Transcript Chapter 9

Electronics
Principles & Applications
Fifth Edition
Charles A. Schuler
Chapter 9
Operational Amplifiers
©1999
Glencoe/McGraw-Hill
INTRODUCTION
• The Differential Amplifier
• The Operational Amplifier
• Determining Gain
• Frequency Effects
• Applications
• Comparators
A differential amplifier driven at one input
+VCC
Noninverted output
Inverted output
C
B
E
C
E
-VEE
B
Both outputs are active because Q1 drives Q2.
Q1 serves as an
emitter-follower
amplifier in this
mode to drive Q2.
+VCC
C
B
Q1 E
Q2 serves as a
common-base
amplifier in this
mode. It’s driven
at its emitter.
C
E Q
2
-VEE
B
A differential amplifier driven at both inputs
Common mode
input signal
+VCC
Reduced output
Reduced output
C
B
E
C
E
-VEE
B
A differential amplifier driven at both inputs
Differential mode
input signal
+VCC
Increased output
Increased output
C
B
E
C
E
-VEE
B
Differential amplifier dc analysis
IRE =
IE =
VEE - VBE
RE
IRE
2
9 V - 0.7 V
= 2.13 mA
=
3.9 kW
VCC
VRL = IC x RL
= 1.06 mA x 4.7 kW
+9 V
= 4.98 V
= 1.06 mA
4.7 kW
RL RL
IC = IE = 1.06 mA
C
B
10 kW
RB
E
3.9 kW
VEE
VCE = VCC - VRL - VE
4.7 kW
= 9 - 4.98 -(-0.7)
C
E
RE
-9 V
= 4.72 V
B
RB
10 kW
Differential amplifier dc analysis continued
Assume b = 200
IB =
IC
b
VB = VRB = IB x RB
1.06 mA
=
200
= 5.3 mA
VCC
4.7 kW
10 kW
RB
+9 V
= 53 mV
RL RL 4.7 kW
C
B
= 5.3 mA x 10 kW
E
3.9 kW
VEE
C
E
RE
-9 V
B
RB
10 kW
Differential amplifier ac analysis
50 mV
50 mV
(50 mV is conservative)
= 47 W
=
rE =
IE
1.06 mA
RL
VCC +9 V
AV(CM) =
RL
2 x RE
AV(DIF) =
2 x rE
4.7 kW
4.7 kW RL RL 4.7 kW
=
4.7 kW
2 x 3.9 kW
=
= 50
2 x 47 W
C
C
= 0.6
B
10 kW
RB
E
3.9 kW
VEE
E
RE
-9 V
B
RB
10 kW
Differential amplifier ac analysis continued
CMRR = 20 x log
AV(DIF)
AV(CM)
VCC
4.7 kW
10 kW
RB
50
= 38.4 dB
0.6
+9 V
RL RL 4.7 kW
C
B
= 20 x log
E
3.9 kW
VEE
C
E
RE
-9 V
B
RB
10 kW
AV(CM) =
A current source can
replace RE to decrease
the common mode gain.
RL
VCC
2 x RE
Replaces this
with a very high
resistance value.
4.7 kW
C
B
10 kW
RL RL 4.7 kW
RB
E
*
C
E
B
RB
10 kW
2 mA
*NOTE: Arrow shows conventional current flow.
A practical current source
IC
9 V - 5.1 V
IZ =
= 10 mA
390 W
IE =
390 W
5.1 V - 0.7 V
2.2 kW
IC = IE = 2 mA
5.1 V
2.2 kW
-9 V
= 2 mA
Differential amplifier quiz
When a diff amp is driven at one input,
the number of active outputs is _____. two
When a diff amp is driven at both inputs, there
is high gain for a _____ signal. differential
When a diff amp is driven at both inputs, there
is low gain for a ______ signal. common-mode
The differential gain can be found by dividing
the collector load by ________. 2rE
The common-mode gain can be found by dividing
the collector load by ________. 2RE
Op amps have
two inputs
Inverting
input
Non-inverting
input
Output
Op-amp Characteristics
•
•
•
•
High CMRR
High input impedance
High gain
Low output impedance
•
•
•
•
Available as ICs
Inexpensive
Reliable
Widely applied
With both inputs grounded through equal
resistors, VOUT should be zero volts.
+VCC
VOUT
-VEE
Imperfections can make VOUT non-zero. The
offset null terminals can be used to zero VOUT.
Dt
DV
741
DV
Slew rate =
Dt
0.5 V
ms
The output of an op amp cannot change instantaneously.
VP
f > fMAX
fMAX =
Slew-rate distortion
Slew Rate
2p x VP
Operational amplifier quiz
The input stage of an op amp is a
__________ amplifier.
differential
Op amps have two inputs: one is inverting
and the other is ________.
noninverting
An op amp’s CMRR is a measure of its ability
to reject a ________ signal.
common-mode
The offset null terminals can be used to zero
an op amp’s __________.
output
The ability of an op amp output to change
rapidly is given by its _________. slew rate
Op-amp follower
AV(OL) = the open loop voltage gain
AV(CL) = the closed loop voltage gain
This is a closed-loop
circuit with a voltage
gain of 1.
RL
It has a high input impedance
and a low output impedance.
Op-amp follower
AV(OL) = 200,000
AV(CL) = 1
The differential input
approaches zero due
to the high open-loop
gain. Using this model,
VOUT = VIN.
VDIF = 0
VIN
VOUT
RL
Op-amp follower
AV(OL) = 200,000
B=1
A
AB +1
VIN
VOUT
The feedback ratio = 1
200,000
@1
AV(CL) =
(200,000)(1) + 1
VIN
VOUT
RL
The closed-loop gain is increased by decreasing
the feedback with a voltage divider.
R1
200,000
AV(CL) =
= 11
(200,000)(0.091) + 1
RF
R1
B=
100 kW
RF + R1
10 kW
=
VIN
VOUT
RL
10 kW
100 kW + 10 kW
= 0.091
It’s possible to develop a different
model for the closed loop gain
by assuming VDIF = 0.
R1
VIN = VOUT x
R1 + RF
RF
Divide both sides by
VOUT and invert:
100 kW
R1 10 kW
VOUT
VDIF = 0
VIN
VOUT
VIN
RF
=1+
R1
RL
AV(CL) = 11
In this amplifier, the assumption VDIF = 0 leads
to the conclusion that the inverting op amp terminal is
also at ground potential. This is called a virtual ground.
Virtual ground
RF
We can ignore the op amp’s input
current since it is so small. Thus:
IR1 = IRF
10 kW
R1
By Ohm’s Law:
1 kW
VIN
VDIF = 0
VIN
R1
VOUT
RL VOUT
VIN
=
=
-VOUT
RF
-RF
R1
= -10
The minus sign designates an inverting amplifier.
Due to the virtual ground, the input impedance
of the inverting amplifier is equal to R1.
Virtual ground
RF
10 kW
R1
1 kW
VDIF = 0
VIN
Although op amp input
currents are small, in
some applications, offset
error is minimized by
providing equal paths for
the input currents.
R2 = R1 || RF = 910 W
This resistor reduces offset error.
A typical op amp has
internal frequency
compensation.
R
Output
C
Break frequency:
1
fB =
2pRC
Bode plot of a typical op amp
Break frequency
120
100
80
60
Gain in dB
40
20
0
1
10 100 1k 10 k 100 k 1M
Frequency in Hz
Op amps are usually operated with negative feedback
(closed loop). This increases their useful frequency range.
AV(CL) =
RF
VIN
RF
=1+
R1
100 kW
=1+
= 101
1 kW
100 kW
R1 1 kW
VOUT
dB Gain = 20 x log 101 = 40 dB
VIN
VOUT
RL
Using the Bode plot to find closed-loop bandwidth:
120
100
Gain in dB
80
Break frequency
60
AV(CL)
40
20
0
1
10 100 1k 10 k 100 k 1M
Frequency in Hz
A 741 op amp slews at
0.5 V
ms
70
V
A 318 op amp slews at
ms
There are two frequency limitations:
Slew rate determines the large-signal bandwidth.
Internal compensation sets the small-signal bandwidth.
The Bode plot for a fast op amp shows
increased small-signal bandwidth.
120
100
80
Gain in dB 60
40
fUNITY
20
0
1
10 100 1k 10 k 100 k 1M 10M
Frequency in Hz
fUNITY can be used to find the small-signal bandwidth.
AV(CL) =
RF
VIN
RF
=1+
R1
100 kW
=1+
= 101
1 kW
100 kW
R1 1 kW
VIN
VOUT
VOUT
318 Op amp
RL
fUNITY
fB =
AV(CL)
10 MHz
=
= 99 kHz
101
Op amp feedback quiz
The open loop gain of an op amp is reduced
with __________ feedback
negative
The ratio RF/R1 determines the gain of the
___________ amplifier.
inverting
1 + RF/R1 determines the gain of the
___________ amplifier.
noninverting
Negative feedback makes the - input of the
inverting circuit a ________ ground. virtual
Negative feedback _________ small signal
bandwidth.
increases
R
Amplitude response
of RC lag circuit
Vout
C
1
fb =
2pRC
fb
0 dB
-20 dB
Vout
-40 dB
-60 dB
10fb
100fb
1000fb
f
R
Phase response
of RC lag circuit
 = tan
-1
Vout -45o
-90o
C
-XC
R
0.1fb
0o
Vout
fb
10fb
f
Interelectrode capacitance and Miller effect
The gain from
base to collector
makes CBC
effectively larger C
BE
in the input circuit.
CBC
R
CMiller = AVCBC
CInput = CMiller + CBE
CMiller
CBE
1
fb =
2pRCInput
Bode plot of an amplifier
with two break frequencies.
50 dB
40 dB
20 dB/decade
30 dB
20 dB
40 dB/decade
10 dB
0 dB
10 Hz
100 Hz
1 kHz
fb1
10 kHz 100 kHz
fb2
Multiple lag circuits:
Vout
R1
C1
R2
C2
R3
0o
Vout
Phase reversal
-180o
Negative feedback becomes positive
C3
f
Op amp compensation
• Interelectrode capacitances create several
break points.
• Negative feedback becomes positive at
some frequency due to cumulative phase
lags.
• If the gain is > 0 dB at that frequency, the
amplifier is unstable.
• Frequency compensation reduces the
gain to 0 dB or less.
Op amp compensation quiz
Beyond fb, an RC lag circuit’s output drops
at a rate of __________ per decade. 20 dB
The maximum phase lag for one RC network
is __________.
90o
An interelectrode capacitance can be effectively
much larger due to _______ effect. Miller
Op amp multiple lags cause negative feedback
to be ______ at some frequency. positive
If an op amp has gain at the frequency where
feedback is positive, it will be ______. unstable
Inverted sum of three sinusoidal signals
RF
10 kW
5 kW
5 kHz
3.3 kW
3 kHz
1 kW
1 kHz
Summing Amplifier
Amplifier scaling:
1 kHz signal gain is -10
3 kHz signal gain is -3
5 kHz signal gain is -2
Difference of two
sinusoidal signals
(V1 = V2)
RF
1 kW
Subtracting Amplifier
(A demonstration of
common-mode rejection)
1 kW
1 kW
1 kW
V1
V2
VOUT = V2 - V1
Active low-pass filter
VIN
-3 dB
Gain
Frequency
fC
VOUT
Active high-pass filter
VIN
-3 dB
Gain
fC
Frequency
VOUT
VIN
VOUT
Active band-pass filter
-3 dB
Gain
Frequency
Bandwidth
VOUT
VIN
Active band-stop filter
-3 dB
Gain
Frequency
Stopband
Integrator
C
R
VIN
V
Slope = s
VOUT
1
Slope = -VIN x
RC
Comparator with a 1 Volt reference
+VSAT
1V
0V
-VSAT
VOUT
VIN
1V
Comparator with a noisy input signal
+VSAT
1V
0V
-VSAT
VOUT
VIN
1V
Schmitt trigger with a noisy input signal
+VSAT
UTP
LTP
Trip points:
R1
VSAT x
R1 + RF
-VSAT
VIN
R1
VOUT
RF
Hysteresis = UTP - LTP
+5 V
Window comparator
4.7 kW
VUL
3V
R1
311
VOUT
R2
4.7 kW
VIN
311
VLL
1 V VOUT is LOW (0 V) when VIN
is between 1 V and 3 V.
Window comparator
VUL
3V
+5 V
311
VOUT
VIN
311
VLL
1V
Many comparator ICs
require pull-up resistors in
applications of this type.
+5 V
Window comparator
4.7 kW
VUL
3V
R1
311
VOUT
R2
4.7 kW
VIN
311
VLL
1V
VOUT is TTL logic
compatible.
Op amp applications quiz
A summing amp with different gains for the
inputs uses _________.
scaling
Frequency selective circuits using op amps
are called _________ filters.
active
An op amp integrator uses a _________ as
the feedback element.
capacitor
A Schmitt trigger is a comparator with
__________ feedback.
positive
A window comparator output is active when
the input is ______ the reference points. between
REVIEW
• The Differential Amplifier
• The Operational Amplifier
• Determining Gain
• Frequency Effects
• Applications
• Comparators