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TS615P 数据表(PDF) 24 Page - STMicroelectronics

部件名 TS615P
功能描述  DUAL WIDE BAND OPERATIONAL AMPLIFIER WITH HIGH OUTPUT CURRENT
PDF  27 Pages
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制造商  STMICROELECTRONICS [STMicroelectronics]
网页  http://www.st.com
标志 STMICROELECTRONICS - STMicroelectronics

TS615P 数据表(HTML) 24 Page - STMicroelectronics

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TS615
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CHOICE OF THE FEEDBACK CIRCUIT
Table 71 : Closed-Loop Gain - Feedback Compo-
nents
INVERTING AMPLIFIER BIASING
In this case a resistance is necessary to achieve a
good input biasing, as R on (fig.30).
This resistance is calculated by assuming the neg-
ative and positive input bias current. The aim is to
make the compensation of the offset bias current
which could affect the input offset voltage and the
output DC component. Assuming Ib-, Ib+, Rin, Rfb
and a zero volt output, the resistance R comes: R
= Rin // Rfb .
Figure 72 : Compensation of the Input Bias
Current
ACTIVE FILTERING
Figure 73 : Low-Pass Active Filtering. Sallen-Key
The resistors Rfb and RG give directly the gain of
the filter as a classical non-inverting amplification
configuration :
Assuming the following expression as the re-
sponse of the system:
the cutoff frequency is not gain dependent and it
comes:
the damping factor comes:
The higher the gain the more sensitive the damp-
ing factor is. When the gain is higher than 1 it is
preferable to use some very stable resistors and
capacitors values.
In the case of R1=R2:
VCC (V)
Gain
Rfb (Ω)
±6
+1
750
+2
680
+4
620
+8
510
-1
680
-2
680
-4
620
-8
510
±2.5
+1
1.1k
+2
1k
+4
910
+8
680
-1
1k
-2
1k
-4
910
-8
680
R
Load
Output
Rfb
Rin
Ib-
Ib+
Vcc+
Vcc-
TS615
+
_
R
Load
Output
Rfb
Rin
Ib-
Ib+
Vcc+
Vcc-
TS615
+
_
+
_
RG
IN
Rfb
910
Ω
TS615
25
Ω
OUT
R1
R2
C2
C1
+
_
RG
IN
Rfb
910
Ω
TS615
25
Ω
OUT
R1
R2
C2
C1
A
V
g1
R
fb
R
g
----------
+
==
T
j
ω
Vout
j
ω
Vin
j
ω
-------------------
g
12
ζ
j
ω
ω
c
-------
j
ω
()
2
ω
c
2
--------------
++
---------------------------------------------
==
ω
c
1
R1R2C 1C2
--------------------------------------
=
ζ
1
2
---
ω
c
C
1
R
1
C
1
R
2
C
2
R
1
C
1
R
1
g
–
++
()
=
ζ
2C
2
C
1
R
fb
R
g
----------
–
2C
1
C
2
------------------------------------
=



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