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OP297GS 参数 Datasheet PDF下载

OP297GS图片预览
型号: OP297GS
PDF下载: 下载PDF文件 查看货源
内容描述: 双通道,低偏置电流精密运算放大器 [Dual Low Bias Current Precision Operational Amplifier]
分类和应用: 运算放大器
文件页数/大小: 12 页 / 267 K
品牌: AD [ ANALOG DEVICES ]
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OP297
PRECISION POSITIVE PEAK DETECTOR
In Figure 11, the C
H
must be of polystyrene, Teflon , or poly-
ethylene to minimize dielectric absorption and leakage. The
droop rate is determined by the size of C
H
and the bias current
of the OP297.
1k
+15V
1N4148
2
1k
8
0.1 F
6
1k
C
H
RESET
1k
2N930
–15V
®
All the transistors of the MAT04 are precisely matched and at
the same temperature, so the I
S
and V
T
terms cancel, giving
2
ln I
IN
=
ln I
O
+
ln I
REF
=
ln
(
I
O
×
I
REF
)
Exponentiating both sides of the equation leads to
I
O
(
I
IN
)
2
=
I
REF
V
IN
3
1/2
OP297
1
5
1/2
OP297
7
V
OUT
Op amp A2 forms a current-to-voltage converter, which gives
V
OUT
= R2
×
I
O
. Substituting (V
IN
/R1) for
I
IN
and the above
equation for
I
O
yields
V
OUT
R
2
 
V
IN
=

I
REF
 
R
1
2
0.1 F
A similar analysis made for the square-root circuit of Figure 14
leads to its transfer function
V
OUT
=
R
2
Figure 11. Precision Positive Peak Detector
SIMPLE BRIDGE CONDITIONING AMPLIFIER
(
V
IN
)(
I
REF
)
R
1
C2
100pF
R2
33k
6
I
O
Figure 12 shows a simple bridge conditioning amplifier using the
OP297. The transfer function is
 ∆
R
R
F
V
OUT
=
V
REF
R
+ ∆
R
R
The REF43 provides an accurate and stable reference voltage for
the bridge. To maintain the highest circuit accuracy, R
F
should
be 0.1% or better with a low temperature coefficient.
15V
R
F
1
2
Q1
3
5
1/2
OP297
7
V
OUT
6
7
Q2
5
8
MAT04E
I
REF
14
13
Q4
12
R3
50k
R4
50k
–15V
V
REF
REF43
2
4
R+
R
3
C1
100pF V+
1/2
OP297
1
V
OUT
V
IN
R1
33k
Q3
10
1
9
2
8
3
1/2
OP297
4
6
8
V–
7
V
OUT
= V
REF
R
R+
R
R
F
R
5
1/2
OP297
4
Figure 13. Squaring Amplifier
R2
33k
C2
100pF
6
I
O
Figure 12. A Simple Bridge Conditioning Amplifier
Using the OP297
NONLINEAR CIRCUITS
Due to its low input bias currents, the OP297 is an ideal log
amplifier in nonlinear circuits such as the square and square-
root circuits shown in Figures 13 and 14. Using the squaring
circuit of Figure 13 as an example, the analysis begins by writing a
voltage loop equation across transistors Q1, Q2, Q3, and Q4.
5
1/2
OP297
7
V
OUT
I
REF
Q1
C1
100pF
V+
V
IN
R1
33k
2
8
6
1
3
7
Q2
5
8
MAT04E
13
14
Q4
12
I
I
I
I
V
T
1
ln
IN
 +
V
T
2
ln
IN
 =
V
T
3
ln
O
 +
V
T
4
ln
REF
I
S
3
I
S
4
I
S
1
I
S
2
Q3
10
9
3
1/2
OP297
4
V–
1
R3
50k
R4
50k
–15V
Figure 14. Square-Root Amplifier
–8–
REV. E