SWITCHING POWER SUPPLY DESIGN_ CONTINUOUS MODE FLYBACK CONVERTER.pdf
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SWITCHING POWER SUPPLY DESIGN:
CONTINUOUS MODE
FLYBACK CONVERTER
Written by Michele Sclocchi
michele.sclocchi@nsc.com
Application Engineer
National Semiconductor
Typical Flyback power supply:
D4
N=18T
C10
220p
R5
10
16V
D2
R1
49.9K
N=18T
N=6T
D5
D3
Q1
20 - 55V
3.3V
D1
10V
R3
10K
R6
200K
C1
1
100V
C2
1
100V
13
9,10,11
C9
68u
4V
C8
68u
4V
C3
0.22
Vin
SW
2
FB
LM5000-3
3
SHDN
R4
6.19K
Byp
FS
SS
COMP GND
12
15
14
1
4,5,6,
7,8,16
R2
1K
R7
10K
C4
0.1
C6
100p
C11
1000p
C5
0.1
C7
0.1
LM5000 FLYBACK CONVERTER
Notes:
Write down the power supply requirements on :
X
xx
:=
Get the results on:
Rsults
xx
:=
This Mathcad file helps the calculation of the external components of a typical continuous
mode switching power supply.
Input voltage:
- Minimum input voltage:
-
6
Vi
min
:=
22
volt
m
sec
:=
10
sec
- Maximum input voltage:
Vi
max
:=
55
volt
- Nominal input voltage:
Vi
nom
:=
36
volt
Output:
-
Nominal output voltage, maximum output ripple, minimum output current, maximum output
current
Vo1
:=
3.3
volt
Vrp1
:=
100
mV
Io1
min
:=
0.250
amp
Io1
max
:=
2.
amp
Io2
min
Vo2
:=
0
volt
Vrp2
:=
120
mV
:=
0.000
amp
Io2
max
:=
0.000
amp
( diode's forward drop voltage)
Vd
fw
:=
0.5
volt
(
)
Io1
min
(
)
Io2
min
Po
min
:=
Vo1
+
Vd
fw
+
Vo2
+
Vd
fw
Po
min
=
0.95
watt
(
)
Io1
max
(
)
Io2
max
Po
max
:=
Vo1
+
Vd
fw
+
Vo2
+
Vd
fw
Po
max
=
7.6
watt
1
fsw
- Switching Frequency:
fsw
:=
300
kHz
T
:=
T
=
3.33 m
sec
- Transformer's Efficiency:
:=
(Guessed value)
- Maximum drop voltage across the switching mosfet during the on time:
- On resistance of the Mosfet:
h
0.90
Rds
on
:=
0.180
ohm
Po
max
h
Vi
min
Vds
on
:=
Rds
on
Vds
on
=
0.07
volt
1) Define Primary/secondary turns ratio: Nps1
Primary/secondary turns ratio can be selected as compromise between maximum voltage
across the switching mosfet and desired max.-min. duty cycle.
- Nominal desired on Duty Cycle:
D
nom
:=
0.24
Ł
Vi
nom
-
Vds
on
ł
D
nom
Nps1
:=
Nps1
=
3
Vo1
+
Vd
fw
1
-
D
nom
The calculated turns ratio can be modified to optimise the windings
- Flyback voltage across the mutual inductance during the off time: Vfm
Vfm
(
)
:=
Vfm
11.35
vol
=
- Maximum voltage across the switching-mosfet:
F
spike
Nps1
Vo1
+
Vd
fw
:=
0.15
(
)
Vi
max
(
)
Vds
max
:=
F
spike
+
1
+
Vfm
Vds
max
=
76.3
volt
Safe factor (assume spikes of 20-30% of Vdc )
To reduce the maximum voltage across the switching mosfet reduce Nps turns ratio by reducing
the desired on-duty cycle
- Slave output turns ratio:
Vfm
Nps2
:=
Nps2
=
22.7
Vo2
+
Vd
fw
2) Maximum and minimum duty cycle : Dmax and Dmin
To maintain the continuous mode of operation the dead time has to be equal zero (Ton+Toff = T),
and to reset the core every cycle, the average voltage on the primary inductance must be equal
zero: ( Vi - Vds ) * Ton = ( Vo + Vd ) * Nps * Toff, where Toff is equal to (T - Ton)
Vfm T
Ton
max
:=
Ton
max
=
1.14 m
sec
(
)
Vi
min
-
Vds
on
+
Vfm
Vfm T
Ton
min
:=
Ton
min
=
0.57 m
sec
(
)
Vi
max
-
Vds
on
+
Vfm
Ton
max
T
Maximum duty cycle
D
max
:=
D
max
=
0.34
Ton
min
T
Minimum duty cycle
D
min
:=
D
min
=
0.17
3) Primary winding: Inductance, peak, AC, RMS current
In continuous mode the duty cycle changes with a change of input voltage. An increase of output
current, will temporary increase the duty cycle until the average primary and secondary currents
increase.
Po
max
Ip
cs
:=
Ip
cs
=
1.13
amp
(
)
h
Vi
min
-
Vds
on
D
max
- Primary average current:
There are several criterias to select the primary and secondary inductances, following are
explained two different solutions: the first one is to select the primary inductance in order to
insure continuous mode of operation from full load to minimum load. (about 1/10-1/20 of the
maximum load).
(3-a)
,
The second alternative criteria, is to calculate primary and secondary inductances by defining
maximum secondary ripple current.
(3-b)
3-a) Select primary inductance for continuous mode of operation at
minimum load:
During the transition from discontinuous to continuous mode, the peak primary current it's about
double the central average current Ipcs(min) .In order to maintain continuous mode at minimum
load the maximul ramp amplitude has to be twice the minimum average current.
- Ramp amplitude:
2 Po
min
D
Ip
a
:=
D
Ip
a
=
0.28
amp
(
)
h
Vi
min
-
Vds
on
D
max
- Primary inductance: dIp= (Vi-Vds)*Ton/Lp
(
)
Ton
max
Vi
min
-
Vds
on
Lp
a
:=
Lp
a
=
88.29 m
H
D
Ip
a
3-b) Primary and secondary inductance for a maximum defined secondary
peak to peak ripple current:
AC core losses, AC winding losses, and output ripple current are directly proportional to the
current ramp amplitude of the primary and secondaries windings. Therefore in high current
application, AC ripple currents could have a predominant role on the overall performance of the
converter, a good compromise between transformer's size and AC currents can be obtained by
selecting the most appropriate secondary ripple current:
- Desired secondary ripple current:
D
Is%
(maximum value / average)
:=
30
%
Io1
max
Is1
cs
:=
Is1
cs
=
3.03
amp
(
)
1
-
D
max
- Ramp amplitude:
D
Is1
b
:=
Is1
cs
D
Is%
D
Is1
b
=
0.91
amp
- Secondary inductance :
(
)
T
(
)
Vo1
+
Vd
fw
-
Ton
max
Ls1
b
:=
Ls1
b
=
9.17 m
H
D
Is1
b
- Primary inductance:
Ls1
b
Nps1
2
Lp
b
:=
Lp
b
=
81.75 m
H
- Ramp amplitude:
(
)
Ton
max
Vi
min
-
Vds
on
D
Ip
b
:=
D
Ip
b
=
0.3
amp
Lp
b
Select primary inductance (3-a) or (3-b):---->
Lp
:=
Lp
b
Lp
=
81.75 m
H
(
)
Ton
max
Vi
min
-
Vds
on
D
Ip
:=
D
Ip
=
0.3
amp
Lp
- Primary average current:
Po
max
Ip
cs
:=
Ip
cs
=
1.13
amp
(
)
h
Vi
min
-
Vds
on
D
max
- Primary peak current:
D
Ip
2
Ip
pk
:=
Ip
cs
+
Ip
pk
=
1.28
amp
- Primary RMS current:
Ø
Œ
º
2
ø
œ
ß
Ł
ł
º
Ł
ł
ß
D
Ip
2
D
Ip
2
1
3
Ip
rms
:=
D
max
Ip
pk
Ip
cs
-
+
Ip
pk
-
Ip
cs
-
Ip
rms
=
0.66
amp
- Primary DC current:
Po
max
Ip
dc
:=
Ip
dc
=
0.39
amp
(
)
h
Vi
min
-
Vds
on
- Primar
y AC(rms) cur
rent:
Ip
rms
2
Ip
dc
2
Ip
ac
:=
-
Ip
ac
=
0.54
amp
-
5
=
4) Secondary winding: Inductance, peak, AC, RMS current
-Master output:
- Primary average current:
Edt
:=
Vi
min
Ton
max
Edt
2.5
·
10
volt sec
Io1
max
Is1
cs
:=
Is1
cs
=
3.03
amp
(
)
1
-
D
max
- Secondary inductance :
Lp
Nps1
2
Ls1
:=
Ls1
=
9.17 m
H
- Ramp amplitude:
(
)
T
(
)
Vo1
+
Vd
fw
-
Ton
max
D
Is1
:=
D
Is1
=
0.91
amp
Ls1
- Secondary peak current:
D
Is1
2
Is1
pk
:=
Is1
cs
+
Is1
pk
=
3.49
amp
- Secondary RMS current:
Œ
º
œ
ß
2
Ł
D
Is1
2
ł
1
3
º
Ł
D
Is1
2
ł
ß
(
)
Is1
rms
:=
1
-
D
max
Is1
pk
Is1
cs
-
+
Is1
pk
-
Is1
cs
-
Is1
rms
=
2.47
amp
- Second
ary AC current:
Is1
rms
2
Io1
max
2
Is1
ac
:=
-
Is1
ac
=
1.45
amp
-First slave output:
- Primary average current:
Io2
max
Is2
cs
:=
Is2
cs
=
0
amp
(
)
1
-
D
max
- Secondary inductance :
Lp
Nps2
2
Ls2
:=
Ls2
=
0.16 m
H
- Ramp amplitude:
(
)
T
(
)
Vo2
+
Vd
fw
-
Ton
max
D
Is2
:=
D
Is2
=
6.92
amp
Ls2
- Secondary peak current:
D
Is2
2
Is2
pk
:=
Is2
cs
+
Is2
pk
=
3.46
amp
- Secondary RMS current:
Œ
º
œ
ß
2
Ł
ł
º
Ł
ł
ß
D
Is2
2
1
3
D
Is2
2
(
)
Is2
rms
:=
1
-
D
max
Is2
pk
Is2
cs
-
+
Is2
pk
-
Is2
cs
-
Is2
rms
=
1.62
amp
- Second
ary AC current:
Is2
rms
2
Io2
max
2
:=
Is2
ac
1.62
am
=
5) Maximum Stress across the output diodes: Vdiode
-Maximum stress voltage on the cathode of diodes
Is2
ac
-
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