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i L_min
i =2.2 A L_min
i L_ave
∆i fall
i L_max
i =2.2 A L_max
i L_ave
∆i fall
This is greater than half of the current fall. By the way, the current will have a maximum and
mimimum value as follows:
i L_ave
i =1.2 A L_ave
V
R N
The flyback converter is in continuous conduction because the fall current is equal to the rise
current when a complete duty cycle of just D for rise and 1-D for fall is checked. In discontinuous
conduction, the fall would be greater than the rise.
Another way to check for continuous conduction is to calculate the average inductor current. If it is
greater than half the inductor current fall, then the system is in continuous conduction. Remember
to reflect the current to the proper side of the transformer.
∆i fall
∆i =2 A fall
V
L
m
N
⋅ ( 1 −D)T
s
V
R
=1.2 A
∆i rise
∆i =2 A rise
V
s
L
m
⋅ DT
s
Checking the continuous conduction assumption, inductor current rise and fall are
V
V =24 V
V
s
D
1 −D
⋅ N
T
s
f s
Assuming a continuous conduction condition, :=
f s
L C := 100 ⋅μF :=30 kHz⋅ m
N R :=20 ohm⋅ := 240 ⋅μH 21
V D :=0.4 := 1
s
:=36 V⋅
Restate the given:
Problem 7.1 The flyback converter of Figure 7-2a has the following parameters: Vs-36V, D=0.4,
N1/N2=1, R=20 ohms, Lm-240uH, C=100uF and the switching frequency is 30kHz.
a. Determine the output voltage.
∆V
pu
− 3 ∆V = × pu
D
R C⋅ f s
c. Output voltage ripple
i Lmax
i =3 A Lmax
V
s
⋅D
( 1 −D)
2 ⋅R
N
V
s
⋅ DT
s
2 L
m
i Lmin
i =1 A Lmin
V
s
⋅D
( 1 −D)
2 ⋅R
N
V
s
⋅D T
s
2 L
m
The text gives formulae:
i Lmax
i =3 A Lmax
I
Lm
∆i rise
and maximjm magnetizing inductor current is
i Lmin
i =1 A Lmin
I
Lm
∆i rise
Therefore, minimum magnetizing inductor current is
∆i fall
∆i =2 A rise
=2 A
We have already found (^) ∆i in the inductor. Rise and fall were equal in this cae,.
I
Lm
I =2 A
Lm
I
s
D
Average magnetizing inductor current is
I
s
I =0.8 A
s
P
s
V
s
Finding average input current
P
s
P
We have an ideal system, no losses, so input power = output power.
P
P =28.8 W
V
2
R
Power out: :=
b. Find the average, maximum, and minimum inductor currents
I
Lsmax
I =2.888 A
Lsmax
I
Ls
∆i Ls
I
Lsmin
I =0.612 A
Lsmin
I
Ls
∆i Ls
Minimum inductor current and maximum inductor current are
∆i Ls
∆i =2.275 A Ls
V
s
N
⋅ V
⋅ DT
s
L
s
Checking the rise portion of the cycle, to verify continuous conduction (rise and fall should give
equal currents in continuous conduction over a complete duty cycle),
∆i Ls
∆i =2.275 A Ls
V
⋅( 1 −D)T
s
L
s
Change in inductor current, using the falling inductor current portion of the cycle,
I
Ls
I =1.75 A
Ls
V
R
b. Determine average, maximum, and minimum current in the inductor.
Average current in the inductor is the same as average current in the load because the capacitor
must have zero average current in steady state switching conditions.
∆V
pu
− 3 ∆V = × pu
1 −D
8 L
s
⋅ ⋅C f s
2 ⋅
Normalized voltage ripple is, from equation 7-33 on page 251 of the text:
V
V =35 V
V
s
N
:= ⋅ ⋅D
We will assume the normal mode of operation: Transformer magnetizing inductor current
discontinuous and output inductor current continuous.
T
s
f s
f := s
R :=20 ohm⋅ C := 100 ⋅μF D :=0.35 :=50 kHz⋅
L
s
L := 200 ⋅μH m
N :=5.0 mH⋅ 21
V := 1
s
:=100 V⋅
Problem 7.6 The forward converer of Figure 7.5a has the following parameters: Vs=100V, N=1,
Lm=5mH, Ls=200uH,R=20ohms, C=100uF, D=0.35, and fs=50kHz.
a. Determine the output voltage and the output voltage ripple.
N
N =1.
V
V
s
⋅D
Then,
D :=0.
If we select a nominal duty cycle of
V
V
s
D N =0.
V
V
s
Therefore, the product of duty cycle and turns ratio is
V
V
s
⋅D N
We know that = ⋅
N
N
N
Let (^) = 1.
T
s
f s
Select a maximum value for N1/N3 of 1.00; the magnetizing inductance is not given, so this
number cannot be checked.
f s
L C := 150 ⋅μF :=40 kHz⋅ s
V := 100 ⋅μH 0
P :=50 V⋅
V :=250 W⋅
s
:=75 V⋅
Restate the given:
Problem 7.7a A forward converter has a source of 75V and a load of 250W at 50V. The output filter
has Ls=100uH and C=150uF. The switching frequency is 40kHz.
Select a duty ratio and transformer turns ratios N1/N2 and N1/N3 to rpvide the requested outuput
voltage and verify continuous current in Lx.
I
sw_max
I =3.028 A
sw_max
I
Lsmax
N
⋅ I
Lm_pk
d. Determine the peak current in the switch (and transformer primary).
This current is the sum of the peak inductor current, reflected to the primary, plus the peak
magnetizing current.
I
Lm_pk
I =0.14 A
Lm_pk
V
s
⋅ DT
s
L
m
c. Determine the peak current in the transformer magnetizing inductance,
Using the rising part of the cycle, the peak magnetizing current is
