Alciatore D.G., Histand M.B. Introduction to Mechatronics and Measurement Systems

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Figure 5.12 Buffer or follower.

–

∞

5.6 Noninverting Amplifier 171

so Equation 5.12 can be written as

out

(5.16)

Using Equations 5.14 and 5.16, the voltage gain can be written as

out

--------

out

---------------------------

out

R+

out

-------------------------------- 1

------

+== =

(5.17)

Therefore, the noninverting amplifier has a positive gain greater than or equal to 1.

This is useful in isolating one portion of a circuit from another by transmitting a

scaled voltage without drawing appreciable current.

If we let R

 0 and R   in the noninverting op amp circuit in Figure 5.10 , the

resulting circuit can be represented as shown in Figure 5.12 . This circuit is known as

a buffer or follower because V

out

 V

. It has a high input impedance and low output

impedance. This circuit is useful in applications where you need to couple to a volt-

age signal without loading the source of the voltage. The high input impedance of

the op amp effectively isolates the source from the rest of the circuit. Lab Exercise 6

explores this characteristic by showing how to tap a voltage in a resistor circuit with-

out affecting the currents in the circuit.

Lab Exercise

Lab 6

Operational

amplifier circuits

■ CLASS DISCUSSION ITEM 5.2

Positive Feedback

Ideally, what would happen to the output of the buffer amplifier shown in Figure 5.12

if V

was applied to the inverting input and feedback was from the output to the non-

inverting input instead?

■ CLASS DISCUSSION ITEM 5.3

Example of Positive Feedback

A good example of positive feedback is the effect Jimi Hendrix used to achieve

when he would move his guitar close to the front of his amplifier speaker. Describe

the effect of this technique and describe what is going on physically.

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THREADED DESIGN EXAMPLE

A.3 DC motor power-op-amp speed controller—Power amp motor driver

The figure below shows the functional diagram for Threaded Design Example A (see

Section 1.3 and Video Demo 1.6), with the portion described here highlighted.

potentiometer

PIC microcontroller

with analog-to-digital

converter

power

amp

motor

A/D D/A

light-

emitting

diode

digital-to-analog

converter

The D/A converter outputs a voltage directly related to the potentiometer position. However,

the D/A converter’s output current is limited and not enough to drive a motor. A power op

amp circuit, configured as a noninverting amplifier, can drive the motor at the higher cur-

rents needed. In effect, the power amp will serve as a buffer between the D/A converter and

the motor. The circuit below shows the components used along with their interconnections.

As shown, the OPA547 can be powered by a bipolar (  ) 9 V supply (instead of a standard

bipolar 15 V supply). In Video Demo 1.6, we show how these voltages can be created with

either a standard laboratory power supply or with two 9 V batteries wired in series. With an

input resistor of 10 kΩ and a feedback resistor of 1 kΩ, from Equation 5.17, the power amp

circuit has a gain of 1.1. Therefore, the voltage from the D/A converter is not amplified very

much, but the circuit is able to source ample current to the motor. Op amp components that

are designed to output significant current are called power op amps. The OP547 is one such

op amp. As with all of the design examples, if you want more information about any of the

components used, see Internet Links 1.4 through 1.6.

1 k

–9 V

10 k

R179-6V

DC motor

J3-1

J3-6

TLC7524C

D/A converter

REF

OPA 547

+9 V

3,4

Video Demo

1.6DC motor

power-op-amp

speed controller

Internet Lin

1.4Threaded

design example

components

1.5Digikey

electronics

supplier

1.6Jameco

electronics

supplier

172 CHAPTER 5 Analog Signal Processing Using Operational Amplifiers

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Figure 5.13 Summer circuit.

–

Figure 5.14 Difference amplifier circuit.

–

5.8 Difference Amplifier 173

5.7 SUMMER

The summer op amp circuit shown in Figure 5.13 is used to add analog signals. By

analyzing the circuit with

= R

(5.18)

we can show (Question 5.8) that

out

= −(V

+ V

)

(5.19)

Therefore, the circuit output is the negative sum of the inputs.

5.8 DIFFERENCE AMPLIFIER

The difference amplifier circuit shown in Figure 5.14 is used to subtract analog

signals (see Lab Exercise 6). In analyzing this circuit, we can use the principle of

superposition, which states that, whenever multiple inputs are applied to a lin-

ear system (e.g., an op amp circuit), we can analyze the circuit and determine the

response for each of the individual inputs independently. The sum of the individual

responses is equivalent to the overall response to the multiple inputs. Specifically,

when the inputs are ideal voltage sources, to analyze the response due to one source,

the other sources are shorted. If some inputs are current sources, they are replaced

with open circuits.

Lab Exercise

Lab 6

Operational

amplifier circuits

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174 CHAPTER 5 Analog Signal Processing Using Operational Amplifiers

The first step in analyzing the circuit in Figure 5.14 is to replace V

with a

short circuit, effectively grounding R

. As shown in Figure 5.15 , the result is an

inverting amplifier (see Figure 5.7 and Class Discussion Item 5.1). Therefore, from

Equation 5.11, the output due to input V

out

------

–=

(5.20)

The second step in analyzing the circuit in Figure 5.14 is to replace V

with

a short circuit, effectively grounding R

, as shown in Figure 5.16 a. This circuit is

equivalent to the circuit shown in Figure 5.16 b where the input voltage is

------------------

(5.21)

because V

is divided between resistors R

and R

–

+ R

)

Figure 5.15 Difference amplifier with V

shorted.

Figure 5.16 Difference amplifier with V

shorted.

–

out

(a) V

shorted

(b) equivalent circuit

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5.9 Instrumentation Amplifier 175

The circuit in Figure 5.16 b is a noninverting amplifier (see Figure 5.10 ). There-

fore, the output due to input V

is given by Equation 5.17:

out

------+

⎝⎠

⎛⎞

(5.22)

By substituting Equation 5.21, this equation can be written as

out

------+

⎝⎠

⎛⎞

------------------

⎝⎠

⎛⎞

(5.23)

The principle of superposition states that the total output V

out

is the sum of the

outputs due to the individual inputs:

out

------

⎝⎠

⎛⎞

–1

------

⎝⎠

⎛⎞

------------------

⎝⎠

⎛⎞

+==

(5.24)

When R

 R

 R, the output voltage is an amplified difference of the input

voltages:

out

------

–()=

(5.25)

This result can also be obtained using the op amp rules, KCL, and Ohm’s law

(Question 5.10).

5.9 INSTRUMENTATION AMPLIFIER

The difference amplifier presented in Section 5.8 may be satisfactory for low-

impedance sources, but its input impedance is too low for high-output impedance

sources. Furthermore, if the input signals are very low level and include noise,

the difference amplifier is unable to extract a satisfactory difference signal. The

solution to this problem is the instrumentation amplifier. It has the following

characteristics:

■ Very high input impedance

■ Large common mode rejection ratio (CMRR). The CMRR is the ratio of the

difference mode gain to the common mode gain. The difference mode gain

is the amplification factor for the difference between the input signals, and the

common mode gain is the amplification factor for the average of the input

signals. For an ideal difference amplifier, the common mode gain is 0, imply-

ing an infinite CMRR. When the common mode gain is nonzero, the output is

nonzero when the inputs are equal and nonzero. It is desirable to minimize the

common mode gain to suppress signals such as noise that are common to both

inputs.

■ Capability to amplify low-level signals in a noisy environment, often a require-

ment in differential-output sensor signal-conditioning applications

■ Consistent bandwidth over a large range of gains

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Figure 5.17 Instrumentation amplifier.

–

176 CHAPTER 5 Analog Signal Processing Using Operational Amplifiers

Instrumentation amplifiers are commercially available as monolithic ICs (e.g.,

Analog Devices 524 and 624 and National Semiconductor LM 623). A single exter-

nal resistor is used to set the gain. This gain can be higher and is more stable than

gains achievable with a simple difference amplifier.

An instrumentation amplifier can also be constructed with inexpensive dis-

crete op amps and precision resistors as illustrated in Figure 5.17 . We analyze this

circuit in two parts. The two op amps on the left provide a high-impedance ampli-

fier stage where each input is amplified separately. This stage involves a moder-

ate CMRR. The outputs V

and V

are supplied to the op amp circuit on the right,

which is a difference amplifier with a potentiometer R

used to maximize the over-

all CMRR.

We first apply KCL and Ohm’s law to the left portion of the circuit to express V

and V

in terms of V

and V

. Using the assumptions and rules for an ideal op amp, it

is clear that the current I

passes through R

and both feedback resistors R

. Applying

Ohm’s law to the feedback resistors gives

– I

(5.26)

and

– I

(5.27)

Applying Ohm’s law to R

gives

– I

(5.28)

To express V

and V

in terms of V

and V

, we eliminate I

by solving Equation 5.28

for I

and substituting it into Equations 5.26 and 5.27. The results are

-----

⎝⎠

⎛⎞

-----

–=

(5.29)

and

-----

– V

-----

⎝⎠

⎛⎞

(5.30)

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Figure 5.18 Ideal integrator.

–+

out

–

5.10 Integrator 177

By analyzing the right portion of the circuit, it can be shown (Question 5.12) that

( 5 .

out

+()

-----------------------------

-----

–=

31)

We can substitute the expressions for V

and V

from Equations 5.29 and 5.30 into

Equation 5.31 to express the output voltage V

out

in terms of the input voltages V

and

. Assuming R

 R

, the result is

out

-----

⎝⎠

⎛⎞

–()=

(5.32)

A design objective for the instrumentation amplifier is to maximize the

CMRR by minimizing the common mode gain. For a common mode input,

 V

, Equation 5.32 yields an output voltage V

out

 0. Hence, the common

mode gain is 0, and the CMRR is infinite if R

 R

. In practice, the resistances

never match exactly. Also, if the temperature varies within the discrete circuit,

resistance mismatches are further exaggerated. By using a potentiometer for R

the designer can minimize the mismatch between R

and R

, resulting in a maxi-

mum CMRR.

The problems of resistor matching with discrete components are avoided by

using a monolithic instrumentation amplifier constructed with laser-trimmed resis-

tors. These amplifiers have a very high CMRR not usually obtainable with discrete

components. In addition, the gain is programmable by selecting an appropriate exter-

nal resistor R

5.10 INTEGRATOR

If the feedback resistor of the inverting op amp circuit is replaced by a capacitor, the

result is an integrator circuit. It is shown in Figure 5.18 . Referring to the analysis

for the inverting amplifier, Equation 5.9 is replaced by the relationship between volt-

age and current for a capacitor:

out

-------------

out

------=

(5.33)

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Figure 5.19 Improved integrator.

–

178 CHAPTER 5 Analog Signal Processing Using Operational Amplifiers

Integrating gives

out

t()

out

τ()dτ

∫

(5.34)

where  is a dummy variable of integration. Since i

out

  i

and i

 V

/ R,

out

t()

– V

τ()dτ

∫

(5.35)

Therefore, the output signal is an inverted, scaled integral of the input signal.

■ CLASS DISCUSSION ITEM 5.4

Integrator Behavior

If a DC voltage is applied as an input to an ideal integrator, how does the output

change over time? What is the output given a sinusoidal input? What would the

result be if a small DC offset were added to the sinusoidal input?

A more practical integrator circuit is shown in Figure 5.19 . The resistor R

placed across the feedback capacitor is called a shunt resistor. Its purpose is to limit

the low-frequency gain of the circuit. This is necessary because even a small DC off-

set at the input would be integrated over time, eventually saturating the op amp (see

Section 5.14 and Class Discussion Item 5.7). The integrator is useful only when the

scaled integral always remains below the maximum output voltage for the op amp.

As a good rule of thumb, R

should be greater than 10 R

Because of the impedance and frequency response of the feedback circuit con-

taining R

and C, the circuit in Figure 5.19 acts as an integrator only for a range of

frequencies. At very low frequencies, the circuit behaves as an inverting amplifier

because the impedance of the feedback loop is effectively R

because the impedance

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Figure 5.20 Differentiator.

–

5.11 Differentiator 179

of C is large at low frequencies. At very high frequencies (i.e.,   1/ R

C), the

output is attenuated to zero because the feedback loop is effectively a short. Lab

Exercise 6 explores and Video Demo 5.5 demonstrates how the response of the inte-

grator changes over a large frequency range.

Any DC offset due to the input bias current (see item A in Section 5.14.1) is

minimized by R

, which should be chosen to approximate the parallel combination

of the input and shunt resistors:

(5.3

-----------------

The reason for this is that the input bias current flowing into the inverting terminal is

a result of the currents through R

and R

, and the input bias current flowing into the

noninverting terminal flows through R

. If the voltages generated by the bias current

are the same, they have no net effect on the output.

5.11 DIFFERENTIATOR

If the input resistor of the inverting op amp circuit is replaced by a capacitor, the

result is a differentiator circuit. It is shown in Figure 5.20 . Referring to the analysis

for the inverting amplifier, Equation 5.8 is replaced by the relationship between volt-

age and current for a capacitor:

(5.3

----------

----

Since i

  i

out

and i

out

 V

out

/ R,

( 5 .

out

----------

–=

38)

Therefore, the output signal is an inverted, scaled derivative of the input signal.

One note of caution concerning using differentiation in signal processing is that

any electrical noise in the input signal will be accentuated in the output. In effect, the

differentiator amplifies the fast-changing noise. Integration, on the other hand, has a

smoothing effect, so noise is not a concern when using an integrator.

Lab Exercise

Lab 6

Operational

amplifier circuits

Video Demo

5.5Op amp

integrator circuit

at different

frequencies

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Figure 5.21 Sample and hold circuit.

–

180 CHAPTER 5 Analog Signal Processing Using Operational Amplifiers

5.12 SAMPLE AND HOLD CIRCUIT

A sample and hold circuit is used extensively in analog-to-digital conversion (dis-

cussed in Chapter 8), where a signal value must be stabilized while it is converted

to a digital representation. The sample and hold circuit illustrated in Figure 5.21

consists of a voltage-holding capacitor and a voltage follower. With switch S closed,

(5.3

out

(t) = V

(t)

When the switch is opened, the capacitor C holds the input voltage corresponding to

the last sampled value, because negligible current is drawn by the follower. Therefore,

(5.4

out

(t − t

sampled

) = V

sampled

)

where t

sampled

is the time when the switch was last opened. Often, an op amp buffer

is also used on the V

side of the switch to minimize current drain from the input

voltage source V

The type of capacitor used for this application is important. A low-leakage

capacitor such as a polystyrene or polypropylene type would be a good choice. An

electrolytic capacitor would be a poor choice because of its high leakage. This leak-

age would cause the output voltage value to drop during the “hold” period.

■ CLASS DISCUSSION ITEM 5.5

Differentiator Improvements

Recommend possible improvements to the differentiator circuit in Figure 5.20 .

Consider the effects of high-frequency noise in the input signal.

■ CLASS DISCUSSION ITEM 5.6

Integrator and Differentiator Applications

Think of various applications for the integrator and differentiator circuits. Consider

how a differential equation could be solved with an analog computer. Also consider

how to convert between sawtooth and square-wave function generator outputs and

how to process position and speed sensor signals.

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