Exam 5: Operational Amplifier Circuits

the circuit shown below, let $\mathrm { R } _ { 1 } = 1 \mathrm { k } \Omega , \mathrm { R } _ { 2 } = 2 \mathrm { k } \Omega , \mathrm { R } _ { 3 } = 6 \mathrm { k } \Omega , \mathrm { R } _ { 4 } = 500 \Omega , \mathrm { R } _ { 5 } = 5 \mathrm { k } \Omega , \mathrm { R } _ { 6 } = 3 \mathrm { k } \Omega , \mathrm { R } _ { 7 } = 1.5 \mathrm { k } \Omega , \mathrm { v } _ { \mathrm { s } } = 1 \mathrm {~V}$ (a) write a node equation at node 1 by summing the currents leaving node 1 . Express $v _ { a }$ as a function of $\mathrm { v } _ { 0 }$ . (b) write a node equation at node 2 by summing the currents leaving node 2 . Express $v _ { n }$ as a function of $v _ { 0 }$ . (c) write a node equation at node 3 by summing the currents leaving node 3 . (d) find the numerical value of $\mathrm { v } _ { 0 } , \mathrm { v } _ { \mathrm { p } } , \mathrm { v } _ { \mathrm { n } } , \mathrm { v } _ { \mathrm { a } }$ . $the circuit shown below, let \mathrm { R } _ { 1 } = 1 \mathrm { k } \Omega , \mathrm { R } _ { 2 } = 2 \mathrm { k } \Omega , \mathrm { R } _ { 3 } = 6 \mathrm { k } \Omega , \mathrm { R } _ { 4 } = 500 \Omega , \mathrm { R } _ { 5 } = 5 \mathrm { k } \Omega , \mathrm { R } _ { 6 } = 3 \mathrm { k } \Omega , \mathrm { R } _ { 7 } = 1.5 \mathrm { k } \Omega , \mathrm { v } _ { \mathrm { s } } = 1 \mathrm {~V} (a) write a node equation at node 1 by summing the currents leaving node 1 . Express v _ { a } as a function of \mathrm { v } _ { 0 } . (b) write a node equation at node 2 by summing the currents leaving node 2 . Express v _ { n } as a function of v _ { 0 } . (c) write a node equation at node 3 by summing the currents leaving node 3 . (d) find the numerical value of \mathrm { v } _ { 0 } , \mathrm { v } _ { \mathrm { p } } , \mathrm { v } _ { \mathrm { n } } , \mathrm { v } _ { \mathrm { a } } .$

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Question 1

Correct Answer:

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(a)
$\frac { 0 - v _ { o } } { R _ { 6 } } + \frac { 0 - v _ { a } } { R _ { 7 } } = 0$
$v _ { a } = - \frac { R _ { 7 } } { R _ { 6 } } v _ { o }$
(b)
$\frac { v _ { n } - 0 } { R _ { 4 } } + \frac { v _ { n } - v _ { o } } { R _ { 5 } } = 0$
$\left( \frac { 1 } { R _ { 4 } } + \frac { 1 } { R _ { 5 } } \right) v _ { n } = \frac { v _ { o } } { R _ { 5 } }$
(c)
$\frac { v _ { p } - v _ { s } } { R _ { 1 } } + \frac { v _ { p } } { R _ { 3 } } + \frac { v _ { p } - v _ { a } } { R _ { 2 } } = 0$
$v _ { p } = v _ { n }$
$\left( \frac { 1 } { R _ { 1 } } + \frac { 1 } { R _ { 3 } } + \frac { 1 } { R _ { 2 } } \right) v _ { p } - \frac { 1 } { R _ { 2 } } v _ { a } = \frac { v _ { s } } { R _ { 1 } }$
$\left( \frac { 1 } { R _ { 1 } } + \frac { 1 } { R _ { 3 } } + \frac { 1 } { R _ { 2 } } \right) \frac { R _ { 4 } } { R _ { 4 } + R _ { 5 } } v _ { o } + \frac { R _ { 7 } } { R _ { 2 } R _ { 6 } } v _ { o } = \frac { v _ { s } } { R _ { 1 } }$ $\begin{array} { l } v _ { o } = \frac { \frac { 1 } { R _ { 1 } } } { \left( \frac { 1 } { R _ { 1 } } + \frac { 1 } { R _ { 3 } } + \frac { 1 } { R _ { 2 } } \right) \frac { R _ { 4 } } { R _ { 4 } + R _ { 5 } } + \frac { R _ { 7 } } { R _ { 2 } R _ { 6 } } } v _ { s } \\\mathrm {~V} _ { \mathrm { n } } = 0.2264 \mathrm {~V} , \mathrm { v } _ { \mathrm { p } } = 0.2264 \mathrm {~V} , \mathrm { v } _ { \mathrm { a } } = - 1.2453 \mathrm {~V} , \mathrm { v } _ { 0 } = 2.4906 \mathrm {~V}\end{array}$

circuit shown below generates output signal vo given by $\mathrm { v } _ { \mathrm { o } } = 10 \mathrm { v } _ { 1 } - 5 \mathrm { v } _ { 2 }$ where $v _ { 1 }$ and $v _ { 2 }$ are two input signals. Find $R _ { 1 } , R _ { 2 } , R _ { 3 }$ , and $R _ { 4 }$ . $circuit shown below generates output signal vo given by \mathrm { v } _ { \mathrm { o } } = 10 \mathrm { v } _ { 1 } - 5 \mathrm { v } _ { 2 } where v _ { 1 } and v _ { 2 } are two input signals. Find R _ { 1 } , R _ { 2 } , R _ { 3 } , and R _ { 4 } .$

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Question 2

Correct Answer:

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One solution: $R _ { 1 } = 4 \mathrm { k } \Omega , R _ { 2 } = 5 \mathrm { k } \Omega , R _ { 3 } = 1 \mathrm { k } \Omega , R _ { 4 } = 9 \mathrm { k } \Omega$ There are many solutions. Superposition principle can be used to show that the answer is correct.

the circuit shown below, let $R _ { 1 } = 1 \mathrm { k } \Omega , R _ { 2 } = 3.6 \mathrm { k } \Omega , R _ { i } = 10 \mathrm { k } \Omega , R _ { 0 } = 600 \Omega , A = 5000 , v _ { t } = 1 \mathrm {~V}$ Find the value of $i _ { t }$ and the output resistance $R _ { \text {out } } = v _ { t } / i _ { t }$ . $the circuit shown below, let R _ { 1 } = 1 \mathrm { k } \Omega , R _ { 2 } = 3.6 \mathrm { k } \Omega , R _ { i } = 10 \mathrm { k } \Omega , R _ { 0 } = 600 \Omega , A = 5000 , v _ { t } = 1 \mathrm {~V} Find the value of i _ { t } and the output resistance R _ { \text {out } } = v _ { t } / i _ { t } .$

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Question 3

Correct Answer:

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$v _ { n } = \frac { \frac { R _ { 1 } R _ { i } } { R _ { 1 } + R _ { i } } } { R _ { 2 } + \frac { R _ { 1 } R _ { i } } { R _ { 1 } + R _ { i } } } v _ { i } = \frac { R _ { 1 } R _ { i } } { R _ { 1 } R _ { 2 } + R _ { 2 } R _ { i } + R _ { 1 } R _ { i } } v _ { i } = 0.2016 \mathrm {~V}$
$i _ { t } = \frac { v _ { t } - v _ { n } } { R _ { 2 } } + \frac { v _ { t } - A \left( 0 - v _ { n } \right) } { R _ { o } } = \frac { v _ { l } - \frac { R _ { 1 } R _ { i } } { R _ { 1 } R _ { 2 } + R _ { 2 } R _ { i } + R _ { 1 } R _ { i } } v _ { l } } { R _ { 2 } } + \frac { v _ { l } + A \frac { R _ { 1 } R _ { i } } { R _ { 1 } R _ { 2 } + R _ { 2 } R _ { i } + R _ { 1 } R _ { i } } v _ { i } } { R _ { o } } = 1.682 \mathrm {~A}$
$R _ { \text {out } } = \frac { v _ { t } } { i _ { t } } = \frac { 1 } { \frac { 1 - \frac { R _ { 1 } R _ { i } } { R _ { 1 } R _ { 2 } + R _ { 2 } R _ { i } + R _ { 1 } R _ { i } } } { R _ { 2 } } + \frac { 1 + A \frac { R _ { 1 } R _ { i } } { R _ { 1 } R _ { 2 } + R _ { 2 } R _ { i } + R _ { 1 } R _ { i } } } { R _ { o } } } = 0.5945$ clear all;
$R 1=1000 ; R 2-3600 ; R 1=10000 ; R 0=600 ; V t=1 ; A=5000 ;$
syms vn
$v n=s 0 \perp v e(v n / R I+v n / R 1+(v n-v t) / R 2, v n)$
$i t-(v t-v n) / R 2+(v t-A *(0-v \cap)) / R Q$
Rout =vt/uti
vh=vpa (vn, 1l)
it=vpa $(i t, 1 l)$
Rout=vpa (Rout, 11)

Answers:
$\mathrm { Vn } =$
$0.20161290323$
it $=$
1.6819959677
Rout $=$
$0.59453174632$