9/15
STV9302A
Application Hints
4.3.1
Centring
When idle, both driver outputs provide i
cm
and the yoke current should be null (R
1
is negligible),
hence:
4.3.2
Peak Current
Scanning current should be I
P
when positive and negative driver outputs provide respectively
i
cm
- i
p
and i
cm
+ i
p
, therefore
and since R
7
= R
2
:
Choose R
1
in the 1
Ω
range, the value of R
2
= R
7
follows. Remember that i is one-quarter of driver
peak-peak differential signal! Also check that the voltages on the driver outputs remain inside
allowed range.
●
Example: for i
cm
= 0.4mA, i = 0.2mA (corresponding to 0.8mA of peak-peak differential
current), I
p
= 1A
Choose R
1
= 0.75
Ω
, it follows R
2
= R
7
= 1.875k
Ω
.
4.3.3
Ripple Rejection
Make sure to connect R
7
directly to the ground side of R
1
.
4.3.4
Secondary Breakdown Diagrams
The diagram has been arbitrarily limited to max VS (35 V) and max I0 (2 A).
Figure 8: Output Transistor Safe Operating Area (SOA) for Secondary Breakdown
i
cm
R
7
⋅
i
cm
R
2
therefore R
7
R
2
=
⋅
=
i
cm
i
–
(
)
R
7
⋅
I
p
R
1
⋅
i
cm
i
+
(
)
R
2
⋅
+
=
I
p
i
-----
2R
7
R
1
-----------
–
=
100
µ
s
10ms
100ms
0.01
0.1
1
10
10
60
100
Volts
Ic
(A)
@ Tcase=25°C
35
