operational amplifiersrhabash/elg4139ln112.pdfto solve ideal op-amp circuit • if the noninverting...
TRANSCRIPT
1
Operational Amplifiers
• Introduction of Operational Amplifier (Op-Amp)
• Characteristics of an Op-Amp
• Comparison of ideal and Non-ideal Op-Amp
• Feedback and Non-Feedback
• Configurations
• Gain; Input Impedance; Output Impedance
• Analysis of ideal Op-Amp Circuits.
2
What is an Operational Amplifier?
• Operational amplifier is an amplifier whose output voltage is
proportional to the negative of its input voltage and that boosts the
amplitude of an input signal, many times, i.e., has a very high
gain.High-gain amplifiers.
• They were developed to be used in synthesizing mathematical
operations in early analog computers, hence their name.
• Typified by the series 741 (The integrated circuit contains 8-pin mini-
DIP, 20 transistors and 11 resistors).
• Used for amplifications, as switches, as filters, as rectifiers, and in
digital circuits.
• Take advantage of large open-loop gain.
• It is usually connected so that part of the output is fed back to the
input.
• Can be used with positive feedback to produce oscillation.
Basics of Differential Amplifier Model
Op Amp is represented by:
A = open-circuit voltage gain
vid = (v+-v-) = differential input signal
voltage
Rid = amplifier input resistance
Ro = amplifier output resistance
The output signal of the amplifier is in
phase with the signal applied at the +
input (non-inverting).
The output signal of the amplifier is 180°
out of phase with the signal applied at the
- input (inverting) terminal.
4
Characteristics of Operational Amplifier
• Very high differential gain
• High input impedance
• Low output impedance
• Wide range of applications:
oscillators, filters and
instrumentation
• Accumulate a very high gain by
cascading multiple stages.
ddo VGV
510 than moresay large,y ver
normally gain aldifferenti : dG
Op-Amp: Power Supply Connections
• Commonly used: dual power supplies
• Common values:
– +15V (+V or V+)
– -15V (-V or V-)
• All output loads connected between output terminal and common ground point
• Usually power supply connections are omitted for ease
Common
ground
point 15 v
15 v
Positive power
supply terminal
Negative power
supply terminal
+
+
-
-
6
The ideal Op-Amp
• Infinite Voltage Gain
– a voltage difference at the two inputs is magnified infinitely
– something like 200,000, means difference between + terminal and terminal is amplified by 200,000!
• Infinite Input Impedance
– no current flows into both inputs
– about 1012 for FET input op-amps
• Zero Output Impedance
– rock-solid independent of load
– roughly to current maximum (usually 5–25 mA)
• Infinitely Fast (Infinite Bandwidth)
– limited to few MHz range
– slew rate limited to 0.5–20 V/s
7
Figure 8.2,
8.3
A Voltage Amplifier Simple Voltage Amplifier Model
inLSinS
Lout
L
inS
inL
Lout
LinLS
inS
inin
AvvvvvRR
R
RR
RAv
RR
RAvvv
RR
Rv
;;
;
8
Op-Amp as an Integrated Circuit
• The integrated circuit operational amplifier evolved soon after
development of the first bipolar integrated circuit.
• The A-709 was introduced by Fairchild Semiconductor in 1965.
• Since then, a vast array of op-amps with improved characteristics,
using both bipolar and MOS technologies, have been designed.
• Most op amps are inexpensive (less than a dollar) and available from a
wide range of suppliers.
• There are usually 20 to 30 transistors that make up an op-amp circuit.
• From a signal point of view, the op-amp has two input terminals and
one output terminal as shown in the following figures.
• The ideal op-amp senses the difference between two input signals and
amplifies the difference to produce an output signal.
• Ideally, the input impedance is infinite, which means that the input
current is zero. The output impedance is zero.
9
Figure 8.4
Operational Amplifier Model Symbols and Circuit Diagram
Op-Amp Stages with Pin-outs of IC741
2
4
7
6
3
11
Ideal Versus Real Op-Amps
Characteristics Ideal Op-Amp Typical Op-Amp
Input Resistance Infinity 106 (bipolar)
109 - 1012 (FET)
Input Current 0 10-12 – 10-8 A
Output Resistance 0 100 – 1000
Operational Gain Infinity 105 - 109
Common Mode Gain 0 10-5
Bandwidth Infinity Attenuates and phases at high
frequencies (depends on slew rate)
Temperature Independent Bandwidth and gain
http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/opampcon.html#c1
12
Single Inputs Op-Amp
• + Terminal : Source
• – Terminal : Ground
• 0o Phase change
• + Terminal : Ground
• – Terminal : Source
• 180o Phase change
13
Double Inputs Op Amp
• Differential input
•
• 0o phase shift change
between Vo and Vd
VVVd
Queation: What Vo should be if,
V1
V2
(A) (B)
Ans: (A or B) ?
14
Distortion / Saturation
• The output voltage never exceeds the DC voltage supply of the Op-Amp.
• Practical Op Amps have limited output voltage and current ranges.
• Voltage: Usually limited to a few volts less than power supply span.
• Current: Limited by additional circuits (to limit power dissipation or protect
against accidental short circuits).
15
A Practical Application: Why Feedback
• Self-balancing mechanism, which allows the amplifier to
preserve zero potential difference between its input
terminals.
• A practical example that illustrates a common application of
negative feedback is the thermostat. This simple
temperature control system operates by comparing the
desired ambient temperature and the temperature measured
by the thermometer and turning a heat source on and off to
maintain the difference between actual and desired
temperature as close to zero as possible.
Impedances
RIN
ROUT
VIN AVIN VOUT
• The amplifier measures voltage across RIN, then generates a voltage
which is larger by a factor A
• This voltage generator, in series with the output resistance ROUT, is
connected to the output port.
• A should be a constant (gain is linear)
Add an input - a source voltage VS plus source impedance RS
Impedances
RIN
ROUT
VIN AVIN VOUT
Note the voltage divider RS + RIN.
VIN=VS(RIN/(RIN+RS)
We want VIN = VS regardless of source impedance
So want RIN to be large.
The ideal amplifier has an infinite input impedance!
VS
RS
Add a load - an output circuit with a resistance RL
Impedances
Note the voltage divider ROUT + RL.
VOUT=AVIN(RL/(RL+ROUT))
Want VOUT=AVIN regardless of load
We want ROUT to be small.
The ideal amplifier has zero output impedance!
RIN
ROUT
VIN AVIN VOUT VS
RS
RL
Feedback Feedback refers to connecting the output of the amplifier to
its input
19
20
Op-Amps for Math
• Inverting
• Non-Inverting
• Summing
• Differencing
• Integrating
• Differentiating
21
Figur
e 8.5
Common Op-Amp Configurations
Inverting Amplifier
SvSR
FRoutv
FRvA
outv
FR
outv
SRvA
outv
SR
Sv
vviniFiSi
iniFR
voutvFi
SR
vSvSi
iniFiSi
;0;
0;;
Voltage Gain of Inverting Amplifier
• The negative voltage gain
implies that there is a 1800 phase
shift between both dc and
sinusoidal input and output
signals.
• The gain magnitude can be
greater than 1 if R2 > R1
• The gain magnitude can be less
than 1 if R1 > R2
•The inverting input of the Op
Amp is at ground potential
(although it is physically
connected to ground) and is said
to be at virtual ground.
0ov22
i1
isv RRs
1
svsi
R
But is= i2 and v- = 0 (since vid= v+ - v-= 0)
and
1
2
svov
R
R
vA
Input and Output Resistances
Rin
vsis
R1 since v0
Rout is found by applying a test current
(or voltage) source to the amplifier
output and determining the voltage (or
current) after turning off all
independent sources. Hence, vs = 0
11i
22ixv RR
But i1=i2
)12
(1ixv RR
Since v- = 0, i1=0. Therefore vx = 0
irrespective of the value of ix .
0out
R
Inverting Amplifier: An Example
• Problem: Design an inverting amplifier for
• Given Data: Av= 20 dB, Rin = 20k,
• Assumptions: Ideal op amp
• Analysis: Input resistance is controlled by R1 and voltage gain is set
by R2 / R1.
and Av = -100
A minus sign is added since the amplifier is inverting.
Av dB
20log
10Av
, Av
1040dB/20dB100
k201 in
RR
AvR
2R1
R2100R
12M
25
Design Example: Design an inverting amplifier with a closed loop
voltage of Av = -5. Assume the op-amp is driven by a sinusoidal soorce, vs
= 0.1 sin t volts, which has a source resistance of Rsr = 1 k and which
supply a maximum current of 5 A. Assume that the frequency is low.
k 100205)1(52y Accordingl
.51
2 and k 19 be should 1
kΩ 206105
1.0
(max)
(max)(min)1 write then weA,5 max)( If
1Therefore .resistance source therepresents
.1 :sresistance twomeans example in this (
RsrRR
RsrR
RvAR
si
svsrRRsi
RsrR
svsisrR
srRRsRsRsR
svsi
26
To Solve Ideal Op-Amp Circuit
• If the noninverting terminal of the op-amp is at ground potential, then
the inverting terminal is at virtual ground. Sum currents at this point,
assuming zero current enters the op-amp itself.
• If the noninverting terminal of the op-amp is not at ground potential,
then the inverting terminal voltage is equal to that at the noninverting
terminal. Sum currents at the inverting terminal node, assuming zero
current enters the op-amp itself.
• For an ideal op-amp circuit, the output voltage is determined from
either step 1 or step 2 above and is independent of any load connected
to the output terminal.
27
Inverting Amplifier with a T-Network
R1
R2 R3
vx
R4
0
0 -
+ vo
i1
i2 i3
i4
vs
)2
3
4
31(1
2
get weequations above theCombing
342;342
)1
2(220
R
R
R
R
R
R
sv
ovvA
R
ovxv
R
xv
R
xviii
R
RsvRixv
28
Design Example: An op-amp with a T-network is to be used as a preamplifier for a
microphone. The maximum microphone output voltage is 12 mV (rms) and the
microphone has an output resistance of 1 k. The op-amp circuit is to be designed
such that the maximum output voltage is 1.2 V (rms). The input amplifier resistance
should be fairly large but all resistance values should be less than 500 k.
k 1.384R and k 40032
k 50 be will)effective 1( resistance total then thek 1 sr and k 49 1set weIf
ncalculatio in the resistance source theof value theinclude should We
5.104
3;8)4
31(8100
81
3
1
2 choose weIf
1
3)4
31(1
2
100012.0
2.1
RR
RRR
R
R
R
R
R
R
R
R
R
R
R
R
R
RvA
vA
29
Figure
8.7
Summing Amplifier
N
n
S
S
Fout
F
outF
S
S
n
FN
n
n
n
n
vR
Rv
R
vi
NnR
vi
iiii
1
21
,...2,1......
...
30
Figu
re
8.8,
8.9
Non-inverting Amplifier Voltage Follower
S
F
S
out
R
R
v
v1
outSvv
Non-inverting Amplifier
The input signal is applied to the non-inverting input terminal.
A portion of the output signal is fed back to the negative input terminal.
Analysis is conducted by relating the voltage at v1 to input voltage vs and
output voltage vo.
Non-inverting Amplifier Voltage Gain, Input Resistance and Output Resistance
Since i-= 0 and
But vid = 0
Since i+=0
21
1ov
1v
RR
R
1v
idvsv
1vsv
1
211
21
svov
1
21svov
R
R
R
RRvA
R
RR
i
sv
inR
Rout is found by applying a test current source to the amplifier output
after setting vs = 0. It is identical to the output resistance of the inverting
amplifier i.e. Rout = 0.
Non-inverting Amplifier: An Example
• Problem: Determine the output voltage and current for the given non-
inverting amplifier.
• Given Data: R1= 3k, R2 = 43k, vs= +0.1 V
• Assumptions: Ideal op amp
• Analysis:
Since i-=0,
Av1R
2R1
143k
3k15.3
voAvvs(15.3)(0.1V)1.53V
A3.33k3k43
V53.1
12
ovoi
RR
34
Design Example: Design a noninverting amplifier with a closed loop gain
of Av = 5. The output voltage is limited to -10 V vo +10 V and the
maximum current in any resistor is limited to 50 A
Answer: R1 = 40 k, R2 = 160 k
The Unity Gain Amplifier or “The Buffer”
Buffer is a special case of the non-inverting amplifier with infinite R1 and zero R2. Hence Av = 1.
It provides an excellent electrical isolation while maintaining the signal
voltage level. The ideal buffer requires no input current and can drive any desired load resistance without loss of signal voltage.
Buffer is used in many data acquisition system applications.
36
Figure
8.10
Differential Amplifier
)(12
1
2
2
21
2
21
1
vvR
Rv
vvRR
Rv
vv
R
vv
R
vv
out
out
Differential Amplifier Using Op-Amp
v v
11
1
v vi
R
01
2
v vi
R
+
-1R
2R
1vov
v
v1i
1i
2v
1R
2R
22
1 2
Rv v
R R
01
1 2
v vv v
R R
2 21 2 2 0
1 2 1 2
1 2
R Rv v v v
R R R R
R R
Differential Amplifier Using Op-Amp
+
-1R
2R
1vov
v
v1i
1i
2v
1R
2R
2 21 2 2 0
1 2 1 2
1 2
R Rv v v v
R R R R
R R
2
2 2 20 1 2 2
1 1 2 1 1 2
R R Rv v v v
R R R R R R
2 2 20 1 2
1 1 2 1
1R R R
v v vR R R R
20 2 1
1
Rv v v
R
Differential Amplifier Applications
• Very useful if you have two inputs corrupted with the same noise
• Subtract one from the other to remove noise, remainder is signal
• Applications: Electrocardiagram to measure the potential difference
between two points on the body
The AD624AD is an instrumentation amplifier - this is a high gain, dc coupled
differential amplifier with a high input impedance and high CMRR (the chip
actually contains a few Op Amps)
http://www.picotech.com/applications/ecg.html
Difference Amplifier: An Example
• Problem: Determine vo
• Given Data: R1= 10k, R2 =100k, v1=5 V, v2=3 V
• Assumptions: Ideal op amp. Hence, v-= v+ and i-= i+= 0.
• Analysis: Using dc values,
Adm
R
2R1
100k
10k10
VoAdmV1V
2
10(53)
Vo20.0 V
Here Adm is called the “differential mode voltage gain” of the difference amplifier.
Finite Common-Mode Rejection Ratio
(CMRR)
A(or Adm) = differential-mode gain
Acm = common-mode gain
vid = differential-mode input voltage
vic = common-mode input voltage
A real amplifier responds to signal
common to both inputs, called the
common-mode input voltage (vic).
In general,
voAdmvidAcmvic
Adm
Admvid
vic
CMRR
CMRR AdmAcm
and CMRR(dB) 20log10
(CMRR)
An ideal amplifier has Acm = 0, but for a
real amplifier,
21id
v
icvv
22id
v
icvv
voAdm(v
1v
2)Acm
v1v
22
voAdm(vid
)Acm(vic
)
Finite CMRR: An Example
• Problem: Find output voltage error introduced by finite CMRR.
• Given Data: Adm= 2500, CMRR = 80 dB, v1 = 5.001 V, v2 = 4.999 V
• Assumptions: Op amp is ideal, except for CMRR. A CMRR in dB of 80 dB corresponds to a CMRR of 104.
• Analysis:
The output error introduced by finite CMRR is 25% of the expected ideal output.
vid5.001V4.999V
vic 5.001V4.999V
25.000V
voAdmvid
vic
CMRR
2500 0.0025.000
104
V6.25V
In the " ideal" case, voAdmvid5.00 V
% output error 6.255.00
5.00100%25%
uA741 CMRR Test Differential Gain
Differential Gain Adm = 5 V/5 mV = 1000
uA741 CMRR Test Common Mode Gain
Common Mode Gain Acm = 160 mV/5 V = 0.032
CMRRAdm
Acm
1000
.032 3.125x104
CMRR(dB) 20log10 CMRR 89.9 dB
Op-Amp circuit Analysis
47
48