Download Designing with A perfect operational amplifier does not exist, but

Designing with A perfect operational amplifier does not exist, but useful functional the ideal
op amp circuits can be designed and operated taking the crudest of simplifications. The ideal’
characteristics are as shown in Table 2
Table 2
‘Ideal’
Parameter
Gain
Input impedance
Bandwidth
Noise
Input offset voltage
Input offset current
‘Realistic’
105-106
105-1012
1 MHz
50nV/Hz
1 mV
1 nA
Infinite
Infinite
Infinite
Zero
Zero
Zero
There are five basic configurations to study and understand. First, there is the basic concept of
an operational amplifier virtual earth to explain. If the input impedance to the op-amp is
extremely large then the input current will be small i.e. nearly equal to zero. If the input
current is zero then the difference voltage between the two input will be zero. As theopen loop
gain of the amplifier is extremely large 105, to get a meaningful voltage out the input voltage
must be extremely small, i.e. a virtual earth. To illustrate this let the output voltage be 10V and
the open loop gain be Av = 105 the input voltage is given by:
Av 
Vo
Vin
 Vin 
Vo
10

 0.0001V or 100V  0V virtual earth
A v 100 000
VA
VB
+
Vo
-
It is the minute (V) difference between these inputs which is amplified. The circuits which
follow adopt the common convention that the power supply connections are omitted for
clarity. They have to be connected, of course, but a separate diagram is often issued to show
how the power supplies are connected to all of the operational amplifiers.
Circuit 1 The Buffer Amplifier
, VIN = VB
VB = V0
Hence VIN = V0.
We have designed a circuit for which the output is the same as the input. So what? Well, this
circuit is most frequently used as a buffer where the high (infinite?) input impedance will not
act as a drain on the source of the signals. For example, a simple photodetector could be
made using a photodiode as Circuit 1 The buffer shown in Figure 5.2.3.amplifier
Incident light on the photodiode increases its reverse leakage current, but even so it is still
only in the nA to pA region. A high input impedance ensures that the tiny amount of current
leaking through the diode is not used to bias the input stage of the circuit connected to it.
Circuit 2 The non-inverting amplifier
The circuits in Figure are identical, but can be drawn either way around. R1 and R2 form a
potential divider so that:
VB
R1

Vo R1  R 2
VA  VB and Vin  VA
VIN
R1

Vo R1  R 2
or, more simply :
Av 
VIN
R1

Vo R1  R 2
The gain is controlled solely by the feedback components.
There are two points to note:
(1)
(2)
Try to keep resistor values between I kQ and 1 MQ Less than 1 kQ can cause problems
with the amplifier inputs trying to source or sink significant amounts of current into the
source resistors; more than 1 MQ can cause problems with noise.
Negative feedback — the feedback resistor is always returned to the negative signal
input.
Example
Find the gain of the circuit of Figure
Circuit 3 The inverting amplifier
In Figure VA is connected to ground and since VB is at the same potential as VA, VB is called a
virtual earth
The current through R1 is given by IR1 where:
I R1 
VIN  VB
R1
Since VB  VA  0 then I R1 
VIN
R1
The input impedance is very large, so not much of this current enters the op amp — most
passes through R2.
The current through R2 is given by IR2 where:
I
VB  Vout
R2
Since VB  VA  0 then I R 2  
Vout
R2
I R 2  I R1 so :
Vin
V
  out
R1
R2
which gives the gain as
Vout
R
 2
Vin
R1
Example.2
20k
Calculate the gain of the amplifier of the circuit in Figure
5.2.8.
~VoUT
_
The gain of the amplifier is given by:
VOUT _ R2 20k
___ _ Ri
10k
2
Circuit 4 The mixer or adder
The circuit in Figure sums the voltages at its inputs. In this circuit, the current through the
input resistors Rl, R2 and R3 sum at the —IN point:
‘IN Vl V2 V3
Rl ± R2 + R3
This current passes through the feedback resistor R4 and the output voltage is given by:
0 — VouT
= ‘F
R4
Il
Vl
Ri
Vi
+ 12 + 13 = ‘F so:
V2 V3 VoUT Rl±R2± R3 R4
Operational amplifiers (1) 109
There is no ‘gain’ term as such. The smaller the input resistor, the larger the input current, and
this gives a greater ‘weighting’ or
importance of a particular input.
Example
1e
In the circuit of Figure 5.2.11, calculate the expected output.
5.2.3
First notice the slightly different construction. Electrically, it is identical to the previous
problem (apart from the fact that there are only two input resistors, that is). This economy of
drawing is quite common and should be recognized as easily as the first method.
Vl
V2
VouT
Ri ~R2 - R3
0.05 0.03 _ VOUT
1000 + 2000 10000
from which:
VoUT = —65 mV
~—
~ ~s. — ~ t b~
j~A
Circuit 5 The differential amplifier or subtractor
VoUT As its name implies, this circuit will output a voltage proportional
to the difference in the two inputs. The basic circuit diagram is
shown
in
Figure
5.2.15.
By
voltage
division
at
the
+IN
input:
R4
VA
—
R3
+
R4
Vl
By
current
flow
at
the
—IN
input:
V2—VB
_
VB—VO
Rl
R2
VOUT
V2
V~
V~
V0
RlR2
R2
R2
V2 V _ V(l±ii~l~~V(R1±R2~\
Rl Ri ~RlR2)8~)
B — RIR2