2. Amplifier Configurations and the Instrumentation Amplifier
2.1 Properties:
The operational amplifier is an integrated amplifier which is manufactured
specifically for the purposes of designing negative feedback amplifiers as shown in
Fig. 1. It is a differential amplifier having two input terminals so that two separate
input voltages can be applied to it. Its main properties are as listed below:
+
Ve
_
AO
V+
VO = AOVe
-
V
Fig. 1 An Operational Amplifier
Property
Open Loop Gain
AO
2 x 105 to 106
Input Resistance
Ri
> 2 MΩ to 1014 Ω
Output Resistance
RO
< 500Ω
Bandwidth
BW
1 – 100 MHz
Output Current
IO max
10 mA typical
Note: A very high input resistance means that the input current demanded by
either input of the op-amp, V+ or V-, is negligible and can be taken as zero for the
purposes of analysis. In addition, if the loop gain, AOβ >>1, the error voltage at the
op-amp input, Ve , can be taken as zero. This means that the inverting and noninverting input terminals can be taken as being at the same potential, i.e. V+ = V-.
This is also useful for the purposes of analysis.
1
2.2
Non-Inverting Amplifier:
The schematic diagram of a non-inverting amplifier is shown in Fig. 2, where the
resistors R1 and R2 are used to provide the negative feedback. A non-inverting
amplifier is one in which the output signal is in the same sense as the input signal.
+
_
Ve
A
I
Vi
V
R2
R1
Vf
β
Fig. 2 Schematic Diagram of a Non-inverting Amplifier
The current flowing into the feedback network is given as:
IO =
VO
(R 1 + R 2 )
But the feedback voltage is:
Vf = I O R 1
Substituting gives:
Vf =
R1
VO
(R 1 + R 2 )
But if V+ = V- then:
Vf = Vi
so that:
Vi =
R1
V
(R 1 + R 2 ) O
2
VO =
(R 1 + R 2 ) V
R1
i
The closed loop voltage gain is then given as:
AV =
VO (R 1 + R 2 )
R
=1+ 2
=
Vi
R1
R1
Unity Gain Buffer:
A special case of the non-inverting amplifier where R1 = ∞ and R2 → 0 is shown in
Fig. 3. In this case full feedback is applied to the amplifier with β = 1 and Vf = VO .
V
+
_
A
V
V
Vf =
Fig. 3 Schematic Diagram of a Unity-Gain Buffer
AV =
VO R 1 (∞ ) + R 2 (0 ) R 1
=
=
→1
Vi
R 1 (∞ )
R1
This provides a buffer amplifier, having a very high input resistance which
demands no current and does not cause any attenuation of an input signal applied.
The output of the op-amp, on the other hand, can supply several mA of current to a
modest load. The unity-gain buffer can provide an interface between a transducer
with significant internal source resistance and a load of similar resistance to avoid
attenuation of the signal, even when voltage gain is not required.
3
2.3 The Inverting Amplifier:
The schematic diagram of the inverting amplifier is shown in Fig. 4. It is possible to
apply the input signal to the inverting input of the op-amp to generate an output
signal which is of opposite sense to the input signal. Since the negative feedback
must also be applied to the inverting terminal of the op-amp, the non-inverting
input is simply connected to ground. In this case with Ve → 0V and V+ = V-, the
inverting and non-inverting inputs can be considered as being at the same potential
so that the inverting terminal can be considered to be a ‘virtual earth’ point,
i.e. V+ = V- → 0V.
R
R1
Ii
If
_
A
V
+
Vi
VO
Fig. 4 Schematic Diagram of the Inverting Amplifier
If the input resistance of the amplifier is taken as infinite, then this means that
whatever current flows into the input resistor, R1 , must also flow through the
resistor in the feedback path, R2 and then into the output terminal of the amplifier.
Ii = If
But:
Vi − V Ii =
R1
and
V - − VO
If =
R2
Then:
Vi − V - V - − VO
=
R1
R2
4
But if the inverting input terminal can be taken as a virtual earth point, V- → 0V so
that:
Vi
V
=− O
R1
R2
So that:
AV =
VO
R
=− 2
Vi
R1
This gives a gain determined by the values of the resistors, R1 and R2 as desired.
Note that if:
Ii =
Vi
R1
Then:
Ri =
Vi
= R1
Ii
This is the input resistance seen by the source, Vi , which is finite and considerably
lower than the input resistance of the op-amp itself.
2.4
The Inverting Summing Amplifier:
This is an amplifier structure which is essentially a multiple version of the
Inverting Amplifier, having several inputs as shown in Fig. 5. Each input is
connected via a corresponding resistor to the inverting input terminal of the opamp. Since this input is considered to have a very high input resistance, it can be
assumed that no current flows into the inverting input. This means that, applying
Kirchoff’s Current Law, all of the input currents from the sources, I1 , I2 and I3
must combine at the inverting input terminal to flow out of this node through the
feedback resistor, Rf . In this case the inverting input terminal of the op-amp is
referred to as a current-summing node. It can still also be considered as a virtual
earth, since Ve → 0V and the non-inverting input terminal is connected to ground.
Then:
I f = I1 + I 2 + I 3
5
R3
I3
Rf
R
If
R2
I2
R1
V3
I1
V2
_
Ve
A
+
V1
VO
Fig. 5 Schematic Diagram of the Inverting Summing Amplifier
so that:
V - − VO V1 − V - V2 − V - V3 − V =
+
+
Rf
R1
R2
R3
But since V- → 0V then:
- VO V1 V2 V3
=
+
+
Rf
R1 R 2 R 3
So that finally:
Rf

Rf
Rf
VO = − 
V1 +
V2 +
V3 
R
R
R
2
3

 1
This shows that this circuit structure essentially adds or sums the input source
voltages V1 , V2 and V3. The overall output is inverted but this can easily be
corrected by following this stage with a single inverting amplifier stage having
unity gain. Each input source voltage is scaled by a coefficient which is determined
by the associated input resistor so that individual input voltages can be scaled by
different factors.
6
2.5
The Difference Amplifier:
The Difference Amplifier can be seen to be a combination of an Inverting
Amplifier and a Non-inverting Amplifier, with the addition of the resistor network
consisting of R3 and R4 on the non-inverting input as shown in Fig. 6.
R2
R1
_
I3
Ve
R3
A
+
V1
V2
V+
VO
R4
Fig. 6 Schematic Diagram of a Difference Amplifier
If the two input sources are independent and do not influence each other, then the
effects of each can be assessed independently and the Principle of Superposition
can be applied to find the overall output. In this case the contribution of the V1
source to the output can be found while making V2 = 0V and then the contribution
of the V2 source to the output can be found with V1 = 0V.
Consider firstly the effect of the V1 source with V2 = 0V. In this case the V2
input is connected to ground, in which case the resistors R3 and R4 can be neglected
as they have no voltage across them and no current flows through them. In this
case the configuration is identical to an inverting amplifier with the V1 source as
input so that:
VO1
R
=− 2
V1
R1
so that
VO1 = −
R2
V1
R1
Secondly, consider the contribution of the V2 source with V1 = 0V. In this
case the V1 input terminal is connected to ground, which means that the input end
of the resistor R2 is connected to ground. This gives a circuit structure similar to
the non-inverting amplifier, but with the additional resistor network consisting of
R3 and R4 at the non-inverting input of the op amp, as shown in Fig. 7.
7
The best approach is to consider the amplifier to be effectively a non-inverting
amplifier as far as the input voltage to its non-inverting terminal, V+ is concerned.
The voltage at the non-inverting terminal, V+, can then be determined in terms of
the input voltage V2.
R2
R1
_
I3
Ve
R3
A
+
V+
V2
VO
R4
Fig. 7 Equivalent Diagram of a Difference Amplifier with V1 = 0V
Then for the voltage at the non-inverting terminal of the op-amp:
VO2
R 2 (R 1 + R 2 )
1
=
+
=
+
V
R1
R1
So that:
VO2 =
(R 1 + R 2 ) V +
R1
Resistors R3 and R4 simply act as a potential divider so that:
V+ =
R4
V2
(R 3 + R 4 )
Then substituting gives:
VO2 =
(R 1 + R 2 )
R1
8
R4
V2
(R 3 + R 4 )
Combining the contributions to the output from each of the input sources gives:
VO = VO1 + VO 2 =
(R 1 + R 2 )
R1
R4
R
V2 − 2 V1
(R 3 + R 4 )
R1
If a choice is made so that R3 = R1 and R4 = R2 then:
VO =
(R 1 + R 2 )
R1
R2
R
V2 − 2 V1
R1
(R 1 + R 2 )
so that finally a difference expression is obtained as:
R2
(V2 − V1 )
VO =
R1
2.6
The Instrumentation Amplifier:
A modification of the difference amplifier can be made which significantly
improves on the performance of the latter and overcomes some of its limitations.
This is known as the standard 3 op-amp instrumentation amplifier and is shown in
Fig. 8 below. The structure around op-amp A3 is that of the difference amplifier
discussed above. The modification is that two input-stage op-amps have been
added, namely A1 and A2, which act as buffer amplifiers having high input
impedance and are configured as non-inverting gain stages but with cross coupling
between them via the resistor, R1.
This configuration can be analysed in a similar manner to the other simpler
circuits, using the principle that the error voltage approaches zero for each of the
individual op-amps and that no current flows into any input terminal. This means
that the potential at the inverting and non-inverting terminals of the op-amps can
be assumed to be the same even if they are not ground. Consequently, the potential
at both input terminals of op-amp A1 in Fig. 8 below can be taken as being at the
potential of input V1, while both input terminals of op-amp A2 can be taken as
being at the potential of input V2. In addition, the current flowing through
resistors, R2A, R1 and R2B can be taken as the same current, i. This can be used in
analysing the input stage of the overall amplifier structure.
9
+
A
_ 1
R3A
V1
R4A
i
VO1
R2A
i
+
_
R1
A3
R2B
i
_
VO
A2
R3B
+
R4B
VO2
V2
Fig. 8 Schematic Diagram of the Standard 3 Op-Amp Instrumentation Amplifier
Using this principle then:
i=
VO1 − V1 V1 − V2 V2 − VO 2
=
=
R 2A
R1
R 2B
Taking:
VO1 − V1 V1 − V2
=
R 2A
R1
VO1
V
V
V
− 1 = 1 − 2
R 2A R 2A R 1 R 1
10
VO1
V
V
V
= 1 + 1 − 2
R 2A R 2A R 1 R 1
VO1  1
1 
1
V1 −
= 
+
V2
R 2A  R 2A R 1 
R1
VO1  R 1 + R 2A
= 
R 2A  R 2A R 1

1
V1 −
V2
R
1

 R + R 2A
VO1 = R 2A  1
 R 2A R 1
 R
VO1 = 1 + 2A
R1


R
V1 − 2A V2
R1


R
V1 − 2A V2
R1

Eq.1
Similarly with:
V1 − V2 V2 − VO 2
=
R1
R 2B
VO 2 − V2 V2 − V1
=
R 2B
R1
VO 2  1
1 
1
V2 −
= 
+
V1
R 2B  R 2B R 1 
R1
so that:
 R
VO 2 = 1 + 2B
R1

11

R
V2 − 2B V1
R1

Eq.2
Examining Eq.1 and Eq.2, if R2A = R2B = R2 then:
 R 
R
VO1 = 1 + 2 V1 − 2 V2
R1 
R1

Eq. 3
 R 
R
VO 2 = 1 + 2 V2 − 2 V1
R1 
R1

Eq.4
Using the expressions obtained for the difference amplifier considered above the
overall output voltage from the third stage op-amp A3 in Fig.8 is given as:
 R + R 4B
VO =  3B
 R 3B
 R 4A

 R 3A + R 4A

R
VO1 −  4B

 R 3B

VO2

Eq. 5
If R3A = R3B = R3 and R4A = R4B = R4 then:
 R + R 4  R 4 
R 
R 

VO1 −  4  VO2 =  4 (VO1 − VO2 )
VO =  3
 R 3  R 3 + R 4 
 R3 
 R3 
Combining This with Eq.3 and Eq.4 gives finally:
 R  R 
VO =  4 1 + 2 (V1 − V2 )
R1 
 R 3 
Eq.6
This shows that the 3 op-amp instrumentation amplifier only amplifies the
difference between the two inputs V1 and V2. This makes it ideal for use with
transducers which have a Wheatstone bridge arrangement in their construction.
For these types of transducer, like the SX50 pressure transducer examined earlier,
the output voltage is the same at both terminals when no input pressure is applied.
This common voltage can be quite large at both terminals, amounting to half of the
bridge supply voltage. However, if the potential is the same at both terminals, the
output voltage from the instrumentation amplifier is zero as required. When input
pressure is applied, the potential at one terminal will increase by a small amount,
while that that at the other terminal will decrease by a small amount. The
12
instrumentation amplifier will detect this small difference in the potentials between
the two terminals and amplify this by the gain determined by the combined 3 opamp stages given by the expression in Eq. 6 above.
13