Topic 4.5.2 – Regeneration
Learning Objectives:
At the end of this topic you will be able to;
 recall and describe the difference between noise and distortion.
 understand that the signal to noise ratio degrades down a transmission
line, and that regeneration restores the original signal.
 recognise, analyse and design inverting and non-inverting Schmitt
trigger circuits to regenerate digital signals.
1
Module ET4 – Communication Systems
Regeneration
In the previous topic we introduced the issues of attenuation, noise and
distortion, and how they might affect a signal during transmission. In summary
the main features of these effects are:
Attenuation : The amplitude of a signal decreases as it passes through the
transmission medium.
Noise: Random signals introduced that affect the original transmitted signal
as it passes through the transmission medium. These are impossible to predict.
Distortion : Systematic changes introduced that affect the original
transmitted signal by the components of the system. These can be predicted
in many cases e.g. clipping, crossover and it is sometimes possible to remove
these at the circuit.
Clearly if a transmission system is to be of any use these three effects need
to be eliminated as much as possible. In the previous topic we illustrated how
this is very difficult when dealing with analogue signals as the actual signal
information is contained in the amplitude of the signal. This susceptibility of
analogue signals to noise and distortion in particular and the development of
modern electronic circuits has led to their recent demise in favour of the
digital signal, which is now replacing analogue signals in many applications
including terrestrial TV broadcasting which has been analogue since the first
broadcast in 1941. By 2012 all TV signals in the UK will be digital. The
remaining part of this topic will therefore be devoted to how digital signals
can be regenerated.
2
Topic 4.5.2 – Regeneration
In topic 4.5.1 we used this diagram to illustrate the regeneration process.
In the first diagram we can see a strong digital signal, ready for transmission.
In the middle we see the same signal after transmission, which is now
attenuated and the effect of noise from the transmission process is clearly
visible.
In the final diagram we see a completely regenerated signal which can either
be passed down the next section of the transmission link or it can be used to
produce the final outcome e.g. a television picture.
We are generally unable to affect what happens during the transmission
medium, and so we have to concentrate on how we can reproduce the original
signal from the noisy input signal. The device responsible for this is similar to
the Schmitt NOT gate or Schmitt inverter we met briefly in Module ET2.
The symbol for the Schmitt inverter is as follows:
A
Q
The Schmitt NOT gate has a unique switching characteristic which is very
different to the standard NOT gate. The following diagrams illustrate the
difference between to two.
3
Module ET4 – Communication Systems
VOUT
VOUT
5
5
4
4
3
3
2
2
1
1
0
0
0
1
2
3
4
5
Switching Characteristic
for a standard NOT Gate
VIN
0
1
2
3
4
5
VIN
Switching Characteristic
for a Schmitt NOT Gate
Looking at the two characteristics you should notice that for a standard NOT
gate operating on a 5V supply that the switching point is at the midpoint of the
supply voltage for an increasing or decreasing input voltage.
Comparing this to the Schmitt characteristic reveals a different situation
altogether. As VIN increases the voltage has to increase above 3V before the
output voltage changes. Once the output has changed however, if the input is
then decreased back to 3V, the output does not change back, as it would in the
normal case, but now the voltage has to fall to below 2V before the output will
go high again.
We have therefore created some hysteresis in the NOT gate with two
distinct switching thresholds. It is these different switching thresholds that
allow the regeneration of the digital signal.
The following example will illustrate how this works.
4
Topic 4.5.2 – Regeneration
Example: A Schmitt inverter has the following switching thresholds:
0V to 5V transition at 2V and
5V to 0V transition at 3.5V.
The Schmitt inverter is connected to the signal coming from a
transmission line. The input signal to the Schmitt inverter is shown
in the graph below. Draw the output signal on the axes provided.
6
Vin/V
4
2
0
time/s
Vout/V
5
0
time/s
We will consider the construction of the output signal in stages so that you will
be able to see how the solution is obtained, once you have mastered the steps
required you will be able to complete this on one graph.
5
Module ET4 – Communication Systems
Step 1 : First we need to mark in the thresholds for the Schmitt inverter. In
this example these were at 2V and 3.5V.
6
Vin/V
4
upper
threshold
lower
threshold
2
0
time/s
Vout/V
5
0
time/s
6
Topic 4.5.2 – Regeneration
Step 2 : Secondly we mark in the points where transitions occur. i.e. when a
rising voltage passes the upper threshold (3.5V) and a falling voltage
passes the lower threshold (2V).
6
Vin/V
4
upper
threshold
lower
threshold
2
0
time/s
Vout/V
5
0
time/s
7
Module ET4 – Communication Systems
Step 3 : Finally we can insert the output graph, now that we have the
transition points. We must remember that the general function of
the Schmitt inverter is to invert the input signal, which means when
the input signal is very low the output is high, and vice versa.
6
Vin/V
4
upper
threshold
lower
threshold
2
0
time/s
Vout/V
5
0
time/s
This example should illustrate the usefulness of the Schmitt inverter in being
able to remove noise from an incoming digital signal and produce a ‘clean’ digital
signal at the output.
There are two limitations with the simple Schmitt inverter
 the switching thresholds are usually a function of the supply voltage,
 they are unable to cope with negative voltage inputs.
As a number of transmission systems use a dual rail system where digital logic
levels might be at ±5V for example, this is a big problem.
8
Topic 4.5.2 – Regeneration
We need a different device that can cope with these issues. This device is
based on the comparator we have already met in ET2 with the condition that
part of the output signal is fed back to the non-inverting input to produce
hysteresis. The circuit is referred to as a Schmitt Trigger. There are two
possible versions of the Schmitt trigger, inverting and non-inverting.
Before we look in detail at how this needs to be connected in a circuit and how
we determine the switching thresholds, let us apply our previous idea to
construct the output graph for a dual rail system.
Example:
An inverting Schmitt trigger has the following switching
thresholds:
9V to -9V transition at -2.4V and -9V to +9V transition at 3.6V.
The inverting Schmitt trigger is connected to the signal coming
from a transmission line.
The input signal to the Schmitt trigger is shown in the graph on the
following page. Draw the output signal on the axes provided.
9
Module ET4 – Communication Systems
Remember the switching characteristics are: 9V to -9V transition at -2.4V and
-9V to +9V transition at 3.6V.
6
Vin/V
4
2
0
time (s)
-2
-4
-6
Vout/V
10
0
time (s)
-10
10
Topic 4.5.2 – Regeneration
Step 1 : Mark in the thresholds for the Schmitt trigger. In this example these
were at -2.4V and 3.6V
6
Vin/V
4
Upper
Threshold
2
0
time (s)
-2
Lower
threshold
-4
-6
Vout/V
10
0
time (s)
-10
11
Module ET4 – Communication Systems
Step 2 : Secondly we mark in the points where transitions occur. i.e. when a
rising voltage passes the upper threshold (3.6V) and a falling voltage
passes the lower threshold (-2.4V).
6
Vin/V
4
Upper
Threshold
2
0
time (s)
-2
Lower
threshold
-4
-6
Vout/V
10
0
time (s)
-10
12
Topic 4.5.2 – Regeneration
Step 3 : Finally we can insert the output graph, now that we have the
transition points. We must remember that as this is an inverting
Schmitt trigger we have to invert the output, which means when the
input signal is very low the output is high, and vice versa.
6
Vin/V
4
Upper
Threshold
2
0
time (s)
-2
Lower
threshold
-4
-6
Vout/V
10
0
time (s)
-10
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Module ET4 – Communication Systems
We will now focus our attention on how to create a Schmitt trigger from a
comparator.
The simplest possible Schmitt trigger we can create using an op-amp
configured as a comparator, is as shown below:
+
Vin
Vout
If the comparator is run on a dual rail supply of ±15V, the output will saturate
at approximately ±13V. We will assume to begin with that the output is in
positive saturation, i.e. Vout = +13V. This voltage will be fed back to the noninverting input. For the output to switch to the negative saturation value, the
voltage at the inverting input to the comparator, Vin will need to go above +13V.
Similarly if the comparator is in negative saturation, i.e. -13V, then for the
output to switch to the positive saturation value, the voltage at the inverting
input to the comparator, Vin will need to go below -13V.
For the example given above you should be able to work out that the switching
thresholds are at the saturation values of the comparator (op amp). The
switching characteristic for the circuit above would be as follows:
Vout/V
+15
-15
+15
-15
14
Vin/V
Topic 4.5.2 – Regeneration
The fact that the output voltage is as high as the saturation values is
important, as this provides a large output signal for sending the signal on to
the next stage of the communications system. However the fact that the
switching thresholds are also at ±13V can cause some difficulties, since the
incoming signal is likely to be attenuated, and may not actually reach either of
the switching thresholds to cause the output to change, which makes this
circuit virtually redundant, as it can only regenerate very strong signals, and if
the signals are strong why would we need to regenerate them.
One way to reduce the switching
threshold voltage is to use a couple of
zener diodes in the circuit. An
example is shown opposite.
+
Vin
Vout
4.3V
4.3V
If you remember from your ET2
module the zener diode in forward
bias, has a voltage drop of 0.7V across
it, whilst in reverse bias, it is the zener voltage that appears across it.
For the two zener diodes in the circuit above, one will always be forward
biased and one always reversed biased, when the comparator tries to reach its
saturation voltage of ±13V, the action of the zener diodes will be to force the
output voltage to remain at ±5V, also keeping the switching threshold to ±5V,
as shown on the characteristic below.
Vout/V
+15
-15
+15
Vin/V
-15
15
Module ET4 – Communication Systems
Using two different zener diodes in the circuit allows the possibility of a
different switching threshold for the positive and negative switching
threshold. An example is shown below.
+
Vin
Vout
6.2V
2.7V
When the output voltage tries to reach positive saturation, the 6.2V zener is
forward biased, therefore the voltage drop across it will be 0.7V. The 2.7V
zener is reversed bias, and therefore the voltage drop across it will be the
zener voltage of 2.7V. Vout will therefore be clamped at 0.7+2.7=3.4V.
When the output voltage tries to reach negative saturation, the 6.2V zener is
reverse biased, therefore the voltage drop across it will be the zener voltage
of 6.2V. The 2.7V zener is forward biased, and therefore the voltage drop
across it will be 0.7V. Vout will therefore be clamped at –(6.2+0.7)=-6.9V.
The characteristic for this circuit therefore looks like this:
Vout/V
+15
-15
+15
-15
16
Vin/V
Topic 4.5.2 – Regeneration
Using zener diodes in this way, would allow switching thresholds at different
values to be determined, however the downside of this system is that the
output voltage is also clamped at the same voltage values, which reduces its
usefulness as a digital signal regenerator.
We have so far considered two designs for inverting Schmitt triggers,
(i)
(ii)
firstly using direct feedback of the output voltage to one of the
inputs which resulted in a high output voltage as required, but
unfortunately high switching thresholds also and it is unlikely that
signals will be able to reach these thresholds after transmission
through a system.
secondly, using zener diodes to reduce the input switching
thresholds, but that has the disadvantage that it also reduces the
output voltage levels which again is not really what we are trying to
achieve in the regeneration process.
A different approach is therefore needed, and one which does not prevent the
output voltage from reaching it’s saturation value.
17
Module ET4 – Communication Systems
Calculations of Schmitt Trigger Thresholds
There are two types of Schmitt trigger, inverting and non-inverting. You will
need to be able to work with both systems, and we will consider both but we
will start with the non-inverting type as this is slightly easier to understand.
Non-inverting Schmitt Triggers
One such solution is as shown below.
0V
+
VIN
10kΩ X 20kΩ
VOUT
Let us assume that this comparator made from an op-amp saturates at ±12V.
So how does the circuit work ?
You need to think back to the basic operation of the comparator, when the
voltage at the non-inverting input is greater than the voltage at the inverting
input the output will be in positive saturation. For the circuit above, this means
when the voltage at point ‘X’ goes above 0V the output will be in positive
saturation.
Similarly if the voltage at point ‘X’ goes below 0V the output will be in negative
saturation since the voltage at the inverting input will now be larger than the
voltage at the non-inverting input.
The voltage at ‘X’ is determined by the value of VIN and the value of VOUT. To
make the Schmitt trigger switch between positive and negative saturation, we
need to change the value of VIN, to make the voltage at ‘X’ pass through 0V.
The voltage required to make the voltage at ‘X’ pass through 0V, will be the
switching thresholds for the Schmitt trigger.
18
Topic 4.5.2 – Regeneration
We will now examine how to calculate the switching thresholds which cause the
Schmitt trigger to switch between it’s two saturation values.
Let us reconsider the circuit again as shown below:
0V
-
+12V
+
VIN
10kΩ X 20kΩ
If the circuit is in the condition shown with VOUT at +12V, then VX must be >0V.
We now need to calculate the value of VIN that will cause VX to fall to 0V since
anything below this will then cause VX to be <0V and the output will be in
negative saturation.
We can re-draw the equivalent circuit as shown below:
IR1
R1
+12V
20kΩ
VX=0V
R2
10kΩ
Note: From diagram VIN < 0V
I R1 
12  0 12

 0.6mA
20k
20k
V10k  0.6mA10k
 6V
VIN  VX  V10k
 0  6  6V
VIN
This calculation shows that the value of VIN that causes the voltage at ‘X’ to
fall to 0V is -6V, in practice the output will switch as this voltage transfers
from -5.999V to -6.001V for example. The comparator will switch to negative
saturation instantly because of the high open loop gain of the amplifier, and
therefore we say that -6V effectively is the lower switching threshold.
19
Module ET4 – Communication Systems
Now for the upper switching threshold. The circuit will be in negative
saturation with VOUT at -12V, then VX must be <0V. We now need to calculate
the value of VIN that will cause VX to rise to 0V since anything below this will
cause VX to be <0V and the output will be in negative saturation.
In this case the output of the Schmitt trigger will be at -12V, so we need to
re-draw the equivalent circuit as shown below:
IR1
R1
-12V
Note: From the diagram VIN > 0V, and IR1 flows from VX
towards the -12V.
20kΩ
I R1 
VX=0V
R2
0  ( 12 ) 12

 0.6mA
20k
20k
V10k  0.6mA10k
 6V
10kΩ
VIN  VX  V10k
 0  6  6V
VIN
This calculation shows that the value of VIN that causes the voltage at ‘X’ to
rise to 0V is +6V, in practice the output will switch as this voltage transfers
from 5.999V to 6.001V for example. The comparator will switch instantly to
positive saturation because of the high open loop gain of the amplifier, and
therefore we say that effectively this is the upper switching threshold.
The characteristic for this Schmitt trigger is therefore as shown below.
Vout/V
+15
-15
20
+15
-15
Vin/V
Topic 4.5.2 – Regeneration
There are a couple of observations to make from this characteristic:
1.
2.
3.
The output voltage of the Schmitt trigger is now at the threshold values.
The switching thresholds are lower than the saturation values.
The action of the Schmitt trigger is non-inverting.
The solution to this problem was derived using the equivalent circuit of a
voltage divider.
An alternative solution can be obtained by considering a balanced current
approach which can be found in some text books. We will now consider the
same problem and obtain the solution using the alternative approach.
Balanced Current Approach.
Let us assume that the output is in positive saturation, at +12V.
-
0V
+12V
+
VIN
10kΩ X 20kΩ
From Module ET2, you should remember that the input impedance of the
comparator is extremely high, and considered to be infinite for the purposes
of calculations. This means that no current will flow from point ‘X’ into the
non-inverting input of the comparator. Therefore the current in the 20k
resistor must be the same as the current in the 10k resistor. So to calculate
the switching thresholds – we need to find the voltage at VIN that causes the
voltage at ‘X’ to be 0V.
We can write down two equations that could be used to calculate the current
through the resistors.
I
 12V  VIN
20k  10k
or
I
12  0
20k
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Module ET4 – Communication Systems
Now these equations can be equated (since the value of I is the same) to give:
12  VIN
12  0

20k  10k
20k
12  30k
12  VIN 
20k
12  VIN  18V
12  18  VIN
VIN  6V
We can carryout a similar analysis for when the output is at -12V to determine
the upper threshold.
 12  VIN  12  0

20k  10k
20k
 12  30k
 12  VIN 
20k
 12  VIN  18V
 12  18  VIN
VIN  6V
Using this method we have arrived at the same solution, as before. In the
examination you can use either method. You will never be asked to use one
particular method so use whichever you are most comfortable with.
22
Topic 4.5.2 – Regeneration
In the previous example the voltage at the inverting voltage was set to 0V.
This is sometimes called the reference voltage (VREF). This resulted in
symmetrical switching thresholds of ±6V. The symmetrical nature was due to
the fact that the voltage at the inverting input was 0V, and the 6V was due to
the resistor values. Now we will consider what happens when the voltage at the
inverting input is not zero. We will analyse this circuit using both the voltage
divider method and the balanced currents approach side by side so that you
can compare the different methods.
A Schmitt trigger circuit is shown in the following circuit diagram.
-
3V
VOUT
+
R2
VIN
X
10kΩ
R1
20kΩ
The comparator saturates at ±9V.
i)
Calculate the value of VIN which causes VOUT to change from +9V to -9V
Method A – Voltage Divider
VOUT = +9V
IR1
R1
20kΩ
R2
10kΩ
VX = 3V
VIN = ?V
Note: From diagram V IN < 3V
I R1 
93
6

 0.3mA
20k 20k
Method B – Balanced Currents
VOUT  VIN VOUT  V X

R1  R2
R1
9  VIN
93

( 20  10) k 20 k
9  VIN
6

30 k
20 k
6  30 k
9  VIN 
20 k
9  VIN  9
9  9  VIN
VIN  0V
V10k  0.3mA  10k
 3V
VIN  VX  V10k
 3  3  0V
23
Module ET4 – Communication Systems
ii)
Calculate the value of VIN which causes VOUT to change from -9V to +9V.
Method A – Voltage Divider
VOUT = -9V
Method B – Balanced Currents
VOUT  VIN VOUT  V X

R1  R2
R1
IR1
R1
20kΩ
R2
10kΩ
 9  VIN
93

(20  10) k
20 k
 9  VIN  12

30 k
20 k
 12  30 k
 9  VIN 
20 k
 9  VIN  18
VX = 3V
VIN = ?V
 9  18  VIN
Note: From diagram VIN > 3V
VIN  9V
3  (9) 12
I R1 

 0.6mA
20k
20k
V10k  0.6mA  10k
 6V
VIN  VX  V10k
 3  6  9V
The characteristic for this Schmitt trigger is therefore as shown below.
Vout/V
+15
-15
+15
-15
24
Vin/V
Topic 4.5.2 – Regeneration
Inverting Schmitt Triggers
Occasionally, we need an inverting Schmitt trigger, in the same way as we had
when we looked at the first two ways of regenerating a signal. Fortunately the
construction of such a system is very straightforward as shown below.
-
VIN
VOUT
+
R2
VREF=3V
10kΩ
X
R1
20kΩ
Careful observation of the circuit will show that the only change has been to
swap the location of VIN and the 3V reference voltage.
The analysis can be performed in exactly the same way, it’s just that the value
of VIN now appears at ‘X’ in the equivalent circuits. Consider the following
example to see how this is carried out.
25
Module ET4 – Communication Systems
An inverting Schmitt trigger circuit is shown in the following circuit diagram.
-
VIN
VOUT
+
R2
VREF= 2V
10kΩ
X
R1
30kΩ
The comparator saturates at ±10V.
i)
Calculate the value of VIN which causes VOUT to change from +10V to -10V
Method A – Voltage Divider
VOUT = +10V
I
R1
30kΩ
VIN = ?V
R2
10kΩ
VR= +2V
Note: From the diagram VIN is
somewhere between +2V and
+10V
10  2
8
I

 0.2mA
30k  10k 40k
V10k  0.2mA10k
 2V
VIN  V10k  2V
 2  2  4V
26
Method B – Balanced Currents
VOUT  VIN VOUT  VR

R1
R1  R2
10  VIN
10  2

30k
30k  10k
10  VIN
8

30k
40k
8  30k
10  VIN 
40k
10  VIN  6
10  6  VIN
VIN  4V
Topic 4.5.2 – Regeneration
ii)
Calculate the value of VIN which causes VOUT to change from -10V to +10V.
Method A – Voltage Divider
VOUT = -10V
Method B – Balanced Currents
VOUT  VIN VOUT  VR

R1
R1  R2
I
R1
30kΩ
 10  VIN
 10  2

30k
30k  10k
 10  VIN  12

30k
40k
 12  30k
 10  VIN 
40k
 10  VIN  9
VIN = ?V
R2
10kΩ
VR= +2V
Note: From diagram -10V < VIN < +2V
 10  9  VIN
2  ( 10 ) 12
I

 0.3mA
30k  10k 40k
VIN  1V
V10k  0.3mA10k
 3V
VIN  2V  V10k
 2  3  1V
The characteristic for this Schmitt trigger is therefore as shown below.
Vout/V
+15
-15
+15
Vin/V
-15
27
Module ET4 – Communication Systems
Up until now we have only considered what happens at the switching
thresholds. In an examination you may be asked to calculate the voltage at
point ‘X’ at some condition other than at the transition point between one
output state and the other. This is much easier calculation to perform as we
do not have to concern ourselves with what happens at the inverting input. The
following example will explain what to do.
A non-inverting Schmitt trigger circuit is shown in the following circuit
diagram. The op-amp saturates at ±9V.
-
3V
VOUT
+
R2
VIN
i)
X
10kΩ
R1
20kΩ
Calculate the voltage at X when VIN = +6V, and VOUT = +9V.
Method A – Voltage Divider
VOUT = +9V
I
R1
20kΩ
R2
10kΩ
VX = ?V
VIN = +6V
Method B – Balanced Currents
VOUT  VIN VOUT  V X

R1  R2
R1
9  VX
96

(20  10)k
20 k
3  20 k
 9  VX
30 k
2  9  VX
V X  9  2  7V
Note: From the diagram VX is somewhere
between +6V and +9V
I
96
3

 0.1mA
30k 30k
V10k  0.1mA  10k
 1V
VX  V10k  VIN
 1  6  7V
Now its time for you to have a go at some examples.
28
Topic 4.5.2 – Regeneration
Student Exercise 1:
1.
The following circuit shows a Schmitt trigger built from a comparator
and a couple of zener diodes. The comparator is connected to a ±10V
supply.
Vin
+
Vout
4.3V
3.9V
a.
Determine the switching thresholds for this Schmitt trigger.
...........................................................................................................................
...........................................................................................................................
b.
Sketch the characteristic for this circuit on the grid below.
29
Module ET4 – Communication Systems
2.
A Schmitt trigger circuit is shown in the following circuit diagram.
-
VIN
VOUT
+
R2
VR= 3V
20kΩ
X
R1
30kΩ
The comparator saturates at ±12V.
i)
Calculate the value of VIN which causes VOUT to change from +12V to -12V
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
ii)
Calculate the value of VIN which causes VOUT to change from -12V to +12V.
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
30
Topic 4.5.2 – Regeneration
iii)
Sketch the characteristic for this Schmitt trigger below.
Vout/V
+15
-15
+15
Vin/V
-15
3.
A Schmitt trigger circuit is shown in the following circuit diagram.
-
1V
VOUT
+
R2
VIN
X
15kΩ
R1
30kΩ
The comparator saturates at ±10V.
i)
Calculate the value of VX, when VIN = +4V and VOUT=+10V
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
31
Module ET4 – Communication Systems
ii)
Calculate the value of VIN which causes VOUT to change from +10V to –10V
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
ii)
Calculate the value of VIN which causes VOUT to change from -10V to +10V.
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
………………………………………………………………………………………………..
iii)
Sketch the characteristic for this Schmitt trigger below.
Vout/V
+15
-15
+15
-15
32
Vin/V
Topic 4.5.2 – Regeneration
Design of Schmitt triggers
So far we have considered how to determine the output for a noisy signal, for
a Schmitt trigger of known switching thresholds and how to determine the
switching thresholds of a Schmitt trigger from a circuit diagram. The only
remaining issues to deal with are:
(i)
(ii)
to suggest suitable switching thresholds for a Schmitt trigger,
given the input waveform,
to determine the resistor values for a Schmitt trigger, built from
a comparator.
Firstly we will consider how to determine switching thresholds for a Schmitt
trigger from an input waveform. Assume that the following input is received
from a transmission system.
6
Vin/V
4
2
0
time/s
The switching thresholds need to be set at a level which can distinguish
between the noise and a transition between logic 1 and 0, and vice versa.
33
Module ET4 – Communication Systems
For the example shown previously, we can select thresholds within the bands
shown below
6
4
Vin/V
Upper Threshold
Lower Threshold
2
0
time/s
So suitable values can fit into the range as follows:
Upper Threshold : Between 3.5V to 4.4V
Lower Threshold : Between 2.6V to 3.4V
Once you have selected your thresholds you might then be asked to carry on
and draw the output graph in the same way that you did at the start of this
topic. The exact output graph will vary in shape depending on where the
switching thresholds are set but for examination purposes this will not matter
as long as the changes in output occur at the nominated switching thresholds.
Finally we will now look at how to determine the resistors needed to setup
specific thresholds in comparator circuits.
34
Topic 4.5.2 – Regeneration
Example 1:
A non-inverting Schmitt trigger is required to have saturation values of ±12V,
and switching thresholds of ±3V. Determine the values of VREF, R1 and R2 in the
circuit below.
-
VREF
VOUT
+
R2
VIN
X
15kΩ
R1
30kΩ
i)
In this case VREF must be equal to 0V because the threshold values
are symmetrical at ±3V.
ii)
Now for the calculation of R1 and R2. We have to consider the
conditions that need to occur to cause a transition. If the output is
in positive saturation, then the input VIN must fall below the lower
switching threshold for the output to switch to negative saturation.
The voltage at point ‘X’ will be 0V.
VOUT = +12V
I
I
12  0 12

and
R1
R1
I
0  ( 3 ) 3

R2
R2
R1
The current I is the same so we can equate
these equations.
VX = 0V
R2
VIN =-3V
12 3

R1 R2
12 R2
 R1
3
4 R2  R1
35
Module ET4 – Communication Systems
An alternative approach is shown below.
From the diagram we can see that
VOUT = +12V
I
VR1  12V
VR2  3V
R1
so
VX = 0V
VR1
VR2
R2

R1
R2
12 R1

3 R2
VIN =-3V
R1  4 R2
We have arrived at the same ratio for the resistors using both methods.
Either method is suitable in the examination.
So all that remains now is to choose suitable values for the resistors. Any
values over 1kΩ in an appropriate ratio would be fine, so for example,
R1 = 4kΩ, R2 = 1kΩ;
R1 = 40kΩ, R2 = 10kΩ;
R1 = 16kΩ, R2 = 4kΩ;
R1 = 8.8kΩ, R2 = 2.2kΩ; etc are all
acceptable answers.
The final circuit could then look like this:
-
0V
VOUT
+
R2
VIN
10kΩ
36
X
R1
40kΩ
Topic 4.5.2 – Regeneration
Example 2:
A non-inverting Schmitt trigger is required to have saturation values of ±13V,
and switching thresholds of +0.8V and +6.5V. Determine the values of R1 and R2
in the circuit below if VREF=3V.
-
VREF=3V
VOUT
+
R2
VIN
X
15kΩ
R1
30kΩ
Now for the calculation of R1 and R2. We have to consider the
conditions that need to occur to cause a transition. If the output is
in positive saturation, then the input VIN must fall below the lower
switching threshold for the output to switch to negative saturation.
The voltage at point ‘X’ will be 3V.
VOUT = +13V
I
I
13  3 10

and
R1
R1
I
3  0.8 2.2

R2
R2
R1
The current I is the same so we can equate
these equations.
VX = 3V
R2
VIN =+0.8V
10 2.2

R1 R2
10 R2  2.2 R1
R2  0.22 R1
Suggestions for R1 therefore could be 10kΩ, making R2 = 2.2kΩ
37
Module ET4 – Communication Systems
Again the solution could have been obtained via the alternative approach.
From the diagram we can see that
VOUT = +13V
I
VR1  10V
VR2  2.2V
R1
so
VX = 3V
VR1
VR2
R2

R1
R2
10 R1

2 .2 R 2
VIN =+0.8V
R2  0.22 R1
Again we have arrived at the same ratio for the resistors.
The final circuit could then look like this:
-
3V
VOUT
+
R2
VIN
2.2kΩ
38
X
R1
10kΩ
Topic 4.5.2 – Regeneration
Example 3:
An inverting Schmitt trigger is required to have saturation values of ±10V, and
switching thresholds of -2.5V and +5V. Determine the values of R1 and R2 in
the circuit below if VREF=2V.
-
VIN
VOUT
+
R2
VREF=2V
X
15kΩ
R1
30kΩ
Now for the calculation of R1 and R2. We have to consider the
conditions that need to occur to cause a transition. If the output is
in positive saturation, then the input VIN must rise above the upper
switching threshold for the output to switch to negative saturation.
The voltage at point ‘X’ must be equal to this upper threshold which
in this case is +5V.
VOUT = +10V
I
I
10  5 5

R1
R1
and
I
52 3

R2
R2
R1
The current I is the same so we can equate
these equations.
VIN = 5V
R2
VREF =+2V
5
3

R1 R2
5 R2  3R1
R2  0.6 R1
Suggestions for R2 therefore could be 10kΩ, making R1 = 6kΩ
39
Module ET4 – Communication Systems
And using the alternative approach we have.
From the diagram we can see that
VOUT = +10V
I
VR1  5V
VR2  3V
R1
so
VIN =+5V
VR1
VR2
R2

R1
R2
5 R1

3 R2
VREF =+2V
R2  0.6 R1
We have arrived at the same ratio for the resistors using both methods.
The final circuit could then look like this:
-
VIN
VOUT
+
R2
2V
10kΩ
X
R1
6kΩ
Note: In an examination you will only be asked to give VREF for symmetrical
switching thresholds which means that VREF will be 0V. In non-symetrical
circuits VREF will be provided.
Now here are some examples for you to do.
40
Topic 4.5.2 – Regeneration
Student Exercise 2:
1.
a.
The following signal is received from a transmission line. Determine
suitable switching thresholds for a Schmitt trigger to regenerate
the original signal.
Upper Threshold = ............................
b.
6
Lower Threshold = ............................
Using your switching thresholds from part (a) complete the output
signal for a non-inverting Schmitt trigger with saturation at ±8V.
Vin/V
4
2
0
time (s)
-2
-4
-6
Vout/V
10
0
time (s)
41
-10
Module ET4 – Communication Systems
2.
A non-inverting Schmitt trigger is required to have saturation values of
±9V, and switching thresholds of ±4V. Determine the values of VREF, R1
and R2 in the circuit below.
-
VREF
VOUT
+
R2
VIN
15kΩ
X
R1
30kΩ
i)
What is the value of VREF ? .......................................................
ii)
Determine the value of R1 and R2 to meet the design specification.
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
R1 = ....................................
42
R2 = .......................................
Topic 4.5.2 – Regeneration
3.
An inverting Schmitt trigger is required to have saturation values of
±15V, and switching thresholds of ±5V. Determine the values of VREF, R1
and R2 in the circuit below.
-
VIN
VOUT
+
R2
VREF
15kΩ
X
R1
30kΩ
i)
What is the value of VREF ? .......................................................
ii)
Determine the value of R1 and R2 to meet the design specification.
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
R1 = ....................................
R2 = .......................................
43
Module ET4 – Communication Systems
4.
A non-inverting Schmitt trigger is required to have saturation values of
±11V, and switching thresholds of -5V and +2.33V. Determine the values
of R1 and R2 in the circuit below if VREF=-1V.
-
VREF=-1V
VOUT
+
R2
VIN
15kΩ
X
R1
30kΩ
Determine the value of R1 and R2 to meet the design specification.
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
.........................................................................................................................................
R1 = ....................................
44
R2 = .......................................
Topic 4.5.2 – Regeneration
Solutions to Student Exercises.
Exercise 1:
1.
The following circuit shows a Schmitt trigger built from a comparator
and a couple of zener diodes. The comparator is connected to a ±10V
supply.
+
Vin
Vout
4.3V
3.9V
a.
Switching thresholds are +0.7+3.9 =+4.6V; -0.7-4.3=-5V
b.
Sketch the characteristic for this circuit on the grid below.
VOUT/V
10
5
5
-10
-5
5
10 VIN/V
-5
-10
45
Module ET4 – Communication Systems
2.
A Schmitt trigger circuit is shown in the following circuit diagram.
-
VIN
VOUT
+
R2
X
VR= 3V
20kΩ
R1
30kΩ
The comparator saturates at ±12V.
i)
Calculate the value of VIN which causes VOUT to change from +12V to -12V
Method A – Voltage Divider
VOUT = +12V
I
R1
30kΩ
R2
20kΩ
VIN = ?V
VR =+3V
Note: From diagram 10V < V IN < 3V
I
12  3
9

 0.18mA
( 30  20 )k 50k
V20k  0.18mA 20k
 3.6V
VIN  3  V10k
 3  3.6  6.6V
46
Method B – Balanced Currents
VOUT  VIN VOUT  VR

R1
R1  R2
12  VIN
12  3

30k
30k  20k
12  VIN
9

30k
50k
9  30k
12  VIN 
50k
12  VIN  5.4
12  5.4  VIN
VIN  6.6V
Topic 4.5.2 – Regeneration
ii)
Calculate the value of VIN which causes VOUT to change from -12V to +12V.
Method A – Voltage Divider
VOUT = -12V
Method B – Balanced Currents
VOUT  VIN VOUT  VR

R1
R1  R2
I
R1
30kΩ
 12  VIN
 12  3

30k
30k  20k
 12  VIN  15

30k
50k
 15  30k
 12  VIN 
50k
 12  VIN  9
VIN = ?V
R2
20kΩ
VR= +3V
Note: From diagram -12V < VIN < +3V
 12  9  VIN
3  ( 12 ) 15
I

 0.3mA
30k  20k 50k
VIN  3V
V20k  0.3mA 20k
 6V
VIN  3V  V20k
 3  6  3V
The characteristic for this Schmitt trigger is therefore as shown below.
Vout/V
+15
-15
+15
Vin/V
-15
47
Module ET4 – Communication Systems
3.
A Schmitt trigger circuit is shown in the following circuit diagram.
-
1V
VOUT
+
R2
VIN
X
15kΩ
R1
30kΩ
The comparator saturates at ±10V.
i)
Calculate the voltage at X when VIN = +4V, and VOUT = +10V.
Method A – Voltage Divider
VOUT = +10V
I
R1
30kΩ
R2
15kΩ
VX = ?V
VIN = +4V
Note: From the diagram VX is somewhere
between +4V and +10V
I
10  4
6

 0.133mA
45k
45k
V15k  0.133mA  15k
 2V
VX  V15k  VIN
 2  4  6V
48
Method B – Balanced Currents
VOUT  VIN VOUT  VX

R1  R2
R1
10  4
10  VX

( 30  15 )k
30k
6  30k
 10  VX
45k
4  10  V X
VX  10  4  6V
Topic 4.5.2 – Regeneration
i)
Calculate the value of VIN which causes VOUT to change from +10V to -10V
Method A – Voltage Divider
VOUT = +10V
I
R1
30kΩ
R2
15kΩ
VX = ?V
Method B – Balanced Currents
VOUT  VIN VOUT  V X

R1  R2
R1
10  V X
10  4

( 30  15 )k
30k
6  30k
 10  V X
45k
4  10  V X
V X  10  4  6V
VIN = +4V
Note: From the diagram VX is somewhere
between +4V and +10V
I
10  4
6

 0.1333mA
45k
45k
V15k  0.1333mA15k
 2V
VX  V10k  VIN
 2  4  6V
ii)
Calculate the value of VIN which causes VOUT to change from -10V to +10V.
Method A – Voltage Divider
VOUT = -10V
IR1
R1
30kΩ
R2
15kΩ
VX = 1V
VIN = ?V
Note: From diagram VIN > 1V
I R1 
1  ( 10 ) 11

 0.366mA
30k
30k
Method B – Balanced Currents
VOUT  VIN VOUT  V X

R1  R2
R1
 10  VIN
 10  1

( 30  15 )k
30k
 10  VIN  11

45k
30k
 11 45k
 10  VIN 
30k
 10  VIN  16.5
 10  16.5  VIN
VIN  6.5V
V15k  0.366mA15k
 5.5V
VIN  VX  V10k
 1  5.5  6.5V
49
Module ET4 – Communication Systems
The characteristic for this Schmitt trigger is therefore as shown below.
Vout/V
+15
-15
+15
-15
50
Vin/V
Topic 4.5.2 – Regeneration
Exercise 2:
1.
Upper Threshold = Between +2V and 3.6V
Lower Threshold = Between -0.4V and 1.8V
6
Vin/V
4
2
0
time (s)
-2
-4
-6
Vout/V
10
0
time (s)
-10
51
Module ET4 – Communication Systems
Assuming thresholds at +3V and +1V the output would be as follows:
6
Vin/V
4
2
0
time (s)
-2
-4
-6
Vout/V
10
0
-10
52
Topic 4.5.2 – Regeneration
2.
A non-inverting Schmitt trigger is required to have saturation values of
±9V, and switching thresholds of ±4V. Determine the values of VREF, R1
and R2 in the circuit below.
-
VREF
VOUT
+
R2
VIN
X
15kΩ
R1
30kΩ
i)
In this case VREF must be equal to 0V because the threshold values
are symmetrical at ±4V.
ii)
Now for the calculation of R1 and R2. We have to consider the
conditions that need to occur to cause a transition. If the output is
in positive saturation, then the input VIN must fall below the lower
switching threshold for the output to switch to negative saturation.
The voltage at point ‘X’ will be 0V.
VOUT = +9V
I
I
90 9

R1
R1
and
I
0  ( 4 ) 4

R2
R2
R1
The current I is the same so we can equate
these equations.
VX = 0V
R2
VIN =-4V
9
4

R1 R2
9 R2
 R1
4
2.25 R2  R1
53
Module ET4 – Communication Systems
Using the alternative approach the solution becomes:
From the diagram we can see that
VOUT = +9V
I
VR1  9V
VR2  4V
R1
so
VX = 0V
VR1
VR2
R2

R1
R2
9 R1

4 R2
VIN =-4V
R1  2.25 R2
So all that remains now is to choose suitable values for the
resistors. Any values over 1kΩ in an appropriate ratio would be fine,
so for example,
R1 = 9kΩ, R2 = 4kΩ;
R1 = 45kΩ, R2 = 20kΩ; etc are acceptable answers.
The final circuit could then look like this:
-
0V
VOUT
+
R2
VIN
20kΩ
54
X
R1
45kΩ
Topic 4.5.2 – Regeneration
3.
An inverting Schmitt trigger is required to have saturation values of
±15V, and switching thresholds of ±5V. Determine the values of VREF, R1
and R2 in the circuit below.
-
VIN
VOUT
+
R2
VREF
X
15kΩ
R1
30kΩ
i)
VREF will be 0V since the switching thresholds are symmetrical,
about 0V.
ii)
Now for the calculation of R1 and R2. We have to consider the
conditions that need to occur to cause a transition. If the output is
in positive saturation, then the input VIN must rise above the upper
switching threshold for the output to switch to negative saturation.
The voltage at point ‘X’ must be equal to this upper threshold which
in this case is +5V.
VOUT = +15V
I
I
15  5 10

and
R1
R1
I
50 5

R2
R2
R1
The current I is the same so we can equate
these equations.
VIN = 5V
R2
VREF = 0V
10 5

R1 R2
10 R2  5 R1
R1  2 R2
Suggestions for R2 therefore could be 10kΩ, making R1 = 20kΩ
55
Module ET4 – Communication Systems
Again using the alternative solution method we have:
From the diagram we can see that
VOUT = +15V
I
VR1  10V
VR2  5V
R1
so
VIN = 5V
VR1
VR2
R2

R1
R2
10 R1

5 R2
VREF =0V
R1  2 R2
The final circuit could then look like this:
-
VIN
VOUT
+
R2
0V
10kΩ
56
X
R1
20kΩ
Topic 4.5.2 – Regeneration
4.
A non-inverting Schmitt trigger is required to have saturation values of
±11V, and switching thresholds of -5V and +2.33V. Determine the values
of R1 and R2 in the circuit below if VREF=-1V.
-
VREF=-1V
VOUT
+
R2
VIN
R1
X
15kΩ
30kΩ
Now for the calculation of R1 and R2. We have to consider the
conditions that need to occur to cause a transition. If the output is
in positive saturation, then the input VIN must fall below the lower
switching threshold for the output to switch to negative saturation.
The voltage at point ‘X’ will be -1V.
VOUT = +11V
I
I
11  ( 1 ) 12

and
R1
R1
I
 1  ( 5 ) 4

R2
R2
R1
VX = -1V
R2
VIN =-5V
The current I is the same so we can equate
these equations.
12 4

R1 R2
12 R2  4 R1
R1  3R2
Suggestions for R1 therefore could be 6kΩ, making R2 = 2kΩ
57
Module ET4 – Communication Systems
Or alternatively.
From the diagram we can see that
VOUT = +11V
I
VR1  12V
VR2  4V
R1
so
VX = -1V
VR1
VR2
R2

R1
R2
12 R1

4 R2
VIN =-5V
R1  3R2
The final circuit could then look like this:
-
-1V
VOUT
+
R2
VIN
2kΩ
X
R1
6kΩ
Now for some past examination style questions.
58
Topic 4.5.2 – Regeneration
Examination Style Questions.
1.
The following circuit diagram shows an op-amp connected as a Schmitt trigger. The op-amp
saturates at +10V and 0V.
-
+2V
VOUT
+
R2
VIN
10kΩ
(a)
X
R1
20kΩ
Is this an inverting Schmitt trigger or non-inverting Schmitt trigger?
......................................................................................................................................................
[1]
(b)
Calculate the voltage at ‘X’ when VIN = 7V, and VOUT= +10V.
....................................................................................................................................................
....................................................................................................................................................
(c)
....................................................................................................................................................
[2]
Calculate the upper switching threshold by calculating the value of VIN that produces a
voltage of +2V at X when VOUT=0V.
....................................................................................................................................................
....................................................................................................................................................
(d)
....................................................................................................................................................
[2]
Calculate the lower switching threshold.
....................................................................................................................................................
....................................................................................................................................................
(e)
....................................................................................................................................................
[2]
Give a use for this circuit in a digital communication system.
....................................................................................................................................................
[1]
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Module ET4 – Communication Systems
2.
A digital signal is transmitted down a copper cable communications link.
The signal is attenuated and subjected to noise in the copper cable.
(a)
(i)
Explain what is meant by the phrase “...the signal is attenuated....”
....................................................................................................................................................
....................................................................................................................................................
....................................................................................................................................................
[1]
(ii)
Identify one possible source of noise.
....................................................................................................................................................
....................................................................................................................................................
....................................................................................................................................................
[1]
(b)
The receiver contains a non-inverting Schmitt trigger to regenerate the signal.
The following graph shows the signal picked up by the receiver.
Give suitable threshold values for the non-inverting Schmitt trigger.
Upper switching threshold = ...................................
Lower switching threshold = ...................................
[2]
60
Topic 4.5.2 – Regeneration
(c)
The next diagram shows an inverting Schmitt trigger. The output of the comparator
saturates at ±10V.
-
VIN
VOUT
+
R2
R1
330kΩ
220kΩ
0V
Calculate the switching thresholds for this circuit.
All calculations must be shown.
...................................................................................................................................................
...................................................................................................................................................
...................................................................................................................................................
...................................................................................................................................................
...................................................................................................................................................
...................................................................................................................................................
...................................................................................................................................................
...................................................................................................................................................
...................................................................................................................................................
...................................................................................................................................................
...................................................................................................................................................
...................................................................................................................................................
...................................................................................................................................................
[4]
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Module ET4 – Communication Systems
3.
62
A Schmitt trigger has the following input/output characteristic, when connected to a +15V/-15V
power supply.
Topic 4.5.2 – Regeneration
(a)
The graph shows the signal applied to the input of the Schmitt trigger.
On the same axes draw the resulting output signal.
(b)
Design a suitable circuit for this Schmitt trigger, based on an op-amp running on a +15V/15V supply.
Calculate suitable values for any resistors used, and mark these values on the circuit
diagram.
...................................................................................................................................................
...................................................................................................................................................
...................................................................................................................................................
...................................................................................................................................................
[4]
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Module ET4 – Communication Systems
4.
The following diagram shows the circuit for an inverting Schmitt trigger.
-
VIN
VOUT
+
R2
R1
X
4V
10kΩ
20kΩ
The op-amp saturates at +10V and 0V.
(a)
Calculate the voltage at X when VOUT = +10V
....................................................................................................................................................
....................................................................................................................................................
(b)
....................................................................................................................................................
[2]
Calculate the voltage at X when VOUT = 0V
....................................................................................................................................................
....................................................................................................................................................
(c)
....................................................................................................................................................
[2]
Write down the switching thresholds of the Schmitt trigger.
Lower switching threshold
...............................................
Upper switching threshold
...............................................
[1]
64
Topic 4.5.2 – Regeneration
(d)
A different inverting Schmitt trigger has switching thresholds at +0.8V and +3.2V. Draw the
output for this new Schmitt trigger if the following analogue signal is applied to the input.
The comparator saturates at ±10V.
[3]
65
Module ET4 – Communication Systems
5.
The following diagram shows the circuit for a Schmitt trigger.
-
4V
VOUT
+
R2
VIN
X
22kΩ
R1
47kΩ
The comparator saturates at +12V and 0V.
(i)
Calculate the voltage at ‘X’ when VIN = +4V and VOUT = +12V.
......................................................................................................................................................
......................................................................................................................................................
[2]
(ii)
Determine the upper switching threshold by calculating the value of VIN which produces a
voltage of +4V at X when VOUT = 0V.
......................................................................................................................................................
......................................................................................................................................................
......................................................................................................................................................
......................................................................................................................................................
[2]
(iii)
Determine the lower switching threshold.
......................................................................................................................................................
......................................................................................................................................................
......................................................................................................................................................
......................................................................................................................................................
[2]
66
Topic 4.5.2 – Regeneration
(b)
A different Schmitt trigger is used as a regenerator for a transmission line. A sinusoidal
waveform used as a test signal gives the output shown on the graph below.
Sketch the characteristic for VOUT against VIN for this Schmitt trigger using the axes
provided.
[3]
67
Module ET4 – Communication Systems
6.
The following diagram shows the circuit for a Schmitt trigger. The comparator saturates at +10V
and 0V.
-
+4V
VOUT
+
R2
VIN
200kΩ
X
R1
400kΩ
(a)
Is this an inverting Schmitt trigger or non-inverting Schmitt trigger?
(b)
....................................................................................................................................................
[1]
Calculate the voltage at ‘X’ when VIN = 7V, and VOUT= +10V.
....................................................................................................................................................
....................................................................................................................................................
(c)
....................................................................................................................................................
[2]
Calculate the value of VIN that causes VOUT to change from 0V to +10V.
....................................................................................................................................................
....................................................................................................................................................
(d)
....................................................................................................................................................
[2]
Calculate the value of VIN that causes VOUT to change from +10V to 0V.
....................................................................................................................................................
....................................................................................................................................................
....................................................................................................................................................
[2]
68
Topic 4.5.2 – Regeneration
(e)
Sketch the characteristic for this Schmitt trigger.
[3]
69
Module ET4 – Communication Systems
7.
(a)
The following circuit diagram shows how a Schmitt trigger circuit can be made using an opamp and two zener diodes.
Draw the switching characteristic of the Schmitt trigger when the op-amp is connected to a
±12V power supply using the axes below.
[3]
70
Topic 4.5.2 – Regeneration
(b)
A different Schmitt trigger is shown in the following circuit diagram.
-
+2V
VOUT
+
R2
VIN
150kΩ
X
R1
300kΩ
The op-amp saturates at ±12V.
(i)
Calculate the voltage at ‘X’ when VIN = 3V, and VOUT= +12V.
....................................................................................................................................................
....................................................................................................................................................
....................................................................................................................................................
[2]
(ii)
Calculate the value of VIN that causes VOUT to change from +12V to -12V.
....................................................................................................................................................
....................................................................................................................................................
....................................................................................................................................................
[2]
(iii)
Calculate the value of VIN that causes VOUT to change from -12V to +12V.
....................................................................................................................................................
....................................................................................................................................................
(c)
....................................................................................................................................................
[2]
Identify a situation where a Schmitt trigger would be needed in a communication system and
what improvement it would make.
....................................................................................................................................................
....................................................................................................................................................
....................................................................................................................................................
[2]
71
Module ET4 – Communication Systems
8.
A Schmitt trigger circuit has the following input/output characteristic when connected to a ±10V
supply.
(a)
(i)
What is the value of VIN that causes VOUT to change from -9V to +9V? ..................
(ii)
What is the value of VIN that causes VOUT to change from +9V to -9V? ..................
[1]
(b)
Design a suitable circuit for this Schmitt trigger based on a comparator running on a ±10V
supply. The comparator output saturates at ±9V. Calculate suitable values for any resistors
used and mark these on the circuit diagram.
[4]
72
Topic 4.5.2 – Regeneration
9.
The following circuit diagram shows a comparator connected as a Schmitt trigger.
-
+3V
VOUT
+
R2
VIN
X
10kΩ
R1
30kΩ
The comparator saturates at +12V and 0V.
(a)
Is this an inverting Schmitt trigger or non-inverting Schmitt trigger?
(b)
....................................................................................................................................................
[1]
Calculate the voltage at ‘X’ when VIN = 7V, and VOUT= +12V.
....................................................................................................................................................
....................................................................................................................................................
(c)
....................................................................................................................................................
[2]
Calculate the value of VIN that causes VOUT to change from 0V to +12V.
....................................................................................................................................................
....................................................................................................................................................
(d)
....................................................................................................................................................
[2]
Calculate the value of VIN that causes VOUT to change from +12V to 0V.
....................................................................................................................................................
....................................................................................................................................................
....................................................................................................................................................
[2]
73
Module ET4 – Communication Systems
Self Evaluation Review
Learning Objectives
My personal review of these objectives:



recall and describe the difference
between noise and distortion.
understand that the signal to noise
ratio degrades down a transmission
line, and that regeneration restores
the original signal.
recognise, analyse and design
inverting and non-inverting Schmitt
trigger circuits to regenerate digital
signals.
Targets:
1.
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
2.
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
74