Chapter 19 Electronics Fundamentals Circuits, Devices and Applications - Floyd © Copyright 2007...

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Chapter Chapter 19 19 Electronics Fundamentals Circuits, Devices and Applications - Floyd © Copyright 2007 Prentice-Hall Chapter 19

Transcript of Chapter 19 Electronics Fundamentals Circuits, Devices and Applications - Floyd © Copyright 2007...

Page 1: Chapter 19 Electronics Fundamentals Circuits, Devices and Applications - Floyd © Copyright 2007 Prentice-Hall Chapter 19.

Chapter 19Chapter 19

Electronics FundamentalsCircuits, Devices and Applications - Floyd © Copyright 2007 Prentice-Hall

Chapter 19

Page 2: Chapter 19 Electronics Fundamentals Circuits, Devices and Applications - Floyd © Copyright 2007 Prentice-Hall Chapter 19.

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Electronics FundamentalsCircuits, Devices and Applications - Floyd © Copyright 2007 Prentice-Hall

Summary

Comparators

Op-amps can be used to compare the amplitude of one voltage with another. Although general-purpose op-amps can be used as comparators, special op-amps are available to optimize speed and add features.

An example of a comparison circuit is shown. The input is compared with a reference set by the voltage-divider. Notice that there is no feedback; the op-amp is operated in open-loop, so the output will be in saturation.

Vin

R1

Vout

+

R2

+V

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Summary

Sketch the output of the comparator in relationship to the input; assume the maximum output is ±13 V.

Vin

R1

Vout+

R2

V = +15 V

10 k

3.9 k

The threshold is +4.2 V. The output is in positive saturation when Vin > +4.2 V Vin

+10 V

10 V

0 V+4.2 V

+13 V

13 V

0 V

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Summary

Show the output of the comparator for the last example if the inputs to the op-amp are reversed.

Vin

R1

Vout

+

R2

V = +15 V

10 k

3.9 k

The threshold is still +4.2 V but now the output is in negative saturation when Vin > +4.2 V. Vin

+10 V

10 V

0 V+4.2 V

+13 V

13 V

0 V

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Summary

Summing amplifier

There are a number of useful applications for the basic inverting amplifier configuration. One is the summing amplifier that uses two or more inputs and one output.

R1VIN1

VIN2

VIN3

VINn

R2

R3

Rn

Rf

+

VOUT

The virtual ground isolates the inputs from each other. Input current from each input is passed to Rf, which develops an output voltage that is proportional to the algebraic sum of the inputs.

Virtual ground

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Summary

Averaging amplifier

An averaging amplifier is a variation of the summing amplifier in which all input resistors are equal. The feedback resistor is the reciprocal of the number of inputs times the input resistor value.

R1VIN1

VIN2

VIN3

R2

R3

Rf

+

VOUT

For example, if there are three input resistors, each with a value of 10 k, then Rf = 3.3 k to form an averaging amplifier.

10 k

10 k

10 k

3.3 k

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Summary

Scaling adder

A scaling adder is another variation of the summing amplifier in which the input resistors are adjusted to weight inputs differently. The input “weight” is proportional to the current from that input.

R1VIN1

VIN2

VIN3

R2

R3

Rf

+

VOUT

Larger resistors will allow less current for a given input voltage, so they have less “weight” than smaller resistors. In the case shown, VIN3 is “weighted” 2 times more than VIN2, which is 2 times more than VIN1.

10 k

5.0 k

2.5 k

10 k

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Summary

Scaling adder

R1VIN1

VIN2

VIN3

R2

R3

Rf

+

VOUT

10 k

5.0 k

2.5 k

10 k

What is VOUT for the scaling adder if all inputs are + 1.0 V?

By Ohm’s law, the currents into Rf are I1 = 0.1 mA, I2 = 0.2 mA and I3 = 0.4 mA.

Using the superposition theorem, the current in Rf is 0.7 mA. From Ohm’s law, VOUT = 7 V

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Summary

Integrators

Mathematical integration is basically a summing process. Within certain limitations, an integrator circuit simulates this process.

The ideal integrator is essentially a summing amplifier with a capacitor in place of the feedback resistor.

R

C

Vin

Vout

In practical circuits, a large value resistor is usually in parallel with the capacitor to prevent the output from drifting into saturation.

Rf

+

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Summary

Integrators

For the ideal integrator, the rate of change of the output is given by

R

C

VinVout

out in

i

V V

t R C

The minus sign in the equation is due to the inverting amplifier. If the input is a square wave centered about 0 V, the output is a negative triangular wave (provided saturation is not reached).

Vin

Vout

0 V

0 V

+

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Summary

R

C

Vin

Vout

5 V

100 μs =2.7 k 33 nF

inout

i

VV t

R C

A 5 kHz square wave with 10 Vpp is applied to a practical integrator. Show the output waveform voltages.

33 nF

2.7 k

270 k

Rf

During the positive input (½ the period), the change in the output is

5.6 V

The feedback resistor (Rf) is large compared to R, so has little effect on the shape of the waveform. In a practical circuit, it will cause the output waveform to center on zero as shown on the following slide.

+

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Summary

R

C

Vin

Vout

33 nF

2.7 k

270 k

Rf

The results of a computer simulation on Multisim confirm the calculated change (5.6 V) in output voltage (blue line).

continued…

+

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Summary

Differentiators

In mathematics, differentiation is the process of finding the rate of change. An idea differentiator circuit is shown. It produces an inverted output that is proportional to the rate of change of the input.

+

CVin

Vout

In practical circuits, a small value resistor is added in series with the input to prevent high frequency ringing.

RinVin

Rf

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Summary

Differentiators

The output voltage for the ideal differentiator is given by

+

Rf

CVin

Vout

Cout f

VV R C

t

The minus sign in the equation is due to the inverting amplifier. If the input is a ramp, the output is a negative dc level for the positive slope and a positive dc level for the negative slope.

Vin

Vout

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Summary

A 1.0 kHz, 10 Vpp triangular wave is applied to a practical differentiator as shown. Show the output in relationship to the input.

When the input has a positive slope, the output is

10 V2.7 k 100 nF

0.5 msC

out f

VV R C

t

Rf

CVin

100 nF

2.7 k

120 +

Vout

Rin

Vin

+5.0 V

5.0 V

0 V

5.4 V

By symmetry, when the input has a negative slope, the output will be +5.4 V.

See next slide for waveforms…0 1 ms 2 ms

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Summary

continued…

The results of a computer simulation on Multisim confirm the calculated output voltages (±5.4 V). The output voltage is the blue line.

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JFET bias circuit

Negative feedback with JFET gain control

Lead-lag circuit

Summary

Oscillators

The feedback oscillators introduced in Chapter 17 and other types of feedback oscillators can be implemented with op-amps. One type of feedback oscillator is called the Wien-bridge oscillator. This circuit is useful for generating low distortion sine waves.

+

R1

R2

R3

Rf

Q1

C1

C2

C3

D1

R4

Vout

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+

R3

Rf

Q1

C3

D1

R4

Vout

Summary

Oscillators

R1

R2

C1

C2

The lead-lag circuit in the Wien-bridge oscillator has a maximum response at the resonant frequency given by

1

2πrf RC

This equation is valid when R’s and C’s in the lead-lag circuit are equal.

The lead-lag circuit response is…

Because the attenuation is ⅓ at fr, the gain of the Wien bridge must set for 3.

Vout

Vout

Vin

fr

f

⅓Vin

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Summary

Wien-bridge oscillator

+

R1

R2

R3

Rf

Q1

C1

C2

C3

D1

R4

Vout

47 nF

47 nF

6.8 k

6.8 k

The frequency is given by

1

2π1

2π 6.8 k 47 nF

rf RC

498 Hz1.0 k 10 k

10 k

1.0 F

What is the frequency of the bridge?

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Summary

Triangular-wave oscillator

A triangular-wave oscillator can be made from a comparator and an integrator. The integrator produces a ramp due to the constant current charging of the capacitor. When the ramp reaches a trip point, the comparator suddenly switches to opposite level and the ramp changes direction.

Vout (triangle)

Vout (square)

Comparator Integrator+

+

R3

R2

R1

C

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Summary

Square-wave relaxation oscillator

The square-wave relaxation oscillator uses a comparator to switch the output based on the charging and discharging of a capacitor.

CVout

+

R3

R2

R1

Vout

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Summary

Active filters

By reversing the resistors and capacitors in the low-pass circuit, a high-pass filter is created. This filter has a gain of 1 at frequencies where f > fc.

Vout

+

R2R1

C2

C1

Vin

Gain (dB)

ffc

30

40 dB/decade+

R2

R1

C2C1

Vin

Gain (dB)

fc

30

40 dB/decade

A filter selects certain frequencies and excludes others. Active filters use op-amps to optimize the frequency response. A 2-pole low-pass filter and its response is shown. The gain for this filter is 1 (0 dB) for f < fc.

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Summary

Voltage regulators

Voltage regulators are made from integrated circuits. A basic series IC regulator has four blocks:

Control element

VOUTVIN

Sample circuit

Error detector

Reference voltage

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Summary

Voltage regulators

A series regulator uses a comparator to compare the output voltage with a reference voltage. The series transistor drops more or less voltage to keep the output constant.

VOUT

+

R3

R2

R1

VIN

Q1

The voltage at the inverting input is forced to be the same as the reference voltage by feedback action. Therefore,

2OUT REF

3

1R

V VR

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Summary

Voltage regulators

VOUT

+

R3

R2

R1

VIN = Q1

What is the output voltage of the series regulator?

2OUT REF

3

1R

V VR

5.1 V

4.7 k

+24 V

3.3 k

6.8 k

6.8 k1 5.1 V

3.3 k

=

15.6 V

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Summary

Voltage regulators

A shunt regulator also has four blocks; it controls the current in the parallel control element. A series resistor drops more or less voltage to keep the output constant.

Control element

VOUTVIN

Sample circuit

Error detector

Reference voltage

R1

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Summary

Voltage regulators

VOUTVIN

+

R3

R2

R1

Q1

R4

Shunt regulators are not as efficient as series regulators, but have the advantage of short circuit protection.

Can you identify each element in this circuit?

Sample circuit

Reference voltage

Control element

Error detector

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Summing amplifier

Averaging amplifier

Scaling adder

Selected Key Terms

An amplifier with several inputs that produces an output voltage proportional to the algebraic sum of the inputs.

A special type of summing amplifier with weighed inputs.

An amplifier with several inputs that produces an output voltage that is the mathematical average of the input voltages.

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Integrator

Differentiator

Active filter

Series regulator

Selected Key Terms

A frequency selective circuit consisting of active devices such as transistors or op-amps combined with reactive (RC) circuits.

A circuit that produces an inverted output that approaches the mathematical integral of the input.

A circuit that produces an inverted output that approaches the mathematical derivative of the input, which is the rate of change.

A type of voltage regulator with the control element in series between the input and output.

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Quiz

1. When an op-amp is configured as a comparator, the gain is equal to

a. 0.

b. 1.

c. a ratio of two resistors.

d. the open-loop gain.

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Quiz

2. The approximate voltage at the inverting input of the op-amp shown is equal to

R1VIN1

VIN2

VIN3

R2

R3

Rf

+

VOUT

10 k

10 k

10 k

3.3 k

a. the average of the input voltages.

b. the sum of the input voltages

c. 0 V

d. OUT

3

V

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Quiz

3. For the scaling adder shown, the input with the greatest weight is

a. VIN1

b. VIN2

c. VIN3

d. they are all equal

R1VIN1

VIN2

VIN3

R2

R3

Rf

+

VOUT

10 k

5.0 k

2.5 k

10 k

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Quiz

4. In a practical integrator, the purpose of the feedback resistor (Rf) is to

a. limit the gain.

b. prevent drift.

c. prevent oscillations.

d. all of the above.R

C

Vin

Vout

+

Rf

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Quiz

5. Assume the top waveform represents the input to a differentiator circuit. Which represents the expected output?

Vin

a.

b.

c.

d.

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Quiz

6. The lead-lag network in a Wien bridge with equal value R’s and C’s attenuates the signal by a factor of

a. 2

b. 3

c. 5

d. 10

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Quiz

7. A Wien-bridge is used to produce

a. sine waves.

b. square waves.

c. triangle waves.

d. all of the above.

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Quiz

8. For the circuit shown, the two outputs (in red) produce

a. sine and square waves.

b. triangle and square waves.

c. sine and triangle waves.

d. sawtooth and triangle waves.

Comparator Integrator+

+

R3

R2

R1

C

Vout

Vout

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Quiz

9. The purpose of the op-amp in the series regulator is

a. to sample the output.

b. to establish a reference.

c. as a control element.

d. error detection.VOUT

+

R3

R2

R1

VIN

Q1

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Quiz

10. An advantage of a shunt regulator is

a. short circuit protection.

b. efficiency.

c. no need for a reference voltage.

d. all of the above.

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Quiz

Answers:

1. d

2. c

3. c

4. b

5. c

6. b

7. a

8. b

9. d

10. a