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of the ampli er: The extremely high input resistance of this ampli er (of the order of megohms to gigohms) permits virtually perfect isolation between source and load, and eliminates loading effects Voltage followers, or impedance buffers, are commonly packaged in groups of four or more in integrated circuit (IC) form The data sheets for one such IC are contained in the accompanying CD-ROM, and may also be found in the device templates for analog ICs in the Electronics WorkbenchTM libraries
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Focus on Computer-Aided Solutions: An Electronics WorkbenchTM simulation of the
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circuit of Figure 129 can be found in the accompanying CD-ROM
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The Differential Ampli er
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R2 i2 R1 i1
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The third closed-loop model examined in this chapter is a combination of the inverting and noninverting ampli ers; it nds frequent use in situations where the difference between two signals needs to be ampli ed The basic differential ampli er circuit is shown in Figure 1210, where the two sources, v1 and v2 , may be independent of each other, or may originate from the same process, as they do in Focus on Measurements: Electrocardiogram (EKG) Ampli er The analysis of the differential ampli er may be approached by various methods; the one we select to use at this stage consists of: 1 Computing the noninverting- and inverting-terminal voltages, v + and v 2 Equating the inverting and noninverting input voltages, v = v + 3 Applying KCL at the inverting node, where i2 = i1 Since it has been assumed that no current ows into the ampli er, the noninvertingterminal voltage is given by the following voltage divider: v+ = R2 v2 R1 + R 2 (1238)
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Figure 1210 Differential ampli er
Part II
Electronics
If the inverting-terminal voltage is assumed equal to v + , then the currents i1 and i2 are found to be i1 = and i2 = and since i2 = i1 the following expression for the output voltage is obtained: vout = R2 v1 1 R2 + v2 + v2 R1 R1 + R 2 R1 (R1 + R2 ) (1242) (1241) vout v + R2 (1240) v1 v + R1 (1239)
vout =
R2 (v2 v1 ) R1
Differential ampli er closed-loop gain
Thus, the differential ampli er magni es the difference between the two input signals by the closed-loop gain R2 /R1 In practice, it is often necessary to amplify the difference between two signals that are both corrupted by noise or some other form of interference In such cases, the differential ampli er provides an invaluable tool in amplifying the desired signal while rejecting the noise Focus on Measurements: Electrocardiogram (EKG) Ampli er provides a realistic look at a very common application of the differential ampli er
Electrocardiogram (EKG) Ampli er
This example illustrates the principle behind a two-lead electrocardiogram (EKG) measurement The desired cardiac waveform is given by the difference between the potentials measured by two electrodes suitably placed on the patient s chest, as shown in Figure 1211 A healthy, noise-free EKG waveform, v1 v2 , is shown in Figure 1212 Unfortunately, the presence of electrical equipment powered by the 60-Hz, 110-VAC line current causes undesired interference at the electrode leads: the lead wires act as antennas and pick up some of the 60-Hz signal in addition to the desired EKG voltage In effect, instead of recording the desired EKG signals, v1 and v2 , the two electrodes provide the following inputs to the EKG ampli er, shown in Figure 1213: Lead 1: v1 (t) + vn (t) = v1 (t) + Vn cos (377t + n )
FOCUS ON MEASUREMENTS
12
Operational Ampli ers
Lead 2 + v2
v1 v 2 (V)
Lead 1 + v1
Electrodes
7 6 5 4 3 2 1 0 1 2 0
04 06 Time (s)
Figure 1212 EKG waveform Figure 1211 Two-lead electrocardiogram
vn(t)
Lead 2 v2
Equivalent circuit for lead 2 R1
Lead 1
+ R1
vn(t)
Vout
Equivalent circuit for lead 1
EKG amplifier
Figure 1213 EKG ampli er
Lead 2: v2 (t) + vn (t) = v2 (t) + Vn cos (377t + n ) The interference signal, Vn cos (377t + n ), is approximately the same at both leads, because the electrodes are chosen to be identical (eg, they have the same lead lengths) and are in close proximity to each other Further, the nature of the interference signal is such that it is common to both leads, since it is a property of the environment the EKG instrument is embedded in On the basis of the analysis presented earlier, then, vout = or vout = R2 (v1 v2 ) R1 R2 [(v1 + vn (t)) (v2 + vn (t))] R1
Thus, the differential ampli er nulli es the effect of the 60-Hz interference, while amplifying the desired EKG waveform
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