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Fig. 9-5 Di erentiating ampli er
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Example 9.6. Find an expression for the output of the inverting di erentiator of Fig. 9-5, assuming the basic op amp is ideal. Since the op amp is ideal, vd % 0, and the inverting terminal is a virtual ground. Consequently, vS appears across capacitor C: iS C dvS dt Hence,
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But the capacitor current is also the current through R (since iin 0). vo IF R iS R RC
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dvS dt
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The insertion of a capacitor in the feedback path of an op amp results in an output signal that is a time integral of the input signal. A circuit arrangement for a simple inverting integrator is given in Fig. 9-6.
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Example 9.7. Show that the output of the inverting integrator of Fig. 9-6 actually is the time integral of the input signal, assuming the op amp is ideal. If the op amp is ideal, the inverting terminal is a virtual ground, and vS appears across R. Thus, iS vS =R. But, with negligible current into the op amp, the current through R must also ow through C. Then 1 1 1 vo iF dt iS dt vS dt C C RC
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CHAP. 9]
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Fig. 9-6 Integrating ampli er
LOGARITHMIC AMPLIFIER
Analog multiplication can be carried out with a basic circuit like that of Fig. 9-7. Essential to the operation of the logarithmic ampli er is the use of a feedback-loop device that has an exponential terminal characteristic curve; one such device is the semiconductor diode of 2, which is characterized by iD Io evD =VT 1 % Io evD =VT
iD + ii +
Li LD
9:14
_ + +
_ R1 + +
Lo
Fig. 9-7 Logarithmic ampli er
A grounded-base BJT can also be utilized, since its emitter current and base-to-emitter voltage are related by iE IS evBE =VT 9:15
Example 9.8. Determine the condition under which the output voltage vo is proportional to the logarithm of the input voltage vi in the circuit of Fig. 9-7. Since the op amp draws negligible current, ii Since vD vo , substitution of (9.16) into (9.14) yields vi RIo e vo =VT Taking the logarithm of both sides of (9.17) leads to ln vi ln RIo vo VT 9:18 9:17 vi iD R 9:16
Under the condition that ln RIo is negligible (which can be accomplished by controlling R so that RIo % 1), (9.18) gives vo % VT ln vi .
OPERATIONAL AMPLIFIERS
[CHAP. 9
FILTER APPLICATIONS
The use of op amps in active RC lters has increased with the move to integrated circuits. Active lter realizations can eliminate the need for bulky inductors, which do not satisfactorily lend themselves to integrated circuitry. Further, active lters do not necessarily attenuate the signal over the pass band, as do their passive-element counterparts. A simple inverting, rst-order, low-pass lter using an op amp as the active device is shown in Fig. 9-8(a).
Mdb R R1 Asymptotic Bode plot 3 db
1/sC
R I1(s) R1
20 log
1 VS (s)
IF (s) _ 20 db/decade 3 Vo (s) slope
0.1 RC 1 RC 10 RC
Fig. 9-8 First-order low-pass lter
Example 9.9. (a) For the low-pass lter whose s-domain (Laplace-transform) representation is given in Fig. 9-8(a), nd the transfer function (voltage-gain ratio) Av s Vo s =VS s . (b) Draw the Bode plot (Mdb only) associated with the transfer function, to show that the lter passes low-frequency signals and attenuates highfrequency signals. (a) The feedback impedance ZF s and the input impedance Z1 s are ZF s R 1=sC R R 1=sC sRC 1 and Z1 s R1 9:19
The resistive circuit analysis of Example 9.2 extends directly to the s domain; thus, Av s (b) Letting s j! in (9.20) gives Mdb  20 log jAv j! j 20 log R 20 log j j!RC 1j R1 ZF s R=R1 Z1 s sRC 1 9:20
A plot of Mdb is displayed in Fig. 9-8(b). The curve is essentially at below ! 0:1=RC; thus, all frequencies below 0.1=RC are passed with the dc gain R=R1 . A 3-db reduction in gain is experienced at the corner frequency 1= 1=RC, and the gain is attenuated by 20 db per decade of frequency change for frequencies greater than 10=RC.
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