1
2
3
Difference amplifier ............................................................................................................................................... 1
Instrumentation amplifier...................................................................................................................................... 4
Integrator and Differentiator................................................................................................................................. 5
1
Difference amplifier
An ideal difference amplifier only amplifies the difference between two signal and
reject any common signals to the two inputs. For example, a microphone system
amplifiers an audio signal applied to one terminal and reject any 50Hz noise
signal or “hum’ existing on both terminals.
The difference amplifier circuit is presented in figure 3.1. To analysis this circuit
we will use superposition theorem and virtual short concept.
This circuit performs following arithmetical function:
Vo = A1V1 – A2V2
where A1 and A2 represents amplifier coefficients corresponding to each input
voltage contribution to output op-amp voltage
R2
0
V1
V2
R1
V+
0
Vo
R3
R4
Figure 3.1 Difference amplifier
Consider for the beginning V2 = 0. In this case we have a typical inverting
configuration (there are no current through R3 and R4). We can apply virtual
ground concept and Vo will be:
Vo1 = - V1
R2
R1
Now we will consider V1 = 0. In this case R3 and R4 form a voltage divider. The
voltage potential in noninverting input denoted as V+ is:
R4
R3 + R4
Applied now the virtual short concept we can determinate output voltage in this
case as:
V+ = V2
R2
) V+
R1
Solving last the equation replace V+ we find:
Vo2 = (1 +
Vo2 = V2 (1 +
R2
R1
)
R4
R3 + R4
Since the output voltage is the sum of the individual terms we have:
Vo = Vo1+Vo2 or
R2
R4
R2
)(
) - V1
R1
R3 + R4
R1
A special property of the ideal difference amplifier is that the output voltage is
zero when V1=V2. An inspection of the final output equation shows that the
condition is met when:
Vo = V2 (1 +
R4
R2
=
R3
R1
The output voltage is then
R2
(V2-V1)
R1
which indicates that this amplifier has a differential gain of Ad=R2/R1. This factor
is the closed-loop differential gain, rather then the open-loop differential gain Aod
of the op-amp itself. Other name for this case is balance difference amplifier.
Vo =
As previously started, another important characteristic of the electronic circuits is
the input impedance. The differential input resistance of the differential
amplifier can be determinate using the circuit shown in figure 3.2. We assume,
for easy calculation, that we have a balance difference amplifier so R1=R3 and
R2=R4. The input impedance is then define as
Ri=
Vi
i
Taking into account the virtual short concept we can write a loop equation as
follows:
Vi = iR1 + iR1 = 2 iR1
Therefore, the input impedance is
Ri = 2R1
R2
Vi
+
R1
Vo
=
R1
R2
virtual short
Figure 3.2 Circuit for measuring differential input resistance of op-amp difference
amplifier.
Now, in the ideal difference amplifier, the output is zero when V1=V2. However
inspect output voltage equation we can see that condition is not satisfied when
R4/R3≠R2/R1. When V1=V2 the input is called a common-mode input signal. The
common-mode input voltage is defined as:
Vcm=(V1+V2)/2
The common-mode gain is then defined as:
Vo
Vcm
Ideally, when a common-mode signal is applied, Vo=0 and Acm=0. A nonzero
common-mode gain may be generating in actual op-circuits. This will be
discussed in later chapter.
Acm =
A figure of merit for a difference amplifier is the common-mode rejection ratio
(CMRR), which is defined as the magnitude of the ratio of differential gain to
common-mode gain. Usually CMRR is expressed in decibels
CMRR =
Ad
Acm
or
Ad
Acm
Ideally, the common mode rejection ratio is infinite (usually we want that to be
also in the non-ideal op-amp circuits). Typical values for CMRR is 80-100 dB.
CMRR (dB)= 20 log10
2
Instrumentation amplifier
We saw in last section that it is difficult to obtain high input impedance and a high
gain in a difference amplifier, using reasonable resistor values. One solution is to
insert a voltage follower between each source and the corresponding input. The
disadvantage of this design is that the gain of amplifier can not easily chanced.
We need to change 2 resistor values and also maintain the ration between the
R2/R1 and R4/R3. Optimally we will like to be able to change the gain by changing
only one resistor. This circuit is shown in figure 3.3 and is called
instrumentation amplifier. We will see the flexibility of such circuit. Note that
two noninverting amplifiers A1 and A2 are used in the input stage, therefore the
input impedance is practically infinite. A difference amplifier is used as output
V1
Vo1=V1+i1R 2
=
A1
R2
0
R4
Vo
i1
R1
R3
A3
R3
0
R2
=
R4
A2
V2
Vo2 =V2 -i 1R2
stage of circuit.
Figure 3.3 Instrumentation amplifier (voltage and currents)
We begin analysis wit virtual short concept. For the followers amplifiers (A1, A2)
the voltage on the inverting terminals are equal with the input voltage V1 and V2.
We can write the current through resistor R1 as:
i1=
V1 – V2
R1
The current i1 is also the current through R2 and the output for amplifier A1 and A2
are, respectively
Vo1 = V1 + i1R2 = (1+
and
R2
R1
) V1 – V2
R2
R1
R2
R2
) V2 – V1
R1
R1
From the previous results, the output of the difference amplifier is given as
Vo2 = V2 - i1R2 = (1+
R4
(Vo2-Vo1)
R3
Finally we obtain the output voltage:
Vo =
R4
2R2
(1+
)(V2 – V1)
R3
R1
We observe that the output voltage is depending on the resistor R1, which can be
easily be varied using a potentiometer, thus providing a variable gain with the
adjustment of only one resistance.
Vo =
3
Integrator and Differentiator
We considered until now the external elements of op-amp configuration as
resistance. Other elements can be used, with different results. We can consider a
general case for an inverting op-amp configuration replacing the external
resistors with generalized impedance.
The output voltage will be:
Z2
Z1
Two special circuits can be developed from this generalized amplifier.
Vo= - Vi
In the first, Z1 corresponds to a resistor and Z2 to a capacitor. The impedance are
then Z1 = R1 and Z2=1/sC2
C2
- Vc +
Vi
Vo
R1
Figure 3.4 Op-amp integrator circuit
The output voltage becomes:
Vo= -
Z2
Vi = - Vi
1
Z1
sR1C2
This equation represents integration in the time domain. If VC is the voltage
across the capacitor at time t=0, the output voltage is:
1
R1C2
Vo= VC -
∫
t
Vi (t’) dt’
0
where t’ is the variable of integration. The equation is the output response
of the integrator circuit, for any input voltage Vi. Note that if Vi(t) is a finite
step function, output Vo will be a linear function of time. The output voltage
will be a ramp function and will eventually saturate at a voltage near either
the positive or negative supply voltage. We will use integrator circuits in the
filter circuits, which is cover in the last chapter.
R2
C1
Vi
Vo
Figure 3.5 Op-amp differentiator circuit
The second generalized inverting op-amp application use a capacitor as Z1
= 1/sC1 and a resistor as Z2 =R2. The output voltage can be write as
Vo= -
Z2
Z1
Vi = - sR2C1Vi
Equation represents differentiation in time domain, as follows:
Vo= - R2C1
dVi(t)
dt
The circuit is therefore a differentiator. A limitation of this circuit is that any
discontinuity in the input voltage Vi will cause a spike in the output voltage.
This circuit is usually avoided in practice.