Transcript Lecture 7

Analog Electronics
Lecture 7
Op-amp Circuits and Active Filters
Electronic Devices, 9th edition
Thomas L. Floyd
Muhammad Amir Yousaf
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Lecture:
 How to compare an analog signal with certain voltage level.
Comparing a noisy signal with certain (reference) level.
Binding an signal to fixed +/- max levels.
Analog to digital converters with comparators.
Adding two analog signals.
Adding weighted signals.
Averaging on analog signals.
 Digital to Analog Converter with weighted additions.
Integrating an analog waveform.
Differentiating analog waveform.
Logarithm on analog signal.
Antilog of analog signal.
Multiplying and diving analog signals.
Converters.
Peak Detectors.
Filters.
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Comparators
 A comparator is a specialized nonlinear op-amp circuit that compares two
input voltages and produces an output state that indicates which one is greater.
 Because of the high open-loop voltage gain, a very small difference voltage
between the two inputs drives the amplifier into saturation.
Zero Level Detection:
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Comparators
Non-Zero Level Detection:
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Noise on Comparator
Noise contaminated signal may cause an unstable output.
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Comparator with Hysteresis
To avoid this, hysteresis can be used.
Hysteresis is incorporated by adding regenerative (positive) feedback, which
creates two switching points:
The upper trigger point (UTP) and the lower trigger point (LTP).
After one trigger point is crossed, it becomes inactive and the other one
becomes active.
VUTP
Vin 0
t
VLTP
+Vout (max)
–Vout(max)
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Output Bounding
Some applications require a limit to the output of the
comparator (such as a digital circuit). The output can be
limited by using one or two Zener diodes in the feedback
circuit.
The circuit shown here is bounded as a positive value equal to
the zener breakdown voltage.
Vin
Ri
0V
+VZ
–
0
+
Electronic Devices, 9th edition
Thomas L. Floyd
–0.7 V
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Comparator Applications
 Flash analog-to-digital converters use
2n-1 comparators to convert an analog
input to a digital value of n bits for
processing.
VREF
R
Vin
(analog)
Op-amp
comparators
+
–
R
+
–
 Flash ADCs are a series of
comparators, each with a slightly
different reference voltage.
R
R
R
 The priority encoder produces an output
equal to the highest value input.
–
(7)
(6)
+
(5)
(4)
–
(3)
(2)
+
(1)
(0)
–
R
+
–
R
Priority
encoder
+
D2
D1
D0
Binary
output
Enable
input
+
–
R
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Summing Amplifier
A summing amplifier has two or more inputs; normally all inputs have
unity gain. The output is proportional to the negative of the algebraic sum
of the inputs.
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Example
Summing Amplifier
What is VOUT if the input voltages are +5.0 V, -3.5 V and +4.2 V and all
resistors = 10 kW?
Rf
R1
VOUT = -(VIN1 + VIN2 + VIN3)
= -(+5.0 V - 3.5 V + 4.2 V)
= -5.7 V
Electronic Devices, 9th edition
Thomas L. Floyd
VIN1
R2
VIN2
10 kW
–
R3
VIN3
VOUT
+
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Averaging Amplifier
An averaging amplifier is basically a summing amplifier with the gain
set to Rf /R = 1/n (n is the number of inputs). The output is the negative
average of the inputs.
What is VOUT if the input voltages are +5.0 V, -3.5 V and +4.2 V? Assume
R1 = R2 = R3 = 10 kW and Rf = 3.3 kW?
Rf
R1
VIN1
R2
VOUT = -⅓(VIN1 + VIN2 + VIN3)
= -⅓(+5.0 V - 3.5 V + 4.2 V)
= -1.9 V
Electronic Devices, 9th edition
Thomas L. Floyd
VIN2
3.3 kW
–
R3
VIN3
VOUT
+
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Scaling Adder
A scaling adder has two or more inputs with each input having a different
gain.
Rf
R1
VIN1
R2
VIN2
–
R3
VIN3
Electronic Devices, 9th edition
Thomas L. Floyd
VOUT
+
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Scaling Adder: D/A Converter
An application of a scaling adder is the D/A converter circuit shown here. The
resistors are inversely proportional to the binary column weights. Because of the
precision required of resistors, the method is useful only for small DACs.
+V
8R
20
Rf
4R
21
2R
–
VOUT
2
2
+
R
23
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
The Integrator
An op-amp integrator simulates mathematical
integration, a summing process that determines total
area under curve.
I
dQ
dt
 C.
Ii  - If
Vin
dVout
 -C.
R
dt
dVout
1 Vin
.
dt
RC
Vout= −
C
R
–
Vin
Ii
dV
dt
IC
Vout
+
Ideal
Integrator
1 𝑡
𝑉 𝑑𝑡
𝑅𝐶 0 𝑖𝑛
The ideal integrator is an inverting amplifier that
has a capacitor in the feedback path. The output
voltage is proportional to the negative integral
(running sum) of the input voltage.
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
The Integrator
Capacitor in the ideal integrator’s feedback is
open to dc.
Rf
C
This implies open loop gain with dc offset.
That would lead to saturation.
The practical integrator overcomes these
issues– the simplest method is to add a
relatively large feedback resistor.
R
Vin
–
Vout
+
Practical
Integrator
Rf should be large enough
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
The Differentiator
R
C
An op-amp differentiator simulates mathematical
differentiation, a process to determine instantaneous
rate of change of a function.
Vin
–
Vout
+
Ideal
Differentiator
The ideal differentiator is an inverting amplifier that has a capacitor in the input
path. The output voltage is proportional to the negative rate of change of the input
voltage.
I=
𝑑𝑄
𝑑𝑡
= 𝐶.
𝑑𝑉
𝑑𝑡
Ii  - If
𝑑𝑉𝑖𝑛
𝑉𝑜𝑢𝑡
𝐶.
=−
𝑑𝑡
𝑅
𝑑𝑉𝑖𝑛
𝑑𝑡
𝑉𝑜𝑢𝑡 = -RC.
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Instrumentation Amplifiers
An instrumentation amplifier (IA) amplifies the voltage difference
between its terminals.
It is optimized for amplifying small differential signals that may be
riding on a large common mode voltages.
o High input impedance
o High CMMR
o Low output offset
Input 1
+
R3
R5
A1
Gain set
–
R1
–
o Low output impedance
R2
Gain set
A3
Output
+
–
R4
A2
Input 2
Electronic Devices, 9th edition
Thomas L. Floyd
+
R6
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Instrumentation Amplifiers
IC of instrumentation amplifier is made up of three op amps and several
resistors.
The gain is set by a single
resistor that is supplied
by the user.
+ Vcm
Vin1Input
1
+
R3
R5
A1
Gain set
The output voltage is the
closed loop gain set by
RG multiplied by the
voltage difference in the
inputs.
Electronic Devices, 9th edition
Thomas L. Floyd
–
R1
–
R2
RG
Gain set
+
Output
+
–
R4
A2
+ Vcm
Vin2Input
2
A3
Vout = Acl (Vin2 - Vin1)
R6
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Instrumentation Amplifiers (IA)
Applications:
oUsed where a quantity is sensed by
a remote sensor e.g. temperature,
pressure transducer and sensed signal
is sent over a long line.
oElectrical noise produces common-mode voltages in the line.
oIA at the end of line amplifies only the small differential
signal and reject the common mode signal
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Example
Instrumentation Amplifiers
An IA that is based on the three op-amp design is the AD622.
The formula for choosing RG is:
RG 
50.5 kW
Av - 1
+V
(7)
What value of RG will set the
gain to 35?
RG 
50.5 kW 50.5 kW

Av - 1
35 - 1
= 1.5 kW
Electronic Devices, 9th edition
Thomas L. Floyd
+IN
(3)
(1)
RG
–IN
(6)
AD622
(8)
(5)
(2)
Output
REF
(Output signal
common)
(4)
–V
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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The Logarithmic Amplifier
A logarithmic (log) amplifier produces an output that
is proportional to the logarithm of the input
Log and antilog amplifiers are used in applications that require:
o Compression of analog input data.
o Linearization of transducers that have exponential outputs.
o Analog multiplication and division, etc
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
The Logarithmic Amplifier
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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The Logarithmic Amplifier
A semiconductor pn-junction in the form of either a diode or the baseemitter junction of a BJT provides a logarithmic characteristic.
Voltage across the diode is proportional to the log of the current in the
diode. Compare data for an actual diode on linear and logarithmic
plots:
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
The Logarithmic Amplifier
When a diode is placed in the feedback path of an inverting op-amp,
the output voltage is proportional to the log of the input voltage. The
gain decreases with increasing input voltage; therefore the amplifier is
I
I
said to compress signals.
F
in
Vin
+V –
F
R1
0V
–
Op-amp
Vout
+
Many sensors, particularly photosensors, have a very large dynamic
range outputs. Current from
photodiodes can range over 5 decades.
A log amp will amplify the small
current more than the larger current to
effectively compress the data for further
processing.
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Example
The Logarithmic Amplifier
For the circuit shown, the equation for Vout is
Vout  -  0.025 V  ln
Vin
(IR is a constant for a given diode.)
I R R1
What is Vout? (Assume IR = 50 nA.)
Vin
Vin
+11 V
Iin
R
1
IF
++ V ––
VFF
1.0R1kW
0V
Vout  -  0.025 V  ln
11 V
 50 nA 1.0 kW
––
Op-amp
Op-amp
VVout
out
++
= -307 mV
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
The Antilog Amplifier
The antilogarithm of a number is the result obtained when the base is raised to
a power equal to the logarithm of that number.
x= 𝑒 𝑙𝑛𝑥
Ii  - If
I R𝑒
Electronic Devices, 9th edition
Thomas L. Floyd
𝑞𝑉𝑖𝑛/𝑘𝑇
=
𝑉𝑜𝑢𝑡
−
𝑅𝐹
𝑉𝑜𝑢𝑡 = - 𝑅𝐹 . IR𝑒 𝑞𝑉𝑖𝑛/𝑘𝑇
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Constant-current source
A constant-current source delivers a load current that remains
constant when the load resistance changes.
IL = Ii
Ri
–
+
VIN
Ii
RL
0V 0A
+
–
A basic circuit in which a stable
voltage source (Vin)
provides a constant current
(Ii) through the input resistor (Ri)
If RL changes, IL remains constant as long as
Vin and Ri are held constant.
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Current to Voltage Converter
A current-to-voltage converter converts a variable input current
to a proportional output voltage.
A specific application of this circuit is where a photoconductive cell
is used to sense changes in light level. As the amount of light
changes, the cur-rent through the photoconductive cell varies
because of the cell’s change in resistance. This change in resistance
produces a proportional change in the output voltage.
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Peak Detector
The circuit is used to detect the peak of the input voltage and
store that peak voltage on a capacitor.
Ri
Vin
+
–
R1
Vout
C
The circuit can be used to detect and store the
maximum value of a voltage surge.
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Charge Sensitive Amplifier
It is used in Radiation detection
Charge on a photon is
accumulated in the capacitor
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Active Filters
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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Basic filter Responses
A filter is a circuit that passes certain frequencies and rejects all others.
The passband is the range of frequencies allowed through the filter.
The critical frequency defines the end (or ends) of the passband and
is normally specified at the point where the response drops -3dB (70.7%)
from the passband response.
Following the passband is a region called the transition region that
leads into a region called the stopband.
Gain
Gain
Gain
f
f
Low-pass
Electronic Devices, 9th edition
Thomas L. Floyd
High-pass
Gain
f
Band-pass
f
Band-stop
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The Basic Low-Pass Filter
The low-pass filter allows frequencies below the critical
frequency to pass and rejects other. The simplest low-pass
filter is a passive RC circuit with the output taken across C.
BW = fc
Gain (normalized to 1)
–3 dB 0 dB
Actual response of a
single-pole RC filter
Passband
–20 dB
Transition
region
–40 dB
–60 dB
0.01 fc
Electronic Devices, 9th edition
Thomas L. Floyd
R
–2
0d
B/
de
Stopband
ca
de
region
BW
Vs
Vout
C
f
0.1 fc
fc
10 fc
100 fc
1000 fc
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
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The Basic Low-Pass Filter
o The ideal response is not attainable by any practical filter.
o Actual filter responses depend on the number of poles,
o Pole, a term used with filters to describe the number of RC circuits
contained in the filter.
o This basic RC filter has a single pole, and it rolls off at -20db/decade
beyond the critical frequency.
o20db/decade means that at a
frequency of 10fc the output will be 20dB(10%) of the input.
o This roll-off allows too much
unwanted frequencies through the
filter
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
The Basic Low-Pass Filter
o Actual filters do not have a perfectly flat response up to the cutoff
frequency.
o More steeper response cannot be obtained by simply cascading the
basic stages due to loading effect.
o With combination of op-amps, the filters can be designed with higher
roll-offs
o In general, the more poles the filter
uses, the steeper its transition region
will be. The exact response depends
on the type of filter and the number of
pole.
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Active Filters
General Active Filters
A single pole active filters
The number of filter poles can be increases with cascading
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
The Basic High-Pass Filter
The high-pass filter passes all frequencies above a critical
frequency and rejects all others. The simplest high-pass
filter is a passive RC circuit with the output taken across R.
Gain (normalized to 1)
–3 dB
0 dB
–20 dB
–40 dB
–60 dB
0.001 fc
Electronic Devices, 9th edition
Thomas L. Floyd
Actual response
of a single-pole
RC filter
0d
–2
Passband
C
de
ca
e
d
B/
Vs
Vout
R
f
0.01 fc
0.1 fc
fc
10 fc
100 fc
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
The Band-Pass Filter
A band-pass filter passes all frequencies between two critical
frequencies. The bandwidth is defined as the difference
between the two critical frequencies fc1 and fc2. The simplest
band-pass filter is an RLC circuit.
R
Vout
Bandwidth B.W= fc2 – fc1
Vs
Center frequency fo= √ fc1 fc2
Quality Factor:
In band pass filters it is ratio of center
frequency to its bandwidth.
Q = fo /B.W
C
L
Vout (normalized to 1)
1
0.707
BW
f
fc1
Electronic Devices, 9th edition
Thomas L. Floyd
f0
fc2
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The Band-Stop Filter
A band-stop filter rejects frequencies between two critical
frequencies; the bandwidth is measured between the critical
frequencies. The simplest band-stop filter is an RLC circuit.
Gain (dB)
0
–3
L
C
Vs
fc1
f0
fc2
Vout
R
f
BW
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Ideal vs Real Filters
In comparison to ideal low pass filters, the
real RC or RLC filters lack the following
characteristics:
Vout (normalized to 1)
1
o Flat passband
o Sharp transition region
oLinear phase response
0.707
BW
f
fc1
f0
fc2
Gain (dB)
0
–3
fc1
f0
fc2
f
BW
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Active Filters
Active filters include one or more op-amps in the design.
One of the three characteristic can be achieved with active
filters:
o Flat band pass with Butterworth
Av
Chebyshev: rapid roll-off characteristic
oSharp roll-off rate with Chebyshev
oLinear phase
response.
Butterworth: flat amplitude response
Bessel: linear phase response
f
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
Electronic Devices, 9th edition
Thomas L. Floyd
© 2012 Pearson Education. Upper Saddle River, NJ, 07458.
All rights reserved.
ID (mA)
ID (mA)
10
8.0
7.0
6.0
1.0
5.0
4.0
0.1
3.0
2.0
0.01
1.0
0
0
0.1 0.2
Electronic Devices, 9th edition
Thomas L. Floyd
0.3
0.4 0.5 0.6 0.7 0.8
VD (V)
0.001
0
0.1 0.2
0.3 0.4 0.5 0.6 0.7 0.8
VD (V)
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