mil = 0.566 mm 1A

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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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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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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


500×(1/0.8)×(1/0.4)×√0.4

500×√0.4

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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



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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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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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Push-Pull Topology Vo ≅ 0.8 Vdc (Ns/Np) Ns can be easily found   Wire size of primary winding is (494 Po)/Vdc



500×(1/0.8)×(1/0.4)×√0.4 1/2 ×

500×√0.4

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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



1mT = 10G

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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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Half-Bridge Topology Np = (20000 Vdc )/( Ae ΔB f) Turns   Core area Ae in cm2


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Half-Bridge Topology Vo ≅ 0.4 Vdc (Ns/Np) Ns can be easily found   Wire size of primary winding is (1397 Po)/Vdc



500×(1/0.8)×(1/0.8)×√0.8 2 ×

Two current pulses per period

500×√0.4

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Full-Bridge Topology Np = (40000 Vdc )/( Ae ΔB f) Turns   Core area Ae in cm2


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Full-Bridge Topology Vo ≅ 0.8 Vdc (Ns/Np) Ns can be easily found   Wire size of primary winding is (699 Po)/Vdc



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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  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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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


Vdc/L (Np/Ns)(Vo+ 1)/L

TON TOFF

0.2T

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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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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


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


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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


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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


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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


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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

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  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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10% drop in inductance

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  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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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

transformer & magnetic design - michael...

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