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i +i
1
y
T
r
C
T
Fig. 21: between t
Then we can evaluate the voltages and current waveforms from t
There are only L and C components in the circuit, therefore we can assume that the current i
sinus shape as shown on figure 19:
i
= I
cosΩt (I
is a constant which will be determined in the following paragraph)
y
2
2
From that circuits we can write:
t
∫
1
=
+
•
VV
(
i
i dt
y
Ct
Ct
t
2
=
−
•
VV
Vcc V
Ct
L
1
=
−
•
VV
V
V
Ct
Cs
Ly
From (3), we obtain :
i
From (2), we obtain :
Then we can deduce from this equation that the supply current passing through the transformer
primary (i
) has a sawtooth waveform, similar to the deflection current (i
l
difference : i
has a DC component (I
l
I
constant calculation:
2
The voltage V
in the transformer primary is given by
L1
V
= V
- V
L1
CC
Ct
The voltage V
in a coil cannot obviously contain a DC component. Therefore, the value of average
LI
Voltage should be 0
T
0
∫
⇒
=
V dt
L
1
0
T
0
∫
⇒
−
(
V
V
CC
0
=
VV
I L
or
Ct
2
=0
by replacing V
value to formula (4), we obtain
Ct
V
T
=
CC
0
I
2
2
Ly
V
Ly
i
i
1
y
V
Cs
and t
Fig. 22: between t
2
3
)
1
2
dV
d i
Ct
1
⇒
= −
L
1
2
dt
dt
t
∫
1
di
y
= −
−
i dt
Ly
y
Cs
dt
t
2
1
=
Ω
−
VV
I Ly
(
Ct
2
Ω
Cs
Ly
1
=
−
I
(
) cos
1
2
Ω
2
L
L Cs
1
1
Ly
Ω =
with
+
C
T
)-
1
0
=
)
0
Ct
1
Ω
−
Ω
(
) sin
t
Ω
y
C
s
i +i
1
y
V
L1
T
r
C
T
V
CC
and t
3
4
to t
2
(1)
(2)
(3)
Ω
) sin
t
Ω
+
t
I
( I
: constant)
1
1
+
L
1
L Ly
1
with T
= 64µs
0
between t
and t
1
2
between t
2
27
3DQDVRQLF
V
Ly
i
y
V
Cs
as presented on figure 19.
4
), but with one important
y
(4)
and t
7
V
L1
i
1
V
CC
has a
y
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