OpAmps

Basic Electric Circuits

Introduction To Operational Amplifiers

Lesson 8

Basic Electric CircuitsOperational Amplifiers

One might ask, why are operational amplifiers included in Basic Electric Circuits?

The operational amplifier has become so cheap in price (often less than $1.00 per unit) and it can be used in so many applications, we present an introductory study early-on in electric circuits.

1


What is an operational amplifier? This particular form of amplifier had the name “Operational” attached to it many years ago.

As early as 1952, Philbrick Operational Amplifiers (marketed by George A. Philbrick) were constructed with vacuum tubes and were used in analog computers.* Even as late as 1965, vacuum tube operational amplifiers were still in use and cost in the range of $75. * Some reports say that Loebe Julie actually developed the operational amplifier circuitry.

2


The Philbrick Operational Amplifier.From “Operational Amplifier”, by Tony van Roon: http://www.uoguelph.ca/~antoon/gadgets/741/741.html


My belief is that “operational” was used as a descriptorearly-on because this form of amplifier can performoperations of

• adding signals

• subtracting signals

• integrating signals, dttx )(

The applications of operational amplifiers ( shortenedto op amp ) have grown beyond those listed above.

3


At this level of study we will be concerned with howto use the op amp as a device.

The internal configuration (design) is beyond basiccircuit theory and will be studied in later electroniccourses. The complexity is illustrated in the followingcircuit.

4


The op amp is built using VLSI techniques. The circuitdiagram of an LM 741 from National Semiconductor isshown below.

5

V+

V-

Vo

Vin(-)

Figure 8.1: Internal circuitry of LM741.Taken from National Semiconductor

data sheet as shown on the web.

Vin(+)


Fortunately, we do not have to sweat a circuit with 22transistors and twelve resistors in order to use the op amp

The circuit in the previous slide is usually encapsulated intoa dual in-line pack (DIP). For a single LM741, the pin connections for the chip are shown below.

Taken from National Semiconductordata sheet as shown on the web.6 Figure 8.2: Pin connection, LM741.


i n ve r t i n g i n p u t

n o n i n ve r t i n g i n p u to u t p u t

V -

V +

The basic op amp with supply voltage included is shownin the diagram below.

7Figure 8.3: Basic op am diagram with supply voltage.


In most cases only the two inputs and the output areshown for the op amp. However, one should keep inmind that supply voltage is required, and a ground.The basic op am without a ground is shown below.

8Figure 8.4: Outer op am diagram.


A model of the op amp, with respect to the symbol, isshown below.

V 1

V 2

_

+

V d R i

R o

A V d

V o

Figure 8.5: Op Amp Model.9


The previous model is usually shown as follows:

R i

R o

A V d

_

+

V d

V 1

V 2

V o

+

_

Figure 8.6: Working circuit diagram of op amp.

10


Application: As an application of the previous model,consider the following configuration. Find Vo as a function of Vin and the resistors R1 and R2.

+

_

R 2

R 1

+

_

+

_

V i n V o

11 Figure 8.7: Op amp functional circuit.


In terms of the circuit model we have the following:

R i

R o

A V i

_

+

V iV i n V o

+

_

+

_

R 1

R 2

ab

Figure 8.8: Total op amp schematic for voltage gain configuration.12


R i

R o

A V i

_

+

V iV i n V o

+

_

+

_

R 1

R 2

ab

Circuit values are:R1 = 10 k R2 = 40 k Ro = 50 A = 100,000 Ri = 1 meg

13


We can write the following equations for nodes a and b.

Eq 8.1

Eq 8.2

14

( )10 1 40

( )5040

i oin i i

i oo i

V VV V Vk meg k

V VV AVk


Equation 8.1 simplifies to;

inio VVV 10012625 Eq 8.3

Equation 8.2 simplifies to;

010410005.4 95 io VxVx Eq 8.4

15


From Equations 8.3 and 8.4 we find;

ino VV 99.3

This is an expected answer.

Fortunately, we are not required to do elaborate circuitanalysis, as above, to find the relationship between theoutput and input of an op amp. Simplifying the analysisis our next consideration.

16

Eq 8.5


For most all operational amplifiers, Ri is 1 meg orlarger and Ro is around 50 or less. The open-loop gain, A, is greater than 100,000.

Ideal Op Amp:The following assumptions are made for the ideal op amp.

i

o

RohmsinputInfinite

RohmsoutputZero

AgainloopopenInfinite

;.3

0;.2

;.1

17

Basic Electric CircuitsIdeal Op Amp:

_

+ ++

++

_

__ _

V i

V 1

V 2 = V 1V o

i 1

i 2

= 0

= 0

(a) i1 = i2 = 0: Due to infinite input resistance.

(b) Vi is negligibly small; V1 = V2.18

Figure 8.9: Ideal op amp.


Find Vo in terms of Vin for the following configuration.

+

_

R 2

R 1

+

_

+

_

V i n V o

19 Figure 8.10: Gain amplifier op amp set-up.


+

_

R 2

R 1

+

_

+

_

V i n V o

a

V i

Writing a nodal equation at (a) gives;

21

)(RVV

RVV oiiin

20

Eq 8.6


21

)(RVV

RVV oiiin

With Vi = 0 we have;

With R2 = 4 k and R1 = 1 k, we have

ino VV 4 Earlierwe got ino VV 99.3

21

Eq 8.7

inVRRV1

20


When Vi = 0 in Eq 8.7 and we apply the Laplace Transform;

1

20

RR

)s(V)s(V

in

Eq 8.8

In fact, we can replace R2 with Zfb(s) and R1 with Z1(s) andwe have the important expression;

)s(Z)s(Z

)s(V)s(V

in

fb

in

0 Eq 8.9

22


At this point in circuits we are not able to appreciate theutility of Eq 8.9. We will revisit this at a later point incircuits but for now we point out that judicious selectionsof Zfb(s) and Zin(s) leads to important applications in

• Analog Filters

• Analog Compensators in Control Systems

• Application in Communications

23


Example 8.1: Consider the op amp configuration below.

+

+

+

_

__

3 VV in

6 k

1 k

V 0

a

Figure 8.11: Circuit for Example 8.1.24

Assume Vin = 5 V


+

+

+

_

__

3 VV in

6 k

1 k

V 0

a

At node “a” we can write;

kV

k)V( in

63

13 0

From which; V0 = -51 V (op amp will saturate)

25

Eq 8.10

Example 8.1 cont.


Example 8.2: Summing Amplifier. Given the following:R fb

R 1

R 2

V 2

V 1V 0

a

Figure 8.12: Circuit for Example 8.2.

fbRV

RV

RV 0

2

2

1

1 Eq 8.11

26


Example 8.2: Summing Amplifier. continued

Equation 8.11 can be expressed as;

2

21

10 V

RR

VRR

V fbfbEq 8.12

If R1 = R2 = Rfb then,

210 VVV Eq. 8.13

Therefore, we can add signals with an op amp.27


Example 8.3: Isolation or Voltage Follower.Applications arise in which we wish to connect one circuitto another without the first circuit loading the second. This requires that we connect to a “block” that has infinite inputimpedance and zero output impedance. An operational amplifier does a good job of approximating this. Considerthe following:

T he" B lo c k "C irc u it 1 C irc u it 2

+

_

+

_V in V out

Figure 8.13: Illustrating Isolation.28


Example 8.3: Isolation or Voltage Follower. continued

C i r c u i t 1 C i r c u i t 2

Th e B l o c k

+

_

+V i n V 0_

Figure 8.14: Circuit isolation with an op amp.

It is easy to see that: V0 = Vin29


Example 8.4: Isolation with gain.

+

_

__

+

+

2 0 k

V in

V in

V0

1 0 k

1 0 k

a

+_

Figure 8.15: Circuit for Example 8.4:

30

Writing a nodal equation at point “a” and simplifying gives;

inVV 20


Example 8.5: The noninverting op amp. Consider the following:

R 0

R fb

V 0V 2_

+

+

_

a

Figure 8.16: Noninverting op am configuration.

31


Example 8.5: The noninverting op amp. Continued

Writing a node equation at “a” gives;

20

0

02

0

02

0

2

1

,

11

0)(

VRR

V

giveswhich

RRV

RV

so

RVV

RV

fb

fbfb

fb

Remember this

32


Example 8.6: Noninverting Input.Find V0 for the following op amp configuration.

+

_

+

+

_

_ 4 V

2 k

6 k

5 k

1 0 k

V 0

a

V x

Figure 8.17: Op amp circuit for example 8.6.

33


Example 8.6: Noninverting Input.

The voltage at Vx is found to be 3 V.

Writing a node equation at “a” gives;

0105

0

k

)VV(k

V xx

orVVV x 930

34

End of Lesson 8

CIRCUITS

Operational Amplifiers

OpAmps

Education

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