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Topic 2.3 – Interface Circuits.
Learning Objectives:
2.3.1 Interfacing to Inputs
At the end of this topic you will be able to;
 compare the merits of transistors, comparators, and Schmitt
inverters as interfaces between analogue sensors and digital systems;
 use the property that a Schmitt inverter gate has two different
input switching levels to draw the output signal for a given
analogue input signal;
 explain why a Schmitt inverter is required to debounce mechanical
switches and analogue sensors connected to a counting systems;
2.3.2 Interfacing to Outputs
At the end of this topic you will be able to;
 select transistors in terms of current gain and collector current;
 apply the following rules to a given transistor switching circuit:
For VIN<0.7V: VBE=VIN and VCE = Supply Voltage,
For VIN>0.7V: VBE=0.7V and VCE = 0V;
 design transistor switching circuits which cause output devices to
respond to information from sensors;
 state that IC=hFEIB until saturation is reached;
 calculate values for resistors and input switching voltages in
transistor switching circuits.
1
GCSE Electronics.
Unit E2 : Applications of Electronics
2.3 Interfacing Inputs and Outputs
To complete our study of logic systems, we need to look at how to interface
inputs and outputs correctly to these systems. Some of the ground work for
this topic was carried in Module E1, where basic transistor circuits and
comparators were introduced. In this chapter we look at interfacing to both
CMOS and TTL logic circuits. The chart below briefly compares the
capabilities of the TTL and CMOS families of IC. Correct interfacing will
ensure successful completion of your practical project if it involves these
families of logic gates.
TTL
Requires a 5V supply
CMOS
Will work off any voltage from
3- 15V
Draws a moderate current
Draws a negligible current
from the power supply
from the power supply
Inputs have to source current No current flows in or out of
when at logic 0
inputs
Unused inputs may be left
All unused inputs have to be
unconnected as they will ‘float’ connected to logic 0 or logic 1
up to logic 1
Output current sufficient to
Output current insufficient to
light up a LED
light up a LED
Time taken for a gate input to Time taken for gate input to
change logic levels should be
change logic levels should be
no more than a few microno more than a few milliseconds
seconds
The ICs are very robust
The ICs are easily damaged by
electrically,
static electricity.
2
Topic 2.3 – Interface Circuits.
2.3.1 Interfacing to inputs
For a logic gate input to recognise the signal applied to it as logic 1 the
voltage level of the signal should be as near as possible to the value of the
positive supply rail. Similarly for the signal to be recognised as logic 0 it
should be as near as possible in value to 0 volts.
If the input signal changes from logic 0 to logic 1 or from logic 1 to logic 0 it
should leap from one to the other as rapidly as possible. This prevents the
logic system behaving in an unpredictable way.
We shall now consider how mechanical switches and input sensors are
connected to the inputs of logic gates.
Connecting Mechanical Switches to a Logic System
Consider the following input sub-systems.
(i)
(ii)
The resistor is referred to as
a pull-up resistor because it
pulls the input up to logic 1
when the switch is open. The
resistor value is not critical. A
10 kΩ resistor is usually used
for TTL circuits and a value up
to 100 kΩ for CMOS circuits.
In this case the resistor is
referred to as a pull down
resistor. The resistor
should normally be less than
or equal to 1kΩ for TTL.
For CMOS circuits,
resistors up to 100kΩ may
be used.
Output normally at logic 1. It changes
to logic 0 when the switch is pressed.
Output normally at logic 0. It changes
to logic 1 when the switch is pressed.
3
GCSE Electronics.
Unit E2 : Applications of Electronics
Connecting sensors to a Logic System
Consider the following light sensing sub-system:
Assume that the light level steadily increases as daylight breaks. This causes
the resistance of the LDR to decrease and the output voltage V0 will increase
as shown in the graph below. The output voltage V0 is an example of an
analogue signal.
The change in output voltage may occur over several hours and the voltage V0
could be too big for logic 0 and too small for logic 1 for a few hours.
This slowly changing voltage would be of little use as an input to a logic gate,
which requires a signal that changes logic state in a few microseconds.
4
Topic 2.3 – Interface Circuits.
Signal Conditioning
Slow changing signals from analogue sensing circuits have to be processed or
conditioned before they can be effectively used with logic gates.
A switching circuit such as a comparator or a transistor could be used to
interface the analogue sensing sub-system to the logic system. See topics
1.7.2 and 1.7.4 of E1 notes.
A better solution is to use a Schmitt inverter as an interface.
Schmitt Inverters
Schmitt inverters are ideal for interfacing input sensors to logic systems.
They are particularly useful for improving rise times by converting slowly
changing analogue signals into signals that change logic state almost instantly.
They have the added advantage of ignoring small changes in the signal
produced by noise in the sensing sub-system.
The action of a Schmitt inverter
Ordinary inverters have a single input voltage level which causes the output
to change state. This voltage level is referred to as the threshold level. A
Schmitt inverter has the same truth table as an ordinary inverter but it has
two switching thresholds rather than one.
The switching threshold for a rising input voltage is higher than that for a
falling input voltage. The actual threshold values differ for different types
of Schmitt inverter.
For example on a 5V supply a TTL 7414 Schmitt inverter has an upper
switching threshold of approximately 1 .7V for rising input voltages and a
lower switching threshold of 0.9V for falling input voltages.
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GCSE Electronics.
Unit E2 : Applications of Electronics
This means that:
 when the input voltage at 0V the output is at logic 1;
 as the input voltage increases, the output remains at logic 1 until the
input voltage reaches 1 .7V;
 then, the output abruptly changes to logic 0 and remains there until the
input voltage drops below 0.9V;
 then the output will abruptly change back to logic 1.
This action is summarised in the following graph
The Schmitt inverter has a dual purpose.
(i)
It converts a slowly changing signal to an abruptly changing signal once
the input threshold voltage has been reached.
(ii)
The gap or ‘deadband’ between the threshold for a rising input voltage
and a falling input voltage prevents minor changes in the signal from a
sensor rapidly switching the output logic level repeatedly between logic
0 and 1.
With a 5V supply a CMOS 40106 Schmitt inverter has an upper switching
threshold of approximately 3.0V for rising input voltages and a lower
switching threshold of 2.2V for falling input voltages. For other supply
voltages the upper switching threshold is approximately 60% of the supply
voltage and the lower threshold is approximately 40% of the supply voltage.
6
Topic 2.3 – Interface Circuits.
Example: Improving the rise time of a light sensing sub-system
The signal from a light sensing unit is applied to a Schmitt inverter.
 In darkness the resistance of the LDR is very high, the input voltage Vin
is nearly 0V and the output is at logic 1.
 Vin increases gradually as the light level increases.
 The output, Q, remains at logic 1 until Vin reaches the upper switching
threshold, of 1.7V.
 Then, the output changes almost instantly to logic 0 and remains there
until Vin falls to 0.9V.
 Then, the output changes almost instantly to logic 1.
Typical behaviour is shown in the graph
opposite.
Notice the rapid transitions between output
logic levels, as the input signal reaches the
threshold voltages.
Minor fluctuations in input voltage are
ignored.
The Schmitt inverter has conditioned the input signal to make it suitable for
a logic system.
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GCSE Electronics.
Unit E2 : Applications of Electronics
Worked Example:
The block diagram shows the design of a security light, which turns on
automatically when it gets dark.
(a)
The Light Sensing Unit gives out a Logic 0 signal in the dark.
(i)
The Schmitt Inverter inverts the signal from the Light
Sensing Unit. Explain what this sentence means.
Answer: The Schmitt Inverter produces a logic 1 output when the
input is logic 0 (i.e. dark) or a logic 0 output when the input is logic
1 (i.e. light). The Schmitt inverter therefore swaps the logic level
from input to output.
(ii)
What is the other use of the Schmitt Inverter in this
system?
Answer: The Schmitt Inverter acts as an interface between a
slowly changing light level and produces a fast changing output
when the threshold light level is reached.
(b)
8
Here is part of a data sheet for a Schmitt Inverter:
Topic 2.3 – Interface Circuits.
The input signal for the Schmitt Inverter is shown below.
Use the axes provided to draw the resulting output signal produced by
the Schmitt inverter.
Step 1 : Mark in the
two switching
threshold values.
In this case at 1V,
and 3V.
Step 2 : Mark in the points at
where switching occurs, i.e. at
3V for a rising voltage and 1V
for a falling voltage.
Step 4 : Add in the output
graph.
Step 3 : Add the output
thresholds for the
Schmitt Inverter. In this
case at 0V and 5V.
Initially the output is high,
because the input is low.
At first switching threshold the
output switches from logic 1 to
logic 0 etc.
9
GCSE Electronics.
Unit E2 : Applications of Electronics
Activity 1:
Connecting a light sensor to a counter
1a.
Set up the circuit below which is used to count the number of people
boarding a fairground ride. A light beam is broken as someone walks
through it.
1b.
Comment on how well the system works as the light level falling on the
LDR changes. Try changing the light level both slowly and quickly.
................................................................................................................................
................................................................................................................................
1c.
You should have observed that the system only counts when the light
level is changed vey quickly. Modify your circuit as follows.
1d.
Is the performance of the system better? Give a reason for any change.
................................................................................................................................
................................................................................................................................
10
Topic 2.3 – Interface Circuits.
Interfacing mechanical switches to count and display systems
Switch Bounce
The contact blades of both mechanical and reed switches tend to ‘bounce’
when the switch is closed.
When the contacts are closed, the output signal is logic 0.
When the contacts bounce open, the pull-up resistor makes the output jump
to logic 1.
When the switch is closed, the bounces make the output change rapidly
between logic 0 and logic 1 several times before settling down at logic 0. This
effect is known as switch bounce.
Switch bounce has little effect in a system consisting only of logic gates. If
the system contains an electronic counter, then switch bounce is highly
undesirable. The counter will jump by several numbers each time the switch is
pressed, as it counts each of the bounces.
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GCSE Electronics.
Unit E2 : Applications of Electronics
Switch De-bouncing
A capacitor, connected across the switch, will suppress but not completely
eliminate the effect of the contact blades bouncing.
Adding a Schmitt inverter will eliminate any effect of switch bounce the
output. Notice that we have swapped over the positions of the switch and
resistor, to take account of the inverting action of the Schmitt Inverter.
The small fluctuations in VOUT2 are ignored by the Schmitt inverter. The
output VOUT3 is said to be “de-bounced”.
The value of capacitor used depends on how fast you want to count. A value in
the range 1 – 10F works well for most applications.
12
Topic 2.3 – Interface Circuits.
Activity 2:
2a.
If you have access to Livewire go to Tools >> Simulation and click on
>> Bounce. For Circuit Wizard go to Project >> Simulation and click on
>> Bounce.
Set up the following counting system .
3b.
Press switch SW1 several times. You will probably get an unexpected
result. Can you give a reason why?
.............................................................................................................................
.............................................................................................................................
3c.
Add a de-bounce circuit to the input using a 10F capacitor and a 40106
Schmitt inverter. (You will also have to make one other change to the
circuit) Comment on the accuracy of the count.
.............................................................................................................................
3d.
Does changing the capacitor value to 1 F have any affect?
.............................................................................................................................
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GCSE Electronics.
Unit E2 : Applications of Electronics
Summary of Interfacing to Inputs
Interface Unit
Transistor Switch
Advantages
Starts to switch on when Vin
reaches a fixed voltage of
0.7V.
Has an inverting action.
Disadvantages
Unable to change the
voltage at which turn on
occurs.
Does not work very well
with a slowly changing
signal, as the output
signal is not necessarily
fast changing.
Comparator
Can be configured as an
inverting or non-inverting
interface.
Switching point can be
varied by the user.
Output voltage swings
between voltage rails almost
instantaneously.
Circuit is more complex
than the transistor or
Schmitt trigger.
Schmitt Trigger
Simple interface for
analogue sensors.
Analogue signal is converted
into digital.
‘Dead band’ between
switching thresholds
provides protection against
rapid on/off switching
caused by small fluctuations
at the threshold point.
Can be used to de-bounce
mechanical switches for use
in counting circuits.
Switching thresholds
can not be changed from
their preset values.
14
Topic 2.3 – Interface Circuits.
2.3.2
Interfacing to Outputs
You met the transistor in Module E1, as a simple switching circuit. We now
develop this further so that you can design your own switching circuits.
The simple transistor circuit we set up in E1 was :
+6V
Lamp
Flying Lead
Collector
Base
Emitter
0V
There are a few things we need to remember:
 a resistor has been added to the base terminal to limit the current
flowing into the transistor base
 the load (a lamp in this case) is connected into the collector circuit.
 With the flying lead connected to 0V the transistor is switched off and
the lamp does not light.
 With the flying lead connected to +6V the transistor switches on and
allows current to flow through the collector and emitter, and the lamp
lights
There were also two basic rules we were given about the transistor.
i. if the voltage between the base and emitter (usually referred to as VBE)
< 0.7V then the transistor will be off;
ii. if the voltage between the base and emitter (VBE) > 0.7V then the
transistor will be on.
15
GCSE Electronics.
Unit E2 : Applications of Electronics
To understand the operation of the transistor, it is necessary to study
the switching action of the transistor in a little more detail. A suitable
circuit for this is shown below.
+6V
LOAD
Collector
1kΩ
10kΩ
Base
VOUT
Emitter
VIN
0V
By adjusting VIN from 0V up to +6V, in small increments, and measuring the
corresponding values of VOUT, a picture of what is happening to the transistor
can be shown. The following graph shows the typical response that is obtained
from a circuit like that shown above.
Transistor off
VOUT (V)
Linear Region
Saturation
6
4
2
0.7V
0
0
1
2
3
4
5
6
VIN (V)
There are three key parts to this graph, which is known as the transistor
voltage transfer characteristic.
16
Topic 2.3 – Interface Circuits.
(i)
Off region
This part of the characteristic when VIN is between 0 and 0.7V shows
when the transistor is completely switched off, no current flows
through the base-emitter junction, no current flows through the
collector, and the voltage across the collector emitter junction of the
transistor (VOUT) is equal to the supply voltage.
(ii)
Linear region
When the voltage VIN increases above 0.7V, a base current starts to
flow. The transistor behaves as a current amplifier and the base current
causes a larger amplified current to flow through the collector and load.
As VIN increases further more current flows into the base and this
allows a further increase in the collector-emitter current. Small changes
in VIN cause corresponding changes in VOUT.
(iii) Saturation
As VIN continues to increase, a point is reached where changes to VIN no
longer cause any change to VOUT, and we say that the transistor is
saturated.
The saturation point is reached just before the voltage across the load
reaches the full voltage of the power supply and the voltage across the
collector-emitter junction of the transistor VOUT is about 0.2V (i.e.
nearly = 0V)
Note:
 Although in practice the base emitter voltage VBE does increase slightly
(up to a maximum of about 0.8V) as VIN is increased above 0.7V, we will
assume it is fixed at 0.7V.
 We have referred to the voltage across the collector-emitter junction
of the transistor as VOUT. It is often referred to as VCE .
17
GCSE Electronics.
Unit E2 : Applications of Electronics
Transistor Switching Circuits
Transistors can operate either as switches or as amplifiers. When we use the
transistor as an amplifier we need to use the linear region and avoid the cutoff and saturation regions
When the transistor is being used as a switch, we operate in the cut-off and
saturation regions of the characteristic, avoiding the linear region.
There are two reasons for avoiding the linear region when designing
transistor switching circuits. Firstly the output device will not work
correctly because the full supply voltage does not appear across the load as
VCE will have a significant value. Secondly because of this value of VCE, and the
current flowing in the collector, power will be used up in the collector-emitter
junction causing the transistor to overheat.
In this course we will only be considering switching circuits, and the following
information will be important.
For VIN<0.7V: VBE=VIN and VCE = Supply Voltage,
For VIN>0.7V: VBE=0.7V and VCE = 0V;
Current Gain (hFE)
In order to design circuits for transistors, there is also an important formula
which needs to be considered. This is the current gain formula for the
transistor. We have mentioned several times that the transistor acts as a
current amplifier. Each transistor has a current gain called ‘hFE’ and this is
defined by the following Current Gain formula.
h FE 
IC
IB
where IC is the collector current, and IB is the base current.
18
Topic 2.3 – Interface Circuits.
For example:
(i)
If IB = 10mA and hFE= 120, what is IC?
h FE 
IC
IB
IC
10mA
I C  120  10mA  1200mA  1.2A
120 
(ii)
If IC = 800mA and hFE = 250, what is IB?
h FE 
IC
IB
250 
800mA
IB
IB 
800mA
 3.2mA
250
Always check after using the current gain formula that IB is smaller than IC
Every transistor has a different hFE value and they can range from 10 to over
800 in value depending on the type of transistor. As far as examination
questions are concerned, you would not be expected to remember the
different values of hFE, you will either be told the hFE value for the transistor,
or you will be able to calculate it from values of IB and IC.
Note:
The formula for current gain is only valid in the linear region. But we stated
earlier that we avoid the linear region when considering switching circuits.
We get around this problem when designing switching circuits by assuming any
calculations performed involving current gain is done at the point where the
transistor switching action is just leaving the linear region and entering the
saturation region. That is at the last possible moment the formula is still
valid.
19
GCSE Electronics.
Unit E2 : Applications of Electronics
Examination questions will always be worded similar to the wording below:
“Determine the value of VIN that will cause the transistor to just saturate”
or
“The transistor is just saturated when the input voltage VIN = 2.5V”
We are now in a position to start looking at some design problems involving
the transistor switching circuits. So first a couple of examples:
20
Topic 2.3 – Interface Circuits.
Example 1
The circuit below contains a transistor with a current gain, hFE = 250. The
circuit switches a warning lamp rated at 6V, 200mA
(a)
Determine the collector current when the lamp is working at its
rated voltage and current and the transistor is just saturated.
Solution: Voltage across lamp = 6V and VCE = 0V therefore
IC = 200mA
(b)
Calculate the base current.
Solution: Now that we know IC, we can find IB, by rearranging the
current gain formula.
hFE 
IC
IB
IB 
IC
200

 0.8mA
hFE 250
21
GCSE Electronics.
Unit E2 : Applications of Electronics
(c)
Calculate the voltage across RB
Solution:
Using Ohm’s Law.
VRB  I B  RB
 0.8mA 1k
 0.8V
(d)
Determine the value of VIN that will cause the transistor to just
saturate.
Solution: Finally to get the voltage for VIN, we just have to add on
the voltage across the base-emitter junction VBE, which is always
0.7V, therefore
VIN  VRB  VBE
 0.8V  0.7V
 1.5V
(e) Complete the following table to show :
(i)
the voltage VBE and VCE for the input voltages VIN given.
(ii)
whether the buzzer will be On or Off .
Solution: The answers are shown in red in the table. Remember:
For VIN < 0.7V: VBE = VIN and VCE = Supply Voltage,
For VIN > 0.7V: VBE = 0.7V and VCE = 0V;
22
Input voltage, VIN
VBE
VCE
Lamp On/Off?
0.2V
0.2V
6V
Off
1.9V
0.7V
0V
On
Topic 2.3 – Interface Circuits.
Example 2
The temperature sensing circuit below contains a transistor with a current
gain, hFE = 400. The circuit switches a warning buzzer on when the
temperature in a greenhouse gets too high. The resistance of the buzzer is
30Ω.
9V
Buzzer
R=30Ω
IC
RB
IB
VOUT
0V
(a)
Calculate the collector current when the transistor is just
saturated.
Solution: When the transistor is just saturated, VOUT will be = 0V.
This means that the whole voltage of the power supply must be
across the buzzer.
IC 
(b)
VSupply
RLoad

9
 0.3 A  300mA
30
Calculate the base current.
Solution:
h FE 
IC
IB
 IB 
IC
300

 0.75mA
h FE 400
23
GCSE Electronics.
Unit E2 : Applications of Electronics
(c)
At a certain temperature the base current is 0.5mA.
We know the transistor was just saturated with base current of
0.75mA so now it is no longer saturated.
(i)
What is the new value of collector current?
Solution:
h FE 
IC
IB
IC
0.5mA
I C  400  0.5mA  200mA
400 
(ii)
What is the new value of the voltage across the buzzer
Solution:
VBuzzer  I C  R Buzzer
 200mA  30
 6V
(d)
When the base current was 0.5mA it was found that the transistor
became very hot and the buzzer was quiet. Suggest a reason why
this happened.
Solution: The transistor is not saturated, and is therefore
operating in the linear region. This results in the transistor
overheating and can permanently damage the transistor.
The voltage across the buzzer is only 6V so it sounds quiet.
24
Topic 2.3 – Interface Circuits.
Example 3
A light sensing unit is connected to a transistor switch, to operate a solenoid
when the light level gets too bright. The solenoid is rated at 6V, 800mA.
The circuit diagram for this system is shown below
The transistor is just saturated and V1 =3V.
(a)
What is the value of voltage VBE?
VBE = 0.7V
(b)
What is the value of voltage VCE?
VCE = 0V
(c)
What is the value of the voltage drop across resistor R?
Voltage drop across R = V1 - VBE = 3 - 0.7 = 2.3V
(d)
What is the value of the collector current when the transistor is
just saturated?
IC = rated current of solenoid = 800mA
25
GCSE Electronics.
Unit E2 : Applications of Electronics
(e)
Calculate the value of the base current if the current gain (hFE) of
the transistor is 200.
Solution: I B 
(f)
Use your answers to (c) and (e) to calculate the ideal value of
resistor R.
Solution: R 
(g)
I C 800mA

 4mA
h FE
200
Voltage drop across R 2.3V

 0.575k  575
IB
4mA
Choose a suitable preferred value for R from the E24 series of
resistors
Solution: We have to choose between 560Ω or 620Ω.
If we choose 620Ω the base current will be less than 4mA
and the transistor will not saturate.
Therefore we choose 560Ω.
Now it’s time for you to have a go!
26
Topic 2.3 – Interface Circuits.
Student Exercise 1
1.
The transistor shown in the following switching circuit has a current
gain, hFE = 120.
The value of VIN is sufficient just to saturate the transistor. Calculate:
(a)
The collector current.
........................................................................................................................................
........................................................................................................................................
(b)
The base current.
........................................................................................................................................
........................................................................................................................................
(c)
The voltage drop across the 2.2k resistor.
........................................................................................................................................
........................................................................................................................................
(d)
The value of VIN.
........................................................................................................................................
27
GCSE Electronics.
Unit E2 : Applications of Electronics
2.
The following circuit is used to test whether transistors are good or
faulty.
(a)
Complete the following table to show the test results for a good
transistor.
BULB
SWITCH, S
(ON/OFF)
OPEN
CLOSED
(b)
After switch S is closed what is the voltage at point X in the
circuit?
………………………………………………………………………………………
(c)
The transistor should just saturate when switch S is closed.
When switch S is closed:
(i)
What is the voltage drop across the base emitter-junction of
the transistor?
............................................................................................................................
28
Topic 2.3 – Interface Circuits.
(ii)
Calculate the voltage drop across the resistor R.
............................................................................................................................
(iii) What is the value of the collector current when the
transistor is saturated?
............................................................................................................................
(iv)
Calculate the value of the base current if the current gain
(hFE) of the transistor being tested is 50.
............................................................................................................................
(d)
(i)
Calculate the ideal value of resistor R.
............................................................................................................................
............................................................................................................................
(ii)
Select a suitable preferred value for R.
........................................................
29
GCSE Electronics.
Unit E2 : Applications of Electronics
3.
The following diagram shows a temperature sensing circuit which
operates a warning lamp when the ambient temperature rises above a
pre-determined value.
12V
R=10Ω
VIN
0V
When the circuit was tested it was found that the lamp was dim when
the temperature rose just above the above the pre-determined value.
The base current was measured and found to be 4mA.
(a)
Calculate the value of the collector current if transistor has a
current gain (hFE) of 125.
........................................................................................................................................
........................................................................................................................................
(b)
Calculate the voltage drop across the lamp if it has a resistance of
10.
...................................................................................................................................
...................................................................................................................................
30
Topic 2.3 – Interface Circuits.
(c)
The performance of this system is improved by adding a Schmitt
Inverter.
(i)
Redraw the circuit with a Schmitt Inverter included.
(ii)
Explain how the Schmitt Inverter improves the performance
of the system.
..............................................................................................................................
..............................................................................................................................
..............................................................................................................................
31
GCSE Electronics.
Unit E2 : Applications of Electronics
4.
The circuit diagram shows a transistor switching circuit
The transistor has a current gain (hFE) of 80
The bulb is rated at 6V, 240mA.
(a)
The transistor is just saturated. Calculate:
(i)
the collector current.
……………………………………………………………………………………………………………………
(ii)
the base current IB;
………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………
(iii) the voltage VB across the base resistor;
………………………………………………………………………………………………………………………
………………………………………………………………………………………………………………………
(iv)
the input voltage V1 from the logic system.
………………………………………………………………………………………………………………………
32
Topic 2.3 – Interface Circuits.
(b)
Complete the table by giving voltages V2 and V3, and the state of
the bulb.
Input voltage
V1
V2
V3
Bulb –
On/Off?
0.3V
5.1V
33
GCSE Electronics.
Unit E2 : Applications of Electronics
Selecting a suitable transistor
A brief look through an electronic component supplies catalogue, e.g. Rapid
Electronics, or Maplin will reveal many pages dedicated to transistors. An
example is shown below.
How do you select the most appropriate transistor for your application?
There are several key points to think about:
i)
ii)
iii)
What is the maximum collector current that your load requires.
This will be written as IC(MAX) in the data tables for the transistor,
(highlighted in yellow).
What current gain, hFE, do you require; shown as DC Current gain in
the table above, (highlighted in blue). Note: that it is given at a
specific value of IC.
Power supply voltage, given as VCEO in the table above, (highlighted
in red).
The application will determine the most appropriate transistor to use.
34
Topic 2.3 – Interface Circuits.
Solutions to Student Exercise 1:
1.
(a)
As the transistor is saturated, then VOUT = 0V, therefore
IC 
9
 0.06 A  60mA
150
(b)
IB 
IC
60

 0.5mA
hFE 120
(c)
V2.2 k  I B  2.2k  0.5mA  2200  1.1V
(d)
To find the value of VIN, add the voltage across the 2.2kΩ resistor
to the voltage drop across the base-emitter junction.
VBE  0.7V  VIN  1.1V  0.7V  1.8V
2.
(a)
SWITCH, S
BULB
(ON/OFF)
OPEN
Off
CLOSED
On
(b)
The voltage at X = 9V.
(c)
(i)
The voltage drop across the base emitter-junction of the
transistor is 0.7V.
(ii)
The voltage drop across the resistor R = 9 – 0.7 = 8.3V.
(iii) The value of the collector current when the transistor is
saturated is 200mA.
35
GCSE Electronics.
Unit E2 : Applications of Electronics
(iv)
h FE 
(d)
IC
IB
50 
200mA
IB
IB 
200mA
 4mA
50
R
9  0.7
8.3

 2075
4mA
4mA
(i)
(iii) A suitable preferred value for R will be 2kΩ.
3.
(a)
h FE 
IC
IB
IC
4mA
I C  125  4mA  1000mA  1A
125 
(b)
Vlamp  I C  R lamp
 1 10
 10V
(c)
36
(i)
Redraw the circuit with a Schmitt Inverter included.
Topic 2.3 – Interface Circuits.
4.
(a)
(ii)
The Schmitt waits until the input voltage reaches the
switching threshold of the Schmitt which then rapidly drives
the transistor from being cut off to fully saturated and
therefore less power is wasted in the transistor.
(i)
240mA
(ii)
h FE 
IC
IB
80 
240mA
IB
IB 
240mA
 3mA
80
(iii)
VB  I B  R B
 3mA  1k
 3V
(iv)
V1  VR  VBE
 3V  0.7 V
 3 .7 V
(b)
Complete the table by giving voltages V2 and V3, and the state of
the bulb.
Input voltage
V1
V2
V3
0.3V
0.3V
6V
Bulb –
On/Off?
Off
5.1V
0.7V
0V
On
O.K. Now for some more practice with some examination style questions.
37
GCSE Electronics.
Unit E2 : Applications of Electronics
Examination Style Questions
1.
The following circuit diagram shows part of a system used to switch on a buzzer.
+ 5V
Vc
Rb
Va
Vb
0V
The transistor is just saturated when the input voltage Va is 2.5V.
(a)
Complete the following table to show :
(i)
(ii)
the voltage Vb and Vc for the input voltages Va given.
whether the buzzer will be On or Off .
Va
Vb
Vc
Buzzer
On/Off?
0.5V
3.5V
[5]
(b)
The transistor in this circuit has a current gain hfe of 150.
The transistor is not saturated when the current through Rb is 0.5 mA.
(i)
Calculate the current through the buzzer.
……………………………………………….…………………………………………. [2]
(ii)
Calculate the voltage VC if the buzzer has a resistance of 40.
………………………………………….………………………………………………
(c)
[2]
What is the purpose of the resistor Rb?
……………..……………………………………………………………………………… [1]
38
Topic 2.3 – Interface Circuits.
2.
The diagram shows the circuit for a transistor switch.
(a)
The smallest value of V1 which causes the transistor to saturate is 1.5V.
When V1 = 1.5V,
…………………………
(i)
what is the value of V2;
(ii)
what is the voltage drop across resistor R?
…………………………
[2]
(b)
The transistor is just saturated.
The transistor has a current gain hFE of 100.
The current through the bulb is 60mA.
(i)
What is the current I?
……………………………………………………………………………….……………..
[1]
(ii)
Hence calculate the resistance of resistor R ?
……………………………………………………………………..………………………..
(c)
……………………………………………………………………………..………………..
[2]
What happens to the current through the bulb if the input voltage V1 is increased still
further?
…………………………………………………………………………..…………………..
…………………………………………………………………………..…………………..
[1]
39
GCSE Electronics.
Unit E2 : Applications of Electronics
3.
The circuit diagram shows a transistor switch used as a transducer driver.
6V
Bulb
Signal
from
logic
system
R = 1k I
B
V1
VB
6V
180mA
V3
V2
0V
(a)
Complete the table by giving voltages V2 and V3, and the state of the bulb when the input
voltage V1 = 0.2V. The transistor is switched off.
Input voltage V1
V2
V3
Bulb – On / Off?
0.2V
[2]
(b)
The bulb is rated at 6V 180mA.
Calculate the power dissipated in the bulb when it is switched on fully.
…………………………………………………………………………………………….
[1]
(c)
A transistor with a current gain (hFE) of 90 is used in this circuit.
The transistor is just saturated.
Calculate:
(i)
the base current IB ;
…………………………………………………………………………………………….
…………………………………………………………………………………………….
[2]
(ii)
the voltage VB across the base resistor;
……………………………………………………………………………………………….
………………………………………………………………………………………………..
[2]
(iii)
the input voltage V1 from the logic system.
………………………………………………………………………………………………..
[1]
40
Topic 2.3 – Interface Circuits.
4.
The following circuit diagram shows part of a system used to monitor the temperature of a car
radiator.
The transistor is just saturated.
(a)
(i)
What is the value of voltage VCE ?
…………………….
(ii)
What is the value of voltage VBE ?
…………………….
[1]
[1]
(b)
The minimum value of VIN which causes saturation is 5V.
(i)
Calculate the current flowing through RB when VIN = 5V.
………………………………………..……………………………………………..
……………………………..………………………………………………………..
[2]
(ii)
Calculate the voltage drop across RB when VIN = 5V.
…………………………………………………..…………………………………..
………………………………………..……………………………………………..
[1]
(iii)
Calculate the resistance of RB.
………………………………………..……………………………………………..
……………………………..………………………………………………………..
[2]
(iv)
Choose a suitable preferred value for resistor RB.
………………………………………………………………………………………..
[1]
41
GCSE Electronics.
Unit E2 : Applications of Electronics
5.
The following circuit diagram shows part of a system used to switch on a lamp .
6V
6V , 60 mA
Rb
V1
V2
V3
0V
The transistor is just saturated when the input voltage V1 is 2.1V.
(a)
Complete the following table to show :


the voltage V2 and V3 for the input voltages V1 given
whether the bulb will be On or Off
V1
V2
V3
Bulb
On/Off?
3.4V
1.0V
[5]
(b)
V1 = 2.1V and the transistor is just saturated .
The transistor in this circuit has a current gain hFE of 200.
The collector current through the lamp is 60 mA.
(i)
Calculate the base current through RB.
(ii)
………………………………..……………………………………………………..
[1]
Calculate the voltage drop across RB .
(iii)
……………………………………………………………………………………….
[1]
Calculate the resistance of RB.
……………………………...………………………………………………………..
……………………………………………………………………………………….
[2]
42
Topic 2.3 – Interface Circuits.
6.
An automatic night-light is used in a baby’s bedroom. Here is the circuit diagram.
The lamp is rated at 6 V, 500 mA and the transistor has a current gain (hFE) of 125
The transistor is just saturated.
(a)
What is the value of V2?
..................................
(b)
What is the value of V3?
..................................
(c)
What is the value of the collector current?
.................................
(d)
Calculate the base current.
[1]
[1]
[1]
.......................................................................................................................................
.......................................................................................................................................
[2]
43
GCSE Electronics.
Unit E2 : Applications of Electronics
7.
The block diagram below shows control system for an automatic door. When the light beam is
broken the door opens for 25 seconds.
Light
Source
Light
Sensor
Schmitt
Inverter
Monostable
Interface
Solenoid
(a)
The solenoid is rated at 12 V, 2 A.
(i)
Choose a suitable device for the interface ………………………………………….
(ii)
Draw the part of the circuit diagram that shows how the monostable is interfaced to
the solenoid.
[3]
(b)
The time T, in seconds, that the monostable output is high can be found from the formula
T = 1.1 RC
(where R is in MΩ and C is in µF)
Calculate the value of resistance required with a 22 µF capacitor to produce the 25 second
delay.
..................................................................................................................................................
..................................................................................................................................................
.................................................................................................................................................
[2]
44
Topic 2.3 – Interface Circuits.
(c)
Here is part of a data sheet for the Schmitt Inverter:
When connected to a 12 V supply:
 Logic 0 = 0 V
 Logic 1 =12 V
 The output changes from logic 1 to logic 0 when a rising input voltage reaches 5V
 The output changes from logic 0 to logic 1 when a falling input voltage reaches 3V
The output signal produced by the light sensor is shown in Graph 1.
Complete Graph 2 to show the signal obtained at the output of the Schmitt inverter.
volts
12
10
8
Graph 1
6
4
2
0
volts
time
12
10
8
Graph 2
6
4
2
0
time
[3]
(d)
Other than inverting the signal from the light sensor, explain why a Schmitt Inverter is
required for this application.
…………………...…………………………………………………….........………………………..
………………………………………………………………………………………………………..
………………………………………………………………………………………………………..
[1]
45
GCSE Electronics.
Unit E2 : Applications of Electronics
8.
A factory production line contains a system that shows the operator how many products have been
packed.
The block diagram of the system is given below.
Light
Sensor
(a)
Schmitt
Inverter
Counting
Sub-system
Packing
Sub-system
The counting sub-system must count up to nine and then reset when the tenth product
passes the light sensor.
This sub-system contains a 4-bit binary counter and an AND gate.
Taking the reset pin to logic 1 can reset the counter.
Bit A of the counter is the least significant bit.
Complete the diagram below to show how the AND gate is connected to the counter to
allow it to reset correctly.
[3]
46
Topic 2.3 – Interface Circuits.
(b)
The Schmitt inverter cleans up the signal produced by the light sensor.
A data sheet gives the following information about a Schmitt Inverter when connected to
9V supply:
 The output changes from logic 1 to logic 0 when a rising input voltage reaches 7 V
 the output changes from logic 0 to logic 1 when a falling input voltage reaches 3 V
 assume logic 0 = 0 V and logic 1 = 9 V
Draw the graph of the waveform obtained at the output of the Schmitt Inverter for the
input waveform shown below.
VIN /volts
12
10
8
6
4
2
0
time /s
VOUT /volts
12
10
8
6
4
2
0
time /s
[4]
47
GCSE Electronics.
Unit E2 : Applications of Electronics
9.
The block diagram shows a high temperature alarm, which switches on when the sensor gets too
hot.
Here is part of the data sheet for the Schmitt inverter:
When connected to 5 V supply:
48

Logic 0 = 0 V

Logic 1 = 5 V

The output changes from logic 1 to logic 0 when a rising input voltage reaches 3 V

The output changes from logic 0 to logic 1 when a falling input voltage reaches 1 V
Topic 2.3 – Interface Circuits.
(a)
The input signal for the Schmitt Inverter is shown below.
Use the axes provided to draw the resulting output signal produced by the Schmitt trigger.
[4]
(b)
What is the advantage of using a thyristor rather than a transistor switch in this system?
………………………………………………………………………………………………………..
………………………………………………………………………………………………………..
………………………………………………………………………………………………………..
[1]
49
GCSE Electronics.
Unit E2 : Applications of Electronics
10.
An electronic control system has recently been installed in a factory to control a production line.
(a)
One section of the control system counts products passing along a conveyer.

As a product passes a sensor a 4-bit binary counter is incremented.

When the last product is passing, the counter is reset and a relay is operated for
30seconds.

The conveyer is stopped whilst the relay is operated.

This allows enough time for another section of the control system to box the
products before the conveyer starts again.

This process is repeated continuously.
A block diagram of this section of the control system is shown below
Light
Source
Product
Light
Sensor
Schmitt
Inverter
Conveyor Belt
Product
4-bit
. Counter R
D C B A
Monostable
Relay
The 4-bit counter can be reset after the correct number of sequence steps by
taking the reset to logic 1. Bit A of the counter is the least significant bit
(i)
How many products are boxed at a time?
…………………….
[1]
(ii)
Explain why it is better to reset counter from the monostable output rather than the
D output of the counter.
………………………………………………………………………………………………
………………………………………………………………………………………………
[1]
50
Topic 2.3 – Interface Circuits.
(b)
A data sheet gives the following information about the Schmitt inverter:

the output changes from logic 1 to logic 0 when a rising input voltage reaches 2 V,

the output changes from logic 0 to logic 1 when a falling input voltage reaches 1 V,

when connected to a 5 V supply assume logic 0 = 0 V and logic 1 = 5 V.
Draw the graph of the waveform obtained at the output of the Schmitt inverter
for the following input waveform produced by the light sensor as products pass along
the conveyer
[3]
VIN /volts
5
4
3
2
1
0
time /s
VOUT /volts
5
4
3
2
1
0
time /s
51
GCSE Electronics.
Unit E2 : Applications of Electronics
11.
A company packs DVDs into boxes, five to a box.
Part of the packing system is shown in the following block diagram
Light
Sensing
Sub-system





(a)
Schmitt
Inverter
Counting
Sub-system
To rest of
packing system
The counting sub-system contains a 3-bit binary counter and an AND gate
The system must reset when the fifth DVD passes the light sensor
Taking the reset pin to logic 1 resets the counter
Bit A of the counter is the least significant bit
The counter is initially reset
Complete the diagram below to show how the AND gate is connected to the counter to
allow it to reset correctly.
[3]
52
Topic 2.3 – Interface Circuits.
(b)
The Schmitt inverter cleans up the signal produced by the light sensor.
Here is part of a data sheet for a Schmitt Inverter:
When connected to 5 V supply:



Logic 0 = 0 V
Logic 1 = 5 V
The output changes from logic 1 to logic 0 when a rising input voltage reaches 3 V

The output changes from logic 0 to logic 1 when a falling input voltage reaches 1 V
The output signal produced by the light sensor is shown in Graph 1.
Complete Graph 2 to show the signal obtained at the output of the Schmitt Inverter.
Graph 1
VIN /volts
5
4
3
2
1
0
time /s
Graph 2
VOUT /volts
5
4
3
2
1
0
time /s
53
GCSE Electronics.
Unit E2 : Applications of Electronics
Self Evaluation Review
Learning Objectives
My personal review of these objectives:



2.3.1 Interfacing to Inputs
Compare the merits of transistors,
comparators, and Schmitt inverters as
interfaces between analogue sensors and
digital systems;
Use the property that a Schmitt inverter gate
has two different input switching levels to
draw the output signal for a given analogue
input signal;
Explain why a Schmitt inverter is required to
de-bounce mechanical switches and analogue
sensors connected to a counting systems;
2.3.2 Interfacing to Outputs
Select transistors in terms of current gain and
collector current;
Apply the following rules to a given transistor
switching circuit:
For VIN<0.7V: VBE=VIN and VCE = Supply Voltage,
For VIN>0.7V: VBE=0.7V and VCE = 0V;
Design transistor switching circuits which
cause output devices to respond to information
from sensors;
State that IC=hFEIB until saturation is reached;
Calculate values for resistors and input
switching voltages in transistor switching
circuits.
Targets:
1.
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
2.
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
54