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3DQDVRQLF
Œ
Functioning and waveform explanation
1.3.2.1.1
From t
to t
1
At the initial moment t
1
is zero (iy=0) and there is no horizontal deflection magnetic field. The capacitor Cs is charged and will
be discharged between t
(Vct=0) between t
and t
1
evaluate the horizontal deflection current i
T
We can write V
= V
cs
Ly
We can write V
= V
Cs
Ly
with
Then
The solution for this equation is :
i
= I
y
The iy oscillating period ( F= 2π√LyCs ) is much higher than the time t2-t1. Then the shape of current
iy is almost a straight line ( with a S-Correction shape)
• From t
to t
and t
(retrace time)
2
3
4
At t
, the transistor stops conducting. All the current circulating through the transistor is out off. Then
2
Ct is charging between t
between t
and t
. Cs continues to discharge until t
3
4
17 can be represented by simplified diagram shows on figure 21 arid figure 22.
:
2
, the electron beam is aimed at the center of a line since the deflection current
and t
. As the transistor Tr is in a conductive state, Ct remains discharged
1
2
. The circuit in figure 17 can be simplified (see figure 20). Then we can
2
.
y
r
Fig. 20 : Between t
= −
VV
Cs
=
VV
Ly
∫
1
t
i dx
Cs
t
1
⇒ +
i
y
sin ωt
0
and t
, reaching its maximum voltage at the instant t
2
3
V
L
Ly
1
V
V
Cs
CC
and t
1
2
t
∫
1
i dx
y
C
S
t
1
di
y
L
y
dt
di
y
+
=
Ly
0
y
dt
2
d i
y
=
LyCs
0
2
dt
1
with ω=
LyCs
and then starts to charge again. The circuit in fig
3
26
Current flow
, and discharging
3
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