Further Uses of Operational Amplifiers
Electricity & Electronics 10:
Further Uses of Operation Amplifiers
AIM
In this second unit on op-amps we are going to investigate some more of their uses. So far we
have only used the inverting input of the op-amp, now it is time to find out about how the
non-inverting input is used.
OBJECTIVES
On completing this unit you should be able to:
• identify circuits where the op-amp is being used in the differential mode.
• state that a differential amplifier amplifies the potential difference between its two
inputs.
• state the differential mode gain equation: Vo = (V2 - V1) Rf / R1
• carry out calculations using the above equation.
• describe how to use the differential amplifier with resistive sensors connected in a
Wheatstone Bridge arrangement.
• describe how an op-amp can be used to control external devices via a transistor.
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Electricity and Electronics
Further Uses of Operational Amplifiers
The Differential Amplifier
If the amplifier is set up in the configuration shown below, it is said to be in the differential
mode.
R f
R 1
-
V 1
R 2
V 0
+
V 2
R 3
0V
There are two input potentials,V1 and V2, one applied to each of the input terminals of the opamp.
There is a feedback resistor, Rf, connected between the output and the inverting input. This
allows control over the gain of the amplifier as it did for the inverting mode.
When the op-amp is used in this mode, it amplifies the difference between the inputs V1 and
V2, with a gain set by the ratio
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Electricity and Electronics
Further Uses of Operational Amplifiers
Example
A differential amplifier is set up as shown below.
For the following values shown, calculate the output voltage, V0.
(a)
V1 = +5.0 V, V2 = +4.8 V
(b)
V1 = -2.0 V, V2 = +4.5 V
(a)
V1 = +5.0 V, V2 = +4.8 V, R1 =R2 = 10 kΩ, Rf = R3 = 100 kΩ, V0 = ?
V0 =
Rf
(V - V 1)
R1 2
V0 =
V0 =
(b)
100
(4.8 - 5.0)
10
- 2.0 V
V1 = -2.0 V, V2 = +4.5 V, R1 =R2 = 10 kΩ, Rf = R3 = 100 kΩ, V0 = ?
R
1
0
0
f
V
=
(
V
V
) V
=(
4
.
5
(
2
.
0
)
)V
=
+
6
5
V
i
n
t
h
e
o
r
y
.
0
2
1
0
0
R
1
0
1
But supply voltage = + 15 V hence output voltage, V0, will saturate at + 15 V
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Electricity and Electronics
Further Uses of Operational Amplifiers
The differential amplifier as part of a monitoring system
+Vs
10 kΩ
Rf
0-10 kΩ
V1
t
10
kΩ
V2
R1
10 kΩ
R2
10 kΩ
Ω
R3
100 kΩ
V0
+
100 kΩ
0V
0V
Wheatstone
Bridge input
Differential
Amplifier
The setting of the variable resistor can be adjusted so as to achieve an output voltage of zero
for a particular temperature setting. The bridge circuit would then be balanced, that is the
potential difference V2 - V1 = 0 V.
The potential, V2, will remain constant as long as the resistance of the variable resistor is not
changed. Any change in temperature will change the potential, V1, and will therefore
produce a potential difference between V2 and V1.
The amplifier will then amplify the difference between V2 and V1, giving an output potential,
V0. The output voltage will increase as the change in thermistor resistance, ∆Rt, increases. The
amplifier is, effectively, amplifying the out-of-balance potential difference from the
Wheatstone Bridge.
Practical Application
The output voltage could be calibrated by placing the thermistor in melting ice (0 oC), then in
boiling water (100 oC), noting the output potential, V0, for each case. The range could then be
divided into 100 equal divisions to give an electronic thermometer over the range 0 - 100 oC.
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Electricity and Electronics
Further Uses of Operational Amplifiers
Control Circuits
A transistor, such as those shown below, can act as an electrical switch.
If the input voltage to these transistors, Vi, is positive, then it switches on allowing a current
to flow between the collector and emitter or source and drain, otherwise it is off.
There are two types of transistor switches:
n -ch a n n el en h a n cem en t M O S F E T
b ip o la r n -p -n tra n sisto r
-V S
+V S
drain
collector
base
gate
source
em itter
Vi
Vi
0V
0V
(i)
A p o s itiv e (+ v e ) p o te n tia l > 1 .8 V is
n e e d e d fo r th e in p u t, Vi , to s w itc h
tra n s is to r o n .
(ii) T h e d ra in is c o n n e c te d to a p o s itiv e
(+ v e ) s u p p ly ra il.
(iii) if th e in p u t p o te n tia l is n e g a tiv e o r
< 1 .8 V th e tra n s is to r is o ff.
(i)
A p o s itiv e (+ v e ) p o te n tia l (> + 0 .7 V )
is n e e d e d fo r th e in p u t, Vi , to s w itc h
tra n s is to r o n .
(ii) T h e c o lle c to r a rm is c o n n e c te d to a
p o s itiv e (+ v e ) s u p p ly ra il.
(iii) If th e in p u t p o te n tia l is n e g a tiv e
(< 0 .7 V ), th e tra n s is to r is o ff.
These transistors can be used with a Wheatstone Bridge/differential amplifier circuit to switch
a device on or off.
Low light-level indicator
+Vs
+Vs
Rf
0-10 kΩ
R1
1 kΩ
R2
1 kΩ
R3
+
1 MΩ
V2
10kΩ
10kΩ
V1
0V
Wheatstone
Bridge input
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Warning
Lamp
1 MΩ
Differential
Amplifier
-5-
V0
Vi
0V
Transistor
Switch and
Output
Electricity and Electronics
Further Uses of Operational Amplifiers
The variable resistor in the Wheatstone Bridge is adjusted so that, at a particular light level,
the output of the op-amp is zero or less so that the transistor is off. In practice, the variable
resistor would be adjusted until the warning lamp is just off. In this situation, V1 ≈ V2.
If the light level falls, the resistance of the LDR will increase causing the voltage V1 to fall.
If V1 falls, then V1 < V2, therefore (V2 - V1) will be positive (+ve).
This will cause the output voltage, V0, to be positive also, switching on the transistor and
the Warning Lamp lights.
If the light level to the LDR increases, then the opposite will be true.
The resistance of the LDR will decrease causing the voltage V1 to increase.
If V1 increases, then V1>V2, (V2 - V1) will be negative causing the output to be negative and
the transistor will not switch on.
The gain of the circuit (about 1000) is deliberately chosen to be large so that a small variation
from the balance point of the Wheatstone Bridge produces a large enough output voltage to
switch the transistor on.
Modifications
This circuit does have its limitations however. If the output device to be used has a high
power rating requiring a current larger than the transistor can safely supply, then a relay
switch must be used with the transistor to switch on an external circuit.
An example of this type of switch is shown below. In this case, a relay can be energised by
the current through the transistor (the collector current) and its contacts can then be used to
switch on a high-power device such as a heater.
+Vs
+Vs
Rf
V2
10K
10K
V1
100K
R1
1K
R2
1K
+
R3
V0 Vi
heater
0V
Wheatstone
Bridge input
relay
-
100K
0-10K
t
230 V
mains
Differential
Amplifier
Transistor Switch
and Output
Notice that as the temperature falls the resistance of the thermistor increases, causing V1 to
fall. this means V1 < V2 giving a positive output voltage V0. The transistor will be switched
on, which will energise the relay switch. The heater will be turned on.
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Electricity and Electronics
Further Uses of Operational Amplifiers
Control Circuit Diagrams
The Wheatstone Bridge arrangement can be shown as two straight potential dividers or as a
diamond arrangement. They are the same circuit, just drawn a different way.
Straight potential dividers
+ Vs
t
to d ifferential
am plifier
10KΩ
10KΩ
V1
V2
0V
Diamond arrangement bridge circuit
V2
+Vs
10 kΩ
t
V1
10 kΩ
0V
to op-amp
+Vs
V2
to op-amp
10 kΩ
t
10 kΩ
V1
0V
All the above circuits are identical examples of the sensor bridge circuit that can be used with
a differential amplifier. It is important to recognise each circuit for what it is and how it
behaves.
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Electricity and Electronics
Further Uses of Operational Amplifiers
16.
17.
18.
The circuit below shows an operational amplifier in the differential mode.
Rf
(a) What is the function of this circuit ?
(b) State the relationship that applies to this
R1
+15 V
circuit, giving the condition for this to
R2
hold.
+
(c) Find the value of VO when
V1
-1 5 V
V0
Rf = 10 MΩ R1 = 10 kΩ
V2
R3
R2 = 10 kΩ R3 = 10 MΩ
V1 = 480 mV V2 = 500 mV.
V 1 (V ) V 2 (V ) R f = R 3 R 1 = R 2 V 0 (V )
Calculate the unknown values in
(k Ω)
(k Ω)
the table for a differential
(a )
2 .5
3 .0
100
10
amplifier circuit.
(b )
1 .5
1 .3
30
5
(c )
0 .4
0 .4
10
100
(d )
6 .0
7 .2
3
6 .6
(e )
4 .5
40
5
-8 .0
(a)
(b)
(c)
(d)
In the circuit below, what value should be chosen for R for it to operate as a
differential amplifier?
Determine the reading on the voltmeter with the slider at position A.
If the contact is moved to a position midway between A and B calculate the
voltmeter reading.
Where should the contact be to produce an output voltage of
(i) -1.25 V
(ii) 0 V.
+ 1 .5 V
100 kΩ
10 kΩ
40 kΩ
A
5 kΩ
-
16 kΩ
+
V
R
B
0 V
19.
The circuit opposite can be
shown to monitor the level
of brightness.
(a)
Explain how it operates,
mentioning why the
voltmeter reading changes.
(b)
What is the purpose of the
variable resistor?
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+9 V
0 -1 0 k Ω
LDR
-8-
10
kΩ
10 kΩ
10 kΩ
V1
10
kΩ
100 kΩ
V2
+
V
100 kΩ
0 V
Electricity and Electronics
Further Uses of Operational Amplifiers
20.
(a)
(b)
(c)
Calculate the gain of the circuit shown.
At what value of (i) V0 (ii) V1 will the transistor switch on ?
Write down a rule which will allow you to predict whether a transistor is on or
off. Remember the polarity of the base-emitter voltage is important.
-Vs
+6 V
1 MΩ
V1
V
100
kΩ
-
c
b
+
V0
e
V
0V
21.
State if the transistors below are switched on or off.
(a)
(b)
(c)
(d)
-5 V
+5 V
-5 V
+5 V
0V
0V
0V
0V
(e)
(f)
(g)
(h)
-5 V
+5 V
-5 V
-5 V
-3 V
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+9 V
-9-
-2 V
-9 V
Electricity and Electronics