Flyback Mathcad Example

8/15/2019 Flyback Mathcad Example

1/13

SWITCHING POWER SUPPLY DESIGN:DISCONTINUOUS MODE FLYBACK CONVERTER

Written by Michele Sclocchi

[email protected] Application Engineer National Semiconductor

Typical Flybac ! "#$p#$ p"%&' (#pply:

N"$&(: Write down the power supply requirements on :X :=!et the results on:"sults :=T)i( Ma$)ca* +il& )&lp( $)& calc#la$i", "+ $)& &-$&',al c".p",&,$( "+ a $ypical*i(c",$i,#"#( ."*& (%i$c)i,/ p"%&' (#pply0

I,p#$ 1"l$a/&:

#Minimum input $oltage: %imin 30 $olt⋅:=

Ma imum input $oltage: %ima 50 $olt⋅:=

O#$p#$:

N sp&

' &

(o &a

) &

(o &b

N sp*

' *

(o *a

) *

(o *b

N sp+

' +

(o +a

) +

(o +b

" sense

( s

" ,a

" ,*" ,+( ,+

( c

" c

( i*( i+- *

" +

' .+

( +%o+

%o*

%o&%input

)M&/00

1sen

comp

,b

agnd pgnd

dr

,a2sync2sd

%in

3Needed i, %in 4 /5%

36utput ,ilter


2/13

Nominal output $oltage7 ma imum output ripple7 minimum output current7 ma imum outputcurrent

%o+ 5 $olt⋅:= %rp+ 10 m%⋅:= 1o+min 0.250 amp⋅:= 1o+ma 5 amp⋅:=

%o* 12 $olt⋅:= %rp* 20 m%⋅:= 1o*min 0.250 amp⋅:= 1o*ma 1 amp⋅:=

%o& 12 $olt⋅:= %rp& 20 m%⋅:= 1o&min 0.250 amp⋅:= 1o&ma 1 amp⋅:=8negati$e9

%d,w 0.6 $olt⋅:= 8 diode s ,orward drop $oltage9( ) ⋅ ( ) ⋅ ( ) ⋅: ( ) ⋅ ( ) ⋅ ( ) ⋅

S%i$c)i,/ F'&2#&,cy: ,sw 150

,sw: =

T'a,(+"'.&'3( E++ici&,cy: η 0.95:=

45 Ma-i.#. S$'&(( ", $)& (%i$c)i,/ ."(+&$ :

6 D&+i,& $)& +lybac 1"l$a/& ac'"(( $)& .#$#al i,*#c$a,c&: V+.

?,b 0.8:= 8?,b is a $alue between + to 5. 9

Nps+%o+ %d,w+

: =

6Ma-i.#. S%i$c)i,/ 1"l$a/& ", $)& (%i$c)i,/6."(+&$:

( ) ( )⋅:%ds ma 103.6 $olt=

Sa,e ,actor 8assume spi


3/13

6C)""(& .i,i.#. *&a* $i.& *#$y cycl&: ' dt 0.1:=

6Ma-i.#. *'"p 1"l$a/& ac'"(( $)& (%i$c)i,/ ."(+&$ *#'i,/ $)& ", $i.&:O, '&(i($a,c& "+ $)& M"(+&$:"ds on 0.06 ohm⋅:=

%ds on η %imin⋅ "ds on⋅: =( )⋅

>on ma⋅ ⋅

%imin %ds on−( ) ?lon min⋅ ⋅

%ima %dson−( ) ?lhe energy stored is: E)p 1p 2⋅

2:= = and 1p

%omin >on ma⋅

)p:=

Edt 8.245 10 5−× $olt sec⋅=

- ;rimary inductance:


4/13

)p1pp<

2: =

)p 9.105µ==

95 S&c",*a'y c#''&,$( a,* $#',( 'a$i"( (&c",*a'y;p'i.a'y5 : N(p4 < N(p7

Nsp+ %,b: = Nsp+ 0.233=

Nsp*%,b

: =Nsp* 0.525=

Nsp&%,b

: =Nsp& 0.525=

Nsp+4.286=

Nsp*1.905=

Nsp*1.905=

6Ma($&' "#$p#$:

# Secondary pea< current:1s+p<

⋅

1 ' ma− ' dt−: =

1s+p< 20.503 amp=

# Secondary "MS current:1s+rms

31 ' ma− ' dt−⋅: =

1s+rms 8.267 amp=

# Secondary A( current:

1s+ac 1s+rms2 1o+ma

2−: = 1s+ac 6.584amp=

# Secondary inductance :s sp p⋅: =

)s+ 0.496µ==

6Fi'($ (la1& "#$p#$:

# Secondary pea< current:1s*p<

1 ' ma− ' dt−: =

1s*p< 4.101amp=

# Secondary "MS current:1s*rms

31 ' ma− ' dt−⋅: =

1s*rms 1.653amp=


1s*ac 1s*rms2

1o*ma2

−: = 1s*ac 1.317 amp=


5/13

# Secondary inductance :s sp p⋅: =

)s* 2.51µ==

6S&c",* (la1& "#$p#$:

# Secondary pea< current:1s&p< ⋅

1 ' ma− ' dt−: =

1s&p< 4.101 amp=

# Secondary "MS current:1s&rms

31 ' ma− ' dt−⋅: =

1s&rms 1.653 amp=


1s&ac 1s&rms2

1o&ma2

−: = 1s&ac 1.317amp=

# Secondary inductance:⋅: =

)s& 2.51µ==

=5 Ma-i.#. S$'&(( ac'"(( $)& "#$p#$ *i"*&(: V*i"*&Ma imum $oltage present on the cathode o, diodes

⋅

%diode+ ma 16.667 $olt=

⋅

%diode* ma 38.25 $olt=

⋅

%diode&ma

38.25 $olt=

Select a diode with %a#c44 %diode.ma 7 and ultra#,ast switching diode

>6a5 O#$p#$ 'ippl& Sp&ci+ica$i",( : O#$p#$ Capaci$"'(# Secondary inductance :

s sp p⋅: =)s* 2.51µ==

>o meet the output ripple speci,ications without using an e ternal )( ,ilter7 the output capacitorsha$e to meet two criteria:# satis,y the standard capacitance de,inition: I?C@*V;*$ where t is the >on time7 % is * B o, theallowable output ripple.# >he Equi$alent Series "esistance 8ES"9 o, the capacitor has to pro$ide less than H B o, thema imum output ripple. 8%rippleC1pea


6/13


7/13

, c

2

'b

α

:=

6ne o, the critical ,actor o, a ,ilter design is the attenuating character at the corner ,requency.

>he damping ,actor 8 eta9 describes the gain at the corner ,requency and the time response o,the ,ilter. As the damping ,actor becomes smaller7 the gain at the corner ,requency becomeslarger.

or many ,ilters7 a damping ,actor o, + and a cuto,, ,requency within about an octa$e o, thecalculated ideal should pro$ide suitable ,iltering.'amping ,actor much greater than + may cause unacceptably high attenuation o, lower,requencies and a damping ,actor much less than .H5H may cause undesired ringing and the ,iltermay itsel, produce noise. 83 octa$eC inter$al that has the ,requency ratio *:+ 9# or each output:#6utput load resistance:"o+

1o+ma: =

#'esired damping ,actor: ζ .7:=

⋅ ⋅ ωc 5 .289 10×sec

=

#1nductance calculated:)+

⋅ ⋅

ωc: =

)+ 2.647µ==

#(apacitance calculated:

(o+b )+ ωc 2⋅: = (o+b 1.35µ=

#1nductance used: )+ used 10 µ=:=

#(apacitance used: (o+b used 50 µ:=

ωcu2 π⋅ )+ used (o+b used⋅⋅

: =

, i 10 0250: =

wi , i 2⋅ π⋅sec⋅: = ⋅

ωn)+ used (o+b used⋅

: = ωnc)+ (o+b⋅

: =


8/13

ζ+2 "o+⋅ )+ used (o+b used⋅⋅

: =

ζ+c2 "o+⋅ )+ (o+b⋅⋅

: =

As+ i

1 i 2⋅ ζ+⋅wi

ωcu⋅+

wi

ωn

2

−

: = As+c i1 i 2⋅ ζ+c⋅

wi

ωc⋅+

wi

ωnc

2

−

: =

Mags+ i 20 log As+ i( )⋅:= Magsc+ i 20 log As+c i( )⋅:=

#(apacitor Selection:>he per,ormance o, a ,ilter critically depends on the capacitor used. Iesides the basic $oltageand capacity requirements7 select capacitors with low ES)7 ,or high ,requency attenuation7 andlow ES"7 ,or mid band attenuation and2or high ripple current capability.#(hohe input capacitor has to meet the ma imum ripple current rating 1p8rms9 and the ma imuminput $oltage ripple ES" $alue.


9/13


10/13

other three important parameters o, a Mos,et are "ds8on97 gate threshold $oltage7 and gatecapacitance.>he switching Mos,et has three types o, losses7 conduction loss7 switching loss7 and gate chargeloss:#C",*#c$i", l"((&( are equal to: 1L*3" losses7 there,ore the total resistance between the sourceand drain during the on state7 "ds8on9 has to be as low as possible.#S%i$c)i,/ l"((&( are equal to: Switching#time3%ds313,requecy. >he switching time7 rise time and,all time is a ,unction o, the gate to drain Miller#charge o, the Mos,et7 -gd7 the internal resistanceo, the dri$er and the >hreshold %oltage7 %gs8th97 the minimum gate $oltage which enables thecurrent through drain source o, the Mos,et.6Ga$& c)a'/& l"((&( are caused by charging up the gate capacitance and then dumping thecharge to ground e$ery cycle. >he gate charge losses are equal to: ,requency -g8tot9 %dr Kn,ortunately7 the lowest on resistance de$ices tend to ha$e higher gate capacitance.Iecause this loss is ,requency dependent7 in $ery high current supplies with $ery large E>s7 withlarge gate capacitance7 a more optimal design may result ,rom reducing the operating ,requency.Switching losses are also e,,ected by gate capacitance. 1, the gate dri$er has to charge a largercapacitance7 then the time the Mos,et spends in the linear region increases and the lossesincrease. >he ,aster the rise time7 the lower the switching loss. Kn,ortunately this causes high,requency noise.

=

Mos,et: airchild -I+5N*5)# '*;A?"ds on 0.3 ohm⋅:=8>otal resistance between the source and drain during the on state9(oss 95 p⋅:=86utput capacitance9-g tot 13 n⋅ coul⋅:=8>otal gate charge9-gd miller 6.1 n⋅ coul⋅:=8!ate drain Miller charge9%gs th 2 $olt⋅:=

8>hreshold $oltage9#(onduction losses: ;condon rms⋅ ma⋅: =

#Switching losses: ;sw8ma 9>urn 6n time:tsw -gd miller

%dr %gs th−⋅: =

sw . × sec=

;sw ma tsw %ds ma⋅ 1pphe a$erage current required to dri$e the gate capacitor o, the Mos,et:⋅

aw g .=


11/13

#>otal losses: ;tot8ma 9;tot ma 2.64watt=

#Ma imum unction temperature and heat sin< requirement:#Ma imum unction temperature desired:> ma 130:=

#Ma imum ambient temperature: >a ma 50:=

#>hermal resistance unction to ambient temperature:θ a

;tot ma: = θ a 30.305

watt=

(elsius

1, the thermal resistance calculated is lower than that one speci,ied on the Mos,et s data sheet aheat sin< or higher copper area is needed.

or E ample ,or a >5#*J& 8'*pahe )M&/00 uses a current mode control scheme. >he main ad$antages o, current mode controlare inherent cycle#by#cycle current limit ,or the switch7 and simple control loop characteristics.Since the )M&/00 has a ma imum duty cycle o, +55B7 and the power supply is designed to wor<in discontinuous mode with a 5B ma imum duty cycle7 the current limit should be designed sothat the pea< short circuit current limit is reached ust be,ore the 5B boundary is reached." sense

1pp< 1.1⋅: =

" sense 0.016 Ω=

445 T'a,(+"'.&' D&(i/,:

>he inductor# trans,ormer should be designed to minimi e the leahe total losses are minimi edwhen core losses and winding losses are appro imately the same $alue.

#(ore selection:>o reduce the core losses7 ,errite#; material is usually the pre,erred material ,or discontinuous


12/13


13/13

#;rimary and secondaryOs inductance:

)p 9.105µ== )s+ 0.496µ== )s* 2.51µ== )s& 2.51µ==

Documents

Flyback Mathcad Example