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JWST499-Cetinkunt
JWST499-c05
ELECTRONIC COMPONENTS FOR MECHATRONIC SYSTEMS 289
Notice that we only need to know two of these parameters. The most commonly
referenced ones are K diff and CMRR. Ideally, the differential gain is the desired gain of the
amplifier (i.e., 1.0or10.0) and the common mode gain is zero. That is, the noise induced
on both lines that are common is rejected totally by the differential amplifier. Hence, an
ideal differential amplifier would have infinite CMRR. In reality, the common-mode gain is
small but finite, hence CMRR is large, but finite. Typical values of CMRR are in the order
4
1
of 80 dB to 120 dB , which means that the amplifier amplifies the differential signal 10 to
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10 times more than the common-mode signal (noise).
The output voltage and input voltage relationship of a differential amplifier can be
expressed as
V = K ⋅ (V − V ) + K ⋅ (V + V )∕2 (5.195)
out diff in1 in2 cm in1 in2
( )
K diff
= K diff ⋅ (V in1 − V in2 ) + ⋅ (V in1 + V in2 )∕2 (5.196)
CMMR
In an ideal differential amplifier, CMMR is infinite, hence K diff ∕CMMR = K cm = 0.0. In
reality it is of course a very small, but finite gain.
The parameters of a real op-amp that are of interest for designers are as follows:
4
7
1. open loop gain (i.e., 10 to 10 range),
2. bandwidth (i.e., 1 MHz range),
6
3. input impedance (i.e., Z = R = 10 Ω range),
in
in
2
4. output impedance (i.e., Z out = R out = 10 Ω range),
5. common mode rejection ratio (CMRR) for differential amplifiers, (i.e., CMRR =
60 dB to 120 dB range)
6. power supply required (i.e. 1.5 VDC to 30 VDC range, unipolar or bipolar),
7. maximum power dissipated internally by the op-amp,
8. maximum input voltage and differential input voltage levels,
◦
9. operating temperature range (commercial rating: 0 to 70 C, industrial rating: −25 to
◦
◦
85 C, military rating: −55 to 125 C).
Example Consider a sensory signal in a medical application, that is an EKG signal,
where the nominal value of the signal is about 10 mV, yet the expected noise in the signal can
vary from 0 V to 100 mV. Let us consider two different ways of transmitting and amplifying
this signal: (1) single-ended signal, and (2) differential-ended signal. In both cases let us
assume that the gain of the op-amp is K diff = 40 dB = 100 V∕V, and the common-mode
5
rejection ratio of the differential op-amp is CMMR = 100 dB = 10 V∕V. We assume that
the noise is induced equally in both conductors when differential-ended signal is used.
If single-ended signal is used, output signal voltage level would vary between
V = K diff ⋅ V i (5.197)
o
V = 100 ⋅ 10 mV = 1V; when V = 10 mV (5.198)
i
o
V = 100 ⋅ 110 mV = 11 V; when V = 10 mV + 100 mV (noise) (5.199)
i
o
Clearly, the single-ended signal and op-amp cannot differentiate the difference between the
actual signal and noise. If a differential-ended signal is used, the output signal voltage level
K dB
1 K dB = 20 log Kor K = 10 20 , i.e., K = 0.01, 0.1, 1, 10, 100 is same as K dB =−40, −20, 0, 20, 40, respectively.
10