Topic 2.5 – Signal Sources.
Learning Objectives:
At the end of this topic you will be able to;
 describe the use of photo-transistors, light dependent resistors and
ntc thermistors in voltage dividing chains to provide analogue signals;
 sketch and interpret response curves for the light dependent
resistors and thermistor;
 calculate suitable values for resistors for use with the above devices;
 use Thévenin’s theorem to draw equivalent circuits and predict the
effect of loading;
 appreciate that the current through a potential divider should be at
least 10 times that drawn at the output;
 explain how a Schmitt inverter can be used to provide signal
conditioning with analogue signal sources and to eliminate mechanical
switch bounce.
1
Module ET2
Electronic Circuits and Components.
Signal Sources.
In our work so far we have described how some of the processing blocks of
an electronic system can be constructed from individual components, e.g. time
delays and pulse generators or ‘clock’s. In this section we will spend some
time looking at how external variations in light or temperature can be
converted into a suitably changing electrical signal that can be processed by
an electrical system.
In the AS course we are going to concentrate on three types of input sensor,
(i) the thermistor, (ii) the light dependent resistor, and (iii) the
phototransistor. We will consider each of these in turn and look at their
characteristic behaviour and how they are used in a circuit. An essential part
of using these components is the ‘voltage divider’ circuit introduced in Topic
2.1. – Page 19, and if you cannot remember how this circuit works then now
would be an appropriate time to do a quick bit of revision of this circuit!
If you are happy to proceed, let us start with the first of our components,
the thermistor.
The thermistor is a two leaded component that changes its resistance
responds to a change in temperature. The symbol for a thermistor is shown
below:
The ‘-t°’ alongside the symbol indicates that this is a negative temperature
coefficient (or n.t.c.) thermistor, which simply means that the resistance of
the thermistor decreases as temperature increases.
A positive temperature coefficient (p.t.c.) thermistor does exist where the
resistance increases as temperature increases, but these will not be
examined as part of this course. The symbol just has a ‘+t°’ alongside it should
you see this in any project books you may look at.
2
Topic 2.5 – Signal Sources.
The characteristic curve for a thermistor, therefore looks like this.
Thermistors come in many different physical packages as shown by the
diagram below:
Irrespective of the package style the behaviour of all of these thermistors is
the same, as temperature rises the resistance of the thermistor falls. The
change in package style does however affect the response time of the
thermistors, the ‘rod’ style thermistor is large and bulky and has the slowest
response time, whilst the tiny ‘glass bead’ style has the fastest response.
Depending on the application different styles of package can be selected but
it is important to remember that from circuit design point of view the
package is not important as long as we know the range of resistance the
thermistor has over the temperature range that it will be used.
3
Module ET2
Electronic Circuits and Components.
A typical data sheet for a thermistor is shown below:
The first two rows in this table show that at 25°C the resistance is 300Ω,
and at 50°C the resistance has fallen to 121Ω. All thermistors are different
so it is important to check their data sheets to determine their
characteristics so that a suitable circuit can be designed to use them
effectively.
Now we will look at how the thermistor is used in an electrical circuit. The
thermistor is used as part of a voltage divider circuit, so that the change in
resistance can be used to produce a change in voltage. This will enable the
output of the sensing circuit to be connected to logic circuits or for switching
outputs as we will see later on. A typical circuit is as follows:
By applying some numbers to the circuit we can determine the value of
voltage we would expect to see on the voltmeter for a number of different
input conditions. We will assume that the thermistor has the following
properties.
4
Topic 2.5 – Signal Sources.
Typical resistances at 0°C = 8kΩ, at 25°C = 2.5kΩ, and at 100°C = 1kΩ.
So to calculate the output voltage at each temperature we will apply the
voltage divider rule for the circuit.
i)
at 0°C.
Voltmeter Reading 
VIN  Rthermistor
R1  Rthermistor
9  8k
1k  8k
72

 8V
9

ii)
at 25°C.
Voltmeter Reading 
VIN  Rthermistor
R1  Rthermistor
9  2.5 k
1k  2.5k
22.5

 6.42V
3.5

iii)
at 100°C.
Voltmeter Reading 
VIN  Rthermistor
R1  Rthermistor
9  1k
1k  1k
9
  4.5V
2

The calculations shown above indicate that as the temperature rises the
voltage across the thermistor falls from a high of 8V at 0°C, to a low of 4.5V
at 100°C.
5
Module ET2
Electronic Circuits and Components.
The following graph shows how the output voltage changes as temperature
changes (blue line) and how the resistance of the thermistor changes for the
same temperature range (pink line).
9
9
8
8
7
7
6
6
5
5
4
4
3
3
2
2
1
1
0
0
10
20
30
40
50
60
70
80
90
Resistance of Thermistor (k ohms)
Voltage (V)
Graph to show voltage across, and resistance characteristic of the thermistor
0
100
Temperature ('C)
Voltage across thermistor
Resistance of thermistor
From the graph we can see that the output voltage falls in a non-linear way
and therefore this type of circuit produces an analogue voltage. However in
this case the range of voltage change is not very good, since the output
voltage only changes by 3.5V (8-4.5V) even though the temperature has
changed by 100°C.
The reason for this is the choice of series resistance, which in this example
was given as 1kΩ. This is equal to the lowest value of resistance that the
thermistor has at 100°C, so at this point the output voltage will only be 4.5V,
half of the supply voltage since both components in the series circuit have
the same resistance, so each will have 4.5V across them.
6
Topic 2.5 – Signal Sources.
This is our first design issue then. To maximise the output voltage range from
this type of sensor, the series resistor used should have a resistance equal or
close to the mid point of the resistance range of the thermistor over the
temperature range it is to be used over. For our example above this is 4.5kΩ.
In the E24 series of resistors we have a choice of 4.3kΩ or 4.7kΩ. We will
consider what happens when the 4.3kΩ resistor is used.
The circuit becomes:
Now for the calculations:
i)
at 0°C.
Voltmeter Reading 
VIN  Rthermistor
R1  Rthermistor
9  8k
4.3k  8k
72

 5.85V
12.3

ii)
at 25°C.
Voltmeter Reading 
VIN  Rthermistor
R1  Rthermistor
9  2.5 k
4.3k  2.5 k
22.5

 3.31V
6.8

7
Module ET2
Electronic Circuits and Components.
iii)
at 100°C.
Voltmeter Reading 
VIN  Rthermistor
R1  Rthermistor
9  1k
4.3k  1k
9

 1.70V
5.3

9
9
8
8
7
7
6
6
5
5
4
4
3
3
2
2
1
1
0
0
10
20
30
40
50
60
70
80
90
Resistance of Thermistor (k ohms)
Voltage (V)
Graph to show voltage across, and resistance characteristic of the thermistor
0
100
Temperature ('C)
Voltage across thermistor
Resistance of thermistor
At first glance it would appear that we have not significantly increased the
range of output voltage, as this has only increased to 4.15V from the 3.5V we
had in our previous example.
However the biggest change is in the minimum and maximum values which are
now above and below the mid range voltage of the power supply. This is very
important as we will see later if we want to connect this type of sensor to a
digital system.
Now it’s time for you to have a go!
8
Topic 2.5 – Signal Sources.
Student Exercise 1:
Determine the voltages across the thermistor in our previous example if a
4.7kΩ resistor had been used instead of the 4.3kΩ.
..................................................................................................................................................
..................................................................................................................................................
..................................................................................................................................................
..................................................................................................................................................
..................................................................................................................................................
..................................................................................................................................................
..................................................................................................................................................
..................................................................................................................................................
..................................................................................................................................................
..................................................................................................................................................
..................................................................................................................................................
..................................................................................................................................................
9
Module ET2
Electronic Circuits and Components.
In our last example we obtained an analogue output voltage that decreased as
the temperature rises. Situations may occur in the design of certain
applications where this is not what is required. It may, in certain situations be
necessary that the voltage increases as the temperature rises.
This is very easy to achieve, requiring only a minor change to our previous
circuit. It is only necessary to transpose the position of the fixed resistor
and the thermistor to achieve the desired result, as shown below.
If we assume that we are still using the same resistor as before, then the
calculations are:
i)
at 0°C.
Voltmeter Reading 
VIN  R1
R1  Rthermistor
9  4.7k
4.7k  8k
42.3

 3.33V
12.7

ii)
at 25°C.
Voltmeter Reading 
VIN  R1
R1  Rthermistor
9  4.7k
4.7  2.5k
42.3

 5.88V
7.2

10
Topic 2.5 – Signal Sources.
iii)
at 100°C.
Voltmeter Reading 
VIN  R1
R1  Rthermistor
9  4.7k
4.7k  1k
42.3

 7.42V
5.7

The graph of the output voltage against temperature will therefore look like
this:
9
9
8
8
7
7
6
6
5
5
4
4
3
3
2
2
1
1
0
0
10
20
30
40
50
60
70
80
90
Resistance of Thermistor (k ohms)
Voltage (V)
Graph to show voltage across, and resistance characteristic of the thermistor
0
100
Temperature ('C)
Voltage across thermistor
Resistance of thermistor
In the examination you will be expected to be able to design temperature
sensing circuits which provided either a falling voltage when temperature
rises as in our very first example or as is the case with this last example a
rising voltage when temperature rises.
You will also be expected to perform calculations on a given circuit at specific
temperatures where the resistance of the thermistor will be given either
directly (e.g. R25°C=5kΩ), or indirectly by providing the resistance
characteristic curve on graph paper from which the resistance at specific
temperatures can be obtained.
11
Module ET2
Electronic Circuits and Components.
Student Exercise 2:
1.
The following circuit has been set up as a simple temperature sensing
circuit.
i)
Describe what happens to the output voltage as the temperature
falls.
.............................................................................................................................
.............................................................................................................................
.............................................................................................................................
.............................................................................................................................
ii)
The thermistor has a resistance of 250kΩ at 0°C, and 750Ω at
100°C. Calculate the reading on the voltmeter at these two
temperature extremes.
.............................................................................................................................
.............................................................................................................................
.............................................................................................................................
.............................................................................................................................
12
Topic 2.5 – Signal Sources.
2.
Complete the circuit diagram below to show a temperature sensing
circuit that provides a rising voltage at the output when the
temperature increases. A fixed resistor of value 22kΩ, and a
thermistor with R25°C=90kΩ, and R70°C=5kΩ are available for this task.
For the completed circuit, calculate the voltage shown on the voltmeter
at 25°C and at 70°C.
.......................................................................................................................................
.......................................................................................................................................
.......................................................................................................................................
.......................................................................................................................................
.......................................................................................................................................
.......................................................................................................................................
.......................................................................................................................................
.......................................................................................................................................
.......................................................................................................................................
.......................................................................................................................................
13
Module ET2
Electronic Circuits and Components.
3.
A thermistor has the following characteristic.
Graph to show the resistance characteristic of the thermistor
1300
1200
1100
1000
Resistance (k.ohms)
900
800
700
600
500
400
300
200
100
0
-50
-40
-30
-20
-10
0
10
20
Temperature ('C)
The thermistor is to be used in a temperature sensor for use by a British
expedition to the North Pole in the Arctic Circle. Temperatures in the arctic
circle vary from -40°C to +5°C.
(a)
Complete the circuit diagram below to show how this thermistor could
be used to make a temperature sensor which produces a falling output
voltage when the temperature rises.
(b)
Select a suitable resistor from the E24 series that will provide an
output voltage of approximately 6V at the mid point of the temperature
range required, and mark this on the circuit diagram.
14
Topic 2.5 – Signal Sources.
(c)
(d)
Determine the resistance of the thermistor at
(i)
-40°C
.......................................
(ii)
+5°C
.......................................
Calculate the output voltage from your sensor at these two extreme
temperatures.
(i)
-40°C
........................................................................................................................................
........................................................................................................................................
........................................................................................................................................
........................................................................................................................................
........................................................................................................................................
(ii)
+5°C
........................................................................................................................................
........................................................................................................................................
........................................................................................................................................
........................................................................................................................................
........................................................................................................................................
15
Module ET2
Electronic Circuits and Components.
We will now turn our attention to the Light Dependent Resistor or LDR. This
is a component as it’s name suggests where it’s resistance changes as the
amount of light falling on the window of the sensor changes. The symbol for
an LDR is as follows:
The LDR comes in a variety of different packages as shown below:
The resistance characteristic for the LDR is shown below:
Resistance (kΩ)
Light Intensity (lux)
16
Topic 2.5 – Signal Sources.
The light intensity is measured in a unit called ‘lux’, but we can see from the
characteristic curve that the resistance of the LDR falls as light intensity
increases. The LDR is used in a circuit in a similar way to the thermistor as
shown below.
In the circuit above, when the LDR is in the dark it will have a very high
resistance, this means that current in the circuit will be small, so the voltage
across R2 will be small, so the voltmeter will show a small voltage.
As light falls on to the LDR it’s resistance decreases, this causes the current
in the circuit to increase. When the current increases the voltage across R2
increases which causes the voltmeter reading to increase.
In this circuit then, the output voltage increases as the light intensity
increases. If we want the opposite effect, i.e. a circuit in which the output
voltage increases as the light intensity decreases (i.e. it becomes dark) then
we simply have to reverse the positions of the LDR and fixed resistor as
shown below.
Make sure you can explain how this circuit works!
17
Module ET2
Electronic Circuits and Components.
If you managed to solve the design questions involving the thermistor, then
you will have no problem solving similar problems involving the LDR. The
circuit analysis will be exactly the same, the only difference will be that
instead of having a characteristic relating to temperature, it will have light
intensity values in Lux. If the characteristic is not provided, you will be given
a value for the resistance in the light and in the dark.
Try these questions for practice.
Student Exercise 3.
1.
Complete the circuit diagram below to show how an LDR and a fixed
resistor can be used to make a light sensor which provides a falling
output voltage when it gets dark.
2.
An LDR is used as a light sensor in the following circuit.
The LDR has a resistance of 1.5MΩ at 10 lux, 400kΩ at 20,000 lux and
3kΩ at 100,000 lux.
18
Topic 2.5 – Signal Sources.
Determine the output voltage Vout, for light intensities of 10 lux, 20,000
lux and 100,000 lux.
……………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………
Now we’ll mix the questions up a bit – thermistors and LDR’s together.
19
Module ET2
Electronic Circuits and Components.
3.
A thermistor has a resistance of 1.0MΩ when at the ice point of water
and 2k when at the steam point of water. It is used in the potential
divider circuit shown below.
Calculate the output voltage when the thermistor is
(a)
at the ice point.
…………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………
(b)
at the steam point.
…………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………
20
Topic 2.5 – Signal Sources.
4.
A light dependent resistor (LDR) has a
resistance of 800k when in the dark and 100
when in full sunlight. It is used in the voltage
divider circuit opposite.
Calculate the output voltage when the LDR is
(a)
in full sunlight
…………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………
(b)
in the dark
…………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………
(c)
How would you alter the circuit to get a large output when the LDR
is in the dark and a small output when it is in full sunlight?
…………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………
21
Module ET2
Electronic Circuits and Components.
Our final input sub-system involves the use of a photo-transistor. A phototransistor is a semiconductor device that responds to ‘photons’ of light which
fall on a tiny slice of silicon inside the device. We will look at the detailed
operation of a transistor Topic 2.6.2, and for now we will treat it like a black
box with a specific function, otherwise the discussion will be overly complex.
The photo-transistor is an enhanced version of the much simpler device called
a photo-diode. The photo-diode has many properties similar to that of an
ordinary diode. The symbol for a photo-diode is as follows:
The photo-diode under forward-bias, i.e. with the anode connected
to the positive of the supply, will conduct conventional current
following the direction of the arrow. The physical appearance of a
photodiode is much like that of an l.e.d. (see photo on the right.)
Photo-diodes are used in reverse bias, just like zener diodes as
discussed in Topic 2.3. Conduction in this reverse direction is
achieved when a photon of sufficient energy strikes the diode, it excites an
electron, and a photocurrent is produced.
Photo-diodes are usually matched with a specific wavelength of light and
special corresponding l.e.d.’s that produce exactly the same frequency of
light, usually in the infra-red part of the spectrum, are used to provide the
input light source for the photo-diodes. These are often used in TV remote
controls for example.
The current that flows through the photo-diode is tiny, and some
amplification is needed if we are to be able to make simple use of this input
device. This device is called the photo-transistor.
22
Topic 2.5 – Signal Sources.
The symbol, and picture of a photo-transistor are as follows:
The photo-transistor contains an inbuilt photo-diode with internal gain. A
phototransistor is in essence nothing more than a bipolar transistor (see topic
2.6.2) that is encased in a transparent case so that light can reach the basecollector junction. The electrons that are generated by photons in the basecollector junction are injected into the base, and this photodiode current is
amplified by the transistor's current gain β (or hfe). Note that while
phototransistors have a higher responsivity for light they are not able to
detect low levels of light any better than photodiodes. Phototransistors also
have slower response times.
The following photograph shows three phototransistors mounted on the laser
receiving part of a CD Player, 3 phototransistors can be seen mounted on the
top of the unit.
23
Module ET2
Electronic Circuits and Components.
The photo-transistor is used in many applications where a light triggered
circuit is required, and applications where an LDR could be used but more
sensitivity is required, such as the TV remote controls we have already
mentioned and also in reflective sensors that are used to detect the presence
of reflective surfaces for use in robotic line following robot vehicles. The
sensors basically incorporate an l.e.d. and a photo-transistor in the same
package as shown below.
The l.e.d. and photo-transistor are mounted at an angle so that the light from
the l.e.d. can be reflected from a reflective surface back onto the phototransistor. This will allow the photo-transistor to conduct, and a current will
flow.
Reflective
Surface
Non-Reflective
Surface
If the surface is non-reflective then the light from the l.e.d. will not reach
the phototransistor and no current will flow through the photo-transistor.
We will now look at how this device is used in a circuit diagram.
24
Topic 2.5 – Signal Sources.
The circuit consists of two main parts:
(i)
The l.e.d. light source and
(ii)
The photo-transistor detector.
Metal foil sheet
passes over rollers
In this example the l.e.d. D1 is shining light onto a metal foil sheet which is
reflected back onto the photo-transistor when the metal foil sheet is in
place.
When the metal foil sheet is in place light is reflected back onto the phototransistor which conducts and the voltage at Vout falls to near 0V.
When the metal foil sheet is absent, no light reflects back onto the phototransistor, which does not conduct and resistor R2 pulls up the voltage at Vout
to near 9V.
Resistor R1 must be chosen to limit the voltage across the l.e.d. to a safe
value, typically 2V, and the current flowing to a safe value typically 7-20mA in
the same way that the resistor was calculated for a normal l.e.d.
25
Module ET2
Electronic Circuits and Components.
In the previous section we have looked at the operation of the voltage divider
circuit and how this could be used in conjunction with a sensing element to
produce a changing output voltage in response to a change in input
temperature for example.
We have also performed calculations on voltage dividers to calculate the
output voltage for certain given conditions. We now have to consider what
happens when another circuit is connected to the voltage divider and how
these circuits are analysed.
To begin with, let’s consider a very simple voltage divider made from just two
fixed resistors.
+6V
R1=1kΩ
VOUT
R2=1kΩ
0V
You should be able to work out that the value of VOUT is 3V, since both
resistors are equal in value. Now consider what happens when a load of 100Ω
is connected to the voltage divider.
+6V
R1=1kΩ
VOUT
R2=1kΩ
0V
26
R3=100Ω
Topic 2.5 – Signal Sources.
Analysis of this circuit is now slightly more complicated as we first have to
calculate the effective resistance of R2 and R3 in parallel before we can apply
the voltage divider formula. The calculation is as follows.
Rp 
R2  R3
R2  R3
1000  100
1000  100
100000

 90.9  91
1100

The effective circuit therefore now becomes:
+6V
R1=1kΩ
VOUT
Rp=91Ω
0V
Now we can calculate the output voltage using the potential divider formula.
VOUT 
VIN  R p
R1  R p
6  91
1000  91
546

 0.5V
1091

The effect of adding this load is to drop the output voltage to 0.5V. The
effect on the circuit is dramatic as it alters the conditions greatly. What if
the load was 1kΩ. We will have to repeat all of the previous calculations.
27
Module ET2
Electronic Circuits and Components.
+6V
R1=1kΩ
VOUT
R2=1kΩ
R3=1kΩ
0V
The calculation is as follows.
Rp 
R2  R3
R2  R3
1000  1000
1000  1000
1000000

 500
2000

The effective circuit therefore now becomes:
+6V
R1=1kΩ
VOUT
Rp=500Ω
0V
Now we can calculate the output voltage using the potential divider formula.
28
Topic 2.5 – Signal Sources.
VOUT 
VIN  Rp
R1  Rp
6  500
1000  500
3000

 2V
1500

The effect on the circuit is not quite as dramatic as before but the voltage
at the output has still dropped by 1V or 33% of what it should be. Let’s try
one more example, with a load of 10kΩ. Yet again we will have to repeat the
calculations performed earlier.
+6V
R1=1kΩ
VOUT
R2=1kΩ
R3=10kΩ
0V
The calculation is as follows.
Rp 
R2  R3
R2  R3
1000  10000
1000  10000
10000000

 909
11000

29
Module ET2
Electronic Circuits and Components.
The effective circuit therefore now becomes:
+6V
R1=1kΩ
VOUT
Rp=909Ω
0V
Now we can calculate the output voltage using the potential divider formula.
VOUT 
VIN  R p
R1  R p
6  909
1000  909
5454

 2.86V
1909

The effect on the circuit is now less significant with the output voltage
dropping by just 0.14V or just 4.6% of what it should be. This gives us a very
basic rule that the load resistance must be at least 10x that of the resistor
in the sensing circuit in order to minimise the effect on the output voltage of
the sensor.
In order to work out these results we have had to carry out a number of
calculations, and whilst they were not particularly difficult, they do take time
particularly when the load resistance is constantly changing, as we have to
recalculate the effective resistance of the parallel circuit before we can
work out the output voltage. In 1883 a French telegraph engineer Léon
Charles Thévenin, proposed a simplification technique which would make the
analysis of these types of circuits much easier. This states that a circuit of
voltage sources and resistors can be converted into a Thévenin equivalent
30
Topic 2.5 – Signal Sources.
circuit. The circuit consists of an ideal voltage source in series with an ideal
resistor. Creating the Thévenin equivalent circuit is very straightforward and
follows a simple procedure.
i. Calculate the open circuit voltage VOC .
ii. Calculate the short circuit current, I SC .
iii. Calculate the output resistance, RO 
VOC
.
I SC
iv. Draw the equivalent circuit including the values of VOC , and RO as
follows.
RO
VOC
Now the equivalent circuit can be used for any load that you may want
to connect to the circuit. Let us revisit our previous example to see how the
Thévenin equivalent circuit would have simplified the process. To refresh
your memory, this is the circuit we started with.
+6V
R1=1kΩ
VOUT
R2=1kΩ
0V
Now we will follow the procedure above to obtain the equivalent circuit.
First step is to calculate the open circuit voltage.
31
Module ET2
Electronic Circuits and Components.
VOC 
VIN  R2
6  1000
6000


 3V
R1  R2 1000  1000 2000
In this example we could have just written down the open circuit
voltage, as both resistors are equal, but this may not always be the case, so a
full treatment has been shown here which will work for any circuit.
Now for step 2, calculate the short circuit current. For this the circuit
becomes:
+6V
R1=1kΩ
VOUT
R2=1kΩ
0V
Resistor R2 is effectively removed from the circuit, and therefore ISC is
as follows.
I SC 
VIN
6

 6mA
R1 1000
Now for RO, which is calculated from RO 
RO 
The equivalent
circuit
therefore
becomes:
32
VOC
I SC
VOC
3
3


 500
I SC 6mA 6  103
RO =500Ω
VOC=3V
Topic 2.5 – Signal Sources.
Now that we have the equivalent circuit we can apply our load resistors
to this circuit. Remember we used load resistors of 100Ω, 1kΩ, and
10kΩ in our previous example. Starting with the 100Ω we have the new
circuit of:
VOUT
RO =500Ω
RL =100Ω
VOC=3V
In this circuit there are no complex parallel circuits to work out we can
use the voltage divider rule directly to determine the value of VOUT.
VOUT 
VOC  RL
3  100
300


 0.5V
RO  RL 500  100 600
Changing the load to 1kΩ is now trivial, the circuit simply becomes:
VOUT
RO =500Ω
VOC=3V
So VOUT is
VOUT 
RL =1kΩ
VOC  RL
3  1000
3000


 2V
RO  RL 500  1000 1500
And finally when RL = 10kΩ
VOUT 
VOC  RL
3  10000
30000


 2.86V
RO  RL 500  10000 10500
33
Module ET2
Electronic Circuits and Components.
Using the equivalent circuit has duplicated all our previous results in a much
easier way. If you are experimenting with a number of different loads then
this method is much quicker.
This circuit also has the advantage that it can be used to determine the
minimum load resistance that can be applied to the circuit given the minimum
acceptable output voltage, or the maximum current that is to be drawn from
the circuit.
Consider the following circuit:
+9V
R1=1.2kΩ
VOUT
R2=1.8kΩ
RL
0V
What is the minimum load resistor RL that can be connected to the voltage
divider so that the voltage does not fall below 5V.
First derive the Thévenin equivalent circuit for the voltage divider.
VOC 
VIN  R2
9  1800
16200


 5.4V
R1  R2 1200  1800 3000
I SC 
RO 
34
VIN
9

 7.5mA
R1 1200
VOC
5.4
5.4


 720
I SC 7.5mA 7.5  103
Topic 2.5 – Signal Sources.
The
equivalent
circuit
therefore
becomes:
VOUT
RO =720Ω
VOC=5.4V
RL
Now if the minimum value of VOUT is to be 5V, this means that a
maximum of 0.4V must be dropped across RO, this allows us to calculate
the current flowing in the circuit.
I MAX 
V
0.4

 0.555mA
RO 720
This current must also flow through RL, which will have a voltage of 5V
across it, therefore the value of RL can be calculated as:
RL 
V
5
5


 9009  9k
I 0.555 mA 0.555  10 3
Therefore the minimum resistance that can be connected as a load
without causing the maximum output voltage to fall below 5V is 9kΩ.
If we consider the same circuit we could ask a different question, which is
solved in a similar way. The question might be; What is the maximum load
current that can be provided which will prevent the output voltage falling
below 3.96V.
Our equivalent circuit will still be the same as shown below:
35
Module ET2
Electronic Circuits and Components.
VOUT=3.96V
RO =720Ω
VOC=5.4V
36
RL
Topic 2.5 – Signal Sources.
We are told in the question that the output voltage should not fall below
3.96V, which means this is the minimum we can expect across RL. If this
is the voltage across RL, then this will leave 5.4V – 3.96V across RO.
I MAX 
V 1.44

 2mA
RO 720
The maximum current which can flow through the load, and maintain an
output voltage of 3.96V is just 2mA.
Now its time for you to have a go:
37
Module ET2
Electronic Circuits and Components.
Student Exercise 4:
1.
The following circuit is used as a voltage source.
12V
18Ω
22Ω
VOUT
0V
(a)
Thévenin’s theorem is used to produce an equivalent circuit.
(i)
Calculate the open circuit voltage, VOC.
.............................................................................................................................
.............................................................................................................................
(ii)
Calculate the short circuit current, ISC.
.............................................................................................................................
.............................................................................................................................
(iii) Calculate the equivalent resistance, RO.
.............................................................................................................................
.............................................................................................................................
(iv)
38
Draw the equivalent circuit.
Topic 2.5 – Signal Sources.
2.
The following circuit is used as a voltage source.
12V
120Ω
180Ω
VOUT
0V
(a)
Thévenin’s theorem is used to produce an equivalent circuit.
(i)
Calculate the open circuit voltage, VOC.
.............................................................................................................................
.............................................................................................................................
(ii)
Calculate the short circuit current, ISC.
.............................................................................................................................
.............................................................................................................................
(iii) Calculate the equivalent resistance, RO.
.............................................................................................................................
.............................................................................................................................
(iv)
Draw the equivalent circuit.
39
Module ET2
Electronic Circuits and Components.
(b)
This voltage source is used to drive two output transducers, each
having a resistance of 500Ω. Show how the equivalent circuit can
be used calculate the output voltage, VOUT, when;
(i)
the two output transducers are connected in series across
the output;
..............................................................................................................................
..............................................................................................................................
..............................................................................................................................
[2]
(ii) the two output transducers are connected in parallel across
the output;
..............................................................................................................................
..............................................................................................................................
..............................................................................................................................
[2]
40
Topic 2.5 – Signal Sources.
3.
The following circuit is used as a voltage source.
10V
100Ω
150Ω
VOUT
0V
(a)
Thévenin’s theorem is used to produce an equivalent circuit.
(i)
Calculate the open circuit voltage, VOC.
.............................................................................................................................
.............................................................................................................................
(ii)
Calculate the short circuit current, ISC.
.............................................................................................................................
.............................................................................................................................
(iii) Calculate the equivalent resistance, RO.
.............................................................................................................................
.............................................................................................................................
(iv)
Draw the equivalent circuit.
41
Module ET2
Electronic Circuits and Components.
(b)
Show how the equivalent circuit can be used to calculate the load
voltage, when a load of 300Ω is connected across the output.
..............................................................................................................................
..............................................................................................................................
42
Topic 2.5 – Signal Sources.
4.
The following circuit is used as a voltage source.
12V
120Ω
360Ω
VOUT
0V
(a)
Thévenin’s theorem can be used to produce an equivalent circuit.
(i)
Calculate the open circuit voltage, VOC.
.............................................................................................................................
.............................................................................................................................
(ii)
Calculate the short circuit current, ISC.
.............................................................................................................................
.............................................................................................................................
(iii) Calculate the equivalent resistance, RO.
.............................................................................................................................
.............................................................................................................................
(iv)
Draw the equivalent circuit.
43
Module ET2
Electronic Circuits and Components.
(b)
Calculate the output voltage when a load of 910Ω is connected
across the output.
..............................................................................................................................
..............................................................................................................................
(c)
Calculate the maximum current that can be taken from the output
of the voltage source if VOUT is not to fall below 8V.
..............................................................................................................................
.............................................................................................................................
44
Topic 2.5 – Signal Sources.
5.
The following circuit is used as a voltage source with the variable
resistor set at 360Ω.
16V
360Ω
120Ω
VOUT
0V
(a)
Thévenin’s theorem is used to produce an equivalent circuit.
(i)
Calculate the open circuit voltage, VOC.
.............................................................................................................................
.............................................................................................................................
(ii)
Calculate the short circuit current, ISC.
.............................................................................................................................
.............................................................................................................................
(iii) Calculate the equivalent resistance, RO.
.............................................................................................................................
.............................................................................................................................
(iv)
Draw the equivalent circuit.
45
Module ET2
Electronic Circuits and Components.
(b)
Show how the equivalent circuit can be used to calculate the load
voltage, when a load of 270Ω is connected across the output.
..............................................................................................................................
..............................................................................................................................
46
Topic 2.5 – Signal Sources.
6.
The following circuit is used as a voltage source.
15V
100Ω
150Ω
VOUT
0V
(a)
Thévenin’s theorem is used to produce an equivalent circuit.
(i)
Calculate the open circuit voltage, VOC.
.............................................................................................................................
.............................................................................................................................
(ii)
Calculate the short circuit current, ISC.
.............................................................................................................................
.............................................................................................................................
(iii) Calculate the equivalent resistance, RO.
.............................................................................................................................
.............................................................................................................................
(iv)
Draw the equivalent circuit.
47
Module ET2
Electronic Circuits and Components.
(b)
Show how the equivalent circuit can be used to calculate the load
voltage, when a load of 25Ω is connected across the output.
..............................................................................................................................
..............................................................................................................................
48
Topic 2.5 – Signal Sources.
Schmitt Inverters.
The last section of this topic deals with some additional uses for the Schmitt
inverter first introduced in Topic 2.2.3 – Astable Circuits. You should
remember that the circuit symbol for this device looks very similar to a NOT
gate, as shown below:
The difference between the Schmitt inverter and the NOT gate is the fact
that a Schmitt inverter has two switching thresholds, as shown by the
following characteristic.
VOUT
5
4
3
2
1
0
0
1
2
3
4
5
VIN
Looking at the characteristic we can see that when the input voltage is at 1V
for example the output is at 5V, demonstrating the inverting action of the
Schmitt inverter. The output remains at 5V until the input reaches 4V, when
the output switches to 0V. The output remains at 0V even if the input voltage
increases further. The transition at 4V is called the Logic 1 to Logic 0
threshold.
When the input voltage is now decreased nothing happens at the 4V threshold
the output remains low, until the input reaches 1.5V when the output returns
to 5V, and remains there even if the voltage is reduced further. The
transition at 1.5V is called the Logic 0 to Logic 1 threshold.
49
Module ET2
Electronic Circuits and Components.
The switching thresholds given here at 1.5V and 4V are not the same for all
Schmitt inverters, they can vary from device to device by the principle of
operation is the same, there are two threshold points separated by a couple
of volts. In exam questions you will either be given a graph like the one we
have just looked at, and you will be asked to work out the thresholds from
the graph; or there will be an extract from the datasheet giving the
appropriate values for the transition points, as shown below:
When connected to a 12V supply:




Logic 0 = 0V
Logic 1 = 12V
The output changes from logic 1 to logic 0 when a rising input voltage reaches 7V
The output changes from logic 0 to logic 1 when a falling input voltage reaches 3.2V
The first additional use of the Schmitt inverter is to interface an analogue
sensor like an LDR or thermistor to a digital circuit. It is needed because
these sensors produce voltages between the two extremes of the power
supplies and for many digital circuits, this causes a problem because they
have difficulty determining whether the input is high or low when it is in the
mid range of the power supply and this may cause the circuit to fail.
Using a Schmitt inverter between the analogue sensor and the logic gate,
ensures that the input to the digital circuit is either going to be high or low.
The typical arrangement is as follows:
Light
Sensor
50
Schmitt
Inverter
Rest of system
Topic 2.5 – Signal Sources.
The following graph shows a typical output from a light sensor, and the
corresponding output from a Schmitt inverter having the thresholds given
earlier, i.e. 1.5V and 4V.
VIN
5
4
3
2
1
0
VOUT
5
4
3
2
1
0
The result of the Schmitt inverter should now be very clear as a highly
variable input voltage is transformed into a very clear digital signal. The
Schmitt inverter has removed the small fluctuations that may occur for
example when a cloud passes overhead and reduces the amount of sun falling
onto a light sensor.
51
Module ET2
Electronic Circuits and Components.
The second application is to de-bounce a mechanical switch. This is essential
for counting circuits as left uncorrected this could result in multiple counts
taking place every time the switch is pressed.
The basic circuit is as follows:
+V
VOUT
R
C
0V
The value of R & C need to be experimented with depending on the counting
application you are using, e.g. frequency of pulses, and whether CMOS or TTL
circuits are being used.
No questions will be set in the examination requiring knowledge of the
values of R & C as this is a practical issue and may affect any project
you might design for ET6, however you will be expected to draw the debouncing circuit shown above in the examination without the need to
suggest component values.
52
Topic 2.5 – Signal Sources.
Solutions to Student Exercises.
Student Exercise 1:
i)
at 0°C.
Voltmeter Reading 
VIN  Rthermistor
R1  Rthermistor
9  8k
4.7k  8k
72

 5.67V
12.7

ii)
at 25°C.
Voltmeter Reading 
VIN  Rthermistor
R1  Rthermistor
9  2.5 k
4.7k  2.5 k
22.5

 3.13V
7.2

iii)
at 100°C.
Voltmeter Reading 
VIN  Rthermistor
R1  Rthermistor
9  1k
4.7k  1k
9

 1.58V
5.7

Student Exercise 2:
1.
i)
When the temperature falls, the resistance of the thermistor
increases, causing a decrease in the current flowing in the circuit,
and therefore the voltage across R1 decreases, so voltmeter
reading falls as temperature falls.
53
Module ET2
Electronic Circuits and Components.
ii)
a)
at 0°C.
Voltmeter Reading 
VIN  R1
R1  Rthermistor
6  10k
250k  10k
60

 0.23V
260

b)
at 100°C.
Voltmeter Reading 
VIN  R1
R1  Rthermistor
6  10k
0.750k  10k
60

 5.58V
10.750

2.
22kΩ
a)
at 25°C.
Voltmeter Reading 
VIN  R1
R1  Rthermistor
6  22k
22k  90k
132

 1.18V
112

54
b)
at 70°C.
Voltmeter Reading 
VIN  R1
R1  Rthermistor
6  22k
22k  5 k
132

 4.88V
27

Topic 2.5 – Signal Sources.
3.
(a)
(b)
Graph to show the resistance characteristic of the thermistor
1300
1200
1100
1000
Resistance (k.ohms)
900
800
700
600
500
400
Mid-point
300
200
100
0
-50
-40
-30
-20
-10
0
10
20
Temperature ('C)
The midpoint of the thermistor temperature range required is 350kΩ.
The nearest preferred value from the E24 series is 360kΩ.
(c)
Determine the resistance of the thermistor at
(i)
at -40°C, Tthermistor = 640kΩ (±5kΩ)
(ii)
at +5°C, Tthermistor = 50kΩ (±5kΩ)
55
Module ET2
Electronic Circuits and Components.
(d)
Calculate the output voltage from your sensor at these two
extreme temperatures.
(i)
At -40°C
Voltmeter Reading 
VIN  Rthermistor
R1  Rthermistor
12  640k
350k  640k
7680

 7.75V
990

(ii)
At +5°C.
Voltmeter Reading 
VIN  Rthermistor
R1  Rthermistor
12  50k
360k  50k
600

 1.46V
410

Student Exercise 3.
1.
56
Topic 2.5 – Signal Sources.
2.
An LDR is used as a light sensor in the following circuit.
(i)
At 10 lux
Vout 
VIN  RLDR
R2  RLDR
9  1500k
12k  1500k
13500

 8.92V
1512

(ii)
At 20,000 lux
Vout 
VIN  RLDR
R2  RLDR
9  400k
12k  400k
3600

 8.73V
412

(iii)
At 100,000 lux
Vout 
VIN  RLDR
R2  RLDR
9  3k
12k  3k
27

 1.8V
15

3.
(a)
at the ice point.
Vout 
(b)
Vin  R2
6  8000
48000


 0.047V  47mV
R1  R2 1000000  8000 1008000
at the steam point.
Vout 
Vin  R2
6  8000
48000


 4.8V
R1  R2 2000  8000 10000
57
Module ET2
Electronic Circuits and Components.
4.
(a)
in full sunlight
Vout 
(b)
in the dark
Vout 
(c)
Vin  R2
6  8000
48000


 5.926V
R1  R2 100  8000 8100
Vin  R2
6  8000
48000


 0.059V  59mV
R1  R2 800000  8000 808000
Swap the positions of the LDR and resistor in the potential divider
so that LDR is on bottom and fixed resistor on top.
Student Exercise 4:
1.
(a)
(i)
(ii)
(iii)
12  22 264

 6.6V
18  22 40
12
I SC 
 0.667 A
18
VOC 
RO 
6 .6
 9 .9 
0.667
(iv)
RO =9.9Ω
VOC=6.6V
58
Topic 2.5 – Signal Sources.
2.
(a)
(i)
(ii)
(iii)
12  180 2160

 7.2V
120  180 300
12
I SC 
 0.1A
120
VOC 
RO 
7.2
 72
0.1
(iv)
RO =72Ω
VOC=7.2V
(b)
(i)
VOUT
RO =72Ω
500Ω
VOC=7.2V
500Ω
VOUT 
7.2  1000 7200

 6.72V
72  1000 1072
(ii)
VOUT
RO =72Ω
VOC=7.2V
VOUT 
500Ω
500Ω
7.2  250 1800

 5.59V
72  250 322
59
Module ET2
Electronic Circuits and Components.
3.
(a)
(i)
(ii)
(iii)
10  150 1500

 6V
100  150 250
10
I SC 
 0.1A
100
VOC 
RO 
6
 60
0 .1
(iv)
RO =60Ω
VOC=6V
(b)
VOUT
RO =60Ω
300Ω
VOC=6V
VOUT 
4.
(a)
(i)
(ii)
(iii)
60
6  300 1800

 5V
60  300 360
12  360 4320

 9V
120  360 480
12
I SC 
 0.1A
120
VOC 
RO 
9
 90
0 .1
Topic 2.5 – Signal Sources.
(iv)
RO =90Ω
VOC=9V
(b)
VOUT
RO =90Ω
910Ω
VOC=9V
VOUT 
(c)
9  910 8190

 8.19V
90  910 1000
If VOUT is not to fall below 8V, then VR  9  8  1V
O
Therefore I R 
O
1
 11.1mA
90
Therefore the maximum current that can be taken from the
output of the voltage source is 11.1mA.
5.
(a)
(i)
(ii)
(iii)
16  120 1920

 4V
360  120 480
16
I SC 
 0.044 A
360
VOC 
RO 
4
 90
0.044
61
Module ET2
Electronic Circuits and Components.
(iv)
RO =90Ω
VOC=4V
(b)
VOUT
RO =90Ω
270Ω
VOC=4V
VOUT 
6.
(a)
(i)
(ii)
(iii)
4  270 1080

 3V
90  270 360
15  150 2250

 9V
100  150 250
15
I SC 
 0.15 A
100
VOC 
RO 
9
 60
0.15
(iv)
RO =60Ω
VOC=9V
62
Topic 2.5 – Signal Sources.
(b)
VOUT
RO =60Ω
25Ω
VOC=9V
VOUT 
9  25 225

 2.65V
60  25 85
Now for some examination style questions:
63
Module ET2
Electronic Circuits and Components.
Examination Style Questions.
1.
The graph below shows the switching characteristic of a Schmitt inverter as the input voltage id
gradually increased from 0V to 2V and then decreased back to 0V.
(a)
Determine the switching threshold for:
(i)
an increasing input voltage; ......................................
(ii)
a decreasing input voltage.
......................................
[2]
64
Topic 2.5 – Signal Sources.
(b)
The Schmitt inverter is used to interface a temperature sensor to a logic system.
Draw the waveform obtained at the output of the Schmitt inverter for the input waveform
shown below.
[3]
65
Module ET2
Electronic Circuits and Components.
2.
The following circuit is used as a voltage source.
12V
240Ω
160Ω
VOUT
0V
(a)
Thevenin’s theorem is used to produce an equivalent circuit.
(i)
Calculate the open circuit voltage, VOC.
..................................................................................................................................................
..................................................................................................................................................
[1]
(ii)
Calculate the short circuit current, ISC.
..................................................................................................................................................
..................................................................................................................................................
[1]
(iii)
Calculate the equivalent resistance, RO.
..................................................................................................................................................
..................................................................................................................................................
[1]
(iv)
Draw the equivalent circuit.
[1]
66
Topic 2.5 – Signal Sources.
(b)
This voltage source is used to drive several output devices, each having a resistance of
100Ω.
Draw the equivalent circuit and use it to calculate the output voltage, VOUT, when;
(i)
two of these 100Ω output devices are connected in series across the output;
................................................................................................................................................
................................................................................................................................................
................................................................................................................................................
[2]
(ii)
two of these 100Ω output devices are connected in parallel across the output;
................................................................................................................................................
................................................................................................................................................
................................................................................................................................................
[2]
67
Module ET2
Electronic Circuits and Components.
3.
The following circuit is used as a voltage source.
12V
30Ω
20Ω
VOUT
0V
(a)
Thevenin’s theorem is used to produce an equivalent circuit.
(i)
Calculate the open circuit voltage, VOC.
..................................................................................................................................................
..................................................................................................................................................
[1]
(ii)
Calculate the short circuit current, ISC.
..................................................................................................................................................
..................................................................................................................................................
[1]
(iii)
Calculate the equivalent resistance, RO.
..................................................................................................................................................
..................................................................................................................................................
[1]
68
Topic 2.5 – Signal Sources.
(b)
(i)
Draw the equivalent circuit with a load resistance of 24Ω connected across its
output terminals.
[1]
(ii)
Use the equivalent circuit to calculate the voltage drop across the 24Ω load.
..................................................................................................................................................
..................................................................................................................................................
[1]
(iii)
Calculate the power dissipated in the load.
..................................................................................................................................................
..................................................................................................................................................
[2]
69
Module ET2
Electronic Circuits and Components.
4.
The following circuit is used as a voltage source.
15V
12Ω
18Ω
VOUT
0V
(a)
Thevenin’s theorem is used to produce an equivalent circuit.
(i)
Calculate the open circuit voltage, VOC.
..................................................................................................................................................
..................................................................................................................................................
[1]
(ii)
Calculate the short circuit current, ISC.
..................................................................................................................................................
..................................................................................................................................................
[1]
(iii)
Calculate the equivalent resistance, RO.
..................................................................................................................................................
..................................................................................................................................................
[1]
70
Topic 2.5 – Signal Sources.
(b)
(i)
Draw the equivalent circuit with a load resistance connected across the output
terminals.
[1]
(ii)
Use the equivalent circuit to calculate the maximum permissible load current to
ensure that the output voltage does not fall below 6.3V.
..................................................................................................................................................
..................................................................................................................................................
..................................................................................................................................................
[2]
71
Module ET2
Electronic Circuits and Components.
5.
A continuous paper feed system in a newspaper printer uses a LED and phototransistor to check
the presence of white paper.
(a)
Complete the following table.
Logic Level
at X
Logic Level
at Y
Paper
Present
Paper
Absent
(b)
Complete the circuit diagram by adding a buzzer which should sound when the system
runs out of paper.
(c)
Explain why it is desirable to include a Schmitt inverter in the circuit?
[2]
..................................................................................................................................................
..................................................................................................................................................
..................................................................................................................................................
[1]
72
Topic 2.5 – Signal Sources.
6.
A Schmitt inverter can be used to condition the signal produced by an analogue sensor.
Here is part of the data sheet for a Schmitt inverter:
When connected to a 10V supply:




Logic 0 = 0V
Logic 1 = 10V
The output changes from logic 1 to logic 0 when a rising input voltage reaches 3V
The output changes from logic 0 to logic 1 when a falling input voltage reaches 1V
The Schmitt inverter is connected as follows in an electronic system:
Light
Sensor
Schmitt
Inverter
Rest of system
The upper graph shows the light sensor output.
Complete the lower graph to show the signal obtained at the output of the Schmitt inverter.
[3]
73
Module ET2
Electronic Circuits and Components.
7.
The following circuit is used as a voltage source.
15V
150Ω
100Ω
VOUT
0V
(a)
Thevenin’s theorem is used to produce an equivalent circuit.
(i)
Calculate the open circuit voltage, VOC.
..................................................................................................................................................
..................................................................................................................................................
[1]
(ii)
Calculate the short circuit current, ISC.
..................................................................................................................................................
..................................................................................................................................................
[1]
(iii)
Calculate the equivalent resistance, RO.
..................................................................................................................................................
..................................................................................................................................................
[1]
74
Topic 2.5 – Signal Sources.
(b)
(i)
Draw the equivalent circuit with a load resistance connected across the output
terminals.
[1]
(ii)
Use the equivalent circuit to calculate the voltage across the load when the load
current is 30mA.
..................................................................................................................................................
..................................................................................................................................................
..................................................................................................................................................
[2]
75
Module ET2
Electronic Circuits and Components.
8.
The graph below shows the switching characteristic of a Schmitt inverter as the input voltage is
gradually increased from 0V to 2.5V and then decreased back to 0V.
Determine the input switching threshold for
(i)
an increasing input voltage
..............................
(ii)
a decreasing input voltage
..............................
[2]
76
Topic 2.5 – Signal Sources.
9.
The following circuit is used as a voltage source.
15V
18Ω
36Ω
VOUT
0V
(a)
Thevenin’s theorem is used to produce an equivalent circuit.
(i)
Calculate the open circuit voltage, VOC.
..................................................................................................................................................
..................................................................................................................................................
[1]
(ii)
Calculate the short circuit current, ISC.
..................................................................................................................................................
..................................................................................................................................................
[1]
(iii)
Calculate the equivalent resistance, RO.
..................................................................................................................................................
..................................................................................................................................................
[1]
77
Module ET2
Electronic Circuits and Components.
(b)
(i)
Draw the equivalent circuit with a load resistance of 48Ω connected across the
output terminals.
[1]
(ii)
Use the equivalent circuit to calculate the voltage drop across the output terminals.
..................................................................................................................................................
..................................................................................................................................................
..................................................................................................................................................
[2]
(c)
A second 48Ω resistance is connected in parallel with the original 48Ω load resistance.
Calculate the new voltage drop across the output terminals.
..................................................................................................................................................
..................................................................................................................................................
..................................................................................................................................................
[2]
78
Topic 2.5 – Signal Sources.
Self Evaluation Review
Learning Objectives
My personal review of these objectives:



describe the use of phototransistors, light depenedent
resistors and ntc thermistors in
voltage dividing chains to provide
analogue signals;
sketch and interpret response
curves for the light dependent
resistors and thermistor;
calculate suitable values for
resistors for use with the above
devices;
use Thevenin’s theorem to draw
equivalent circuits and predict the
effect of loading;
appreciate that the current through
a potential divider should be at
least 10 times that drawn at the
output;
explain how a Schmitt inverter can
be used to provide signal
conditioning with analogue signal
sources and to eliminate mechanical
switch bounce.
Targets:
1.
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
2.
………………………………………………………………………………………………………………
79
Module ET2
Electronic Circuits and Components.
………………………………………………………………………………………………………………
80