Flyback converter calculations

items Description values

1 Minimum input voltage(!"#()"#) ) 60V
2 Maximum input voltage (!"#()*+) ) 75V
3 Output power(=! ) 100W
4 Input power(=0# ) 100W
5 Output voltage (!! ) 15V
6 Diode voltage (!1 ) 0.7V
7 Switching frequency 50kHz
8 Efficiency(>) 90%
9 Voltage ripple∆!2 2%
10 operating flux density@) 0.25T
11 Regulation, α 1%

• Voltage gain of Flyback converter, is shown in Equation []

!! D *$
= $ )$ )
!"# 1 − ( *%

• Calculation when Time period:

1 1
T= = = 20 × 10'( = 2012
F 50 × 10&

• Assuming Maximum duty ratio, ()*+ ,

()*+ = 0.5

• Minimum duty ratio, ()"# , is shown in Equation []

!"#()"#) 60
()"# = 4 5 (()*+) = $ ) × 0.5 = 0.4
!"#()*+) 75

• Maximum on time,9!# , is shown in Equation [].

9!# = :()*+ = 20 × 10'( × 0.5 = 10 × 10'( = 1012

• Maximum off time,9!// , is shown in Equation [].

 9!// = :(1 − (34#) = 20 × 10'( × (1 − 0.4) = 12 × 10'( = 1212

• Output current A!56 , is shown in Equation []

=! 100
A!56(7"#) = = = 6.67B
!2 15

• Minimum output power, =!(7"#) is shown in Equation

=!(7"#) = A!56(7"#)(!1 + !2 ) = 6.67 × (0.7 + 15) = 104.72D

• Minimum input power, ="#(7"#) is shown in Equation

=!56(7"#) 104.72
="#(7"#) = = = 116.36D
> 0.9

• Inductance, L, is shown in Equation []

%
H!in(max) ((min) I : (75 × 0.4)% × 20 × 10'(
G= = = 77.3 × 10'( = 77.31J
2=in ( min ) 2 × 116.36

2 of 77.3N
• Finding Maximum, Minimum, delta Inductor currents A8)*+ , A8)"# , ∆A8

Figure:Inductor current
 • The inductor delta current, ∆A8 , is shown in Equation []
H:!"#()"#) (()*+) I (20 × 10'( × 60 × 0.5)
∆A8 = = = 7.76B
G 77.3 × 10'(
• Maximum Input current A"#(7*+) , is shown in Equation [].

="# 100
A"#(7*+) = = = 1.67B
!"#(9"#) 60

• Minimum Input current A"#(7*+) , is shown in Equation [].

="# 100
A"#(7"#) = = = 1.33B
!"#(9*+) 75

• The inductor peak current, A(:;) , is shown in Equation [].

A"#(7*+) ∆A8 1.67 7.76
A(:;) = $ )+ =$ )+$ ) = 7.22B
()*+ 2 0.5 2

• inductor currents A8 is shown in Equation []

1 1
A8 = HA8(7"#) + A8(7*+) I = (0.295 + 8.055) = 4.175B
2 2

• Average source current is related to average inductor current by

=! = ="#
!! A! = !"# A"#
!!
A"# = ( ) A!
!"#
D *$
A"# = $ ) $ ) A!
1 − ( *%

Results is

A"# = A< × (
A"#
A8 =
(
• Maximum and minimum inductor currents I=(74>) , I=(7?@) is shown in Equations []

∆A8 A"#(7*+) !"#(7"#) ()*+ : 1.67 7.76

A8(7*+) = A8 + = + = + = 8.055B
2 ()"# 2G) 0.4 2
∆A8 A"#(7*+) !"#(7"#) ()*+ : 1.67 7.76
A8(7"#) = A8 − = − = − = 0.295B
2 ()"# 2G) 0.4 2

• Output capacitor N!56 is shown in Equations []

(9*+ × A!56 0.5 × 6.67
N!56 = = = 0.2223RO
OA × ∆!2 × PQ 50 × 10& × 0.02 × 15
 Transformer calculations

• The skin depth in centimeters is:

6.62 6.62
S= = = 0.0296WR
TU √50 × 10&

• the wire diameter is:

XYZ[ \Y]R[9[Z = 2 × S = 2 × 0.0296 = 0.0592WR

• the bare wire areaBB is:

^ × (XYZ[ \Y]R[9[Z)% ^ × (0.0592)%

BB = = = 0.00275 WR%
4 4

• Calculate the primary peak current, A:(CD)

2 × =!(7"#) × : 2 × 104.72 × 20 × 10'(
A:(CD) = = = 7.76B
> × !"#(9"#) × 9!# 0.9 × 60 × 10 × 10'(

• Calculate the primary rms current, A:(E)F)

6 $I×$I#$
!"
A:(E)F)G A:(CD) × _ &H = 7.76 × _&×%I×$I#$ = 3.17B

• Calculate the equivalent input resistance, `"#(KL5"M)

%
H!"#(9"# I (60)%
`"#(KL5"M) = = = 30.94 ≈ 31 QℎR2
="#(7"#) 116.36

• Calculate the energy-handling capability in watt-seconds, w-s.

*
8×0%('() NN.&×$I#$ ×(&.$N)*
cd[Zef = %
= %
= 0.000388 X − 2
 • Opera ting flux density@7PQ is typically a defined input parameter; for ferrite cores it is
generally between 0.2T and 0.3T. Calculating the average of @7PQ equation:

0.2 + 0.3
@7?@(?RS) = = 0.25:
2

• Calculate the electrical conditions,gK .

gK = 0.145 × =!56 × @) % × 10'T = 0.145 × 100 × 0.25% × 10'T = 0.0000906

• Calculate the core geometry, Kg. See the design specification for, window utilization
factor, g5

cd[Zef % (0.000388)%
gU = = = 0.00167 = 0.00167(1.35) = 0.00226 WRV
gK ∝ 0.0000906 × 1.0

• Select, from Chapter 3, an EE core comparable in core geometry, Kg.

Core number = EE-187
Manufacturer = Ferroxcube
Material = 3C85
Magnetic path length, MPL = 4.01 cm
Core weight, Wtfe = 4.4 grams
Copper weight, Wtcu = 6.8 grams
Mean length turn, MLT = 3.8 cm
Iron area, Ac = 0.226cm2
Window Area, Wa = 0.506 WR%
Area Product, Ap = 0.114 WRT
Core geometry, Kg = 0.0027WRV
Surface area, At = 14.4 WR%
Core Permeability = 2500
Winding Length, G = 2.080 cm
▪ Calculate the current density, J, using a window utilization, g5 = 0.29.

2(cd[Zef)(10T ) 2 × (0.000388)(10T )
i= = = 938.9 ]Rj/WR%
@) g5 B: 0.25 × 0.29 × 0.114

• Calculate the primary wire area, BCW(X)

A:Y)F 3.17
BCW(X) = = = 0.00338WR%
i 938.9

• Calculate the required number of primary strands, l#:

BCW(X) 0.00338
l#: = = = 2.086 o2[ 2
#25(n]Z[ ]Z[]) 0.00162

• Calculate the number of primary turns, *: . Half of the available window is primary,
D*: /2. Use the number of strands, l#: , and the area for #26.
D* 0.506
D*: = = = 0.253 WR%
2 2
g5 D*: 0.29 × 0.253
*: = = = 15.09 o2[ 15 9oZd2
3(#25(n]Z[ ]Z[])) 3(0.00162)
• Calculate the required gap, pU .
0.4^*: % BZ (10'[ ) q=G 0.4^ × 15% × 0.226 × (10'[ ) 4.01
pU = − = −
G 1) 77.3 × 10'( 2500
= 0.00662WR

▪ Calculate the equivalent gap in mils.

RYp2 = (WR)(393.7) = (0.00662)(393.7) = 2.606

• Calculate the fringing flux factor, F

pU 2t 0.00662 2 × 2.080
O =1+ ln 4 5 = 1 + ln $ ) = 1.0897
TBZ pU √0.226 0.00662

• Calculate the new number of turns, *#: , by inserting the fringing flux, F.
 pU G 0.00662 × 77.3 × 10'(
*#: = u = u = 12.85 = 13 9oZd2
0.4^BZ O(10'[ ) 0.4^ × 0.226 × 1.0897(10'[ )

• Calculate the peak flux density, @:;

0.4^*#: O(A:(:;) )(10'T ) 0.4^ × 13 × 1.0897 × 7.76 × (10'T )
@:; = = = 1.679 9[2p]
q=G 4.01
pU + v 1 w 0.00662 + v2500w
)

• Calculate the primary, the new µΩ/cm

Ω µΩ/cm 1062
(d[X)µ = = = 531
cm l#: 2

• Calculate the primary winding resistance, `:

Ω
`: = qG:H*#: I $µ ) × 10'( = 3.8 × 13 × 531 × 10'( = 0.0262 QℎR2
cm

• Calculate the primary copper loss, =:

=: = A % : × `: = (3.17)% × 0.0262 = 0.263 X]992

• Calculate the secondary turns,*FI$ .

*#: (!I + !1 )(1 − ()*+ ) 13(15 + 0.7)(1 − 0.5)
*FI$ = = = 3.4 = 3 9oZd2
H !: ()*+ I ( 60 × 0.5)

• Calculate the secondary peak current, pFI$(:;)

2AI$ 2 × 6.67
pFI$(:;) = = = 26.68]Rj2
(1 − (7?@ ) (1 − 0.5)
• Calculate the secondary rms current, pFI$(Y)F) .

(1 − (7?@ ) (1 − 0.5)
pFI$(Y)F) = pFI$(:;) u = 26.68 × u = 10.89 ]Rj2
3 3

• Calculate the secondary wire area, B\WI$(X)

10.89
B\WI$(X) = = 0.0116 WR%
938.9

• Calculate the required number of secondary strands, l#FI$ .

B\WI$(X) 0.0116
l#FI$ = = = 7.16 o2[ 7
XYZ[] 0.00162
• Calculate the, lI$ , secondary, μΩ/cm.
Ω
Ω µ cm 1062
(lI$ )µ = = = 152
cm l#FI$ 7

• Calculate the winding resistance, `FI$

Ω
`FI$ = qG:(*FI$ ) $µ ) × 10'( = 3.8 × 3 × 152 × 10'( = 0.00173QℎR
cm

• Calculate the secondary copper loss, =FI$ .

=FI$ = A % F!$ × `FI$ = (10.89)% × 0.00173 = 0.205 X]992

• Calculate the secondary turns, *FI% .

*#: (!I% + !1 )(1 − ()*+ ) *#: (+0.7)(1 − 0.5)
*FI% = = = 9oZd2
H !: ()*+ I ( 60 × 0.5)
• Calculate the secondary peak current, pFI%(:;)
2AI%
pFI%(:;) = = ]Rj2
(1 − (7?@ )

• Calculate the secondary rms current, pFI$(Y)F) .

(1 − (7?@ )
pFI%(Y)F) = pFI%(:;) u = ]Rj2
3
• Calculate the secondary wire area, B\WI$(X)
pFI%(Y)F)
B\WI%(X) = = WR%
i

• Calculate the required number of secondary strands, l#FI% .

B\WI%(X)
l#FI% =
XYZ[]

• Calculate the, lI% , secondary, μΩ/cm.

Ω µΩ/cm
(lI% )µ =
cm l#FI%

• Calculate the winding resistance, `FI%

Ω
`FI% = qG:(*FI% ) $µ ) × 10'( = QℎR
cm
• Calculate the secondary copper loss, =FI% .

=FI$ = A % F!% × `FI% = X]992

• Calculate the window utilization,g5 .

[9oZd2] = H*: l#: I[jZYR]Zf]

[9oZd2] = (*FI$ l#FI$ )[2[WQ\]Zf]
[9oZd2] = (*FI% l#FI% )[2[WQ\]Zf]
*6 = 9oZd2#26
*6 BB
g5 =
D*
• Calculate the total copper loss,=^5 .

=^5 = =: + =FI$ + =FI%GB*66F

• Calculate the regulation, α, for this design.

=^5
α= × 100 = %
=I
• Calculate the ac flux density, @*^ .
A:(:;)
0.4^*#: O $ 2 ) (10'T )
@*^ = = 9[2p]2
q=G
pU + v 1 w
)
• Calculate the watts per kilogram, WK.

X]99
Dg = 4.855(10'V )(U)$.(& (@*^ )%.(% =
€YpQeZ]R
• Calculate the core loss, =/K .

=/K = (Dg)D6/K × 10'& = X]992

• Calculate the total loss, core =/K and copper =^5 , in watts, =∑ .

=∑ = =/K + =^5 = X]992

• Calculate the watt density, ψ

=∑ X]992
•= =
B` WR%

• Calculate the temperature rise, :Y , in, °C.

:Y = 450(•)I.[%( = °N
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