Transcript SECTION 12 - Electrical engineering

Magnetics
Courtesy of Dr. Slobodan
Cuk
SIMPLE DEVICE MODEL (PIECEWISE LINEAR)

 sat
i
i
Slope L
i
v
PP
S


v
L
S
L = N2 P
Saturation switch closes when   S. Current is then limited only by the source
Impedance and sacond order monidealities such as winding resistance,
leakage inductance etc. This is only an approximate modul, since in saturation,
  1 curve still has a nonzero slope (due to  0 H term), hence a finite
inductance (althoughvery small) is still present.
SATURATION OF FLUX LINKAGES

= 
vdt
v(t)

 
=
 (t 2 ) -  (t 1 )
v
t1
t2
i
t
AREA UNDER VOLTAGE
WAVEFORM (VOLT SEC)
corresponds to
VERTICAL 
DISPLACEMENT
Flux linkage  is the total flux as seen by the winding. For N turns and with  core flux.
  N
From Faraday’s law the induced voltage is
v 
d
dt
Hence, flux linkages are integral of induced voltage
   vdt
t2
Finally,
   vdt   t  / tt12   t 2    t1 
t1
PERIODIC WAVEFORMS AND SATURATION

 (t)
v (t)

s
+V
VT s
t
-V
T
t
Slope L
2
i
s
For a periodic waveform with equal volt sec area in positive and negative direciton, equal and opposite
direction flux excursions are obtained in the steady state. Thus, indeally there is no “core creep” problem,
that is walking toward saturation. However, to prevent saturation in steady-state.
VT2
  max  2NB sS
2
where Ts switching period or
V N  4 B sSfs
1
where f s 
is switching frequency. This transformer equation shows how
T
many volts turn a given core can support without saturation at the frequency
SIZE AND WEIGHT OF MAGNETIC COMPONENTS
Basic equation:
V
 4 BS Sf S
N
Fundamental reason for increasing switching frequency in switched mode power converters.
1. By Increase of fs smaller cores with smaller cross-section S can be used with no saturation.
Also, smaller number of turns require smaller window area.
2.
Better ferromagnetic materials with higher Bs lead directly to smaller, more compact designs.
For example, iron with Bs = 1.6T compared to ferrite with Bs = 0.3T would result in more compact and
smaller design.
INCREASE OF THE DRIVE VOLTAGE AMPLITUDE
v(t)
0
t
v(t)
 - flux linkages
0
0
t
i
Symmetrical and alternative (both positive and negative) voltage excitation is assumed and only
positive excursion illustrated here. Also for simplicity of demonstration a square-loop core
material is assumed.
AIR GAPS IN MAGNETIC CIRCUITS
index m-material;g-gap
i
v
i
Ni
g
Ni
m
g
-
Lm
v
Lg
m
magnetic circuit with air-gap
AMPERE’S Circuital Law
Ni 

 Hd
Hd 
material
Rm 
airgap
Ni  Hmm  Hgg
Ni 
reluctance model
Reluctance definitions
m
g
, Rg 
S
 0S
Pm 
From reluctance model:
Bm
Bg
m 
g

0

Ni
Rg  Rm
2

N2
 Leff 
i
Rg  Rm
Ni   Rm  Rg
Uniform Flux
 assumed
1
Rm
Leff  N2
and
i
  N  N
Rg  Rm
m g
Ni 

S  0S
device (inductance)model
Permeance definitions

Pg 
PmPg
Pm  Pg
 
so
Leff  N2 Pm H N2 Pg
so
Leff  Lm // Lg
1
Rg

PRACTICAL IMPLICATIONS FOR SWITCHED-MODE
POWER CONVERSION
In Majority of the applications
Rm  Rg
hence
or
m
R
m
 R 0 S

g
0 S
 g
For example even for m  100g but since r  1000 the above inequality is satisfied (0.1 << 1).
Thus, the overall magnetic circuit is abmoniated by the air-gap
reluctance Rg, and often reluctance of magnetic path is neglected to result in:
i
Leff  Lg 
N  0S
g
2
+
v
Lg
Lm (neglected)
CONCLUSION: The change of the air-gap significantly alters the inductance.
However, decrease of the inductance value by increase of air
gap is compensated by increase in dc current capacity.
PRACTIAL CORE MATERIALS FOR
INDUCTOR WITH LARGE DC BIAS
r
= 60, 125 etc
s
CORE

B
Bs = I T
MATERIAL
PROPERTY
sl ope
PROPERTY
sl ope
 e ff
Ni
EQUIVALENT
TO 
iron powder
nonmagnetic
binder
When iron powder is mixed with the NONMAGNETIC material as binder,
the powdered iron core.is obtained. The binder act as the distributed air-gap.
The relative proportion of the binder determines the effective inductance. However,
the manufacturers of these cores like to specify the core by its GO, =125 etc.)
Note however, that permeability of magnetic part is still r=10,000 (as also Bs =IT or higher).
Similarly, for mollypermalloy (MMP) coves.
H
DC CURRENT CAPABILITY AND AIR-GAP
s

At no-gap or with small
air gap the inductor can
support only a small DC
current without saturation
Small DC Current
i
Note that from s = Li
for any air-gap value
Hence, higher peak
current are followed by
lower inductance since s is
constant for given core.
DC CURRENT CAPABILITY AND AIR-GAP

s
With large air-gap the inductor
can support a LARGE DC current
without. Note that the relative
ripple  i Idc remains unchanged
saturation.
Large DC current
i
Note that from
 sLi
for any air-gap value.
Hence, higher peak current
are followed by lower inductance
since  sis constant for given core.
INDUCTORS WITH LARGE DC BIAS
CURRENT- AN ALTERNATIVE


Fm
SQUARE-LOOP
MAGNETIC MATERIAL

Fg
+
AIR-GAP
(LINEAR MAT)
=
F= Fm + Fg
"SHEARED" HYSTERISIS LOOP
OF MATERIAL WITH AIR-GAP
A graphical illustration demonstrating how air gap DOMINATES the overall magnetics
structure. Airgap in square loop materials causes “shearing” effect, yielding linearized
hysterisis characteristics. Graphically two magnetic potentials add.
F  Fm    Fg
NI  NI     NI gap
(material)
Low-loss square-loop material can be used for large DC bias inductors.
HYBRID CORE - THE CASE OF A POWERFUL AIR-GAP


NO GAP
m
m
Rc
Rm
i
+
-
Ni

ONLY AIR-GAP
g
 g
Rc
i
Rg
+
-
m
 g
i
2
Ni
HYBRID CORE-COMBINED
m
i
1

g
Rc
Rm
Rg
+
-
i
1
i
2
i
Ni
CONCLUSION: AIR-GAPS CAN DRASTICALLY CHANGE i CHARACTERISTICS
PRACTICAL APPLICATIONS OF HYBRID CORE

TOROIDAL VERSION OF
HYBRID CORE
SQUARE-LOOP MAT
VOLT
EXCITATION
at starting
primary
secondary
INNER CORE - SQUARE-LOOP UNGAPPED
OUTER CORE - GAPPED
In switching converters with isolation transformers made
on SQUARE-LOOP material - the saturation can occur during
start-ups by Volt-sec excitation above remnant magnetism point.
Simple cure is to use HYBRIC CORE’S as above.
Commercial line made by Magnetic, Inc.
i
SATURATION IN SWITCHING CONVERTERS


t
i
t
i
current spikes like this
indicate approach to saturation
I
Typical inductor current waveform in switching converters:
dc currents superimposed triangular current ripple i
INDUCTOR DESIGN VIA AREA PRODUCT Ap
a)
b)
c)
dc current I
i
relative ripple L
I
efficiency (copper loss)  J
J - current density (A/cm2)
Typically
A
A
J ~ 250 2  1000 2
cm
cm
AREA PRODUCT Ap APPROACH:
To fully utilize flux capability:
  LI  N 
LI
BS
t
Mean length
per turn
Electrical specifications:
W
window area
core
cross-section
s
S


Slope L
(1)
I
To fully utilize window area:
KW 
NI
J
Magnetic
path
length 
(2)
where K is empirical window utilization
factor (between 0.3 - 0.7)
From (1) and (2)
LI 2
Ap  WS 
BJk
t
Ip
i
i
DESIGN STEPS IN AREA PRODUCT APPROACH
1.
Pick the core with area product closest to but higher than on calculated by use of
Ap  LI 2 BJk
Record its W and S. (Cores come only in one combination of W and S - possible
limitations if optimum design is sought)
2.
Determine the number of turns N by use of N  LI BS
3.
Determine the wire cross-section
4.
Using mean length per turn data, find the copper losses Pcu.
5.
Finally, calculate the air gap
g
Aw  I J
and wire size.
2

N
S
0
by use of g 
L
INDUCTOR DESIGN VIA CORE GEOMETRY APPROACH
Here instead of current density J, the copper losses Pcu are taken as a design parameter c
This is more realistic criterion in practice, since designer allocates total Watts lost in windings.
DERIVATION:
Nt 2
N 2 tI2
kW
Pcu  RI  
I 
sin ce Aw 
Aw
kW
N
2
where t is mean length per turn of the winding and   1.724 10 6 cm is resistivity
of copper at 25C.
LI
N

To fully utilize flux capability:
(1)
BS
To fully utilize window area W:
N 2 tI2
Pcu  
kW
(2)
Substitution of (1) into (2) and rearranging results in:
 
WS 2  LI 2
Kg 
 2
t
kB Pcu
2

core geometry (m5)
geometrical parameters
only
(3)

electrical and magnetic
parmaters
DESIGN STEPS VIA CORE GEOMETRY Kg
1.
Pick the core with the core geometry product Kg closest but higher than
one calculated from
2
  LI2 
Kg 


Pcu  B 
For the chosen core record its W, S, and t.
LI
BS
2.
Determine the number of turns from N 
3.
Determine the wire cross-section Aw = k W/N and wire size
4.
2

N
S
0
Calculate the air gap g by use of g 
L
B
TRANSFORMER FUNDAMENTALS
N1 turns
N2 turns
i1
v2
primary
N1 N2
i1
i2
v1

vertical
R
i2
v1
v2
R
secondary
Two winding transformer
Additional of one more winding to the core results
in a very useful device-transformer
The two windings are electrically isolated but
coupled magnetically through mutual flux .
DOT CONVENTION: The mutual flux  generates
positive voltages at the dot marked terminals.
Primary current flows into dot marked terminal and
secondary out of dot marked terminal
Ampere turns N2i2 ARE SUBTRACTED from
Ampere turns N1i1 (Lenz’s Law)
Ideal transformer model
From Faraday’s Law:
v 1  N1d dt
v
v 2  N2 d dt

2
v1

N2
N1
From Ampere’s Law:
Hcm  N1i 1  N2 i 2
Assuming core with extremely steep
B-H loop  Hc  0
 N1 i 1  N 2 i 2 
i 2 N1

i1 N2
EVEN MORE REALISTIC MODEL...
Note that the core function (mutual coupling ) can be represented by
magnetizing inductance either on primary or on secondary.
COUPLED INDUCTORS MODEL
i
1
L
i
L 1 - aL M
2
L1 
L
2
-
1
a
L
M
i
V1
a : 1

L
2
V2
V1
i
1
aL M
2
V2
GENERAL MODEL : a ANY NUMBER!
Coupled inductor equations
di
di
v 1  L1 1  L M 2
dt
dt
L 1
N1
N2
v2  LM
N1 :
L
L 2
N2
M
di1
di
 L2 2
dt
dt
"PHYS ICAL" MODEL :
LM - MUTUAL INDUCTANCE
a=N
1
/ N
a - arbitrary number
Note that mutual inductance LM and magnetizing inductance
Lm are different in general and related by:
2
M
TRANSFORMER SIZING
S
W
Intuitive argument:
V~S
VI ~ WS
or

I~W
P ~ Ap
Proportional
N1
N2
AREA PRODUCT Ap
To fully utilize the core capability:
N1 
Transformer specifications:
V1
4 BmSfs
N2 
V2
4 BmSfs
(1)
(square wave, V1 primary, V2 - second)
To fully utilize window area:
kW   N1I1  N2 I 2  J
a) desired power P
b) copper losses (regulation)
(2)
Substitution of (1) in (2) and rearranging
In the area product Ap approach
requirement b) is translated into
current density J(design parameter).
In kg approach actual copper losses
Pcu are specified.
Ap  WS 
OR
Ap 
V1 I1  V2 I 2
4 BS kJfS
P
2BS kJfs
(3)
(4)
DESIGN STEPS
1.
Pick the core with Ap closest but higher than the one calculated by use of
Ap  P  2BS kJfs
2.
Determine the number of primary and secondary turns by use of
N1  V1 4 BSSfs
N 2  V2 4 BSSfs
Awz  I 2 J
3.
Calculate the wire cross-sections Aw 1  I1 J
the wire sizes
4.
Using mean length per turn t, calculate the primary and secondary copper
losses and evaluate the transformer efficiency.
5.
Evaluate the magnetizing inductance by use of Lm  N12S m , and estimate
the magnetizing current.
and determine
CORE GEOMETRY Kg APPROACH
Total copper losses:
Pcu  R1I12  R 2 I 22  2
Substitution of
N1  V1 4 B s fSS

t
N12 I12  N22 I 22
kW

and N2  V2 4 BS fSS and rearrangement yields 25C
2
kWS2
  P 

P2
Kg 


 
t
4 Pcu  BS fS 
4 Pcu BS2 fS2
where   1.724  10 6 cm
is resitivity of copper at 25
DESIGN STEPS (Kg APPR0ACH)
1.
Choose the core with Kg closest but higher than on calculated by use of

P2
Kg 
4 BS2 fS2 Pcu
2.
.
Record core S, W, and t.
Determine the number of turns from
N1  V1 4 BS fSS
N2  V2 4BS f S S
3.
Calculate the wire cross-sections AW1 and AW2 and determine the wire sizes
4.
Evaluate the magnetizing inductance by use of
Lm  N12S m