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F5.1 OPERATIONAL AMPLIFIERS
An operational amplifier (op-amp) is a voltage amplifier which amplifies the difference between the
voltages on its two input terminals. Op-amps often require a dual balanced d.c. power supply, e.g.
± 15V. However, they will work over a wide range of supply voltages from ±5V to ±15V. The
power supply connections are often omitted from circuit diagrams for simplicity. The diagram
below shows the typical connections for an op-amp.
–
VV
+
V+
V
V+
out
0V
+
-V
s
+Vs
V
out
©IKES07
The + input terminal is known as the non-inverting input and the – input terminal is known as the
inverting input terminal.
The output voltage is given by
V out  AV   V  
where A is the voltage gain.
An op-amp is assumed to behave ideally and the general properties of an ideal op-amp include: the voltage gain is very large (typically, 106 at low frequencies),
 the maximum output voltage is equal to the power supply voltage, (in practice it is about 2V
less),
 it has infinite input resistance so no current passes into the input terminals, (typically 109,
so there is an input current of a few nano-amps),
 the output impedance is zero so it can supply any required current, (in practice many opamps are designed to limit the current to approximately 15mA),
 the output voltage is zero when the two inputs are equal, (in practice there is a small offset
voltage which needs a variable resistor to balance out).
General Amplifier Definitions
An amplifier is designed to produce an output voltage or current which is a magnified copy of the
input voltage or current. When power amplification occurs the extra power is provided by the
external power supply. The peak power output is limited by the power supply voltage. The gain is
calculated by the ratio of the output quantity to the input quantity.
voltage gain =
output voltage ( Vout )
input voltage ( Vin)
power gain =
output power ( Pout )
input power (Pin)
F5.2 VOLTAGE AMPLIFIERS
This section shows how operational amplifiers can be used as voltage amplifiers.
The Inverting Amplifier
For most amplifiers, the voltage gain of an op-amp is too large to be of practical use. It is not
possible to adjust the open loop voltage gain of an op-amp to enable them to be used in normal
amplifier circuits so Negative Feedback is used to reduce the overall gain of the circuit. With
negative feedback a proportion of the output signal is 'fed-back' to the input so that it cancels out
some of the input signal and reduces the apparent voltage gain of the whole circuit. With an opamp, negative feedback is the only way of controlling the voltage gain of the circuit. The voltage
gain of the op-amp itself is unchanged but the overall voltage gain of the circuit is reduced
significantly. The simplest example of this is the Inverting Amplifier.
The circuit diagram of an inverting amplifier is shown below. The power supply connections are
not shown, but it is assumed that the circuit is operating from a dual voltage power-supply
e.g. +15V. Rf joins the output to the inverting input and provides the negative feedback.
Rf
_
R1
Vin
+
Vout
0V
With an inverting amplifier, as the input voltage increases, the output voltage decreases and vice
versa. This is represented in the diagram of an oscilloscope display shown below.
input
output
In this diagram, both the input and output traces are to the same scale and so it can be seen that the
output voltage is four times the size of the input voltage. Therefore the amplifier giving these traces
has a voltage gain of 4 but the output voltage is inverted compared to the input voltage.
The Non-inverting Amplifier
The circuit for a non-inverting voltage amplifier using an op-amp is very similar to the circuit for
the inverting amplifier. The only difference is that the two inputs are reversed i.e. the signal is
connected to the non-inverting input of the op-amp and the free end of resistor R1 is connected to
0V. This is shown in the circuit diagram below. Again, the power supply connections are not
shown, but it is assumed that the circuit is operating from a dual voltage power-supply
e.g. +15V.
+
_
Rf
Vin
Vout
R1
0V
Rf joins the output to the inverting input and provides the negative feedback for the circuit, so
reducing the overall voltage gain for the circuit.
With a non-inverting amplifier, as the input voltage increases, the output voltage increases and vice
versa. This is represented in the diagram of an oscilloscope display shown below.
input
output
Again the output voltage is four times the size of the input voltage. Therefore the amplifier giving
these traces has a voltage gain of 4 but the output voltage is not inverted compared to the input
voltage.
Since the input voltage is applied directly to the op-amp in this circuit, the input resistance of this
amplifier will be very large (109). This should be compared with the input impedance of the
inverting amplifier where the input resistance is equal to R1.
F5.3 VOLTAGE GAIN CALCULATIONS
Calculating the voltage gain of amplifiers made with op-amps is relatively straight forward.
For an op-amp, the output voltage, Vout, is given by
V out  AV   V  
Vout cannot exceed the voltage of the power supply. So if A is 106 and the power supply voltage is
±10V, then, if the output voltage is to be less than the power supply voltage, the difference in
voltage between the two input terminals will be
6
10  10 V   V  
 V   V   
10
10
6
 10  5  10V
The Inverting Amplifier
For the inverting voltage amplifier the non-inverting input terminal of the op-amp is connected to
0V. This means that the inverting op-amp input terminal will be within 10µV of 0V and so will be
virtually at 0V, i.e. it is a Virtual Earth Point.
Virtual Earth Point
Rf
_
R1
+
Vin
Vout
0V
This means that the input voltage, Vin, appears across the input resistor R1.
V in
R1
Since one end of Rf is also connected to the virtual earth point, the output voltage Vout appears
across the feedback resistor Rf.
V
So using Ohm's Law, The current through R f  out
Rf
Assuming that the input resistance of the op-amp is so large that no current passes into its input
terminals, then the current passing through R1 must pass through Rf..
So using Ohm's Law,
The current through R1 
V
V in
  out
R1
Rf
The negative sign comes from the current passing from the virtual earth point to the output.

Rearranging gives

V out
R
 f
V in
R1
Therefore the voltage gain of the inverting amplifier is determined solely by the two external
resistors and the negative sign indicates that the amplifier is inverting.
It should be noted that the voltage gain of the op-amp itself (open loop gain) has not been altered; it
is still very large at low frequencies. But the voltage gain of the overall circuit (closed loop gain)
has been reduced significantly. Also it is important to realise that the equation is only valid when
the open-loop gain is significantly greater than the voltage gain of the whole circuit. At high
frequencies the voltage gain of the circuit will decrease, since most op-amps are frequency
compensated to ensure that they are stable and do not oscillate.
The Non-inverting Amplifier
+
_
Rf
Vin
Vout
R1
0V
When considering the voltage gain of the non-inverting voltage amplifier, it is important to
remember that the output voltage of an op-amp is equal to the differential input voltage multiplied
by the open loop voltage gain.
V out  AV   V  
Since the open loop gain of the op-amp is very large at low frequencies, the voltage at the positive
input terminal and the voltage at the negative input terminal of the op-amp will be almost identical,
so long as the output has not saturated at the power supply voltage. Therefore the voltage at the
negative input terminal of the op-amp will be the same as that of the junction of the two resistors Rf
and R1. These two resistors form a voltage divider as shown below.
Vout
Rf
VR1
0V
Vout causes a current to pass through Rf and R1. This current can be found from Ohm's Law.
V out
I
R f  R1
This current causes a voltage to appear across R1, which is the voltage at the inverting input of the
op-amp. Using Ohm's Law,
V   I  R1
Substituting for I
 V 
V out
 R1
R1  R f
But this will also be equal to the input voltage Vin since the two op-amp input terminals must have
almost the same voltage, if the op-amp is not to saturate.
 Vin 

Vout  R1
( Rf  R1)
Vout ( Rf  R1)
R

1 f
Vin
R1
R1
The non-inverting amplifier has the same limitations at high frequencies as the inverting amplifier
owing to the frequency compensation of the op-amp, but it does have the advantage of having a
very large input impedance.
F5.4 THE OP-AMP AS A COMPARATOR
The output voltage of an op-amp is given by
Vout = A( V  - V -)
where A is the open loop voltage gain of the op-amp (106) and (V+-V-) is the voltage difference
between the two inputs of the op-amp.
Since the op-amp has a very large open loop gain, A, only a very small difference between V+ and
V- is needed for the output voltage, Vout, to equal the supply voltages. Once this has occurred,
increasing the voltage difference between V+ and V- will not cause any further increase in Vout and
the op-amp is said to be saturated.
The transfer characteristic for an op-amp is shown below, Vs is the power supply voltage.
Vout / volts
saturation
+ Vs
V+ > VV+ - V- /V
V- > V+
saturation
- Vs
©IPK07
This characteristic enables the op-amp to compare the two voltages on its input terminals.
If V+ is greater than V- the output saturates at the positive supply voltage. If V- is greater than V+
the output saturates at the negative supply voltage. This principle can be used to compare two
voltages, a reference voltage and a varying input voltage.
V   V   V out   V s
V   V   V out   V s
An op-amp can be used as a comparator either with a dual power supply or with a single power
supply. Both are considered below, as each have their own specific problems when used with real
op-amps.
As an example of the use of a comparator circuit, consider an electronic fire alarm system.
A thermistor is a resistor made from semiconductor materials whose resistance changes with
temperature. The symbol for a thermistor is shown below
The most common type of thermistor has a negative temperature coefficient (ntc) which means that
as the temperature increases, the resistance of the thermistor decreases.
The circuit diagram below shows how the thermistor can be used with an op-amp comparator to
sound an alarm if the temperature goes above a set temperature.
+Vs
R2
thermistor
V+
VR1
Diode
+
–
+
-
Buzzer
0V
R3
©IPK07
–Vs
The circuit operates from a dual power supply, +Vs and -Vs. Resistors R2 and R3 set the voltage at
the op-amp inverting input terminal, which gives the reference voltage for the comparison with the
non-inverting input terminal. If R2 is equal to R3, then the reference voltage will be 0V, half way
between +Vs and -Vs.
Resistor R1 forms a voltage divider with the thermistor. As the temperature of the thermistor
increases, its resistance decreases, and so the voltage at the non-inverting input of the op-amp will
increase. If V+<V-, then the output of the op-amp will be at –Vs. The diode prevents the buzzer
being damaged by being powered the wrong way round.
When V+>V-, the output of the op-amp will switch to +Vs and the buzzer will switch on sounding
the alarm.
The temperature at which the alarm sounds can be altered by changing the value of any of the
resistors, but the most convenient arrangement is to make R2 equal to R3, (e.g. 10k) and then use a
variable resistor for R1.
A similar circuit could be used to provide warning lights for a freezer. The buzzer would be
replaced by two LEDs (Light Emitting Diodes), one red and the other green.
+Vs
R2
thermistor
V+
VR1
+
–
red
green
0V
1k 
R3
©IPK07
–Vs
When the temperature of the thermistor is high, the output of the op-amp will be positive and the
red LED will light. When the temperature of the thermistor is low, the output of the op-amp will be
negative and the green LED will light. The 1k resistor limits the current passing through the
LEDs to a safe level.
If op-amps were ideal, then both of these circuits could be easily converted to operate from a single
power supply. Unfortunately, many real op-amps suffer from the problem that their outputs do not
saturate at the power supply voltages but usually approximately two volts less. This can be a real
nuisance when the output of the op-amp is connected to a transistor or even a red LED, since both
will be switched on permanently, irrespective of the output state of the op-amp, unless precautions
are taken to prevent this from happening.
The circuit diagram below shows the fire alarm circuit modified to operate from a single power
supply.
+Vs
R2
thermistor
V+
VR1
+
–
red LED
+
-
Buzzer
R3
©IPK07
0V
When the temperature is low, the output of a real op-amp will also be approximately 2V. This is
sufficient to make some buzzers sound, and so a red LED can be connected in series with the buzzer
to prevent this problem. A red LED has a forward voltage of approximately 2V, and so when the
output of the op-amp is low, there will now be almost no voltage across the buzzer.
The modified circuit diagram for the freezer alarm is shown below.
+Vs
1k 
R2
thermistor
V+
V-
+
green
–
red
R1
R3
1k 
©IPK07
0V
The two diodes provide an additional voltage drop of 0.7V, which will ensure that each LED has
insufficient voltage available to light when it is not required.
F5.5 The Schmitt trigger
When any signal is transmitted from one place to another it will suffer attenuation, (become
weaker), dispersion (spread out) and it will gather noise and interference.
Consider the diagram below which shows a digital signal as transmitted and the same signal as
received after passing through a transmission medium.
transmission medium
©IPK07
transmitted
received
If the received signal was just amplified, as shown below, the original waveform would not be
restored and it would be unsuitable for use as a digital signal.
amplifier
©IPK07
received
amplified
If the received signal was passed through a comparator, as shown below, the noise would cause
stray pulses which would again make it unsuitable as a digital signal.
comparator
©IPK07
received
compared
In order to restore the digital signal it is necessary to use a circuit called a Schmitt trigger. This
circuit uses positive feedback to create two voltage switching levels instead of the single voltage
switching level of a comparator. This produces hysteresis, the voltage level needed to make the
output go high is different from the voltage level needed to make it go low. As a result the circuit
can be set so that it will ignore the majority of the noise on the signal and so not produce stray
pulses.
Schmitt trigger
©IPK07
received
regenerated
The circuit diagram for an inverting Schmitt trigger operating from a single power supply is shown
below.
+12V
10k
V– _
V+
+
R 10k
Vin
Vout
10k
©ikes1001
0V
In this circuit the op-amp functions as a comparator but the reference voltage, produced by the two
10k resistors connected across the power supply, changes with the value of Vout. Assuming an
ideal op-amp, if Vout is 0V, then the feedback resistor, R, will effectively be in parallel with the
lower 10k resistor as shown in the left diagram below.
+12V
10k
+12V
10k
V+
10k
V+
R 10k
©ikes1001
5k
0V
©ikes1001
0V
The two parallel resistors form a 5k resistor as shown in the diagram above on the right. Using
the voltage divider formula
12  5 60
V 

 4V
10  5 15
In order for the output to go high, Vin must go below this voltage. When it does, V– will be less
than V+ and so Vout will become 12V. The feedback resistor, R, is now effectively in parallel with
the upper 10k resistor and so, using the technique above, the reference voltage at V+ will be 8V.
In order for the output to go low, Vin must go above this voltage. This circuit therefore has two
switching levels (4V and 8V) which are separated by 4V and is less susceptible to the effects of
noise than a comparator circuit.
F5.6 Op-amp problems.
Please see the Student Work Book for suitable tasks and problems together with appropriate responses.