Obsolete Product(s) - Obsolete Product(s)
Vertical Section
AN1290
24/62
STMicroelectronics Confidential
For the calculation of vertical frequency, we can neglect this discharge time and find:
with I
0
= 45 µA (typical).
Nevertheless, there is a ±20% spread on this current, so that the same spread must be expected on
the free-running frequency (not taking into account the capacitor spread).
When a sync pulse is received, the system switches to synchronized mode. In this mode, the
vertical sawtooth amplitude is maintained constant in relation to the frequency by an AGC loop, as
described below.
The front edge of the first Vsync pulse opens switch NoS and closes switch S for a 13µs sampling
duration. C
s
will then charge to the voltage present at that moment on the amplifier output.
The discharge of C
o
is only triggered at that moment. At the same time, the discharge threshold
switches from 5V to 7V.
Every subsequent sync edge will trigger the sampling and then the discharge of C
0
, so that C
s
always stores the previous sawtooth peak value. V(C
s
) is then converted to current I'
0
by a high-
impedance converter using the following formula:
with R = 18 k
Ω
.
Since C
o
is charged by I
0
-I'
0
, there is a feedback effect. If the last sampled voltage was higher than
5V, the charge current will decrease, and conversely, so that next peak value will be closer to 5V. At
the end of the process, the voltage on C
s
will be 5V + (R x I'
0
), and the sawtooth peak voltage:
with R
1
= 19 x R
2
.
5.1.3
Frequency Range and Precision
Value I'
o
may vary between -3xI
o
/4 and +2xI
o
, so that the total charge current presents a total range
of: I
o
/4 to 3xI
o
.
Therefore, the frequency range may reach a ratio of 12. Nevertheless, the upper and lower limits are
subject to spread and to thermal drift, which considerably reduces the really usable range.
Therefore the guaranteed autosync frequency range is only 50 to 185Hz, with C
o
= 150 nF. This
includes a ±5% spread for C
o
.
F
v
I
0
Cx V
D
----------------
=
= 100 Hz (typical)
V C
s
(
)
5
–
(
)
R
------------------------------
I’
0
=
5V
Rx I
0
'
x
R
2
R
1
R
2
+
----------------------
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