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 1

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

2

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 3

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. 4

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 5

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 6

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 7

Transformer Core Geometries RM or “Square” cores   Compromise between a pot and an EE core   Good EMI performance   No narrow notch   Unimpeded airflow

8

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 9

Transformer Core Geometries LP cores   Designed for low profile applications   Minimizing leakage inductance with long centre legs

10

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 11

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 12

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 13

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 14

Hysteresis Curve At 100oC, complete saturation occurs at a flux density of 3200 G 1mT = 10G At 100oC, residual flux density is about 900 G

3F3

15

Core Losses Mainly due to hysteresis loss; expressed in mW/cm3 Peak flux density 3F3

16

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 thirdquadrant operation) by various manufacturers   For unipolar circuits, e.g., forward, divide by 2 17

Table of Core Losses

Page A.1 18

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 19

3F3

Peak Flux Density Selection   How to detect the occurrence of saturation in practice?

Saturation starts

Current Waveform

20

Maxi. Transformer Core Output Power Forward Topology Po = (0.5 Bmax f Ae Ab)/Dcma Watts   Peak flux density Bmax in Gauss 1mT = 10G   Switching frequency f in kHz 1A/m = 0.0126 Oersted   Core area Ae in cm2   Bobbin winding area Ab in cm2   Dcma in circular mils per rms ampere (Inverse of current density) 21

Maxi. Transformer Core Output Power 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 1A

√500 = 22.3 mil = 0.566 mm

22

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 23

Maxi. Transformer Core Output Power What is space factor ?   The fraction of bobbin winding area occupied by the winding of primary and secondaries 4mm

Why SF is 0.4? 4mm 24

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) 25

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 26

Primary & Secondary Windings Forward Topology Vo ≅ 0.4 Vdc (Ns/Np) 500×(1/0.8)×(1/0.4)×√0.4 Ns can be easily found   Wire size of primary winding is (988 Po)/Vdc in circular mils 500×√0.4   Wire size of secondary winding is 316 Io in circular mils Assume that duty cycle is 0.4 and efficiency is 0.8 27

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) 28

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 29

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 30

Primary & Secondary Windings Push-Pull Topology Vo ≅ 0.8 Vdc (Ns/Np) Ns can be easily found 1/2 × 500×(1/0.8)×(1/0.4)×√0.4   Wire size of primary winding is (494 Po)/Vdc in circular mils 500×√0.4   Wire size of secondary winding is 316 Io in circular mils Assume that duty cycle is 0.4 and efficiency is 0.8 31

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 1mT = 10G   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) 32

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 HalfBridge 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) 33

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 34

Primary & Secondary Windings Half-Bridge Topology Two current pulses per period Vo ≅ 0.4 Vdc (Ns/Np) Ns can be easily found 2 × 500×(1/0.8)×(1/0.8)×√0.8   Wire size of primary winding is (1397 Po)/Vdc in circular mils 500×√0.4   Wire size of secondary winding is 316 Io in circular mils Assume that duty cycle is 0.4 and efficiency is 0.8 35

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 36

Primary & Secondary Windings Full-Bridge Topology Two current pulses per period Vo ≅ 0.8 Vdc (Ns/Np) 500×(1/0.8)×(1/0.8)×√0.8 Ns can be easily found   Wire size of primary winding is (699 Po)/Vdc in circular mils 500×√0.4   Wire size of secondary winding is 316 Io in circular mils Assume that duty cycle is 0.4 and efficiency is 0.8 37

Table of Maxi. Output Power

Page A2-5 38

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. 39

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 40

Flyback Transformer (Magnetics)   However in Flyback, no mmf balance during ON time therefore cores can reach saturation easily if they are ungapped 41

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! 42

Flyback Transformer (Magnetics) Turn ratio (Np/Ns)   voltage stress on power switch Vstress = 1.3 Vdcmax + (Np/Ns)(Vo + 1) Voltage spike due to leakage inductance

Choose Vstress with 30% margin Vstress = 0.7 Vbreakdown 43

Flyback Transformer (Magnetics) Find TONmax from

Vdc/L

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

0.2T

For DCM, dead time of 0.2T

TON TOFF

  TONmax= (Np/Ns)(Vo+ 1)(0.8 T)/(Vdc+(Np/Ns)(Vo+ 1))

For CCM,   TONmax= (Np/Ns)(Vo+ 1)T/ Vdc

44

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 45

Flyback Transformer (Magnetics) Primary inductance 2)/T P = (0.8 × 0.5L I omax p p DCM   Lp = (Vdc TONmax)2/(2.5 T Pomax) CCM   Lp = (Vdc TONmax)2/(2.5 T Pomin) Inductance at mode boundary

Assume that efficiency is 0.8 46

Primary & Secondary Windings 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 47

Primary & Secondary Windings 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

48

Primary & Secondary Windings 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 49

Primary & Secondary Windings 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 50

Primary & Secondary Windings 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 51

Primary & Secondary Windings Alg = Al(l/µ)/(lair + l/µ)

NIsat = (Bsat (lair + l/µ))/(0.4 π) 52

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 53

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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 54

Powdered Molypermalloy (MPP) Cores 10% drop in inductance

55

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 56

Powdered Molypermalloy (MPP) Cores

NI = H lm/(0.4 π)

Page A6-8

Nmax

57

Lmax

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 58

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 59

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 60

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 1A/mm2   f is frequency in Hz   µ is permeability in H/m 0.36A/mm2   σ is the conductivity in S/m   For copper, d = 2.1mm at 1kHz and 0.067 mm at 1MHz 61

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 62

Table of Rac/ Rdc (Sine-Wave Currents)

63

Table of Rac/ Rdc (Square-Wave Currents)

64

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 65

Skin Effect   Skin effect is more profound for square-wave currents because they have a lot of highfrequency harmonics

66

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 67

Proximity Effect   Proximity effect is more profound when the no. of winding layers is large A 3-layer secondary winding in a EE core

0A

–1A –2A

1A

2A

3A

68

Proximity Effect   Primary/secondary without interweaving Sec-- Layer 2 Sec-- Layer 1 Higher Rac/ Rdc

Pri-- Layer 2 Pri-- Layer 1 mmf Core centre leg

69

Proximity Effect   Primary/secondary with interweaving Sec-- Layer 2 Pri-- Layer 2 Lower Rac/ Rdc

Sec-- Layer 1 Pri-- Layer 1 mmf Core centre leg

70