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Unformatted text preview: G. Rizwni, Principles and. Applications of Electrical Engineering Problem solutions, Chapter 8 b. The input voltage of the first stage can be determined in terms of the source voltage using voltage division:
Re
  ”3  + R 7 7
A“: Ezmm=mmﬂn_s__149_9._.,;_=144_3=20mgm[144.3]=43.13d3
VS vn vs vn vs ' 0.3 +7.7 8.3 Solution: Known quantities: Figure P85. Find: What approximations are usually made about the voltages and cunents shown for the ideal operational
ampliﬁer [opamp} model. Analysis: Bo‘l Solution:
Known quantiﬁes:
Figure P8.6. Find:  . ‘ What approximations are usually made about the circuit components and parameters shown for the ideal
Operational ampliﬁer [opamp] model. Analysis: n. goo p, a: 00 re :30. —————————w———————___.___._____m__—_~___ 8.5 V5 (ehllgk = 3m. _. 51k .. ‘1:
V"V3\[EE]' a (“Q . C31)“ Amp 1% :1 \JOHQQ ﬂFOl,\OW'\f gag Voljgage mm “1L” ampU‘JC 7: 13V , %\\*’l§€_ \ngo [VouagxQ
\\ H \ CLQVTJ'SS \WA'PU'E 012 019*60MKP) \Jot‘xcﬁQ (Wk — W1??? (1130
= \3V ' . ESL
\3543’ : QJQCSJA 3 \EW
V0: \au+\&v':30V L241, : = {0px 1‘ L36, ﬂincﬂ ("MFA “~ 0 (\Aeqjéyﬂ'a3 Lio/L: 30"”; : 3A $MUL C’rCn’vult :0 GOLE‘JUVGMP) 8,10: G. Rlzzoni, Principles and Applications of Electrical Engineering Problem solutions. Chapter 3 Section 8.2: The Operational Amplifier 8.24 ' Seiuiion: .
Known Quantities:
The circuit shown in Figure P81 V
Find: _ ' §
The resistor R, that will accompiish the nominal gain requirement, and stare what the maximum and
minimum values (1fo can be. Will a Standard 5 percent tolerance resistor be adequate to satisfy this requirement? Analysis: For a nuninverting ampiiﬁer the voltage gain is given by: i
=————=~———=—+l ; =16=~——+1 => R =—=1k.0.  i A” vb. R, 11, Am” a, ‘ 16 1 E 1
To ﬁnd the 1113inme and minimumRS we note that R, 0: ——, so to ﬁnd the maximumRs We consider 15m ' 
+———~—=1.02m.
"m 16(10.02)—1 ' 15m
mm=——.———_=9 12. 1
(161+ 0112)— 1 80 Since a. standard 5% tolerance Ikﬂ resistor has mistame 950 < R < 10502511111de resistor will [33?
sufﬁce In this application. am the minimum Av: R Conversely, to ﬁnd the minimum Rs we consider the maximumAp: R 8.2K — 33k '
Lad AU: "'— ——— ‘1: “ ——__ :"270g 11% ’ 12L 3 .5 “$312+L393uo \3
US [Ml . \‘Jk— 02mm) (A ”Wk : 331<~L33kx00 \QK + 03110 I): ’ouﬁk
l 008: 301 7k
133?. =33. (cl  225 * 8.26 Rs
.. , Um}?
Jr
2W ’1'“ A 
' . ‘ \hm: wHSgsimfr
I ' 20 ”VA IVA ““ Vaujf
0* KC  V‘t (1e A '  = M
U L C", 0 RS \olc
.. I A 3 faintly}
 VA“; \Mﬁ‘) =7 \onSanch .. 00*” —Uou£¥
 R's . \OK
\30 4' : ._.. \OV‘UD'WDS 'SmmJC) + 10+“) sSimhﬁ
Re.
TO Aumxnoéﬁ. TX. 3.9};  \bkL‘Q +10 f: O
W $0M 14w is =3 123mm
3 ‘ 8.28 2" Solution: Known quantities: 1
1393:): —(2sin 031t+4sin (92:4. Ssinw3r+ lﬁsinmg) V, RF =5 k9.  1
Design an inverting summing ampliﬁer to obtain vm (t) and demine the require some resistam. ‘! , Analysis: ' ' i
The inverting summing ampliﬁer is shown in the folluwing ﬁgme. :
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i
i By superposition and by sclecnng R5: = 3%, vs, a 51!! (qt, I =1...4. the output voltagc is vm=—£——ivg= —22‘vg= ;2'sinm,t that coincides with the dashed output voltage.
So. the required some msistors are . ‘ 1 14: R Iii R,.=§=2.,5mxﬂ=f=1.2sm, Rﬁ= R 7‘ =625£2,R 5 _ 3
8 “=11; =3.129
W . Wald: sin”; “Um‘ki ““039 R‘ —__ 0‘10”“ 83411 ‘ VORcm‘e 01m \‘—h.in\)u'\' '3 V“) I%N\¢Q \Aeqn'a \Juwcwxe, (LONﬁS Dyvoxmp \n’PA i% ¢o Luau?— LID, Lainey; thug MAO gram930).. . 1h: 3 A3 \bhg (19; “$50 (Lowe/VJ“ an“ Row  Améue, 40 W JtA’Muqlx L‘d) «AWMQOIQ “QmeoXJOi‘ax M “\W,
\Q \JgLO ‘ LUWQNJQ wmﬁi Hm up ~\5\\voug\\ ”0/7, %T M W341 Jr0 .. L61) M “Me/12w: MéQ lawmug‘k
(KM . \‘mmaghovg ask mi“ wk \ E
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E 80%2 L V E. r Sanding: Known quantiﬁes: ' The circuit shown in Figurc P821
Find: ' ‘
Show that the voltage VM is proportional to the current gcnmtad by the Cats solar cell. Show that the
transimpedancc of the circuit Van/I, is —R. Analysis: 011
Assuming an ideal op~ampz 11"" .=. v" = 0 21> Vm = —RIS
V i
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E 1
a
1'
,
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1
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, R . The transimpbdance is gchn by R. .___. ii
S All Isqmgwbg
Is ‘19 ‘ n—ﬁor 8.16 8 .L‘Ll G. Rizzoni. Principlm and Applications of Electrical Engineering Problem solutions. Chapter 8 v0 =_.R£ +§L 5'.
R52 R51 1
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3 BS1 G. Rizzoni, Principles and Applications of Electrical Enginaering Problem solutions. Chapter 8 In fact. the output voltage is R31
V a) =————————V '
0U ) R 1+ .iszR 5U”) . Vo(jq£} R2 1
13. H0 E0  =—————
0) var» ) R 1+jmcirzz c. The gain in decibel is obtaiiwd by evaluating 1H,, (ice)! at (0:0. is. [H,(j0)Lm = 20Log%= 201033221 =7'_55 dB. The cutoﬂ’ {raquency is SST—““Hﬁ Solution: Known quantities: . Forthe circuit shown in Figure P8.40: C = 0.47 11F R = 2.2169 R2 = .6816} Rt. 21m ‘
Find: a. An expression in standard form fur the vultage ﬁequency response function. b. Ther gain in dwibeis' m the passbmd,tl1at'1s, at the frequmcies being passed by tho ﬁlter. and mi: cutoff
frequency. Analysis: 3. From“ rigure 3. 21 itresults that the ampliﬁer in Figure P8. #0' Is aiompass ﬁlter Th: voltage frequency response function is . V (1w) R 1'
HAW) V,(jw) R, 1+ jay CR2 ' b. The gain in decibel is obtained by evaluating [H11 jag] iii «i=0, Le. . R2 68 
H o = Lo———=20Lo ——=29.3dB.
I ”(1)!” 20' th g2.2
The cumﬁ'fmquﬂncy is .
mo=L=———l_—=31289iadjs CR2 0.47 10‘9 68 103 8.13 FY‘OM U} '. EH ...
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