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OP297_03 Datasheet, PDF (8/12 Pages) Analog Devices – Dual Low Bias Current Precision Operational Amplifier
OP297
PRECISION POSITIVE PEAK DETECTOR
In Figure 11, the CH must be of polystyrene, Teflon®, or poly-
ethylene to minimize dielectric absorption and leakage. The
droop rate is determined by the size of CH and the bias current
of the OP297.
All the transistors of the MAT04 are precisely matched and at
the same temperature, so the IS and VT terms cancel, giving
( ) 2 ln IIN = ln IO + ln IREF = ln IO × IREF
Exponentiating both sides of the equation leads to
2
8
1/2
VIN
1k⍀
OP297
3
RESET
1k⍀
1N4148
+15V
0.1␮F
1
1k⍀
6
1/2
7
1k⍀
OP297
5
CH
0.1␮F
2N930
–15V
VOUT
Figure 11. Precision Positive Peak Detector
SIMPLE BRIDGE CONDITIONING AMPLIFIER
Figure 12 shows a simple bridge conditioning amplifier using the
OP297. The transfer function is
VOUT
=
VREF


R
∆R
+ ∆R


RF
R
The REF43 provides an accurate and stable reference voltage for
the bridge. To maintain the highest circuit accuracy, RF should
be 0.1% or better with a low temperature coefficient.
15V
REF43
4
VREF
R + ⌬R
RF
2
1/2
1
3 OP297
VOUT
( ) IO =
IIN 2
IREF
Op amp A2 forms a current-to-voltage converter, which gives
VOUT = R2 × IO. Substituting (VIN/R1) for IIN and the above
equation for IO yields
VOUT
=


R2
IREF




VIN
R1


2
A similar analysis made for the square-root circuit of Figure 14
leads to its transfer function
( )( ) VOUT = R2
VIN IREF
R1
R1
33k⍀
VIN
C2
100pF
R2
33k⍀
2
1
Q1
3
C1
100pF
V+
2
8
1/2
OP297
3
4
6
7
Q2
5
1
6
1/2
7
IO
OP297
5
VOUT
MAT04E
8
Q3 9
10
14
IREF Q4 13
12
R3
50k⍀
R4
50k⍀
6
8
1/2
7
OP297
5
4
⌬R
RF
VOUT = VREF R + ⌬R
R
Figure 12. A Simple Bridge Conditioning Amplifier
Using the OP297
V–
–15V
Figure 13. Squaring Amplifier
R2
33k⍀
C2
100pF
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.
VT1 ln


IIN
IS1


+ VT2
ln


IIN
IS2


=
VT
3
ln


IO
IS3


+VT 4 ln


IREF
IS 4


R1
33k⍀
VIN
6
1/2
7
IO
OP297
5
VOUT
IREF
C1
100pF
V+
Q1 1
3
7
6 Q2
5
MAT04E
8
Q3 9
10
13
14
Q4
12
2
8
1/2
1
OP297
3
4
R3
50k⍀
R4
50k⍀
V–
–15V
Figure 14. Square-Root Amplifier
–8–
REV. E