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Execute hEx9_12.CIRi and use the Probe feature of PSpice to plot the gain magnitude Mdb as shown in Fig. 9-11. The low-frequency gain magnitude is seen to have the value predicted by the results of Example 9.9. ! 10 103 Mdbj! 0 20 log R=R1 20 log 20 db 1 103
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The gain magnitude has decreased from the low-frequency value of 20 db to 17 db (drop of 3 db) at the corner frequency fc 159:1 Hz. Clearly, the gain magnitude decreases by 20 db per decade of frequency for values of high frequency. Hence, the characteristic of Fig. 9-8(b) is veri ed.
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9.1 For the inverting ampli er of Fig. 9-2: (a) Show that as AOL ! 1; vd ! 0; thus, the inverting input remains nearly at ground potential (and is called a virtual ground). (b) Show that the current feedback is actually negative feedback.
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(a) By KVL around the outer loop, vS vo i1 R1 iF RF Using (9.1) in (1), rearranging, and taking the limit give
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AOL ! 1
vd
AOL ! 1
i1 R1 iF RF vS 0 AOL
(b) The feedback is negative if iF counteracts i1 ; that is, the two currents must have the same algebraic sign. By two applications of KVL, with vd % 0, i1 vS vd v % S R1 R1 and iF vo vd vo % RF RF
But in an inverting ampli er, vo and vS have opposite signs; therefore, i1 and iF have like signs.
(a) Use (9.4) to derive an exact formula for the gain of a practical inverting op amp. (b) If R1 1 k; RF 10 k; Rd 1 k; and AOL 104 , evaluate the gain of this inverting ampli er. (c) Compare the result of part b with the ideal op amp approximation given by (9.5).
(a) Rearranging (9.4) to obtain the voltage-gain ratio gives Av  vo AOL vS 1 R1 =RF 1 AOL R1 =Rd
(b) Substitution of the given values yields Av (c) From (9.5), Avideal so the error is 9:979 10 100% 0:21% 9:979 Note that Rd and AOL are far removed from the ideal, yet the error is quite small. RF 10 R1 104 9:979 1 1=10 1 104 1=1
A di erential ampli er (sometimes called a subtractor) responds to the di erence between two input signals, removing any identical portions (often a bias or noise) in a process called commonmode rejection. Find an expression for vo in Fig. 9-12 that shows this circuit to be a di erential ampli er. Assume an ideal op amp.
Since the current into the ideal op amp is zero, a loop equation gives v1 vS1 Ri1 vS1 R By voltage division at the noninverting node, v2 R1 v R R1 S2 vS1 vo R R1
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In the ideal op amp, vd 0, so that v1 v2 , which leads to vo R1 v vS1 R S2
Thus, the output voltage is directly proportional to the di erence between the input voltages.
i1 1
L0 LS
iin _
R1 0
Fig. 9-12 Di erential ampli er
Fig. 9-13 Unity follower
Find the input impedance Z1 of the inverting ampli er of Fig. 9-2, assuming the basic op amp is ideal.
Consider vS a driving-point source. Since the op amp is ideal, the inverting terminal is a virtual ground, and a loop equation at the input leads to vS i1 R1 0 so that Z1 vS R1 i1
The unity-follower ampli er of Fig. 9-13 has a voltage gain of 1, and the output is in phase with the input. It also has an extremely high input impedance, leading to its use as an intermediatestage (bu er) ampli er to prevent a small load impedance from loading a source. Assume a practical op amp having AOL 106 (a typical value). (a) Show that vo % vS . (b) Find an expression for the ampli er input impedance, and evaluate it for Rd 1 M (a typical value).
(a) Writing a loop equation and using (9.1), we have   1 vS vo vd vo 1 AOL vS vS from which 0:999999vS % vS vo 1 1=AOL 1 10 6 (b) Considering vS a driving-point source and using (9.1), we have vS iin Rd vo iin Rd AOL vd iin Rd 1 AOL v Zin S Rd 1 AOL % AOL Rd 106 106 1 T iin
Find an expression for the output vo of the ampli er circuit of Fig. 9-14. amp. What mathematical operation does the circuit perform
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