MODULE TITLE : ELECTRONICS TOPIC TITLE : AMPLIFIERS LESSON ... · 12/3/2015 · ADJUSTABLE GAIN...

MODULE TITLE : ELECTRONICS

TOPIC TITLE : AMPLIFIERS

LESSON 3 : LINEAR APPLICATIONS OF OP AMPS

EL - 3 - 3

© Teesside University 2012

Steve

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________________________________________________________________________________________

INTRODUCTION________________________________________________________________________________________

In this lesson, we start to use the operational amplifier as a functional block.

We have moved on from individual transistor circuits.

We shall investigate a number of circuits which can perform linear analytic

functions on signals, such as amplifying, adding, subtracting, filtering,

integrating and differentiating.

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YOUR AIMS________________________________________________________________________________________

At the end of this lesson, you should be able to:

• analyse simple op amp circuits

• design simple active filters.

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STUDY ADVICE________________________________________________________________________________________

You may find it quite useful to make a separate summary of all the op amp

circuits we investigate, together with just their function and basic equation; it

acts as an aide memoire.

The most common error in applying the formulae derived in this lesson is

carelessness with signs. Remember that, if I ask you for the output voltage of

an amplifier, or the gain of an amplifier, the sign is part of the answer!

1

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Steve

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________________________________________________________________________________________

BASIC APPLICATIONS________________________________________________________________________________________

In the previous lesson, we introduced the basic ideal operational amplifier. In

this lesson, we shall be building up a library of applications. We shall assume

a perfect amplifier; i.e. one which has infinite gain, no input current and

infinite bandwidth. In practice of course, no amplifier is ideal and, in the next

lesson, we shall try to see what effects the imperfections cause.

Remember the basic formula for the gain of the op amp in inverting mode.

INVERTER

FIG. 1

This is a special case of the basic application we looked at in the last lesson;

here the two resistors R1 and R2 are the same.

V

V

R

Ro

in

= = −– 1 ........................................ ( )2

R

Vin

R

Vo

0 V

–+

V

V

R

Ro

in

= – 2

1

................................................ ( )1

2



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ADJUSTABLE GAIN CIRCUITS OR SCALERS

In many cases, we may want adjustable gain. Of course, one can replace either

R1 or R2 by a variable resistance, but the following circuits have some

advantages. In these circuits, we take an adjustable fraction, x, of the input or

the output. Obviously, x is less than or equal to 1.

FIG. 2(a) and (b)

From FIGURE 2(a), we obtain the following relationship.

This is just an application of the fundamental gain rule of the op amp.

For FIGURE 2(b), assume the point V1 is a virtual earth.

V

R

xV

R

V

V x

R

R

in o

o

in

– –

–

0 0

1

1 2

2

1

=

= ........................................... ( )4

V

Vx

R

Ro

in

= – 2

1

........................................... ( )3

(a) (b)

R

Vin

Vo

0 V

–+

R1

R2

xVin

0 V

Vin Vo

0 V

–+

R1

R2

V1

RxVo

0 V

3



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These formulae assume that R1 and R2 are much greater than the potentiometer

resistance R.

In Figure 2(b), what happens if x is set to 0?

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________________________________________________________________________________________

If x → 0, gain tends to A, the gain of the op amp (not infinity in practice!).

From a practical point of view, it is often desirable to prevent x going to 0 by

including a fixed resistor in series with the potentiometer.

WORKED EXAMPLE 1

In the circuit of FIGURE 3, find the exact gain.

FIG. 3

Vin Vo

0 V

–+R1

0 V

R2

100 kΩ

1 kΩ

R310 kΩ

R4

100 kΩ

4



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We could use equation (4) if x can be deduced. Now, an approximate answer

would give x as , but this does not take into account the loading

effect of R2. Because V– is a virtual earth, R2 is effectively in parallel with R4.

This leads to the following expression for x.

Hence, gain is calculated from equation (4) as follows.

Put in the values given in FIGURE 3.

If we had used x as , the gain would have been calculated as –1010.

This is quite a useful circuit, because we have achieved a high gain without

using very high resistance ratios, or making R1 very low.

R

R R4

3 4+( )

Gain = ⎛⎝⎜

⎞⎠⎟

+( ) + ×( )–

1010

10 10 10 10 10

10

5

4

5 3 5 3 5

3 ××⎛

⎝⎜⎞

⎠⎟

=

10

1020

5

–

Gain =+

=+( ) +

–

–

R

R

R R R

R R

R

R

R R R R R

R

2

1

3 4 2

4 2

2

1

3 4 2 4 2

4 RR2

xR R

R R R=

+4 2

3 4 2

R

R R4

3 4+( )

5



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SUMMER

FIG. 4

Since input current = feedback current, we can write the following equation.

The circuit gives the inverted algebraic sum of all input signals and can, of

course, be extended to any number of inputs.

For Vin1 = +1 V, Vin2 = –2 V, Vin3 = +3 V, R2 = 10 kΩΩ, Ri1 = 1 kΩΩ, Ri2 = 1.5 kΩΩ and

Ri3 = 10 kΩΩ, find Vo.

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The value of Vo is +0.33 V.

VR

RV

R

RV

R

RVo

iin

iin

iin= + +

⎡

⎣⎢

⎤

⎦⎥– 2

11

2

22

2

33 ................. ( )5

Vin1

Vo

0 V

–+

R2

Vin2

Vin3

Ri1

Ri2

Ri3

6



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DIFFERENTIAL AMPLIFIER

FIG. 5

Assume that the amplifier is ideal.

Note that V1 is not a virtual earth, since V2 is not 0 V. V1 is here virtually equal

to V2.

V V

R

V V

Rin o1 1

1

1

2

– –= ..............................................

.

( )

–

6

2 2

3

2

4

V V

R

V

Rin = .....................................................

.......................

( )7

1V V= ................................. ( )8

Vin1

Vo

0 V

–+

R2

Vin2

R4

R1

R3

V1

V2

7



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Rearrangement of equation (7) allows us to determine V2.

We could have deduced this straight from the diagram. How?

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Because it is just a voltage divider.

Rearrangement of equation (6) yields the equation below.

Since V1 = V2, and substituting for V2 from equation (9), we can obtain an

equation with V1 and V2 eliminated.

This can then be arranged to express the output voltage in terms of the input

voltages.

V

R

V

RV

R R R

R R R Rin o

in1

1 22

4 1 2

3 4 1 2

= ++( )

+( )–

V

R

V

RV

R Rin o1

1 21

1 2

1 1= + +⎡

⎣⎢

⎤

⎦⎥

V

RV

R R

R V V R R

VR

in

in

2

32

3 4

4 2 2 4 3

24

1 1= +⎡

⎣⎢

⎤

⎦⎥

= +( )

=VV

R Rin2

4 3+ ........................................... ( )9

8



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As we would expect, the output depends on both Vin1 and Vin2.

Choice of the resistors can considerably simplify this equation.

If

then the equation simplifies to that below in equation (11).

So now we have a differential amplifier with a well defined gain.

If Vin1 + 3 V, Vin2 = +4 V, R1 = 1 kΩΩ and R2 = R3 = R4 = 10 kΩΩ, find Vo.

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The value of Vo is –8 V.

VR

RV Vo in in= ( )– –2

11 2 .............................. ( )11

R

R

R

R1

2

3

4

=

VR

RV V

R

R

R

R

o in in=+

⎛⎝⎜

⎞⎠⎟

+⎛⎝⎜

⎞⎠⎟

⎡

⎣

– –2

11 2

1

2

3

4

1

1

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

........... ( )10

9



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NON-INVERTING AMPLIFIER

FIG. 6

From equation (10) and putting Vin1 = 0, R3 = 0 and replacing R4 by an open

circuit (∞), we obtain the following simple relationship.

This gives us a non-inverting amplifier; i.e. the gain is positive. The minimum

gain is 1! With the inverting amplifier, we can get gains less than 1.

BUFFER

FIG. 7

From equation (12), with R1 = ∞ and R2 = 0, we have the special case of a

buffer.

V Vo in= 2 ............................................. ( )13

Vo–+Vin2

V

V

R

Ro

in2

2

1

1= + ............................................. ( )12

Vo–+

R2

Vin2

R10 V

10



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REVIEW OF CIRCUITS EXAMINED

We have built up a repertoire of circuits which can:

• give high programmable negative or positive gains

• add or subtract multiple inputs

• provide buffer amplifiers with gains of +1 or –1.

________________________________________________________________________________________

FREQUENCY DEPENDENT CIRCUITS________________________________________________________________________________________

INTRODUCTION

FIG. 8

When we consider sine waves, operational amplifiers can be used as active

filters to provide amplification or attenuation and/or phase-shift for selected

frequency ranges of the input signal. The usual advantages of operational

amplifiers are retained.

These advantages are that the gain is:

• defined accurately by the ratio of impedances

• independent of the load.

Vin

Vo–+

Z2

Z1

0 V

11



Steve

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Equation (14) below gives the general relationship.

G(jω) is the gain expressed as a frequency dependent complex number, and

can be plotted on a Bode Diagram.

In general, the output is a sine wave with the same frequency as the input, but

with a different amplitude and displaced in phase. By choosing Z1 and Z2

appropriately, we can design low-pass, high-pass and band-pass filters. The

filters can be cascaded to give second and higher order filters, discussion of

which is beyond the scope of this module.

Let us compare a basic first-order low-pass filter with a similar active filter.

LOW-PASS FILTER

(a) Passive Filter

The circuit is shown in FIGURE 9(a).

where

and

Z

R

C

RC

R

CR

Z R R

L

L

L

L

s

2

1

1 1=

+=

+

= +

j

jj

ω

ωω

V

V

Z

Z Zo

s

=+2

2 1

V

VG

Z

Zo

in

j

jj

j

j ........

ωω

ωωω

( )( ) = ( ) =

( )( )– 2

1

......... ( )14

12



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So

(b) Active Filter

The circuit is shown in FIGURE 9(b).

FIG. 9 Passive and Active Filter

Vo–+

C

R1Rs

R2

Vs

RL

Source Filter Load

Rs

Vs

RLC

Source Filter Load

R

Vo

Z2

V

V

RC

R R

Z

Z

R

R R

o

s s

s

=+

=⎛⎝⎜

⎞⎠⎟

=+( ) +

––

–

2

1

2

1

2

1

1

1

j

j

ω

ωωCR2( )

V

V

R

CRR

CRR R

R

R

o

s

L

Ls

L

L

=+( )

++ +

⎧⎨⎩

⎫⎬⎭

=

–

–

112

2jj

ωω

++ +( ) ++⎡⎣ ⎤⎦

+ +( )⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪R R

CR R R

R R RsL s

L s

1jω

13



Steve

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We now draw the conclusions shown in FIGURE 10.

FIG. 10

It is clear that active filters can be designed so that they are largely

independent of source and load impedances, and that they can be designed to

have any desired gain.

In general, active filters can be designed to use smaller values of capacitance

(cheaper) and to avoid the use of inductors which are heavy and bulky, and

often lead to electrical interference and audible noise.

DC Gain

InputResistance

Passive Active

Always < 1

Depends on Rs and RLas well as R and C

Depends only onC and R2

Normally it iseasy to chooseR1 » Rs

NormallyRs < R < RLotherwise there areloading problems asRs + RL changes.

But this is sometimesdifficult to achieve.

Bandwidth

=1

CR

Depends on load RL

Can be

programmedby R2

Rs + R1( )

14



Steve

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However, active filters can only be used for signal conditioning; they cannot

be used in the filtering of high power circuits.

INTEGRATOR

This is the first of two special frequency dependent circuits which will now be

investigated.

FIG. 11

Since input current = feedback current, use of the basic relationship for a

capacitor gives the following equation.

Integrate both sides with respect to t.

VV

RCto

in=⌠

⌡

⎮⎮⎮⎮

– d ...................................... ( )15

V

RC

V

tin o– –0 0

=( )d

d

Vin

Vo–+

C

R

0 V

15



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Hence, this circuit will perform integration of time varying signals, with a gain

constant of RC. One problem with an integrator is that any small errors, e.g.

voltage offset, will be integrated by the amplifier, which will therefore slowly

head towards positive or negative saturation. A pure integrator cannot

therefore be used alone. A modification which limits low frequency gain is

shown in FIGURE 12. The resistance RI is chosen to be a high value, typically

1 MΩ.

FIG. 12

In the frequency domain, the gain is given by equation (16).

As ω increases, the gain falls at the rate of 20 dB/decade. Unity gain (0 dB)

occurs where ωCR = 1. The gain tends to infinity (or in practice the d.c. gain

of the amplifier) as ω tends to 0. This is shown in FIGURE 13, which also

shows the effect of introducing RI; this makes the circuit into a low-pass filter.

V

V RCo

in

j

j j .....................

ωω ω

( )( ) = −1

........................... ( )16

Vin

Vo–+

C

R

RI

0 V

16



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FIG. 13

DIFFERENTIATOR

FIG. 14

Vin

Vo–+

C

R

0 V

0

Gain(dB)

Effect of RI

–20 dB/decade

Frequencydecade

10m 10m+1

ω = 1CR

Log10ω

17



Steve

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Assuming an ideal op amp, the following equation holds.

This represents an ideal differentiator, but it is rarely used in this form.

In the frequency domain, the gain is given by equation (18)

Note that the gain increases as frequency!

This property which the differentiator has of amplifying any high frequency

signal means that any 'noise' in the input is also greatly amplified; and the

higher the noise frequency, the greater the gain. A modification to reduce the

gain at high frequencies is shown in FIGURE 15, which also shows the

frequency response.

G CRj j ..............................ω ω( ) = − ...................... ( )18

CV

t

V

R

V RCV

t

in o

oin

d

d

Sod

d ...............

=−

= − ................................. ( )17

18



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FIG. 15

Feedback Impedancej j

Input I

= =+

RC

R

C R22

2

2 2

11ω ω

mmpedancej

j

j

Soj

j

= + =+

( )

RC

C R

C

V

Vo

in

11

1 1

1

1 1

ωω

ω

ωωω

ωω ω( ) =

+( ) +( )–j

j j .........

C R

C R C R1 2

1 1 2 21 1................. ( )19

Vin

Vo–+

0 V

R1C1

C2

R2

0

Gain(dB)

+20 dB/decade

Response withoutR1 and C2

ω = 1C1R2

Log10ω

–20 dB/decade

ω = 1C2R2 ω = 1

C1R1

R1= 0

19



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At high frequencies, the gain falls, because the two terms in the denominator

dominate.

An interesting aside is that, if R1 = 0, the response is as shown above in

FIGURE 15. The gain is flat from ω = to infinity (at least

theoretically).

So the circuit then becomes a high-pass filter. As ω becomes much greater

than , the gain tends to .C

C1

2

1

2 2C R

1

2 2C R

20



Steve

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NOTES________________________________________________________________________________________

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________________________________________________________________________________________

SELF-ASSESSMENT QUESTIONS________________________________________________________________________________________

1. Find the output voltage Vo of the following circuits. REMEMBER

THAT SIGNS ARE IMPORTANT!

Vo–+

0 V

10 kΩ

1 kΩ

–1 VVo

–+

0 V

15 kΩ

1 kΩ–0.5 V

Vo–+

0 V

100 kΩ

–100 V

Vo–+

0 V

15 kΩ1 kΩ–0.5 V

1.5 kΩ

12 kΩ

10 kΩ

+2 V

–4 V

–3 V

(a) (b)

(c) (d)

100 kΩ1MΩ

100 kΩ

VX

22



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2. What is the gain in each of the following circuits?

3. Design an amplifier circuit which gives an output of +10 V when the

input is –2V, and has an output of 0 V when the input is –6 V.

Hint: Sketch the gain line, get an equation for the straight line and then

match the equation with one of the circuits we have developed. You can

use one or two amplifiers and you will need a reference voltage.

Vo–+

0 V

10 kΩ

1 kΩ

Vo–+

0 V

10 kΩ

1 kΩ

Vo–+

0 V

100 kΩ

Vo–+

0 V

20

(a) (b)

(c)

(d)

100 kΩ

10 kΩ

10 kΩ

0 V

Vin

10 kΩ

0 V

Vin

1 kΩ

10 kΩ

20 kΩ20 kΩ

20 kΩ10 kΩ20 kΩ

Vin

VX VY

100 kΩ

–+

0 VVin

VX

10 kΩ

100 kΩ

23



Steve

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4. Find the peak output of the amplifier shown below.

Hint: How do you find the maximum of a signal?

5.

Show that the expression for gain in the circuit above is

GCR

CRj

j

jω

ωω

( ) =+1 2

1

Vo

0 V

–+R1

Vin

R2C

R2 sin β

R

Vo

0 V

–+3 sin 2β

2R

24



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6. Sketch the shape of the outputs of (i) an op amp integrator and (ii) a

differentiator to the following input wave-forms.

Hint: Remember the definitions of integration and differentiation.

7. Design an active low-pass filter with a low frequency gain of –15 and a

bandwidth of 400 radians/second.

8. Obtain the gain expression for the circuit opposite. Show that the

magnitude of the gain is +1 for all frequencies.

Hint: you can use a modified gain expression from one of the circuits we

have already derived, or you can obtain the gain from first principles.

0A

A

T T

t

0A t

0

A

tA

Sine wave

(a)

(b)

(c)

25



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Vin Vo

0 V

–+

R

C

R

R

26



Steve

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________________________________________________________________________________________

SELF-ASSESSMENT QUESTIONS________________________________________________________________________________________

1. (a) Apply equation (1) for the gain of an inverting amplifier.

(b) Apply equation (12) for the gain of a non-inverting amplifier.

(c) Apply the Summer equation.

(d) It is easy to see that

V V Vo x x= − ⎛⎝⎜

⎞⎠⎟ = −100

100

Vo = ⎛⎝⎜

⎞⎠⎟ −( ) + ⎛

⎝⎜⎞⎠⎟ ( ) + ⎛

⎝⎜⎞⎠

151

0 5151 5

21512

.. ⎟⎟ −( ) + ⎛

⎝⎜⎞⎠⎟ −( )⎡

⎣⎢⎤⎦⎥

= − + − −[ ]

41510

3

7 5 20 5 4 5. .

== − 3V

VR

RVo in= +

⎛⎝⎜

⎞⎠⎟

= +⎛⎝⎜

⎞⎠⎟ ( ) =

1

1101

1 11

2

12

– – V

VR

RVo in=

⎛⎝⎜

⎞⎠⎟

= ⎛⎝⎜

⎞⎠⎟ ( ) = +

–

– – . .

2

1

151

0 5 7 5V

27



Steve

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We have to find Vx. Because V1 is a virtual earth, the input circuit

becomes a voltage divider, in which the resistance of the two 100 kΩresistors in parallel is 50 kΩ. This is shown below.

Hence,

2. (a) We can use the differential amplifier formula, putting Vin1 = 0.

VR

RV

R

R

R

R

o in= −+

⎛⎝⎜

⎞⎠⎟

+⎛⎝⎜

⎞⎠⎟

⎡

⎣

⎢⎢⎢

2

12

1

2

3

4

0

1

1

–

⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

= ++⎛

⎝⎜⎞⎠⎟

+( )

= +

101

110

1 1

5 5

2

2

V

V

in

in.

Vo = + 4 76. V

Vx = ×+ ×

−( )

= −

50 1010 50 10

100

4 76

3

6 3

. V

1 MΩ

100 kΩ

–100 V

100 kΩ

0 V

Vx

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Steve

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(b) Use the formula derived in Worked Example 1.

(c) This is an extension of the circuit of Question 1(d).

We now find by a similar process. See the circuit below.

20 kΩ

0 V

Vx

Vin

20 kΩ 10 kΩ

20 kΩ 20 kΩ

V

Vx

in

V

V

V

V

o

y

y

x

= −

=Ω Ω

Ω Ω + Ω

=

1

12

20 k 20 k

20 k 20 k 10 k

treati[ nng the circuit as a divider]

Gain =+( ) +

=+( ) +

–

–

R

R

R R R R R

R R2

1

3 4 2 4 2

4 2

101

10 1 10 100

10

120= −

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Steve

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The overall gain is calculated as follows.

(d) For the inverting amplifier,

For the differential amplifier,

V V V

V V

V

o x in

in in

i

= − −( )

= − −⎡⎣ ⎤⎦ −( )

= − −

10

10 10

10 11 nn

inV

( )

= + 110

V Vx in= − 10

V

V

V

V

V

V

V

Vo

in

o

y

y

x

x

in

=

= −( ) ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

= −

112

13

166

V

Vx

in

=Ω Ω + Ω Ω( )

Ω Ω + Ω Ω(20 k k 20 k 20 k

20 k k 20 k 20 k

10

10 )) + Ω

=

20 k

13

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3. The two input/output conditions define two points on a straight line, the

graph of which is sketched overleaf; the equation of the line is

The gain is positive.

There are many possible solutions.

Solution by Adapting a Differential Amplifier

From equation (10), we can see that the input in this case must be Vin2, because

of the positive gain requirement.

Try putting R3 = 0 and R4 as an open circuit. Then use equation (10).

VR

RV V

R

R

R

R

o in in=⎛⎝⎜

⎞⎠⎟

+ +⎛⎝⎜

⎞⎠⎟

⎛

⎝⎜⎞

⎠⎟

=

2

11

1

2

2

1

111

2

1

1⎛⎝⎜

⎞⎠⎟

+ +⎛⎝⎜

⎞⎠⎟

⎛

⎝⎜⎞

⎠⎟V V

R

Rin in

Vin

Vo

+10

–6 –2

V Vo in= +2 5 15.

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Making this identical with the target equation gives two conditions.

The second of these conditions gives = 1.5.

Choose R2 = 15 kΩ and R1 = 10 kΩ from the E12 series to satisfy this

condition.

The first condition then gives the following equation.

Hence,

The circuit is shown below.

–10 V 10 kΩ

Vo–+

Vin

15 kΩ

R1

R2

Vin1 10= − V

− ( ) = +1 5 151. Vin

R

R2

1

−⎛⎝⎜

⎞⎠⎟

= + + =R

RV

R

Rin2

11

2

1

15 1 2 5and .

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Solution by Using Two Amplifers

The first amplifier gives a gain of –2.5.

The second is a summer with a gain of –1.

4. This is a fairly difficult question. First get an expression for the output.

Differentiate with respect to β.

d

d

[Usin

Vo

ββ β

β β

= −

= − ( )

2 3 2

2 3 2 12

cos – cos

cos – cos – gg double angle formula]

Vo = − 2 1 5 2sin – . sinβ β

V V

V

V

V

o x

x

in

in

= − ( )

= − +

= − −( ) +

= + +

–

.

.

15

15

2 5 15

2 5 15

Vo–+

–+

100 kΩ10 kΩ

0 V

10 kΩ

0 V 10 kΩ–15 V

Vx

39 kΩ1 kΩVin

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The peak of the output will be where = 0.

Solve this quadratic for cos β.

Hence,

Not all these are maxima; if you sketch the wave-forms of 2sinβ and

1.5sin2β and add them, you will see that the solutions we want are given

by ±55.9°.

Hence,

5. The complex gain is given by –

In this case, we have the following values of Z1 and Z2.

Hence, G

RC

R

CR

CRj

j j

jω

ω ωω

( ) =+

⎛⎝⎜

⎞⎠⎟

=+( )

– –2

1

2

1

11

Z RC

Z R2 2 1 11= + =

jand

ω

Z

Z2

1

Vo (max)

V

At

= − ( ) ×( )

= −

2 55 9 1 5 2 55 9

3 05

sin . – . sin .

.

ββ = − = +55 9 3 05. , .Vo (max) V

β = ± ° ± °55 9 153 3. .or

cos

. .

β = − ± −

= −

2 4 7212

0 56 0 89or

2 6 3 02cos cos –β β+ =

d

d

Vo

β

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6. Integration Response Differentiation Response

The total area under a graph The slope of a graph

(Remember the – sign!)

(a)

t t

∞

(b)

t t

∞

(c)

t

0

+cos

t0

–cos

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7. The gain of an active low-pass filter is given by the equation below (see

notes).

We require a d.c. gain of –15. Hence = 15.

We also require that ωCR2 = 1 when ω = 400.

It is best to make C small and R large, so try C = 10 nF.

Therefore,

Select R2 = 270 kΩ as the nearest value in the E12 series.

Finally,

Select R1 = 18 kΩ, which is a standard value in the E12 series.

8. We can adapt the differential amplifier formula, equation (10), by putting

V V V R R R R RCin in in1 2 2 3 4= = = = = =, and

1j1 ω

RR

12

3

15270

1015

18= = × = kΩ

R28 1

400 10 250= ×( ) =− −kΩ

R

R2

1

GR

R CRj

jω

ω( ) = =

+– 2

1 2

11

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Then equation (10) simplifies as follows.

Therefore, the magnitude of the gain = = 1

Alternatively, we can work from first principles.

and

and also

So,

V V

V V V Vin o

1 2

1 1 1

=

=– – [since no current enters oop amp]

j

Rj

j

V

V

C

CCR

V V

in

o

2

1

11

1

2

=

⎛⎝⎜

⎞⎠⎟

+=

+

=

ω

ωω

11

22

2

1

–

–

–

V

V V

V

CRV

in

in

inin

=

=+ jω

1

1

2

2

+ ( )+ ( )

ω

ω

CR

CR

VR

RV V

R

RCRo in inj

jω

ω( ) = − −

+⎛⎝⎜

⎞⎠⎟

+

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥

1

1 ⎥⎥⎥

= + −+

⎡

⎣⎢

⎤

⎦⎥

=+

–

–

VCR

CR

VCR

C

in

in

1 21

11

jj

jj

ωω

ωω RR

⎡

⎣⎢

⎤

⎦⎥

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Finally,V

V

CR

CR

CR

CR

o

in

=− +( )

+

=+

2 1

1

11

j

j

jj

ωω

ωω

–

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________________________________________________________________________________________

SUMMARY________________________________________________________________________________________

The operational amplifier has proved to be a versatile implement in the design

of a wide range of linear circuits.

The lesson has developed the theory of d.c. amplifier circuits for the:

• inverter

• scaler

• summer

• differential amplifier

• non-inverting amplifier

• buffer.

We have also investigated frequency dependent circuits for the:

• low-pass filter

• integrator

• differentiator.

Many more circuits are possible and, indeed, a comprehensive treatment of

active filters alone could easily run to a full module. This lesson has provided

the groundwork.

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MODULE TITLE : ELECTRONICS TOPIC TITLE : AMPLIFIERS LESSON ... · 12/3/2015 · ADJUSTABLE GAIN...

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Transcript of MODULE TITLE : ELECTRONICS TOPIC TITLE : AMPLIFIERS LESSON ... · 12/3/2015 · ADJUSTABLE GAIN...