transformer for a flyback converter is used as the converters inductor as well as an isolation transformer.
Variables and acronyms [ ] Universal constants
Permittivity of free space
μ
o
{\displaystyle \mu_o}
−1 m−1 )
μ
o
=
4
π
10
−
7
{\displaystyle \mu_o = 4\pi 10^{-7}}
−1 m−1 )
Wire variables:
ρ
{\displaystyle \rho}
I
t
o
t
{\displaystyle I_{tot}}
RMS winding currents (A)
I
m
,
m
a
x
{\displaystyle I_{m,max}}
I
R
M
S
{\displaystyle I_{RMS}}
P
c
m
{\displaystyle P_{cm}}
A
c
{\displaystyle A_c}
2 )
Xformer/inductor design parameters
n
1
,
n
2
{\displaystyle n_1, n_2}
L
m
{\displaystyle L_m}
L
{\displaystyle L}
K
u
{\displaystyle K_u}
B
m
a
x
{\displaystyle B_{max}}
Core parameters
EC35, PQ 20/16, 704, etc , Core type (mm)
K
g
{\displaystyle K_g}
5 )
K
g
f
e
{\displaystyle K_{gfe}}
x )
A
c
{\displaystyle A_c}
2 )
W
A
{\displaystyle W_A}
2 )
M
L
T
{\displaystyle MLT}
l
m
{\displaystyle l_m}
l
{\displaystyle l}
l
g
{\displaystyle l_g}
μ
{\displaystyle \mu}
−1 m−1 )
μ
r
{\displaystyle \mu_r}
μ
=
μ
o
μ
r
{\displaystyle \mu = \mu_o \mu_r}
Acronyms RMS: root-mean-squared -
x
rms
=
⟨
x
2
⟩
{\displaystyle x_\text{rms} = \sqrt{ \langle x^2 \rangle} \,\!}
⟨
…
⟩
{\displaystyle \langle \ldots \rangle}
arithmetic mean )
MLT: mean length turn
AWG: American wire gauge Initial calculations [ ] Variables
V
o
{\displaystyle V_{o}}
V
i
n
{\displaystyle V_{in} }
V
D
{\displaystyle V_D}
V
R
d
s
{\displaystyle V_{Rds}}
N
{\displaystyle N}
D
{\displaystyle D}
Calculate turns ratio
V
o
+
V
D
V
i
n
−
V
R
d
s
=
1
N
∗
(
D
m
a
x
1
−
D
m
a
x
)
{\displaystyle \frac{ V_o + V_D }{ V_{in} - V_{Rds} } = \frac{ 1 }{ N } * \left ( \frac{ D_{max} }{ 1 - D_{max} } \right )}
Diode
Rectifier:
V
D
=
0.8
V
{\displaystyle V_D = 0.8V}
Schottky diode:
V
D
=
?
{\displaystyle V_D = ?}
Inductance calculations [ ] The inductance of the transformer,
L
m
{\displaystyle L_m}
Say you want a current ripple 50% of average current.
Δ
i
=
0.5
∗
I
{\displaystyle
\Delta i = 0.5 * I
}
Solve for
L
m
{\displaystyle L_m}
let
n
=
n
2
n
1
{\displaystyle n = \frac{n_2}{n_1}}
I
=
n
D
′
I
l
o
a
d
{\displaystyle
I=\frac{n}{D'}I_{load}
}
Δ
i
=
n
I
l
o
a
d
2
D
′
{\displaystyle
\Delta i = \frac{nI_{load}}{2D'}
}
L
m
=
V
g
D
T
s
2
Δ
i
{\displaystyle
L_m = \frac{V_g D T_s}{2 \Delta i}
}
L
m
=
μ
A
c
n
1
2
l
{\displaystyle
L_m=\frac{\mu A_c n_1^2}{l}
}
The permittivity of free-space is so much larger than the permittivity the transformer material, that the magnetic path length,
l
{\displaystyle l}
l
g
{\displaystyle l_g}
l
=
l
g
{\displaystyle l = l_g}
L
m
=
μ
o
A
c
n
1
2
l
g
{\displaystyle
L_m=\frac{\mu_o A_c n_1^2}{l_g}
}
Solve for
n
{\displaystyle n}
Minimize total power loss:
P
t
o
t
=
P
f
e
+
P
c
u
{\displaystyle P_{tot} = P_{fe} + P_{cu}}
P
f
e
=
K
f
e
Δ
B
β
A
c
l
m
{\displaystyle P_{fe} = K_{fe} \Delta B^\beta A_c l_m}
B
a
c
=
L
m
Δ
i
n
1
A
c
{\displaystyle B_{ac} = \frac{L_m \Delta i}{n_1 A_c}}
β
{\displaystyle \beta}
K
f
e
{\displaystyle K_{fe}}
Core calculations [ ] Core selection [ ] Variables
P
F
e
{\displaystyle P_{Fe}}
W
{\displaystyle W}
B
s
a
t
{\displaystyle B_{sat}}
T
{\displaystyle T}
B
m
a
x
{\displaystyle B_{max}}
T
{\displaystyle T}
Δ
B
{\displaystyle \Delta B}
T
{\displaystyle T}
B
a
c
{\displaystyle B_{ac}}
A
w
{\displaystyle A_w}
c
m
2
{\displaystyle cm^{2}}
A
e
{\displaystyle A_e}
c
m
2
{\displaystyle cm^{2}}
A
P
{\displaystyle AP}
c
m
4
{\displaystyle cm^4}
K
u
{\displaystyle K_u}
N
P
{\displaystyle N_P}
N
S
{\displaystyle N_S}
N
B
{\displaystyle N_B}
μ
o
{\displaystyle \mu_o}
μ
o
=
2
π
10
−
7
{\displaystyle \mu_o = 2 \pi 10^{-7}}
Material specifications
Grade
B
s
a
t
{\displaystyle B_{sat}}
Specific Power Losses @100 °C [W/cm3]
Manufacturer
B2
0.36
P
F
e
=
1.15
∗
10
−
5
∗
Δ
B
2.26
∗
f
s
w
1.11
{\displaystyle P_{Fe} = 1.15 * 10^{-5} * \Delta B^{2.26} * f_{sw}^{1.11}}
THOMSON
3C85
0.33
P
F
e
=
1.54
∗
10
−
7
∗
Δ
B
2.62
∗
f
s
w
1.54
{\displaystyle P_{Fe} = 1.54 * 10^{-7} * \Delta B^{2.62} * f_{sw}^{1.54}}
PHILIPS
N67
0.38
P
F
e
=
8.53
∗
10
−
7
∗
Δ
B
2.54
∗
f
s
w
1.36
{\displaystyle P_{Fe} = 8.53 * 10^{-7} * \Delta B^{2.54} * f_{sw}^{1.36}}
EPCOS (ex S+M)
PC30
0.39
P
F
e
=
1.59
∗
10
−
6
∗
Δ
B
2.58
∗
f
s
w
1.32
{\displaystyle P_{Fe} = 1.59 * 10^{-6} * \Delta B^{2.58} * f_{sw}^{1.32}}
TDK
F44
0.4
P
F
e
=
2.39
∗
10
−
6
∗
Δ
B
2.23
∗
f
s
w
1.26
{\displaystyle P_{Fe} = 2.39 * 10^{-6} * \Delta B^{2.23} * f_{sw}^{1.26}}
MMG
Calculate minimal AP needed
A
P
m
i
n
=
10
3
∗
(
L
p
∗
I
P
r
m
s
Δ
T
1
2
∗
K
u
∗
B
m
a
x
)
1.316
{\displaystyle AP_{min} = 10^3 * \left ( \frac{ L_p * I_{Prms} }{ \Delta T^{ \frac{1}{2} } * K_u * B_{max} } \right )^{1.316}}
c
m
4
{\displaystyle cm^4}
B
m
a
x
{\displaystyle B_{max}}
B
s
a
t
{\displaystyle B_{sat}}
B
s
a
t
>
0.3
T
{\displaystyle B_{sat} > 0.3T}
B
m
a
x
=
0.25
T
{\displaystyle B_{max} = 0.25T}
Δ
T
=
T
m
a
x
−
T
a
m
b
{\displaystyle \Delta T = T_{max} - T_{amb}}
Generally
T
m
a
x
=
100
C
{\displaystyle T_{max} = 100C}
T
a
m
b
=
30
C
{\displaystyle T_{amb}=30C}
Using
K
u
=
0.3
{\displaystyle K_u=0.3}
Calculate minimum number of primary and secondary turns
N
P
−
m
i
n
=
L
p
∗
I
p
k
∗
10
4
B
m
a
x
∗
A
e
{\displaystyle N_{P-min} = \frac{ L_p * I_{pk} * 10^4 }{ B_{max} * A_e }}
N
S
−
m
i
n
=
N
P
−
m
i
n
N
{\displaystyle N_{S-min} = \frac{ N_{P-min} }{ N }}
Calculate actual number of turn on the primary and secondary to be used.
N
S
{\displaystyle N_S}
N
S
−
m
i
n
{\displaystyle N_{S-min}}
N
P
=
N
∗
N
S
{\displaystyle N_P = N * N_S}
Calculate air gap
l
g
=
μ
o
∗
N
P
2
∗
A
e
∗
10
−
2
L
p
{\displaystyle l_g = \frac{ \mu_o * N_P^2 * A_e * 10^{-2} }{ L_p }}
Current calculations [ ] Variables
I
p
k
{\displaystyle I_{pk}}
I
m
i
n
{\displaystyle I_{min}}
Δ
I
p
p
{\displaystyle \Delta I_{pp}}
I
p
k
−
I
m
i
n
{\displaystyle I_{pk} - I_{min}}
Peak current
I
p
k
=
(
I
o
u
t
−
m
a
x
N
)
∗
(
1
1
−
D
m
a
x
)
+
Δ
I
L
2
{\displaystyle I_{pk} = \left ( \frac{ I_{out-max} }{ N } \right ) * \left ( \frac{ 1 }{ 1 - D_{max} } \right ) + \frac{ \Delta I_L }{ 2 }}
DC current
I
d
c
=
D
I
p
k
+
I
m
i
n
2
{\displaystyle I_{dc}=D \frac{I_{pk}+I_{min}}{2}}
RMS current
I
r
m
s
=
D
(
(
I
p
k
+
I
m
i
n
)
+
1
3
(
I
p
k
+
I
m
i
n
)
2
)
{\displaystyle I_{rms}=\sqrt{ D \left ((I_{pk}+I_{min}) + \frac{1}{3} (I_{pk}+I_{min})^2 \right )}}
AC current
I
r
m
s
=
I
r
m
s
2
−
I
d
c
2
{\displaystyle I_{rms}=\sqrt{ I_{rms}^2 - I_{dc}^2 }}
Power Loss [ ]
P
t
o
t
=
P
f
e
+
P
c
u
{\displaystyle P_{tot} = P_{fe} + P_{cu}}
References [ ] 
