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OP193 Datasheet, PDF (12/16 Pages) Analog Devices – Precision, Micropower Operational Amplifiers
OP193/OP293/OP493
A Single-Supply Instrumentation Amplifier
Designing a true single-supply instrumentation amplifier with
zero-input and zero-output operation requires special care. The
traditional configuration, shown in Figure 31, depends upon
amplifier A1’s output being at 0 V when the applied common-
mode input voltage is at 0 V. Any error at the output is multi-
plied by the gain of A2. In addition, current flows through
resistor R3 as A2’s output voltage increases. A1’s output must
remain at 0 V while sinking the current through R3, or a gain
error will result. With a maximum output voltage of 4 V, the
current through R3 is only 2 µA, but this will still produce an
appreciable error.
R1
R2
20k
1.98M
R1
R2
20k
1.98M
+5V
V+
A1
1/2 OP293
–IN
V–
+5V
10k
Q1
VN2222
+IN
R3
R4
20k
1.98M
Q2
+5V
V+
A2
1/2 OP293
V–
VOUT
–IN
+IN
+5V
V+
A1
R3
20k
1/2 OP293
V–
IS I N K
R4
1.98M
+5V
V+
A2
1/2 OP293
V–
VOUT
Figure 31. A Conventional Instrumentation Amplifier
Figure 32. An Improved Single-Supply, 0 VIN, 0 VOUT
Instrumentation Amplifier
A Low-Power, Temperature to 4–20 mA Transmitter
A simple temperature to 4–20 mA transmitter is shown in Fig-
ure 33. After calibration, this transmitter is accurate to ± 0.5°C
over the –50°C to +150°C temperature range. The transmitter
operates from +8 V to +40 V with supply rejection better than
3 ppm/V. One half of the OP293 is used to buffer the VTEMP
pin, while the other half regulates the output current to satisfy
the current summation at its noninverting input:
One solution to this problem is to use a pull-down resistor. For
example, if R3 = 20 kΩ, then the pull-down resistor must be
less than 400 Ω. However, the pull-down resistor appears as a
fixed load when a common-mode voltage is applied. With a 4 V
common-mode voltage, the additional load current will be 10 mA,
which is unacceptable in a low power application.
IOUT
+ VTEMP × (R6 +
R2 × R10
R7)
–
VSET


R2 + R6 + R7
R2 × R10 
The change in output current with temperature is the derivative
of the transfer function:
Figure 32 shows a better solution. A1’s sink current is provided
by a pair of N-channel FET transistors, configured as a current
mirror. With the values shown, sink current of Q2 is about
340 µA. Thus, with a common-mode voltage of 4 V, the addi-
tional load current is limited to 340 µA versus 10 mA with a
400 Ω resistor.
∆IOUT
=
∆VTEMP (R6 + R7)
∆T
∆T
R2 × R10
REF-43BZ
VIN 2
VOUT 6
VTEMP 3 R1 10kΩ
GND 4
2
8
1
1/2 OP293
3
4
R2
VTEMP 1kΩ
R3
R5
100kΩ 5kΩ
R4
20kΩ
VSET
ZERO
TRIM
R6
SPAN TRIM
3kΩ
R7
5kΩ
6
1/2 OP293 7
5
R8
1kΩ
R9
100kΩ
1N4002
V+
+8V TO +40V
2N1711
ALL RESISTORS 1/4W, 5% UNLESS OTHERWISE NOTED
R10
100Ω
1%, 1/2 W
IOUT
RLOAD
Figure 33. Temperature to 4–20 mA Transmitter
–12–
REV. A