Transcript Document

electronics fundamentals
circuits, devices, and applications
THOMAS L. FLOYD
DAVID M. BUCHLA
chapter 19
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
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.
+V
An example of a comparison circuit
is shown. The input is compared
with a reference set by the voltagedivider. Notice that there is no
feedback; the op-amp is operated in
open-loop, so the output will be in
saturation.
Electronics Fundamentals 8th edition
Floyd/Buchla
R1
Vin
Vout
+
R2
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
Sketch the output of the comparator in relationship to the input;
assume the maximum output is ±13 V.
+10 V
The threshold is +4.2 V. The output is in
positive saturation when Vin > +4.2 V
+4.2 V
Vin
0V
V = +15 V
R1
10 kW
Vin
-10 V
+13 V
+
Vout
0V
R2
3.9 kW
-13 V
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
Show the output of the comparator for the last example if the inputs
to the op-amp are reversed.
The threshold is still +4.2 V but now
the output is in negative saturation
when Vin > +4.2 V.
+10 V
+4.2 V
Vin
0V
V = +15 V
-10 V
+13 V
R1
10 kW
+
-
Vin
Vout
0V
R2
3.9 kW
-13 V
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
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.
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.
VIN1
VIN2
VIN3
VINn
Rf
R1
R2
-
R3
VOUT
+
Rn
Virtual ground
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
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.
For example, if there
are three input
resistors, each with a
value of 10 kW, then
Rf = 3.3 kW to form an
averaging amplifier.
Electronics Fundamentals 8th edition
Floyd/Buchla
VIN1
VIN2
VIN3
R1
Rf
10 kW
R2
3.3 kW
10 kW
R3
+
VOUT
10 kW
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
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.
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.
Electronics Fundamentals 8th edition
Floyd/Buchla
VIN1
VIN2
VIN3
R1
Rf
10 kW
R2
10 kW
5.0 kW
R3
+
VOUT
2.5 kW
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
Scaling adder
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
VIN1
VIN2
VIN3
R1
Rf
10 kW
R2
10 kW
5.0 kW
R3
+
VOUT
2.5 kW
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
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.
Vin
Rf
C
In practical circuits, a large
value resistor is usually in
parallel with the capacitor
to prevent the output from
drifting into saturation.
R
-
Vout
+
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
Integrators
For the ideal integrator, the rate of change of the
output is given by Vout  - Vin
t
Ri 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).
C
Vin
Vin
0V
R
+
Electronics Fundamentals 8th edition
Floyd/Buchla
Vout
0V
Vout
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
A 5 kHz square wave with 10 Vpp is applied to a
practical integrator. Show the output waveform voltages.
During the positive input (½ the period), the change
in the output is
V
5V
Vout  - in t  100 μs = 5.6 V
Ri C
 2.7 kW  33 nF
Rf
270 kW
C
Vin
R
2.7 kW
33 nF
+
Electronics Fundamentals 8th edition
Floyd/Buchla
Vout
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.
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
continued…
The results of a computer
simulation on Multisim confirm
the calculated change (5.6 V) in
output voltage (blue line).
Rf
270 kW
C
Vin
R
2.7 kW
33 nF
+
Electronics Fundamentals 8th edition
Floyd/Buchla
Vout
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
Differentiators
In mathematics, differentiation is the process of finding
the rate of change. An ideal differentiator circuit is
shown. It produces an inverted output that is
proportional to the rate of change of the input.
In practical circuits, a small value
resistor is added in series with the input
to prevent high frequency ringing.
Vin
Rin Vin
Rf
C
-
Vout
+
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
Differentiators
The output voltage for the ideal differentiator is given
by
 VC 
Vout  -   R f 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.
Rf
Vin
Vin
Vout
Electronics Fundamentals 8th edition
Floyd/Buchla
C
-
Vout
+
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
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
V
Vout  -  C
 t

 10 V 
R
C


  2.7 kW 100 nF  -5.4 V
 f
0.5
ms



By symmetry, when the input has a negative slope, the
Rf
output will be +5.4 V.
Rin
Vin
+5.0 V
Vin
2.7 kW
C
120 W 100 nF
-
Vout
+
0V
-5.0 V
0
Electronics Fundamentals 8th edition
Floyd/Buchla
1 ms
2 ms
See next slide for waveforms…
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
continued…
The results of a computer
simulation on Multisim confirm
the calculated output voltages
(±5.4 V). The output voltage is
the blue line.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
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.
Lead-lag circuit
Negative feedback with
JFET gain control
JFET bias circuit
Electronics Fundamentals 8th edition
Floyd/Buchla
Rf
C1
-
R1
Vout
+
D1
Q1
R2
C2
R3
R4
C3
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
Oscillators
The lead-lag circuit in the Wien-bridge oscillator has a
maximum response at the resonant frequency given by
1
Vin
The lead-lag circuit
fr 
response is…
2πRC
Rf
C
1
Vout
This equation is valid when
R’s and C’s in the lead-lag
circuit are equal.
Because the attenuation is
⅓ at fr, the gain of the
Wien bridge must set for 3.
Electronics Fundamentals 8th edition
Floyd/Buchla
-
R1
Vout
Vout
⅓Vin
+
D1
Q1
R2
C2
R3
R4
fr
f
C3
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
Wien-bridge oscillator
What is the frequency of the bridge?
The frequency is given by
fr 

1
2πRC
1
2π  6.8 kW  47 nF 
 498 Hz
Electronics Fundamentals 8th edition
Floyd/Buchla
Rf
10 kW
C1
47 nF
R1
6.8 kW
-
Vout
+
D1
Q1
R2
6.8 kW
C2
47 nF
R3
1.0 kW
R4
10 kW
C3
1.0 mF
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
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.
C
Vout (square)
+
Comparator
Electronics Fundamentals 8th edition
Floyd/Buchla
R1
Vout (triangle)
R2
R3
+ Integrator
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
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.
R1
Vout
C
Vout
+
R2
R3
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
Active filters
A
selectsthe
certain
frequencies
and excludes
others.
Byfilter
reversing
resistors
and capacitors
in the low-pass
Active
use op-amps
optimize
thefilter
frequency
circuit, filters
a high-pass
filter is to
created.
This
has a gain
response.
A 2-pole where
low-pass
of 1 at frequencies
f > filter
fc. and its response is
shown. The gain for this filter is 1 (0 dB) for f < fc.
Gain (dB)
C1
R1
V Vin
R1
--
R2
++
in
C1
C2
R2
Vout
00
-3
-3
-40 dB/decade
-40 dB/decade
C2
f
fc fc
* -40dB/decade means a 40dB decrease per 10 fold change in frequency
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
Voltage regulators
Voltage regulators are made from integrated circuits. A
basic series IC regulator has four blocks:
Control
element
VIN
Reference
voltage
Electronics Fundamentals 8th edition
Floyd/Buchla
Error
detector
VOUT
Sample
circuit
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
Voltage regulators
A series regulator can use a comparator to compare the
output voltage with a reference voltage. The series
transistor drops more or less voltage to keep the output
constant.
Q1
VIN
Therefore,
atthe
thethe
When VOUTvoltage
exceeds
goes
below
inverting
is forced
reference input
voltage,
the to be
the
samehappens,
as
the exceeds
reference
inverting
reverse
input
causingthe
voltage
by feedback
action
non-inverting
the
op-amp
to input
turn on
voltage,
causing the
transistor
Q1,
op-amp
and the
to turn


R
2 the
off
output
transistor
voltage
Q1,
to
pull
and
VOUT  1 
V
output
voltage
voltage
up.
to R
dip. REF

3
Electronics Fundamentals 8th edition
Floyd/Buchla

VOUT
R1
+
-
R2
R3
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
Voltage regulators
What is the output voltage of the series
regulator?
VIN =
 R 
+24 V
VOUT  1  2 VREF
R1
 R3 
4.7 kW
 6.8 kW 
 1 
 5.1 V
 3.3 kW 
= 15.6 V
Q1
VOUT
+
5.1 V
Electronics Fundamentals 8th edition
Floyd/Buchla
R2
6.8 kW
R3
3.3 kW
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
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.
R1
VIN
VOUT
Reference
voltage
Error
detector
Control
element
Sample
circuit
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
Voltage regulators
Shunt regulators are not as efficient as series
regulators, but have the advantage of short circuit
protection.
R1
VIN
Can you identify
each element in
this circuit?
Reference
voltage
Electronics Fundamentals 8th edition
Floyd/Buchla
VOUT
R2
-
Error
detector
Q1
+
Control
element
R3
Sample
circuit
R4
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
Selected Key Terms
Summing An amplifier with several inputs that produces
amplifier an output voltage proportional to the algebraic
sum of the inputs.
Averaging An amplifier with several inputs that produces
amplifier an output voltage that is the mathematical
average of the input voltages.
Scaling adder A special type of summing amplifier with
weighed inputs.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
Selected Key Terms
Integrator A circuit that produces an inverted output that
approaches the mathematical integral of the
input.
Differentiator A circuit that produces an inverted output that
approaches the mathematical derivative of the
input, which is the rate of change.
Active filter A frequency selective circuit consisting of
active devices such as transistors or op-amps
combined with reactive (RC) circuits.
Series A type of voltage regulator with the control
regulator element in series between the input and output.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
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.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
Quiz
2. The approximate voltage at the inverting input of the
op-amp shown is equal to
a. the average of the input voltages.
b. the sum of the input voltages.
c. 0 V
d. VOUT
3
VIN1
VIN2
VIN3
R1
Rf
10 kW
R2
3.3 kW
10 kW
R3
-
VOUT
+
10 kW
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
Quiz
3. For the scaling adder shown, the input with the greatest
weight is
R1
R
f
VIN1
a.
VIN1
b. VIN2
c.
VIN3
VIN2
VIN3
10 kW
R2
5.0 kW
R3
10 kW
+
VOUT
2.5 kW
d. they are all equal.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
Quiz
4. In a practical integrator, the purpose of the feedback
resistor (Rf) is to
a. limit the gain.
Rf
b. prevent drift.
C
c. prevent oscillations.
d. all of the above.
Vin
R
+
Electronics Fundamentals 8th edition
Floyd/Buchla
Vout
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
Quiz
5. Assume the top waveform represents the input to a
differentiator circuit. Which represents the expected output?
Vin
a.
b.
c.
d.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
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
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
Quiz
7. A Wien-bridge is used to produce
a. sine waves.
b. square waves.
c. triangle waves.
d. all of the above.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
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.
C
Vout
+
Comparator
Electronics Fundamentals 8th edition
Floyd/Buchla
R1
Vout
R2
R3
+ Integrator
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
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.
Q1
VIN
VOUT
R1
+
-
R2
R3
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
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.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 19
1
Quiz
Answers:
Electronics Fundamentals 8th edition
Floyd/Buchla
1. d
6. b
2. c
7. a
3. c
8. b
4. b
9. d
5. c
10. a
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.