Download 2.2.1 RC Networks Word Document | GCE AS/A

Topic 2.2.1 – RC Networks.
Learning Objectives:
At the end of this topic you will be able to;
 Explain how capacitors can be used to form the basis of timing
circuits;
 Calculate the value of the time constant for an RC circuit using
T  RC ;
 Sketch the charge and discharge curves for voltage and current;
 Select and use the following formulae relating to the voltage across a
capacitor as a function of time:
VC  VS (1 e
VC  VS e
t
RC
t
RC
) for a charging capacitor
for a discharging capacitor
 Realise that:
VC = 0.5VS after 0.69RC
VC  VS after 5RC for charging capacitors
VC  0V after 5RC for discharging capacitors.
1
Module ET2
Electronic Circuits and Components.
RC Networks.
In this section we are going to investigate the basic circuit elements relating
to timing circuits. In order to do this we have to introduce a new component
called the capacitor.
The capacitor is essentially a device for storing small quantities of electric
charge. It is made from two metal plates separated by an insulator. We can
represent this as shown in the following diagram.
Metal
Plates
Insulator
When connected into a circuit as shown below we observe some unusual and
often unexpected results.
When pupils see this circuit for the first time, having looked at the
construction of the capacitor, they often predict that in the configuration
shown above with the switch at position A, that the lamp BL1 will not light,
because there is an insulator between the plates of the capacitor, and with an
effective break in the circuit current cannot possibly flow!
However when connected together lamp BL1 actually does come on briefly but
then goes off again. So what exactly is happening.
2
Topic 2.2.1 – RC Networks.
Well, when the battery is first connected, some electrons move from the
negative of the battery to the bottom plate of the capacitor. The presence
of these electrons on the bottom plate of the capacitor repels a similar
number of electrons from the top plate which return to the battery through
the lamp which causes it to glow. It looks like a current is flowing, even
though electrons are not crossing the insulating layer in the capacitor.
The electrons keep building up on the bottom plate until there are just as
many on this plate as are being generated at the negative terminal of the
battery. With two equal negative charges repelling each other, no other
electrons can flow from the negative terminal onto the plate of the capacitor,
so current stops flowing and the lamp goes out.
When the switch is moved to position B, the battery is disconnected from the
circuit, but when contact is made with B, the lamp BL2, comes on for a short
while before going out. We can explain this by thinking about the state of the
capacitor when the switch is moved from contact A. The bottom plate has a
large negative charge, whilst the top plate has a positive charge, since
electrons have been repelled by the lower plate.
When contact is made with terminal B, a circuit is made between the two
plates, through the bulb. The negative charges rush around the circuit to the
positive charges on the top plate, causing the bulb to light as it passes
through it. The bulb goes out after a very short time because there is only a
small amount of electrons on the bottom plate, once these have moved to the
top plate the capacitor cannot produce any extra electrons, as it is not a
battery and so current stops and the light goes out.
Moving the contact back to A charges the capacitor again, and a short flash
from the bulb is observed, before it goes out, and the process starts again.
It is always worth remembering that a capacitor can only store the
charge it is given, it cannot under any circumstances generate its own
charge, and therefore must not be compared to a battery or power
supply.
3
Module ET2
Electronic Circuits and Components.
In this example, the charging, and discharging of the capacitor occurred
almost instantly, as the only resistance in the circuit is provided by the lamps,
which is only about 100Ω. In electronics terms this is a really small
resistance. As it stands the capacitor does not look like a very useful device
in timing circuits, unless the time delays required are very small.
We can however improve the usability of the capacitor by slowing down the
rate at which charge arrives at the capacitor by using a resistor. The basic
timing circuit is shown below:
The following graph shows what happens to the voltage across the capacitor
when the switch is closed.
Switch
closed
here
From the graph you should observe that voltage across the capacitor does not
rise in a linear manner. The increase is rapid at the start, when there is no
charge on the capacitor, but slows towards the end as the charge arrives
more slowly at the capacitor plate. The graph shows that the voltage across
the capacitor eventually reaches the supply voltage some time after the
switch has been closed.
4
Topic 2.2.1 – RC Networks.
Now consider the same circuit with the resistance decreased to 1kΩ.
The voltage across the capacitor increases
more rapidly, since the resistance is lower, so
charge builds up faster.
And when the resistance is increased to 100kΩ we obtain the following graph
of the voltage across the capacitor.
The voltage across the capacitor increases very
slowly, in this example not even reaching the
supply voltage in the time shown. The larger
resistor slows down the rate of charging.
5
Module ET2
Electronic Circuits and Components.
So far we have looked at examples of a capacitor charging, we will now look at
what happens when a capacitor is discharged through a resistor.
Switch
closed
here
Switch
opened
here
Here we can see that when the switch is closed the capacitor charges
instantly, as there is no resistance to slow down the build up of charge on the
plates. However when the switch is opened the charge on the capacitor slowly
drops as current flows through the resistor. Once again changing the size of
the resistor can change the rate at which the charge leaves the capacitor.
The following graph for example shows the response when the resistance is
increased to 68kΩ.
6
Topic 2.2.1 – RC Networks.
Now that we understand the basic operation of the capacitor, it is time to
introduce some more mathematical information that will allow us to calculate
specific time delays.
The first thing to consider are the units of Capacitance, these are called
Farads, ‘F’. The Farad is a very large unit and it is much more common in
electronics circuits to see much smaller units such as
1
F  10 6 F
1,000,000
1
 The nanofarad (nF) =
F  109 F
1,000,000,000
1
 The picofarad (pF) =
F  1012 F
1,000,000,000,000
 The microfarad (μF) =
For example:
1000pF = 1 nF = 0.001μF
You will have to become familiar with converting from one unit to another,
recognising either the written units of μF, nF and pF or their standard form
multipliers i.e. x10-6, x10-9, or x10-12. Indeed in order to perform the
calculations involving capacitors you should be comfortable handling numbers
in standard form, and have access to a scientific calculator.
One of the most common calculations you will perform as part of the design
process for timing circuits is something called the Time Constant of the
circuit.
This is very easily defined as
Time Constant = R x C
where R is in Ohms and C is in Farads, the Time Constant will be in seconds.
7
Module ET2
Electronic Circuits and Components.
Examples:
1.
Calculate the time constant of a circuit containing a 10kΩ resistor and a
470μF capacitor.
Time Constant  R  C
 10k  470 F
 10  103  470  10 6
 4.7 seconds
2.
Calculate the time constant of a circuit containing a 22kΩ resistor and
a 330nF capacitor.
Time Constant  R  C
 22k  330nF
 22  103  330  10 9
 7.26  10 3 seconds
 7.26ms
Student Exercise 1:
Calculate the time constant for the following combinations of resistor and
capacitor.
i)
R = 4.7kΩ, C = 1000μF.
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ii)
R = 82kΩ, C = 15μF.
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8
Topic 2.2.1 – RC Networks.
iii)
R = 56kΩ, C = 33nF.
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iv)
R = 470kΩ, C = 8.2nF.
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v)
R = 2.2MΩ, C = 33μF.
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vi)
R = 91kΩ, C = 0.22μF.
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vii)
R = 1.8MΩ, C = 470pF.
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viii) R = 240kΩ, C = 33nF.
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9
Module ET2
Electronic Circuits and Components.
When we looked at the earlier examples of how capacitors behave in a circuit
we saw that the voltage across the capacitor rises non-uniformly as time goes
on.
The curve followed by the rising (or falling) voltage can be described
mathematically as an exponential curve, and it can be shown that the
following two mathematical expressions can be used to determine the voltage
across a capacitor at any particular time.
VC  VS (1  e
VC  VS e
t
RC
t
RC
)
for a charging capacitor
for a dischargin g capacitor
Where VC is the voltage across the capacitor, VS is the supply voltage, t is the
time in seconds, and RC is the time constant of the circuit.
‘e’ is the exponential function, usually found on a scientific calculator above
the natural logarithm ln function key as ex. Unfortunately calculators deal
with calculations in different ways and the way in which you enter the
numbers into the calculator affects how the calculation is done. You must
ensure that you know how to use your calculator correctly. Practice with the
worked examples to be sure you get the same answers.
(Note: there is no requirement to be able to prove how these formulae
are derived, nor do you need to remember them as they will be printed
on the formula sheet at the front of each examination paper.)
Note : More information about the different types of capacitors available,
and their properties will be found in the Supplementary Pack of Notes which
are non-examinable called “Practical Circuit Assistance.”
10
Topic 2.2.1 – RC Networks.
Examples:
1.
Find the voltage across the capacitor, after 0.05s and then after 0.25s,
assume that before the switch is closed there is no charge on the
capacitor.
Solution:
be
The circuit is a charging circuit, and therefore the formula required will
VC  VS (1 e
t
RC
)
It is often good practice to calculate the Time Constant before using
this equation as this simplifies the numbers to be put into the equation, and
hence into the calculator.
Time Constant  R  C
 1k  100 F
 1 103  100  10 6
 0.1 seconds
Now to calculate the voltage across the capacitor.
i)
at 0.05s
ii)
VC  VS (1  e
 9(1  e
t
RC
0.05
)
0.1
)
 9(1  0.6065)
 3.5415V
at 0.25s
VC  VS (1  e
 9(1  e
t
RC
0.25
)
0.1
)
 9(1  0.0821)
 8.2611V
11
Module ET2
Electronic Circuits and Components.
2.
Calculate the voltage across the capacitor in the following circuit, 3s
and 6s after the switch is opened.
Solution:
The circuit is a discharging circuit, and therefore the formula required
will be
VC  VS e
t
RC
Once again we will calculate the Time Constant before using this
equation as this simplifies the numbers to be put into the equation, and hence
into the calculator.
Time Constant  R  C
 22k  330 F
 22  103  330  10 6
 7.26 seconds
Now to calculate the voltage across the capacitor.
i)
at 3s
ii)
VC  VS e
t
 6e
RC
3
7.26
 6  0.6615
 3.969V
12
at 6s
VC  VS e
t
 6e
RC
6
7.26
 6  0.4376
 2.6256V
Topic 2.2.1 – RC Networks.
Student Exercise 2:
1.
Find the voltage across the capacitor, after 0.5s and then after 8s,
assume that before the switch is closed there is no charge on the
capacitor.
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Module ET2
Electronic Circuits and Components.
2.
Calculate the voltage across the capacitor in the following circuit, 15s
and 40s after the switch is opened.
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14
Topic 2.2.1 – RC Networks.
So far we have looked at how to calculate the voltage across the capacitor
after a specific time has passed. Sometimes we may need to calculate the
time taken for a specific change in voltage to occur. Consider the following
examples.
1.
In the circuit below the switch is closed at time t = 0, calculate the time
taken for the voltage across the capacitor to reach 6V. The capacitor is
initially uncharged.
First we calculate the time constant, as before
Time Constant  R  C  56k  330F  56  103  330  106  18.48 seconds
Now we need to find the time from the charging equation
VC  VS (1  e
6  9(1  e
t
t
RC
)
18.48
)
6
t
 (1  e 18.48 )
9
0.67  1  e
e
e
Now we take the natural log ' ln' of both sides
ln( e

t
t
t
t
18.48
18.48
 1  0.67
18.48
 0.33
18.48
)  ln( 0.33)
t
 1.109
18.48
t  1.109  18.48
t  20.48 s
So the voltage will reach 6V, 20.48 seconds after the switch is closed.
15
Module ET2
Electronic Circuits and Components.
For those of you that are not taking A-Level Mathematics, this last example
may have proved to be a bit too difficult given all the re-arrangements
needed to the equation. In electronics it is more important that you are able
to determine the answer than it is to understand the mathematical process
going on in the background.
Therefore on the examination paper you will find an additional formula, which
will allow you to calculate this time directly. For a charging capacitor as in
this case the formula is:
 V
t   RC ln  1  C
 V0



So a possible solution to the previous problem would be:
 V
t   RC ln  1  C
 V0



 6
t  18.48 ln  1  
 9
t  20.48 s
If the capacitor is discharging the there is a different formula to use which
is as follows:
V
t   RC ln  C
 V0



We will now look at a discharging example, and solve this using the simplified
method.
16
Topic 2.2.1 – RC Networks.
2.
In the circuit below, the switch SW1 is closed which charges up the
capacitor. Determine the time taken for the voltage across the
capacitor to fall from 9V to 2V, when the switch is opened.
First we calculate the time constant, as before
Time Constant  R  C  220k  10F  220  103  10  106  2.2 seconds
Now we need to find the time from the discharging equation
V
t   RC ln  C
 V0
2
t  2.2 ln  
9
t  3.3 s



So the voltage will reach 2V, 3.33 seconds after the switch is opened.
Using this method, the solution is obtained very quickly, without the
need for extensive mathematics. Remember you can use either method in
the examination, the most important thing is to obtain the correct
answer.
17
Module ET2
Electronic Circuits and Components.
Quite often we want to know how long it takes the capacitor to reach the
half-way point in the charging or discharge cycle but this calculation can be
very time consuming.
So let us consider the charging circuit to start with, at what time does the
voltage across the capacitor reach half of the supply voltage. We will not put
any values in for R and C at the moment. The equation becomes:
VC  VS (1  e
t
RC
)
VS
t
 VS (1  e RC )
2
VS
t
 1  e RC
2VS
1
t
 1  e RC
2
1
t
e RC  1 
2
1
t
ln( e RC )  ln( )
2
t

 0.693
RC
t  0.693 RC
This means that for any charging circuit, the time taken to reach half of the
supply voltage is 0.69RC, and is completely independent of the supply voltage,
depending only on the value of the capacitor and resistor.
This provides a very simple way of determining this point on the graph and is
summarised as follows.
VC  0.5VS after 0.69 RC
18
Topic 2.2.1 – RC Networks.
What about a discharging capacitor ?
We can perform a similar calculation using the discharging formula, as shown
below:
VC  VS e
t
RC
VS
t
 VS e RC
2
VS
t
 e RC
2VS
1
t
 e RC
2
1
t
e RC 
2
1
t
ln( e RC )  ln( )
2
t

 0.693
RC
t  0.693 RC
This result is exactly the same as for a charging capacitor, which is really
good news as we do not have to remember another formula.
We now have a very simple way of determining the time taken to reach half
of the supply voltage. It would be nice if there was a similar result we could
use to determine the time taken to reach maximum charge when VC = VS and
full discharge when VC = 0V.
The problem with the exponential function is that the maximum charge is
never actually reached in theory, because the curvature of the line gets
smaller and smaller, never quite reaching the maximum value.
For our purposes, it would be a reasonable approximation to determine how
long it would take for the voltage to rise to 99% of VS in the case of a
charging circuit and fall to 1% of VS in the case of discharging circuit.
We can use a similar argument to the one we have just used, as follows.
19
Module ET2
Electronic Circuits and Components.
For a charging capacitor
VC  VS (1  e
t
0.99VS  VS (1  e
t
0.99VS
t
 1  e RC
VS
0.99  1  e
e
ln( e
t
t

RC
RC
t
For a discharging capacitor
RC
)
VC  V S e
t
RC
)
0.01V S  V S e
t
)  ln( 0.01)
t
  4 .6
RC
t  4.6 RC
RC
0.01V S
t
 e RC
VS
0.01  e
RC
 1  0.99
RC
e
ln( e
t
t

RC
RC
t
RC
 0.01
)  ln( 0.01)
t
  4 .6
RC
t  4.6 RC
Once again we see that the results are identical, and as a general rule of
thumb electronics engineers will consider a capacitor to be full charged, or
fully discharged after a time period = 5RC. We can summarise these results
as follows:
VC  VS after 5 RC for charging capacitors
VC  0V after 5 RC for dischargin g capacitors
We are now in a position to be able to quickly sketch a charging or discharging
curve for a circuit, by only making a few simple calculations.
Example :
For the circuit shown opposite, sketch the
charging characteristic to show how the
voltage across the capacitor changes with
time.
20
Topic 2.2.1 – RC Networks.
Step 1 : Calculate the time constant of the circuit.
Time Constant  R  C
 100k  100 F
 100  103  100  10 6
 10 seconds
Step 2 : Calculate the time taken to reach half of the supply voltage.
Time 1  0.69  VS
2
 0.69  10
 6 .9 s
Step 3 : Calculate the time taken to reach VS.
TimeVS  5 RC
 5  10
 50 s
Step 4 : Plot these points on graph and join up with a smooth curve.
10
8
Step 3
6
Step 2
4
2
0
0
10
20
30
40
50
60
time (s)
21
Module ET2
Electronic Circuits and Components.
Student Exercise 3:
Now try this discharging example.
Step 1 : Calculate the time constant.
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Step 2 : Calculate the time to reach half the supply voltage.
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Step 3 : Calculate the time to reach 0V.
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Step 4 : Plot the points on the graph paper and join with a smooth curve.
22
Topic 2.2.1 – RC Networks.
2.
In the following circuit, the capacitor is initially uncharged and the
switch is closed at time t = 0. How long will it take for the voltage
across the capacitor to rise to 9V.
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23
Module ET2
Electronic Circuits and Components.
3.
In the following circuit, the switch is closed to charge the capacitor
instantly. Calculate how long after the switch is released for the output
voltage to fall from 6V to 2V.
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24
Topic 2.2.1 – RC Networks.
Solutions to Student Exercises.
Student Exercise 1:
i)
R = 4.7kΩ, C = 1000μF.
Time Constant  R  C
 4.7k  1000 F
 4.7  103  1000  10 6
 4.7 seconds
ii)
R = 82kΩ, C = 15μF.
Time Constant  R  C
 82k  15 F
 82  103  15  10 6
 1.23 seconds
iii)
R = 56kΩ, C = 33nF.
Time Constant  R  C
 56k  33nF
 56  103  33  10 9
 1.848  10 3 seconds
 1.848ms
iv)
R = 470kΩ, C = 8.2nF.
Time Constant  R  C
 470k  8.2nF
 470  103  8.2  10 9
 3.854  10 3 seconds
 3.854 ms
25
Module ET2
Electronic Circuits and Components.
v)
R = 2.2MΩ, C = 33μF.
Time Constant  R  C
 2.2 M  33 F
 2.2  106  33  10 6
 72.6 seconds
vi)
R = 91kΩ, C = 0.22μF.
Time Constant  R  C
 91k  0.22 F
 91 103  0.22  10 6
 0.02002 seconds
vii)
R = 1.8MΩ, C = 470pF.
Time Constant  R  C
 1.8 M  470 pF
 1.8  106  470  10 12
 846  10 6 seconds
 846 s
viii) R = 240kΩ, C = 33nF.
Time Constant  R  C
 240k  33nF
 240  103  33  10 9
 7.92  10 3 seconds
 7.92ms
26
Topic 2.2.1 – RC Networks.
Student Exercise 2:
1.
Time Constant  R  C
 100k  22 F
 100  103  22  10 6
 2.2 seconds
i)
at 0.5s
ii)
VC  VS (1  e
 6(1  e
0.5
t
2.2
RC
VC  VS (1  e
)
 6(1  e
)
 6(1  0.7967)
 1.2198V
2.
at 8s
8
2.2
t
RC
)
)
 6(1  0.0263)
 5.8422V
Time Constant  R  C
 56k  470 F
 56  103  470  10 6
 26.32 seconds
i)
at 15s
ii)
VC  VS e
 12  e
t
15
at 40s
RC
VC  VS e
26.32
 12  e
 12  0.5656
 6.7872V
t
40
RC
26.32
 12  0.2188
 2.6256V
27
Module ET2
Electronic Circuits and Components.
Student Exercise 3.
Step 1 : Calculate the time constant.
Time Constant  R  C
 30k  220 F
 30  103  220  10 6
 6.6 seconds
Step 2 : Calculate the time to reach half the supply voltage.
Time 1  0.69  VS
2
 0.69  6.6
 4.554 s
Step 3 : Calculate the time taken to reach 0V.
Time0V  5 RC
 5  6.6
 33 s
Step 4 : Plot the points on the graph paper and join with a smooth curve.
6
4
2
0
28
0
10
20
30
40
50
time (s)
Topic 2.2.1 – RC Networks.
2.
Time Constant  R  C
 10k  100 F
 10  10 3  100  10 6
 1 second
For the mathematician:
VC  VS (1  e
t
9  12(1  e
t
9
 (1  e t )
12
0.75  1  e t
e t  1  0.75
e t  0.25
ln( e t )  ln( 0.25)
RC
1
For the non mathematician
)
)
 V
t   RC ln  1  C
 V0
9 

t  1ln  1  
 12 
t  1.386 s



 t  1.386
t  1.386 s
So the voltage will reach 9V, 1.386 seconds after the switch is closed.
29
Module ET2
Electronic Circuits and Components.
3.
Time Constant  R  C
 100k  22 F
 100  103  22  10 6
 2.2 seconds
For the mathematician
VC  VS e
t
For the non-mathematician
RC
t
2  6e 2.2 )
2
t
 e 2 .2
6
0.33  e
e
ln( e
t
t

2 .2
2 .2
t
2 .2
 0.33
)  ln( 0.33)
V
t   RC ln  C
 V0
2
t  2.2 ln  
6
t  2.44 s



t
 1.109
2 .2
t  1.109  2.2
t  2.44 s
So the voltage will reach 2V, 2.44 seconds after the switch is opened.
30
Topic 2.2.1 – RC Networks.
The following section contains some past examination questions on RC Timing Circuits.
1.
The following circuit is used to provide a time delay.
(a)
Calculate the time constant of the circuit.
............................................................................................................................
(b)
............................................................................................................................
[1]
When the switch S is momentarily closed the capacitor charges up to 15V.
Calculate the output voltage 5 seconds after the switch is released.
............................................................................................................................
............................................................................................................................
(c)
............................................................................................................................
[2]
Calculate the time taken after the switch is released for the output voltage to
fall from 15V to 3V.
............................................................................................................................
............................................................................................................................
............................................................................................................................
............................................................................................................................
[2]
31
Module ET2
Electronic Circuits and Components.
2.
(a)
The capacitor shown in the circuit below is initially discharged. Switch S1 is
closed at time t = 0.
(i)
Calculate the time constant of the circuit.
............................................................................................................................
............................................................................................................................
[1]
(ii)
Calculate the output voltage after 2s.
............................................................................................................................
............................................................................................................................
............................................................................................................................
[2]
(iii) Calculate the time taken for VOUT to reach 5V.
............................................................................................................................
............................................................................................................................
............................................................................................................................
............................................................................................................................
[2]
32
Topic 2.2.1 – RC Networks.
(b)
The capacitor shown in the circuit below charges up to 12V when switch S2 is
closed.
Calculate the time taken, after the switch is released, for the voltage across
the capacitor to fall from 12V to 3V
............................................................................................................................
............................................................................................................................
............................................................................................................................
............................................................................................................................
............................................................................................................................
............................................................................................................................
[3]
33
Module ET2
Electronic Circuits and Components.
3.
The capacitor shown in the following circuit is initially discharged. The switch S is
opened at time t = 0.
(a)
Calculate the time constant of the circuit.
............................................................................................................................
(b)
............................................................................................................................
[1]
Determine the time taken for VOUT to reach 5V.
............................................................................................................................
............................................................................................................................
(c)
............................................................................................................................
[2]
Calculate the output voltage 20 seconds after the switch is opened.
............................................................................................................................
............................................................................................................................
............................................................................................................................
............................................................................................................................
............................................................................................................................
[2]
34
Topic 2.2.1 – RC Networks.
4.
The capacitor shown in the following circuit is initially discharged. Switch S is
closed at time t = 0.
(a)
Calculate the time constant of the circuit.
............................................................................................................................
(b)
............................................................................................................................
[1]
Determine the time taken for VOUT to reach 3V.
............................................................................................................................
............................................................................................................................
(c)
............................................................................................................................
[2]
Calculate the output voltage after 5s.
............................................................................................................................
............................................................................................................................
............................................................................................................................
............................................................................................................................
[2]
35
Module ET2
Electronic Circuits and Components.
5.
(a)
The capacitor shown in the circuit below is initially uncharged. S1 is closed at
time t = 0.
Calculate:
(i)
the time constant of the circuit.
............................................................................................................................
............................................................................................................................
[1]
(ii) the time taken for VOUT to reach 2V.
............................................................................................................................
............................................................................................................................
............................................................................................................................
[2]
(iii) the output voltage after 0.5s.
............................................................................................................................
............................................................................................................................
............................................................................................................................
36
Topic 2.2.1 – RC Networks.
(b)
Complete the following graph by showing the output from the circuit when
square pulses are applied at its input.
[2]
37
Module ET2
Electronic Circuits and Components.
6.
The following circuit is used to provide a time delay. The capacitor charges up to
12V when switch S is momentarily closed.
(i)
Calculate the time constant of the circuit.
............................................................................................................................
(ii)
............................................................................................................................
[1]
Calculate the output voltage 3 seconds after the switch is released.
............................................................................................................................
............................................................................................................................
(iii)
............................................................................................................................
[2]
Calculate the time taken after the switch is released for the output voltage to
fall from 12V to 4V.
............................................................................................................................
............................................................................................................................
............................................................................................................................
............................................................................................................................
[2]
38
Topic 2.2.1 – RC Networks.
7.
The capacitor shown in the following circuit is initially discharged.
(a)
Calculate the time constant of the circuit.
............................................................................................................................
(b)
............................................................................................................................
[1]
Switch S is closed at time t = 0.
(i)
Determine the time taken for VOUT to reach half the supply voltage.
............................................................................................................................
............................................................................................................................
............................................................................................................................
[2]
(ii) Calculate the output voltage 2s after the switch is closed.
............................................................................................................................
............................................................................................................................
............................................................................................................................
[2]
39
Module ET2
Electronic Circuits and Components.
Self Evaluation Review
Learning Objectives
My personal review of these objectives:



Explain how capacitors can be used
to form the basis of timing circuits;
Calculate the value of the time
constant for an RC circuit using
T  RC ;
Sketch the charge and discharge
curves for voltage and current;
Select and use the following
formulae relating to the voltage
across a capacitor as a function of
time:
VC  VS (1 e
VC  VS e
t
RC
t
RC
) for a charging
capacitor
for a discharging
capacitor
Realise that:
VC = 0.5VS
VC  VS
VC  0V
Targets:
1.
after 0.69RC
after 5RC for
charging capacitors
after 5RC for
discharging
capacitors.
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
2.
………………………………………………………………………………………………………………
………………………………………………………………………………………………………………
40