Download Module 4

4
AMPLIFIERS
By the end of this unit you should be able to:
OBJECTIVES
1.
Define or explain the meaning of the following terms:
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Unit 4: Amplifiers
ac amplifier
bandwidth
buffer amplifier
closed-loop gain
comparator
current gain
dc amplifier
decibel
equivalent circuit
gain-bandwidth product
hysterisis in a comparator
ideal op-amp
input resistance
inverting amplifier
negative feedback
non-inverting amplifier
open-loop gain
operational amplifier
output resistance
power gain
saturation
summing amplifier
tuned amplifier
virtual ground/earth
voltage follower
voltage gain
2.
Understand the basic operation of amplifier circuits and perform
simple calculations on voltage gain, current gain and power gain in
an amplifier, given the appropriate data.
3.
Understand the operation of operational amplifier circuits for the
amplification, summation, and comparison of voltage signals.
4.
Calculate the voltage gain and input resistance of simple inverting,
non-inverting and voltage - follower operational amplifier circuits,
given the appropriate data.
5.
Calculate the output of simple operational amplifier summing
circuits, given the appropriate data.
6.
Sketch the output of operational amplifier comparator circuits for
given input signals.
4-1
7.
INTRODUCTION
Design simple inverting, non-inverting and summing operational
amplifier circuits given the appropriate data.
An amplifier is a very signifcant components in a telecommunications
system. Its basic function is to take an electronic signal of small
amplitude and create a similar version of it with a much greater
amplitude, so that it is less corruptible by noise that naturally occurs
within any communications system.
In the last Unit we studied the bipolar junction transistor (BJT) and the
metal-oxide-semiconductor field-effect transistor (MOSFET) and
described how these devices could provide signal amplification. We
also considered simple, complete, single-stage amplifier circuits based
on both devices. It is possible to extend such amplifier circuits to have
several stages, and consequently greater signal gain, as well as other
desirable amplifier characteristics such as high input resistance and low
output resistance. Such amplifiers can be manufactured on a tiny piece
of silicon and supplied as single integrated circuit (IC) package.
AMPLIFIER
EQUIVALENT CIRCUITS
To understand or use amplifiers, we do not need to have a detailed
knowledge of the complete circuitry inside the amplifier, but rather we
can consider the amplifier as a black box, which can be represented by
the equivalent network shown in Figure 4.1
Ro
Io
Ii
+
+
Figure 4.1
Equivalent Circuit of a
Voltage Amplifier
Vi
-
+
Ri
-
Gvo Vi
Vo
-
When the voltage signal to be amplified, Vi is connected to the input
terminals of the amplifier, a current Ii flows. From Ohm’s law, the ratio
Ri = Vi / Ii is referred to as the input resistance of the amplifier. Thus
the input side of the amplifier can be modelled as a resistance Ri
comnnected between the two input terminals. The ability of the
amplifier to produce an amplified signal requires a source of controlled
energy and this is represented by the voltage generator GvoVi , where Gvo
denotes the open-circuit voltage gain of the amplifier, ie the gain when
there is no load connected to the amplifier output terminals. In practice,
when the amplifier is connected to a load (other electronic circuitry), the
voltage gain will be reduced slightly from its open-circuit value. This is
modelled by the inclusion of the component Ro which represents the
output resistance of the amplifier. In the open-circuit condition the
cuurent Io = 0 and the output voltage Vo = GvoVi . When the amplifier
output is connected to other electronic circuitry, the current Io is not zero
and some of the amplified voltage will be lost across the output
resistance Ro , just as we saw how some voltage is lost across the internal
resistance of a battery in Unit 1.
In most practical amplifiers the input resistance is relatively high and the
output resistance is relatively low. Ideally Ri would be infinitely large,
thereby reducing the input current and the input power to zero, and Ro
would be zero minimising the output power loss in Ro.
4-2
Communications Technology 1
Figure 4.2 shows an amplifier, represented by its equivalent circuit with
an input signal source Vs and internal resistance Rs driving a resistive
load RL. The circuit may be easily analysed as follows.
Figure 4.2
Equivalent Circuit of a
Voltage Amplifierwith Input
Signal Source and Output
Load
Rs
+
Ro
Ii
+
+
Vi
Vs
Io
Ri
-
-
Gvo Vi
Vo
RL
-
Ii =
Vs
Ri
and Vi = Ii Ri =
.Vs
Rs + Ri
Rs + Ri
Similarly
Vo =
RL
.GvoVi
Ro + RL
Therefore the overall voltage gain Gv is
Vo
Ri
RL
=
.Gvo .
Vs Rs + Ri
Ro + RL
Gv =
The output signal current is
Io =
GvoVi
G IR
= vo i i
Ro + RL Ro + RL
Therefore the current gain Gi is
Gi =
Io
G R
= vo i
Ii
Ro + RL
The power gain Gp is given by
Gp =
V2 / R
I2R
VI
signal power into load
= o2 L = o2 L = o o = Gv Gi
signal power into amplifier Vi / Ri
Vi Ii
Ii Ri
The voltage gain of an amplifier is often expressed in logarithmic units
called decibels, or dBs ie
GvdB = 20 log10 Gv = 20 log10
Vo
Vs
In practice a number of amplifiers may be connected in cascade as shown
in Figure 4.3 for two amplifiers in which the output of the first amplifier
is connected to the input of the second. Thus the second amplifier acts
as a load for the first. The overall voltage gain Gvo is
Gvo =
Unit 4: Amplifiers
Vo Vo1 Vo
=
.
Vi
Vi Vo1
4-3
Amplifier 1
Figure 4.3
Amplifiers in Cascade
Vi
Amplifier 2
Vo1
Vo
Expressing this in decibels gives
V V 
V
V
dB
Gvo
= 20 log10  o1 . o  = 20 log10 o1 + 20 log10 o
V
V
V
V
 i
o1 
i
o1
Thus the overall voltage gain in decibels is equal to the sum of the
individual voltage gains in decibels. This is a very useful result and can
be extended to any number of amplifiers in cascade.
SAQ 1
FREQUENCY
RESPONSE
An integrated circuit amplifier has an open-circuit voltage
gain of 60dB, an input resistance of 90kΩ and an output
resistance of 50Ω. Its input is connected to a 1mV (rms)
signal source with an internal resistance of 10kΩ and its
output to a 450Ω resistive load. Determine the rms value of
the output voltage across the load.
So far we have been assuming that the gain of an amplifier is constant no
matter what the frequency of the input signal is. Unfortunately practical
amplifiers do not amplify all input signal frequencies to the same degree.
This often results from a natural practical limitation of the amplifier but
may also be due to the fact that the amplifier has been specifically
designed to amplify a small range of frequencies only. The general
shape of the frequency response of an ampifier which uses capacitors at
the input and output of the amplifier is shown in Figure 4.4(a). Such
amplifiers are referred to as capacitively coupled or ac amplifiers (see,
for example, Figures 3.24 and 3.29). For low-frequency input signals,
the presence of the input and output coupling capacitors causes the
amplifier frequency response to fall off as shown. The frequency at
which the voltage gain of the amplifier has fallen to 0.707Gv (-3 decibels)
is referred to as the low-frequency cut-off of the amplifier fL. At high
frequencies the amplifier gain decreases also due to internal transistor
capacitances. At the high-frequency end, the freqency at which the gain
is down 3dBs on its nominal value is referred to as the high-frequency
cut-off fH. The frequency difference fH- fL is referrred to as the bandwith
of the amplifier and defines the range of input signal frequencies over
which the amplifier can provide the specified gain.
It is possible to construct an amplifier which doesn’t use input and output
coupling capacitors. Consequently such an amplifier doesn’t have a
low-frequency cut-off and for low-frequency input signals the gain
remains constant right down to 0 Hz, as shown in Figure 4.4(b). Another
common type of amplifer is the so-called tuned amplifier which is
common in radio circuits and is designed to amplify a very narrow range
of frequencies only. A typical tuned amplifier response is shown in
Figure 4.4(c).
4-4
Communications Technology 1
gain
bandwidth
of tuned
amplifier
(b) dc amplifier
Gv
0.707Gv
(a) ac amplifier
1
10
10 2
10 3
(c) tuned amplifier
10
4
10 5
Gain (db)
db
Gv
Frequency (Hz)
(log scale)
3db
bandwidth of ac amplifier
Figure 4.4
Frequency Response of an
AC, DC and Tuned
Amplifiers
NEGATIVE FEEDBACK
bandwidth of dc amplifier
1
10
10 2
10 3
10
4
10 5
Frequency (Hz)
(log scale)
Another important technique that is used in amplifier design is negative
feedback . A detailed study of this topic is beyond the scope of this unit,
but basically feedback is a process whereby a signal (voltage or current)
is taken at the output side of th amplifier and a fraction of it is fed back
and subtracted from a signal on the input side of the amplifier. This
process is used mainly to alter the gain of the amplifer but also alters its
input resistance, its output resistance and its frequency response.
Figure 4.5 shows a basic amplifier with an open-loop (non-feedback)
voltage gain Gv = A in which feedback is applied such that a fraction β
of the output voltage across the load is fed back and subtracted from the
input voltage. The circuit may be easily analysed as follows.
Amplifier
+
Vin
+ Vf -
+
Vi
-
Figure 4.5
Amplifier with Negative
Feedback
Io
+
Gv =A
RL
-
Vo
β
Feedback
Network
The feedback signal Vf is
V f = βVo
Vin = Vi + V f = Vi + βVo
Vo = AVin = A(Vi + βVo )
Vo − AβVo = AVi
⇒ the voltage gain with feedback Gvf =
Unit 4: Amplifiers
Vo
A
=
Vi 1 − Aβ
4-5
If the magnitude of (1 - Aβ) is greater than 1 then the magnitude of Gvf
is less than that of A and the feedback is said to be negative, otherwise
the feedback would be positive and the amplifier output could potentially
oscillate, ie generate a sinusoid at a particular frequency . The easiest
way to ensure the feedback is negative is to use an amplifier which
provides a 180°phase shift between input and output, which means that
A will be negative and the magnitude of (1 - Aβ) is guaranteed to be
greater than unity. Thus in this case the gain with feedback can be written
as
Gvf =
Vo
A
1
=
=
1
Vi 1 + Aβ
A+β
If the term 1/A is much smaller than the quantity β, as is generally the
case, then the gain with feedback is given approximately by:
Gvf ≈
1
β
This is an important result because it indictes that in a feedback amplifier
the gain is determined essentially by the feedback network and not the
amplifier itself. We will see examples of this in the next section.
To conclude this section on negative feedback let’s consider its effect on
the frequency response of an amplifier. Figure 4.6 shows the typical
response of an ac amplifer without negative feedback. Its gain is Gv =
A and its upper and lower frequencies are fH and fL respectively. When
feedback is applied the gain falls to a value Gvf which is much smaller
than Gv = A, and the characteristic becomes much flatter. It follows that
the new lower and upper cut-off frequencies fLf and fHf are further apart
and the bandwidth of the feedback amplifier is increased. In fact the
product of gain and bandwidth ( the gain-bandwidth product ) in an
amplifier is a constant. Thus if b denotes the bandwith of the original
amplifier and bf denotes the bandwidth of the feedback amplifier then we
can say that
Gv × b = Gvf × b f
Gain
Gain without
Feedback
Gv = A
b
0.707A
bandwidth without feedback
Figure 4.6
Effect of Negative Feedback
on Amplifier Frequency
Response
4-6
Gain with
Feedback
Gvf
0.707Gvf
bf
bandwidth with feedback
f Lf
fL
f Hf
Frequency (Hz)
Communications Technology 1
OPERATIONAL
AMPLIFIERS
There are certain types of amplifier which have the general capability of
amplifying the difference between two input signals. When such an
amplifier is manufactured on a tiny piece of silicon and supplied as a
single integrated circuit (IC) package, it is generally referred to as an
operational amplifier or op-amp. The term ‘operational’ derives from
the fact that an op-amp can electronically perform such mathematical
operations as addition, multiplication and integration.
Fortunately, it is not necessary to understand the detailed circuitry inside
an op-amp in order to analyse and design circuits which contain op-amp
devices. We may therefore treat the op-amp as a ‘black box’, as shown
in Figure 4.7.
+ VS
Inverting Input
Figure 4.7
Circuit Symbol for an
Operational Amplifier
Positive Supply
v2
-
v1
+
Non-Inverting Input
vo
Negative Supply
- VS
Firstly for an op-amp to work two DC power supplies, ±VS , must be
connected to the device. These supplies are typically in the range ±5V
to ±15V. To simplify circuit diagrams containing op-amps, the power
supply connections are often omitted; but it is important to remember
that they must be connected for the amplifier to function.
There are two separate input terminals: the non-inverting input terminal
(marked +) and the inverting input terminal (marked -). When a
voltage V1 is applied to the non-inverting input, an amplified voltage Vo
= +AV1 appears at the output terminal, where A is the voltage gain of the
amplifier. The + sign indicates that the input and output voltages are in
phase.
If a voltage V2 is applied to the inverting input, an amplified voltage Vo
= -AV2 appears at the output terminal. The - sign indicates that the input
and output voltages are 180∞ out of phase (antiphase).
If voltages V1 and V2 are applied simultaneously then the output voltage
Vo is given by the expression:
Vo = +AV1 - AV2 = A(V1 - V2 )
That is, the input signal V2 is subtracted from the input signal V1 and the
result of the subtraction is amplified. This signal differencing and
amplification function is very useful in many practical applications. For
example, when an electrocardiogram (ecg), which is a signal indicatiing
the electrical activity of the heart, is acquired by placing a pair of
electrodes on the human body (chest), one electrode is the reference
electrode and the ecg signal potential at the other electrode is with
respect to this reference electrode. The ecg signal is very small, of the
order only of a few microvolts, and requires a high level of amplification.
Unfortunately both electrodes pick up high levels of other electrical
Unit 4: Amplifiers
4-7
activity in the human body which appears as high amplitude-noise
signals on both electrodes, thus masking the desired signal on the ecg
electrode. This is where an op-amp can prove extremely useful. The
reference electrode is fed to the inverting input of the op-amp and the ecg
signal electrode is fed to the non-inverting input. Since the noise on both
electrodes is virtually the same, the differencing action of the op-amp
causes the noise to be cancelled and the small ecg signal is amplified by
the high gain of the op-amp.
We can draw a ‘black box’ equivalent circuit of an op-amp as shown in
Figure 4.8 and summarise its main properties as follows:
.
A very high voltage gain, called the open-loop gain. This is
typically in the range 105 - 106 at DC and low frequencies, but
decreases with frequency as shown in Figure 4.9.
.
A very high input resistance Ri between the inverting and noninverting input terminals, which is typically of the order 1012Ω.
.
A very low output resistance Ro which is typically 100Ω.
+
V1
Ro
Vd = V1 - V2
Figure 4.8
‘Black Box’ Representation
of an Op-Amp
Vd
Ri
V2
Vo
AVd
_
Open-Loop
Voltage Gain
10 6
10 5
10 4
10 3
10 2
10
Figure 4.9
Open-loop voltage gain as a
function of frequency
4-8
1
0
10
10
2
10
3
10
4
10
5
10
6
10
7
10
8
Frequency (Hz)
Communications Technology 1
THE IDEAL OP-AMP
The above properties of an op-amp - high voltage gain, high input
resistance and low output resistance - imply that it is close to being an
ideal amplifying device. In fact, the analysis and design of op-amp
circuits is much simplified if ideal devices are assumed. The properties
of an ideal op-amp are:
.
Infinite open-loop gain (A = ∞).
.
Infinite input resistance (Ri = ∞). That is, the op-amp does not draw
any current from any source connected to it.
.
Zero output resistance (Ro = 0 Ω). That is, the output acts as an ideal
voltage source.
The equivalent circuit of an ideal op-amp is shown in Figure 4.10.
v0 = A (v1 - v 2 )
+
Figure 4.10
Equivalent circuit of an ideal
op-amp
INVERTING
AMPLIFIER
v2
v1
The very high gain of an op-amp (typically 105 - 106) is far too high for
many applications and some method of controlling the gain is required.
This is achieved using negative feedback as described above, in which
resistors are connected around the op-amp in such a way as to reduce the
gain. Consider the circuit shown in Figure 4.11.
i2
i1
v in
R1
P v2
v1
+
R2
v0
v0
-R 2
=
v in
R1
Figure 4.11
Op-Amp inverting amplifier
With the resistors R1 and R2 connected, the gain G = Vo /Vin is called the
closed-loop gain and will always be less than the open-loop gain, A,
which is the gain without any resistors connected. V1 and V2 denote the
values of the voltages at the non-inverting and inverting input terminals
of the op-amp. Since the non-inverting input terminal is connected to 0V
or ‘ground’, V1 = 0.
Vo = - AV2
where A is the open-loop gain of the amplifier. If the DC supply voltages
±VS applied to the op-amp are ±15V, and the op-amp gain A = 105, then
the maximum AC output voltage swing VO (max) from the amplifier is
30V and the maximum input voltage swing at point P, V2 (max) is given
by:
Unit 4: Amplifiers
4-9
V2 (max) =
Vo (max) 30
= 5 = 300 µV ( ±150 µV )
A
10
This is a very small voltage. In fact, if we assume an ideal op-amp in
which A = ∞, then V2 = 0 and the point P is virtually at ground or earth
potential. Thus, the point P is often referred to as a virtual earth or
virtual ground.
The current i1 through R1 is given by:
i1 =
Vin − V2 Vin
=
, since V2 = 0
R1
R1
Since the input resistance of the op-amp is so high, a negligibly small
current flows into the inverting input terminal and therefore all of the
current i which flows through R1 also flows through R2.
V o = V 2 - i2 R 2
= -i1R2, since V2 = 0 and i1 = i2
The gain G is given by:
G=
Vo −i1 R2
R
=
=− 2
Vin
i1 R1
R1
The gain of the circuit is therefore simply the ratio of the two resistors
o
connected around the op-amp. The minus sign indicates that a 180
phase shift exists between the input voltage Vin and the output voltage,
Vo. Hence the term ‘inverting amplifier’.
The input resistance Rin of the amplifier is given by:
Rin =
SAQ 2
Vin
Vin
=
= R1
i1
Vin / R1
An inverting operational amplifier circuit has R2 = 33 kΩ and
R1 = 3.3 kΩ. Calculate the voltage gain of the amplifier.
SAQ 3
Design an operational amplifier circuit which has a gain of 100 and an input resistance of 1 kΩ.
4-10
Communications Technology 1
NON-INVERTING
AMPLIFIER
Consider the circuit of Figure 4.12, which is similar to the previous
circuit, except in that the input voltage source is now directly connected
to the non-inverting input terminal and the input end of the resistor is
connected to ground.
i
R2
P
i
R1
v in
vin
v2 v1
+
v0
Figure 4.12
Op-amp non-inverting
amplifier
v0
v0
v in
= (1 +
R2
)
R1
An expression for the voltage gain of the circuit can again be calculated,
by assuming that no current flows into the amplifier input terminals and
that the voltage between the inverting and non-inverting input terminals
is zero.
For zero voltage between the inverting and non-inverting input terminals,
the voltage at point P must equal the input voltage Vin.
i=
Vin
R1
Vo = Vin + iR2 = Vin +
∴ G=

Vin
R 
R2 = Vin 1 + 2 
R1
R1 

Vo
R
= 1+ 2
Vin
R1
Because the gain is positive, the output voltage is in phase with the input
voltage and hence the name ‘non-inverting amplifier’. In this case, the
input resistance Rin of the amplifier is equal to the very high input
resistance of the op-amp, between the non-inverting and inverting
terminals.
SAQ 4
A non-inverting operational amplifier circuit has R1 = 1 kΩ
and R2 = 10 kΩ. Calculate the voltage gain of the amplifier.
SAQ 5
Design an operational amplifier circuit with a gain of +50, the
smallest resistor used being 1 kΩ.
VOLTAGE FOLLOWER
(BUFFER) AMPLIFIER
Consider the simple circuit of Figure 4.13. Again assuming an ideal opamp, with zero voltage difference between the inverting and noninverting input terminals, we can say that
Vo = Vin or
Unit 4: Amplifiers
Vo
=1
Vin
4-11
Figure 4.13
Voltage Follower or Buffer
Amplifier
v0
v in
+
Thus this circuit provides a gain of +1 and the output voltage simply
‘follows’ the input voltage - hence the term ‘voltage follower’. The
input resistance of the circuit is the high differential input of the
operational amplifier. This property of the circuit, together with the low
output resistance of the op-amp, makes it useful for converting a high impedance source to a low - impedance one. When it is used in this way
it is often referred to as a buffer amplifier.
SUMMING AMPLIFIER
It is often required to have a circuit which can add two or more voltages
together. Consider the circuit of Figure 4.14 which is capable of adding
two voltage signals V1 and V2 and amplifying the result. Assuming an
ideal op-amp.
i1 =
v
V1
v
, i2 = 2 and i = o
R
R
RF
i1
i
R
P
i2
v1
R
RF
+
v0
v2
Figure 4.14
Summing Amplifier Circuit
From Kirchoff’s Laws:
i + i1 + i2 = 0, or i = −(i1 + i2 )
∴
Vo
V V
= − 1 + 2 
 R R
RF
⇒ Vo = −
RF
(V1 + V2 )
R
The output signal, V0, is the sum of the input signal V1 and V2 amplified
by the factor - RF/R. More than two signals can be summed by having
additional input resistors, R. It is also possible to multiply the input
signals by different gain factors by having unequal input resistors R and
hence different gain factors RF/R.
4-12
Communications Technology 1
SAQ 6
In the circuit of Figure 4.14, R = 10 kΩ, RF = 10 kΩ. Derive
an expression for the output voltage for the following input
voltages.
(a) V1 = 2 sin (400πt), V2 = 3 sin (400πt)
(b) V1 = + 3V (dc), V2 = 4 sin (200πt)
SAQ 7
OP-AMP VOLTAGE
COMPARATOR
Design an op-amp summing amplifier which adds three input
signals V1, V2 and V3 such that V0 = -(V1+2V2+4V3 ). The
smallest resistor to be used is 1 kΩ.
Consider an operational amplifier connected in the open-loop
configuration with input signals V1 and V2 connected to the inverting
and non-inverting inputs as shown in Figure 4.15. From our previous
study we know that the output voltage in this configuration is given by
V0 = A(V1 - V2)
where A is the open-loop gain of the amplifier which normally has a
value in the range 105 - 106. The voltage difference between the inputs
V1 and V2 is amplified and appears at the output. Since the value of A
is so large a voltage difference of only a few hundred microvolts between
V1 and V2 causes the amplifier output to saturate, i.e. have a value close
to the supply voltage - either + VS or - VS. When V1 is slightly greater
than V2, then V0 ≅ + VS and when V1 is slightly less than V2, V0 ≅
- VS. Thus the circuit can be used as a comparator to determine whether
V1 is greater than or less than V2.
Suppose in a voltage comparator V1 is a triangular waveform and V2 is
a dc voltage as shown in Figure 4.16(a). The output voltage waveform
is as shown in Figure 4.16(b).
When the instantaneous value of the triangular signal, V1, is less than the
d.c signal, V2, the output signal has value close to - VS and when V1 is
greater than V2, the output signal has a value close to + VS. Thus V0
repeatedly switches between approximately -VS and +VS producing a
pulse waveform. The mark - to - space ratio, ie the ratio of the high-level
voltage to low-level voltage in each period, of the output waveform
depends on the switching threshold which is determined by the value of
the d.c voltage V2 .
+Vs
+
Figure 4.15
Op-Amp Voltage
Comparator Circuit
Unit 4: Amplifiers
v2
v1
v0
-Vs
4-13
v1
vin
v2
+v s
Figure 4.16
Comparator Input and
Output Signals
t
(b) Comparator
Output Signals
v0
-vs
SAQ 8
t
(a) Comparator
Input Signals
In a simple voltage comparator circuit the power supply
voltages are ± 15V. Input signal V1 is a triangular voltage
waveform with an amplitude of 10V and a period of 4
seconds. Calculate the mark-to-space ratio when V1 is a dc
voltage of:
(i) 0V
(ii) -5V
(iii) +5V
(iv) +12V
A disadvantage of the simple comparator configurations above is that
their behaviour is erratic in the presence of noise. If the input voltage has
superimposed noise, and has an amplitude close to Vref the output may
be forced to change states several times. This erratic behavior may be
suppressed by introducing hysteresis into the comparator. Hysteresis is
a phenomenon in which the transition point is different when switching
from the high-to-low state as compared with switching from the low-tohigh state. Comparators with hysteresis are often referred to as Schmitt
triggers. Figure 4.17 indicates the circuitry of an inverting Schmitt
Trigger.
Vo
V+
Sat
Vi
_
Vo
R1
Figure 4.17
(a) Circuitry of a Schmitt
Trigger,
(b) Output characteristic.
4-14
-Vt
Vt
Vi
_
R2
(a)
V Sat
(b)
Communications Technology 1
Resistors R1 and R2 form a voltage divider, establishing a voltage at the
noninverting terminal Vt, proportional to Vo.
Vt =
R2
Vsat
R1 + R2
when, Vo=+Vsat, the noninverting input = Vt
Vo=-Vsat, the noninverting input = -Vt
When Vi is negative, Vo = +Vsat and V+ = Vt. In order for the output to
change signs, Vi must exceed the value of Vt. To return to the previous
state the input voltage must reach a value of Vi < -Vt, causing the output
to return to +Vsat. By choosing a sufficiently large value of Vt the effects
of noise at the transition points may be minimised. Care however must
be taken to ensure that Vt does not become too large, in which case the
accuracy may be degraded for certain applications. The rectangle on the
input-output curve is referred to as a hysteresis loop. Note, it is
important to label the curve with arrows to indicate directions of change.
It is also possible to develop a non-inverting Schmitt trigger, which is
based on the same principles as above; however this circuit will have a
slightly different hysteresis loop.
SAQ 9
Design an inverting Schmitt Trigger having thresholds close
to +/- 50mV based on supply voltages to the op-amp of +/15V. The exact value of Vt is not critical in this application.
Assume op-amp saturation voltages of +/- 14V.
1.
An amplifier is an electronic device for increasing the amplitude of
a signal, and may be characterised by its voltage gain, its input
resistance, its output resistance, and its frequency response.
2.
Negative feedback is used to alter the gain of an amplifier but also
affects its input resistance, its output resistance and its frequency
response.
3.
An operational amplifier is a high gain, BJT, FET or BJT/FET
amplifier with a high input resistance and low output resistance. It
is normally supplied as a single integrated (IC) package.
4.
An ideal operational amplifier has infinite gain, infinite input
resistance and zero output resistance.
5.
The voltage gain G of the inverting amplifier circuit of Figure 4.11
is given by the expression:
SUMMARY
G=−
R2
R1
The input resistance Rin = R1 . The input and output voltages are out
o
of phase by 180 .
Unit 4: Amplifiers
4-15
6.
The voltage gain G of the non-inverting amplifier circuit of Figure
4.12 is given by the expression:
G = 1+
R2
R1
The input resistance Rin = ∞ . The input and output voltages are in
phase.
4-16
7.
A voltage follower or buffer amplifier has a high input impedance,
low output impedance and a gain of +1.
8.
A summing amplifier circuit may be used to add two or more
voltages together - in different ratios if required.
9.
A voltage comparator indicates whether the instantaneous value
of one input signal is greater than or less than that of another.
Communications Technology 1
ANSWERS TO SAQS
SAQ 1
dB
Gvo
= 20 log10
Gv =
Vo
= 60 ⇒ Gvo = anti log10 3 = 1000
Vi
Vo
Ri
90
RL
=
.Gvo .
=
× 1000 ×
Vs Rs + Ri
Ro + RL 10 + 90
450
= 0.9 × 1000 × 0.9 = 810
50 + 450
Vo = Gv Vs = 810 × 1mV = 0.81V (rms)
SAQ 2
Voltage Gain G = −
R2
33 × 10 3
=−
= −10
R1
3.3 × 10 3
SAQ 3
Input Resistance Rin = R1
Rin = 1 kΩ => R1 = 1 kΩ
Voltage Gain G = −
R2
= −100
R1
or R2 = 100 x R1 = 100 x 1 kΩ
= 100 kΩ.
Therefore, the circuit is as shown in Figure 4.18.
100k Ω
v in
1k Ω
v0
+
Figure 4.18
OP-AMP Circuit for SAQ 3
Unit 4: Amplifiers
4-17
SAQ 4

R 
10
Voltage Gain G = 1 + 2  = 1 +  = 11

R
1

1
SAQ 5

R 
Voltage Gain G = 1 + 2  = 50
R1 

⇒
R2
= 49 or R2 = 49 R1
R1
Let R1 = 1 kΩ
R2 = 49 x R1 = 49 x 1 kΩ = 49 kΩ.
The circuit is shown in Figure 4.19.
49k Ω
1k Ω
Figure 4.19
OP-AMP Circuit for SAQ 5
v0
v in
+
SAQ 6
(a)
Vo = −
RF
10
(V1 + V2 ) = − 10 (2 sin 400πt + 3 sin 400πt ) = −5 sin 400πt
R
(b) Vo = −1(3 + 4 sin 200ωt ) = −3 − 4 sin 200ωt
4-18
Communications Technology 1
SAQ 7
The basic summing circuit is shown in Figure 4.20.
Let R3 = 1 kΩ => RF = 4 kΩ.
∴
R2 = 2 kΩ and R1 = 4 kΩ.
v1
R1
RF
v2
R2
+
v3
R3
v0
Figure 4.20
SAQ 8
Consider the sketch of Figure 4.21.
The mark -to-space ratios are as follows:
(i) 1 : 1
(ii)
T
3T
:
= 1: 3
4
4
(iii)
3T
T
:
= 3 :1
4
4
(iv) Output voltage always low (≅ -15V)
Voltage
(volts)
V2
+15
+12V
+10
V1
+5
+5V
0
-5
Figure 4.21
0V
T/4
T/2
3T/4
T
t
-5V
-10
-15
Unit 4: Amplifiers
4-19
SAQ 9
Vt =
R2
Vsat
R1 + R2
50 × 10 −3 =
R2
.14
R1 + R2
0.05( R1 + R2 ) = 14 R2
⇒ 0.05 R1 = 13.95 R2
⇒ R1 = 279 R2
Let R2 = 100Ω and R1 = 27.9kΩ (28kΩ)
4-20
Communications Technology 1