ECE 300

ideal ap-amp

C v s - \I, - - a k a

ECE 300 Assuminy idea1 op-amps I

Chapter 5, Solutiou 5.

- v ;+Ava+ (&+Ro) I=0

v6 = \I,

- v j + (R i+Ro+R iA ) I=0

I= v i

Ro + (l+A)Ri

-Avd -R{ I+vo=0

vn= Avd+ Rn l = (Ro+ R ,A t l = (Ro+R 'A) t '

R o . r ( l + A )R ,

u. l= \q j IL = loo+loaxl95.ro 'vt Ro +( l+A)Ri 100+(1+105)

rne rnofno: ,__--.10, = -:::::' = 0.9999990

u. r0 ' ,1 100.001

(1 )

But

(2)

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Chaoter 5. Solution 6.

+

But

(Ro+&)R+v i+Ava=0

va = RiI,

v i+Go+Ri+RiA) I=0

Ro +G+A)Rt

-Avd-RoI+v .=0

v"=Ava+RoI=(Ro+RA) I

Substituting for I in (l),

I n^+n,e )lKo + { r + A)K, /(50 + 2x106 x2xlo' ).10r50 + (1+ 2x10'tr2x10'

- 200,000x2x106 ,,- 200,001x2x106

vo = {492925-s!Y

(1)

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Chapter 5, Problem 9Determine v, for each of the opamp circuits in Fig. 5.48.

: kf]

lV

--€-

(b)

Figure 5.48 for Prob. 5.9

Chapter 5, Solution 9.(a) Let v. and vb be respectively the voltages at the inverting and noninverting

terminals of the op amp

vt=vb=4V

At the inverting terminal,

l.A = 4:,uo -----> ve = !!2k

+

Since v" = vb = 3V,

-vb+ 1 +vo=0 - - - - - | Yo=Y6- | =2 !

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Chapter 5, Problem 13

Find v, ard i, in the circuit of Fig. 5.52.

Figue 5.52 for Prob. 5.11

ChaDter 5. Solution 13.

By voltage divisioir,

';

q0v ,= : f1 )=0 .9V-

100

50 v^v h = - v ^ = -

150 " 3

But vu = v6 --->

io = i1 a i2 = 5', i= = 0.27nA + 0.0l8rnA = 2EE-EA- rok t50kPROPRIETARY MATERIAL. @ 200? The Mccmw-Hill Companies, Inc. AI rights reserved. Nq!3dof this Marual rnay be displayed. reoroduced or dbwritten permission of the publisher. or used bevo fieperndtted bv Mccraw-Hil for their individual course p!9saratj9!-II-yqu,gq-a-!Eds$-udl&]U!-!444rdvou are usrnq it wilhout pE{!qi!qb&

5=o.s3

-> v"=ZE

Chapter 5, Solution 16

kt currents be in mA and resisl"nces be in k Q . At node a,

= -11'7OnA= -14.28ltA

I _., uft:-\L -v"

I o ' ,' '$v'

0.5-v , v - -v -

5 " = - 1 0 - - - - ) t = 3 v o - v o

But DY-8 //' l0'\"

v , = v r = - v ^ + t , = - v ,

Substituting (2) into (1) gives \-_ _,-,1088 ' ' ' 1 ,4

Thus,

"5

i - vo -vb +vo-va =Q.61y - y - 1= g .61 . ]9u_ - r - , . 9 { * ! * - * r . r , ,' 2 i0 " ' 8 " " 4 t4

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(1 )

(2)

10kQ

Chapter 5, Solution 18.We tempomrily rcmove the 20-kO rcsistor. To find Vr!, we coNider the circuit below.

10 kf,)

This is an inverting ampliJier.1nL

V^=- : : Qm\A= -1OmV2k

To find Rr!, we note that the 8-kO resistor is aqoss the output ofthe op amp which isacting like a voltage source so the only resistance seen looking itr is the 12-kO resistor.

The Thevenin equivalent with the 20-kO resistor is $hown below.

12kct

-10 mV

I = -10r'/(12k + 20k) = 0.3125x10j A

p = rh = 10.:tzsxto{)'?x20x103 = 1.9531 nw

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Chapter 5, Problem 29

Detemine the voltage gain volti of the op alnp circuit in Fig. 5.67.

Figure 5.67

Chapter 5, Solution 29

&' R, +R,

B.ut v"=vb

Orv-&,,&

&t . = - t -" 4+R,

-

R" R,

R, +Rz fir +R) "

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"Titten p9!4iqsion of the pubfisher ff

permitted bv Mccraw-Hill for their hdividuat course orepaiatiol If vou are a student usvou are usinq it without permission.

Chapter 5, Problem 3l

For fte circujl ir Fig. 5.69. f iod i,.

Figure 5.69

Chapter 5, Solution 31.

After convefing the cuFent source to a voltage source, the citcuit is as shown below:

At node 1,

At node 2,

(1 )

(2)

t t _ w-i----1, ]l----j-c+.L. _______+ 48 = 7vr - 3vu

3612

w _ n' r _ ' " _ ' . _ " = i , - _ _ - _ _ + v l = 2 v o6 t )

From (1) and (2),48

" 11

i = --! = 727.2tt[OK

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D kti

Chapter 5, Problem 40.

Find vo ir terms ofvr, v2, ard v3, in the circuit ofFig. 5.77.

Figure 5.77 For Prob. 5.,t0.

Chapter 5, Solution 40.

Applying KCL at node a, where node a is the input to the op amp.

f1+.f2R-Ya.?=o or va= (vl + v2 r v3Y3

vo=(1 + Rl/R v. = (1 + R'/Rr)(vr + vz + vrV3.

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