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The preceding Focus on Measurements introduces the concept of so-called common-mode and differential-mode signals The desired differential-mode EKG signal was ampli ed by the op-amp while the common-mode disturbance was canceled Thus, the differential ampli er provides the ability to reject commonmode signal components (such as noise or undesired DC offsets) while amplifying the differential-mode components This is a very desirable feature in instrumentation systems In practice, rejection of the common-mode signal is not complete: some of the common-mode signal component will always appear in the output This fact gives rise to a gure of merit called the common-mode rejection ratio, which is discussed in Section 126 Often, to provide impedance isolation between bridge transducers and the differential ampli er stage, the signals v1 and v2 are ampli ed separately This technique gives rise to the so-called instrumentation ampli er (IA), shown in Figure 1214 Example 123 illustrates the calculation of the closed-loop gain for a typical instrumentation ampli er
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Figure 1214 Instrumentation ampli er
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EXAMPLE 123 Instrumentation Ampli er
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Determine the closed-loop voltage gain of the instrumentation ampli er circuit of Figure 1214
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Solution
Known Quantities: Feedback and source resistances Find:
AV =
vout v1 v 2
(1243)
Assumptions: Assume ideal op-amps Analysis: We consider the input circuit rst Thanks to the symmetry of the circuit, we
can represent one half of the circuit as illustrated in Figure 1215(a), depicting the lower half of the rst stage of the instrumentation ampli er We next recognize that the circuit of
12
Operational Ampli ers
Figure 1215(a) is a noninverting ampli er (see Figure 128), and we can directly write the expression for the closed-loop voltage gain (equation 1236): A=1+ R2
R1 2
2R2 R1
Each of the two inputs, v1 and v2 , is therefore an input to the second stage of the instrumentation ampli er, shown in Figure 1215(b) We recognize the second stage to be a differential ampli er (see Figure 1210), and can therefore write the output voltage after equation 1242: vout = RF RF (Av1 Av2 ) = R R 1+ 2R2 R1 (v1 v2 ) (1244)
from which we can compute the closed-loop voltage gain of the instrumentation ampli er: AV = vout RF = R (v1 v2 ) 1+ 2R2 R1
R R1 2 Av2 v1 + R Av1 RF R2 +
Vout
Figure 1215(a)
Figure 1215(b)
Comments: This circuit is analyzed in depth in 15 Focus on Computer-Aided Solutions: An Electronics WorkbenchTM simulation of the
circuit of Figure 1214 can be found in the accompanying CD-ROM
Multisim
Because the instrumentation ampli er has widespread application and in order to ensure the best possible match between resistors the entire circuit of Figure 1214 is often packaged as a single integrated circuit The advantage of this con guration is that the resistors R1 and R2 can be matched much more precisely in an integrated circuit than would be possible using discrete components A typical, commercially available integrated circuit package is the AD625 Data sheets for this device are provided in the accompanying CD-ROM Another simple op-amp circuit that nds widespread application in electronic instrumentation is the level shifter Example 124 discusses its operation and its application The following Focus on Measurements illustrates its use in a sensor calibration circuit
Part II
Electronics
EXAMPLE 124 Level Shifter
Problem
The level shifter of Figure 1216 has the ability to add or subtract a DC offset to or from a signal Analyze the circuit and design it so that it can remove a 18-VDC offset from a sensor output signal
vsensor RS + Vref
Solution
Known Quantities: Sensor (input) voltage; feedback and source resistors Find: Value of Vref required to remove DC bias
vout VS
RF = 220 k ; RS = 10 k
Schematics, Diagrams, Circuits, and Given Data: vS (t) = 18 + 01 cos( t); Assumptions: Assume ideal op-amp
Figure 1216 Level shifter
Analysis: We rst determine the closed-loop voltage gain of the circuit of Figure 1216 The output voltage can be computed quite easily if we note that, upon applying the principle of superposition, the sensor voltage sees an inverting ampli er with gain RF /RS , while the battery sees a noninverting ampli er with gain (1 + RF /RS ) Thus, we can write the output voltage as the sum of two outputs, due to each of the two sources:
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