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so that the output voltage may be computed to be vout = RF R + RF R va RF + R + R RF v R b (158)
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Note that if the term R in the denominator were zero, the same result would be obtained as in Example 124: vout = (RF /R)(va vb ); however, because of the resistor mismatch, there is a corresponding mismatch between the gains for the two differential signal components Further and more important if the original signals, va and vb , contained both differential-mode and common-mode components: va = va dif + vcom such that va = A(va dif + vcom ) vb = A(vb dif + vcom ) (1510) vb = vb dif + vcom (159)
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then the common-mode components would not cancel out in the output of the ampli er, because of the gain mismatch, and the output of the ampli er would be given by vout = RF R + RF R A(va dif + vcom ) RF A(vb dif + vcom ) RF + R + R R (1511)
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resulting in the following output voltage: vout = vout, dif + vout, com with vout, dif = RF and vout, com = RF RF = R R + RF R Avcom RF + R + R RF Avcom R R + RF R Ava dif RF + R + R RF Avb dif R (1513) (1512)
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The common-mode rejection ratio (CMRR; see Section 126) is given in units of dB by CMRRdB = Adif Adif = 20 log10 Acom vout, com /vcom Adif
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RF R R+RF RF +R+ R
(1515) 1 A
= 20 log10
where Adif is the differential gain (which is usually assumed equal to the nominal design value) Since the common-mode gain, vout, com /vcom , should ideally be zero, the theoretical CMRR for the instrumentation ampli er with perfectly matched resistors is in nite In fact, even a small mismatch in the resistors used would dramatically reduce the CMRR, as the Check Your Understanding exercises at the end of this subsection illustrate Even with resistors having 1 percent tolerance, the maximum CMRR that could be attained for typical values of resistors and an overall gain of 1,000 would be only 60 dB In many practical applications, a requirement for a CMRR of 100 or 120 dB is not uncommon, and these would demand resistors of 001 percent tolerance (see Check Your Understanding Exercise 153) It should be evident, then, that the discrete design of the IA, employing three op-amps and discrete resistors, will not be adequate for the more demanding instrumentation applications
EXAMPLE 151 Common-Mode Gain and Rejection Ratio
Problem
Compute the common-mode gain and common-mode rejection ratio (CMRR) for the ampli er of Figure 1516
Solution
Known Quantities: Ampli er nominal closed-loop gain; resistance values; resistor
tolerance
15
Electronic Instrumentation and Measurements
Find: vout, com /vcom , CMRRdB Analysis: The common-mode gain is equal to the ratio of the common-mode output
signal to the common-mode input: vout, com = 100 vcom 11 1 = 01815 1102
The CMRR (in units of dB) can be computed from equation 1515 where Adif = A and therefore, CMRR = Adif Acom = 20 log10
RF = 100 R
Adif = 20 log10 vout, com /vcom
Adif
RF R R+RF R+RF + R
1 A
= 20 log10
10 1
11 1102
100 = 5482 dB 1 10
Comments: Note that, in general, it is dif cult to determine exactly the level of resistor mismatch in an instrumentation ampli er One usually makes reference to manufacturer data sheets, such as those you will nd in the accompanying CD-ROM
The general expression for the CMRR of the instrumentation ampli er of Figure 1516, without assuming any of the resistors are matched, except for R2 and R2 , is
CMRR =
(RF /R)(1 + 2R2 /R1 ) Adif = RF RF RF +R Acom 1
R RF RF +R
(1516)
and it can easily be shown that the CMRR is in nite if the resistors are perfectly matched Example 151 illustrated some of the problems that are encountered in the design of instrumentation ampli ers using discrete components Many of these problems can be dealt with very effectively if the entire instrumentation ampli er is designed into a single monolithic integrated circuit, where the resistors can be carefully matched by appropriate fabrication techniques and many other problems can also be avoided The functional structure of an IC instrumentation ampli er is depicted in Figure 1517 Speci cations for a common IC instrumentation ampli er (and a more accurate circuit description) are shown in Figure 1518 Among the features worth mentioning here are the programmable gains, which the user can set by suitably connecting one or more of the resistors labeled R1 to the appropriate connection Note that the user may also choose to connect additional resistors to control the ampli er gain, without adversely affecting the ampli er s performance, since R1 requires no matching In addition to the pin connection that permits programmable gains, two additional pins are provided,
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