clc5602 National Semiconductor Corporation, clc5602 Datasheet - Page 10

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clc5602

Manufacturer Part Number
clc5602
Description
Dual, High Output, Video Amplifier
Manufacturer
National Semiconductor Corporation
Datasheet

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This example illustrates a lowpass filter with Q = 0.707
and corner frequency f
chosen to achieve a maximally flat, Butterworth
response. Figure 14 indicates the filter response.
http://www.national.com
V
in
0.1 F
For R
Gain K 1
Corner frequency
Q
Q
Figure 12: Lowpass Filter Topology
c
-15
-21
-3
-9
3
5k
158
5k
Figure 14: Lowpass Response
(3 K)
R
Figure 13: Design Equations
RC
1
R C
R C
1
1
1
2 2
1 1
R
1M
158
2
100pF
R
1.698k
2
R
C
R
R and C
Frequency (Hz)
2
g
f
c
R C
R C
= 10MHz. A Q of 0.707 was
1 2
2 1
10M
R
c
0.1 F
3
2
1
g
1
CLC5602
-
+
+5V
1/2
R R C C
C
4
(1 K)
8
1 2 1 2
2
100M
0.1 F
1k
C
R
1
C
1
f
1
R C
R C
2 2
1 1
0.1 F
100
V
o
10
V
Differential Line Driver With Load
Impedance Conversion
The circuit shown in the Typical Application schematic
on the front page and in Figure 15, operates as a
differential line driver. The transformer converts the load
impedance to a value that best matches the CLC5602’s
output capabilites. The single-ended input signal is
converted to a differential signal by the CLC5602. The
line’s characteristic impedance is matched at both the
input and the output. The schematic shows Unshielded
Twisted Pair for the transmission line; other types of lines
can also be driven.
Set up the CLC5602 as a difference amplifier:
Make the best use of the CLC5602’s output drive
capability as follows:
where R
ance, V
maximum Output Current.
Match the line’s characteristic impedance:
Select the transformer so that it loads the line with a
value very near Z
impedance of the CLC5602 also affects the match. With
an ideal transformer we obtain:
in
R
t1
R
g1
Return Loss
+
CLC5602
-
max
1/2
Figure 15: Differential Line Driver wtih
R
eq
g2
R
f1
is the transformed value of the load imped-
is the Output Voltage Range, and I
V
d/2
Load Impedance Conversion
R
t2
V
V
in
d
+
CLC5602
-
R
o
1/2
R
20 log
m
2
f2
over frequency range. The output
-V
R
R
n
R
d/2
1
L
m
eq
10
R
R
R
R
R
m/2
m/2
Z
R
R
R
eq
g1
f1
n
o
2 V
eq
eq
L
2
I
max
1:n
Z
max
2
o 5602
Z
R
o
R
g2
f2
UTP
Z
j
o
max
,dB
I
o
is the
R
L
V
+
-
o

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