ST TDA911 Series, Примечание к применению

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STMicroelectronics Confidential 43/62
AN1290 Geometry and Focus Control Section
or,
The output must go to 40 V when input is 8.5 V, therefore:
Combining both equations:
or,
Returning to first equation:
We determine:
Now we must choose the value of one resistor out of R17, R18 and R19, the others follow:
Choosing R17 = 270 k
Ω
entails R18 = 68 k
Ω
, R19 = 270 k
Ω
(nearest normalized values).
Þ
Please notice that the R17/R18 divider applies a bias voltage of 2.4 V to the feedback input,
which is slightly higher than the “plateau” voltage which corresponds to the maximum H-amplitude.
This is mandatory to ensure that the amplifier output voltage will be approximately 0 regardless of
the frequency when the H-size is set for maximum amplitude.
Þ
Since this bias voltage and the voltage from the IC derive from different sources, there may exist
some offset between these voltages. This offset might be non-negligible compared to the total E/W
voltage from IC (mainly at a low H frequency, when frequency tracking is used). As a consequence,
it might be impossible to properly adjust the horizontal amplitude at a low frequency. A convenient
countermeasure is to implement a bias adjustment (sub-HSize) with a DAC and a resistor (470k
Ω
should be convenient). The adjustment is shown in Figure 17 .
Þ
Take care to provide a pull-down resistor to always keep the output transistor conductive on pin
24! (Not needed with the TDA9112A)
Þ
You will find on the ST Evaluation Board a similar application; except that the signal from pin 24 to
the input passes through an attenuator (R
34
constitutes a resistive divider with R
33
//R
37
). The loss
in gain has been more than compensated by increasing the value of R
19
. This leads to a set of
values (to be adapted to specific applications):
R
34
= 1 k
Ω
, R
33
= 4.7 k
Ω
, R
37
= 27 k
Ω
, R
17
= 270 k
Ω
, R
18
= 39 k
Ω
, R
19
= 270 k
Ω
V
s
2
-------
R
17
1
R
19
-----------
1
R
17
-----------
1
R
18
-----------
+ +
è ø
æ ö
⋅
=
40
R
19
-----------
8.5
1
R
19
-----------
1
R
17
-----------
1
R
18
-----------
+ +
è ø
æ ö
×
V
s
R
17
-----------
– =
40
R
19
-----------
8.5
V
s
2 R
17
⋅
-------------------
×
V
s
R
17
-----------
– =
R
17
R
19
-----------
V
S
8.5
8.5
--------
2
-------- 1
–
40
------------------
⋅
0.975 = =
V
s
2
------- R
17
1
R
19
-----------
1
R
17
-----------
1
R
18
-----------
+ +
è ø
æ ö
⋅
=
1
R
17
R
18
-----------
R
17
R
19
-----------
+ + =
1
R
17
R
18
----------- 0.975
+ + =
R
17
R
18
----------- 4.025
=
12
R
18
R
17
R
18
+
----------------------------
⋅
2.4V
=