Since the gain A ≈ ∞, v
Operational Amplifiers
for Basic Electronics
http://cktse.eie.polyu.edu.hk/eie209
by Prof. Michael Tse
January 2005 Where do we begin?
We begin with assuming that the op-amp is an ideal element satisfying the following conditions: Output resistance = 0 (perfect output stage) Input resistance = ∞ (perfect input stage) Differential voltage gain = ∞
v v v v Av i i ± o o i
- – – –
- – Since the gain A ≈ ∞, v ≈ 0 if v is infinite,
i o
the two input terminals have same potential if v is infinite
o
a “virtual” short-circuit exists between the two input terminals
- –
- + – + – + – + –
Ω; A
8
9
10
11
12
13
14
GBW = 4MHz (gain-bandwidth) SR = 13V/µs
5 CMRR = 100dB
=100dB = 10
VOL
12
The 347 IC op-amp
= 10
in
R
7 Manufacturer listed spec:
6
5
4
3
2
1
single-ended output The 347 is a Quad JFET input op-amp using biFET technology.
output stage
V+ V– The basics
An op-amp is a very high gain differential amplifier. In almost all applications (except in comparator and Schmitt trigger), feedback is used to stabilize the gain.
TWO GOLDEN RULES:
RULE 1:
attempts
The output to do whatever is necessary to make the voltage difference between the two inputs zero.
RULE 2: The inputs draw no current. Example
Consider the following op-amp circuit. What is the voltage gain?
R
2 Apply the Golden Rules:
i x R
1
0V try
It first says that the output will to set
- – v i i x
- will try to make the –ve input 0 V because the +ve input is 0 V.
- – –
- – +
- – +
- –1V
- – +
- – + sq.
- – +
- – v o This is the voltage follower
- v i
- – 2
- This is the summing amplifier.
- – v
- 2
- – i
- This is the integrator.
- – v i
- This is the differentiator (theoretically). In practice, this circuit won’t work!!!
- – out
- 1
- – 2
- t
- – 2
- – + A R Assume the op-amp is powered by 2 ±10V, and now v = +10V. R out
- –10R 1
- – + out v
- R
- –10R
- R
- +10 v in v out –10
- – + –10V 90k What are the upper and lower trip points? 10k W – 8V + W Practical considerations Finite input currents
- – –
- – +
- – +
- – load current I , must be o
- + I x
- – v i
- still 0V because there is no
- – v
i
- i b
- For example, for the non-inverting amplifier, if we set
- – R
itself in order to make the difference
v
o itbetween the inputs zero. That means,
Then, it says that the current flowing into the inputs are zero.
Therefore, This is the inverting amplifier. Warnings
The Golden rules sometimes do not apply. NOTE CAREFULLY that Golden Rule 1 says that “the output attempts to…”. The output attempts, but it may fail to do what it wants to do!
Do Golden rules apply in the following circuits?
2 x x Other examples (where Golden rules work) R
2 Applying the Golden rules, we get
R
1
v i This is the non-inverting amplifier.
Here, simply
More examples
R f
R 1 v
1 R
2 v
R 3 v
3
R
2 R
1
1 R 1 o v v
R 2 This is the difference amplifier. Other examples (where Golden rules work)
More examples
C R v
R C
v o
Examples (where Golden rules do not work)
2
by more than 0.0001 V, the output will swing to –10 V.
1
is larger than v
2
2. If v
by more than 0.0001 V, the output will swing to +10 V.
is larger than v
v
1
1. If v
Since the voltage gain typically exceeds 100,000, the inputs must be within a fraction of a millivolt in order to prevent the output from swinging all the way to extreme positive or negative. It is assumed that the supply voltages are +10 V and –10 V and that the gain is 100,000.
Comparator
2 v
1 v
The output cannot make the two inputs equal!!! Golden Rule 1 fails!!! Examples (where Golden rules do not work) Comparator
The output cannot make
v
the two inputs equal!!!
v out
Golden Rule 1 fails!!!
v
But this simple comparator suffers from a problem if the input signals have noise! The output may switch (jump up and down) when the signals are close to each other.
Examples (where Golden rules do not work) v
1 Comparator v
5V
2 v
1 v out v
v out
±
5V Suppose v = 5V (constant) and v is an input.
2
1 This circuit is supposed to compare v with 5V.
1 But if v has noise, the output may jump when v is near 5V.
2
2 Examples (where Golden rules do not work) Schmitt Trigger — a better comparator
How does it work? v in v out
Obviously, v must be less than v : A 1
in
10R 1
v in A < = v R 1 + R 2 What happens if v moves just above in
10R /(R +R )? Clearly, v falls to
1
1 2 out –10V because of comparator action.
Therefore, v drops to –10R /(R +R ), A
1
1
2
and v must be greater than v : A
in
v in A > = v R 1 + R 2 Examples (where Golden rules do not work) Schmitt Trigger
R 1 R 2 v in A We have a situation similar to hysteresis.
upper trip point =
10R
1 R
1
2
lower trip point =
1 R
1
2 t t
10R
1 R + R 1 210R
1 R + R 1 2 –Examples (where Golden rules do not work) Schmitt Trigger
10V v in v out
Very small currents are in fact needed to bias the op-amp input stage. Circuits that have no DC path to inputs won’t work! None of these works!
C
v i v i v o v o x x
C Practical considerations Offset in integrator
The op-amp integrator is very easily saturated if there is a small lack of symmetry in the input signals. This is because the error gets integrated quickly and the output will soon move towards the maximum voltage.
v i C
C
In practice we need a discharge path to prevent saturation. Usually R has to be big enough, so that the discharge rate becomes insignificantly slow compared to the signal frequency.
R
The op-amp will make sure that the voltage across
R is also equal to v , which is fixed! + v R R –
Therefore the current flowing down R must be
LOAD
I o R
which is the load current. Thus, this circuit provide a constant current source for the load.
Note: the load is floating for this case! Applications Current source for grounded load Again v is fixed by the voltage divider. R
V cc R
The op-amp will make sure that the voltage at the – lower end of R is also equal to v , which is fixed! R + Therefore the current flowing down R must be
v R
I o LOAD
which is very close to the load current (if base current is small and op-amp draws very small current). Thus, this circuit provide a constant current source for the grounded load. Applications Current source for grounded load
(voltage controllable) Here, v is controllable/adjustable by v . R IN
V cc R
The current flowing down R, which is close to the
R 2
V I cc cc x - (V - R ) 2 I o = v R
R v IN +
I o R v 2 – IN
= LOAD R R 1 R 1 Thus, this circuit provide a controllable constant current source for the grounded load.
Other non-ideal behaviour Example of input bias current
Problem: Since i flows into b R
2
both inputs, the negative input side will have a slightly negative
R
1
dc voltage even when v = 0, i
i b v whereas the positive input side is o
i b
resistor there! Therefore, v ≠ 0, i i.e., some unwanted offset !
R
2 R
1
i Practical solution: b v o
R
||R
1
2 Other non-ideal behaviour Input offset voltage
Due to imperfect symmetry, some voltage has to be applied to the input to get the output to zero. Typical value ≈ 5 mV.
The input offset voltage is a function of temperature (due to temperature drift of device parameters).
Practical solution: In many applications, the dc gain is not needed. We can simply drop the dc gain to 1.
= 2kΩ
1 C
= 4.7µF
1 R R
2
1 the cutoff frequency is approx 17 Hz. C
1 Summary We have studied the basics of op-amps, and some applications.
Basic rules of op-amp circuit analysis Some practical considerations Some applications