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CLC428 Datasheet(PDF) 6 Page - National Semiconductor (TI)

[Old version datasheet] Texas Instruments acquired National semiconductor.
Part # CLC428
Description  Dual Wideband, Low-Noise, Voltage Feedback Op Amp
PDF  8 Pages
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Manufacturer  NSC [National Semiconductor (TI)]
Direct Link  http://www.national.com
Logo NSC - National Semiconductor (TI)

CLC428 Datasheet(HTML) 6 Page - National Semiconductor (TI)

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Figure 6
Positive Peak Detector
The CLC428's dual amplifiers can be used to implement a
unity-gain peak detector circuit as shown in Figure 7.
Figure 7
The acquisition speed of this circuit is limited by the
dynamic resistance of the diode when charging Chold. A
plot of the of the circuit's performance is shown in Figure
8 with a 1MHz sinusoidal input.
Figure 8
A current source, built around Q1, provides the necessary
bias current for the second amplifier and prevents satura-
tion when power is applied. The resistor, R, closes the loop
while diode D2 prevents negative saturation when Vin is
less than Vc. A MOS-type switch (not shown) can be used
to reset the capacitor's voltage.
The maximum speed of detection is limited by the delay
of the op amps and the diodes. The use of Schottky diodes
will provide faster response.
Adjustable or Bandpass Equalizer
A "boost" equalizer can be made with the CLC428 by
summing a bandpass response with the input signal, as
shown in Figure 9.
Figure 9
The overall transfer function is shown in Eq. 5.
V
V
R
KR
R
s2Q
ss
Q
1
out
in
b
a
b
o
2
o
o
2
=
+
F
HG
I
KJ ++ −
ch
ω
ω
ω
Eq. 5
To build a boost circuit, use the design equations Eq. 6 and
Eq. 7.
RC
2
Q
,2C R ||R
1
Q
2
o
a
b
o
==
ωω
ch
Eq. 6,7
Select R2 and C using Eq. 6. Use reasonable values for
high frequency circuits - R2 between 10Ω and 5kΩ, C
between 10pF and 2000pF. Use Eq. 7 to determine the
parallel combination of Ra and Rb. Select Ra and Rb by
either the 10
Ω to 5kΩ criteria or by other requirements
based on the impedance Vin is capable of driving. Finish
the design by determining the value of K from Eq. 8.
Peak Gain
V
V
R
KR
out
in
o
a
==
ω
ch 2
2
1
Eq. 8
Figure 10 shows an example of the response of the circuit
of Figure 9, where fo is 2.3MHz. The component values
are as follows: Ra =2.1kΩ, Rb =68.5Ω, R2 =4.22kΩ, R
=500
Ω, KR =50Ω, C =120pF.
Figure 10
Q1
http://www.national.com
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