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Operational Amplifiers
forBasic Electronics
http://cktse.eie.polyu.edu.hk/eie209
byProf. Michael Tse
January 2005
C.K. Tse: Operational Amplifiers 2
Where do we begin?
We begin with assuming that the op-amp is an ideal element satisfying thefollowing conditions:
Output resistance = 0 (perfect output stage)Input resistance = ∞ (perfect input stage)Differential voltage gain = ∞
Since the gain A ≈ ∞, vi ≈ 0 if vo is infinite,the two input terminals have same potential if vo is infinitea “virtual” short-circuit exists between the two input terminals
+
–
+vi–
vo
+vi–
Avi±+vo–
C.K. Tse: Operational Amplifiers 3
The 347 IC op-amp
output stage
+–
single-ended output
The 347 is a QuadJFET input op-ampusing biFETtechnology.
1234567
Manufacturer listed spec:Rin = 1012Ω; AVOL=100dB = 105
CMRR = 100dBGBW = 4MHz (gain-bandwidth)SR = 13V/µs
141312111098
+–
+–
+–
+–
V+ V–
C.K. Tse: Operational Amplifiers 4
The basicsAn op-amp is a very high gain differential amplifier. In almost all applications(except in comparator and Schmitt trigger), feedback is used to stabilize thegain.
TWO GOLDEN RULES:
RULE 1:The output attempts to do whatever is necessary to make the voltagedifference between the two inputs zero.
RULE 2:The inputs draw no current.
C.K. Tse: Operational Amplifiers 5
ExampleConsider the following op-amp circuit. What is the voltage gain?
R1
R2
–
+
vivo
Then, it says that the current flowing intothe inputs are zero.
ix
ix
Apply the Golden Rules:
It first says that the output will try to setitself in order to make the differencebetween the inputs zero. That means, itwill try to make the –ve input 0 Vbecause the +ve input is 0 V.
0V
Therefore,
This is the inverting amplifier.
C.K. Tse: Operational Amplifiers 6
WarningsThe Golden rules sometimes do not apply. NOTE CAREFULLY that GoldenRule 1 says that “the output attempts to…”. The output attempts, but it mayfail to do what it wants to do!
+
–
Do Golden rules apply in the following circuits?
+
–
–
+–
+
–
+
+
–
x2 xsq.
–1V
C.K. Tse: Operational Amplifiers 7
Other examples (where Golden rules work)
R1
R2
–
+vi
vo
Applying the Golden rules, we get
This is the non-inverting amplifier.
–
+vi
Here, simply
This is the voltage follower.
C.K. Tse: Operational Amplifiers 8
Other examples (where Golden rules work)More examples
R2
Rf
–
+
v2
vo
This is the summing amplifier.
–
+v2
This is the difference amplifier.
R1
R3
v1
v3
R1
R1
v1
R2
R2
C.K. Tse: Operational Amplifiers 9
Other examples (where Golden rules work)More examples
R–
+
vi
vo
This is the integrator.
–
+
This is the differentiator (theoretically).In practice, this circuit won’t work!!!
vi
R
C
C
C.K. Tse: Operational Amplifiers 10
Examples (where Golden rules do not work)
+
–outv
1v
2v
Comparator
Since the voltage gain typically exceeds 100,000, the inputs must be withina fraction of a millivolt in order to prevent the output from swinging all theway to extreme positive or negative. It is assumed that the supply voltagesare +10 V and –10 V and that the gain is 100,000.
1. If v1 is larger than v2 by more than 0.0001 V, the output willswing to +10 V.2. If v2 is larger than v1 by more than 0.0001 V, the output willswing to –10 V.
The output cannot makethe two inputs equal!!!Golden Rule 1 fails!!!
C.K. Tse: Operational Amplifiers 11
Examples (where Golden rules do not work)
+
–outv
1v
2v
Comparator
But this simple comparator suffers from a problem if the input signals havenoise! The output may switch (jump up and down) when the signals areclose to each other.
The output cannot makethe two inputs equal!!!Golden Rule 1 fails!!!
C.K. Tse: Operational Amplifiers 12
Examples (where Golden rules do not work)
+
–outv
1v
2v
Comparator
Suppose v2 = 5V (constant) and v1 is an input.This circuit is supposed to compare v1 with 5V.
But if v2 has noise, the output may jump when v2 is near 5V.
±5V
v1
5V v2
t
vout
C.K. Tse: Operational Amplifiers 13
Examples (where Golden rules do not work)
Schmitt Trigger — a better comparator
–
+outv
R1
R2
vin
A
How does it work?
Assume the op-amp is powered by±10V, and now vout = +10V.
Obviously, vin must be less than vA:
What happens if vin moves just above10R1/(R1+R2)? Clearly, vout falls to–10V because of comparator action.Therefore, vA drops to –10R1/(R1+R2),and vin must be greater than vA:
†
vin <10R1
R1 + R2= vA
†
vin >–10R1
R1 + R2= vA
C.K. Tse: Operational Amplifiers 14
Examples (where Golden rules do not work)
Schmitt Trigger
–
+outv
R1
R2
vin
A
We have a situation similar to hysteresis.
†
upper trip point = 10R1R1 + R2
†
lower trip point = –10R1R1 + R2
t
t
vin
vout
+10
–10
10R1R + R1 210R1
R + R1 2
–
C.K. Tse: Operational Amplifiers 15
Examples (where Golden rules do not work)
Schmitt Trigger
What are the upper and lower trip points?
–
+outv
vin
10V
–10V
10kW
90kW
8V+–
C.K. Tse: Operational Amplifiers 16
Practical considerations
Finite input currents
Very small currents are in fact needed to bias the op-amp input stage. Circuitsthat have no DC path to inputs won’t work!
None of these works!
vo
–
+
vi
C
vo
–
+
vi
C
x x
C.K. Tse: Operational Amplifiers 17
Practical considerationsOffset in integrator
The op-amp integrator is very easily saturated if there is a small lack ofsymmetry in the input signals. This is because the error gets integrated quicklyand the output will soon move towards the maximum voltage.
–
+
vi
C
C
–
+
In practice we need a discharge path toprevent saturation. Usually R has to bebig enough, so that the discharge ratebecomes insignificantly slow comparedto the signal frequency.
R
C.K. Tse: Operational Amplifiers 18
ApplicationsCurrent source
+–
RIo
LOAD
vR
We see that vR is fixed by the voltage divider.
The op-amp will make sure that the voltage acrossR is also equal to vR, which is fixed!
Therefore the current flowing down R must be
which is the load current.
Thus, this circuit provide a constant currentsource for the load.
Note: the load is floating for this case!
C.K. Tse: Operational Amplifiers 19
ApplicationsCurrent source for grounded load
–+
R
Io
LOAD
vR
Again vR is fixed by the voltage divider.
The op-amp will make sure that the voltage at thelower end of R is also equal to vR, which is fixed!
Therefore the current flowing down R must be
which is very close to the load current (if basecurrent is small and op-amp draws very smallcurrent).
Thus, this circuit provide a constant currentsource for the grounded load.
Vcc
C.K. Tse: Operational Amplifiers 20
ApplicationsCurrent source for grounded load(voltage controllable)
–+
R
Io
LOAD
vR
Here, vR is controllable/adjustable by vIN.
The current flowing down R, which is close to theload current Io, must be
Thus, this circuit provide a controllable constantcurrent source for the grounded load.
Vcc
+–
vIN
R2
†
Io =Vcc - (Vcc - R2Ix )
R
=R2
R1
vIN
R
R1
Ix
C.K. Tse: Operational Amplifiers 21
Other non-ideal behaviourExample of input bias current
R1
R2
–
+
vivo
ib
ib
Problem: Since ib flows intoboth inputs, the negative inputside will have a slightly negativedc voltage even when vi = 0,whereas the positive input side isstill 0V because there is noresistor there! Therefore, vi ≠ 0,i.e., some unwanted offset!
R1
R2
–
+
vivo
ib
ibPractical solution:
R1||R2
C.K. Tse: Operational Amplifiers 22
Other non-ideal behaviourInput offset voltage
Due to imperfect symmetry, some voltage has to be applied to the input to get theoutput to zero. Typical value ≈ 5 mV.
The input offset voltage is a function of temperature (due to temperature drift ofdevice parameters).
Practical solution: In many applications, the dc gain is not needed. We cansimply drop the dc gain to 1.
For example, for the non-inverting amplifier, if we set+
–R1 = 2kΩC1 = 4.7µF
the cutoff frequency isapprox 17 Hz.
R1R2
C1
C.K. Tse: Operational Amplifiers 23
Summary
We have studied the basics of op-amps, and some applications.
Basic rules of op-amp circuit analysisSome practical considerationsSome applications