Download Relationship between charge and pd

Electricity
Current, voltage, power and resistance
What you should already know
From National 5 you should understand and be able to use the terms charge,
conductor, insulator, electric field, current, circuit, series, parallel and potential
difference.
You should also be able to identify and draw circuit symbols and their correct position
in circuit diagrams.
You will have already used the relationships Q = It, E w = QV, V = IR and P = IV.
Circuit rules
Series
Parallel
Current
IS  I1  I 2  I 3
I P  I1  I 2  I 3
Voltage
VS  V1  V2  V3
VS  V1  V2  V3
Resistance
RS  R1  R2  R3
1
1
1
1
 

RP R1 R2 R3
In a circuit, when the switch is closed the free electrons in the conductor experience a
force that causes them to move. Electrons will tend to drift away from the negatively
charged end towards the positively charged end.
Current is the rate of flow of charges (electrons) and is given by the flow of charge
(coulombs) per second:
I
Q
t
The energy required to drive this current around the circuit is provided by the
battery. The electrical energy that is supplied by the source is transformed into
other forms of energy in the components that make up the circuit.
Potential difference is defined as the energy transferred per unit charge.
𝑉=
𝐸𝑤
𝑄
Where E W is the work done, ie the energy transferred, therefore:
1 volt (V) = 1 joule per coulomb (J C –1 ).
When energy is being supplied by a source to the circuit, the voltage is referred
to as an electromotive force (emf).
When energy is transformed into another form of energy by a component in the
circuit, the voltage is referred to as a potential difference (pd).
In the supply
Energy is given to the charges passing
through the supply.
Chemical energy  electrical energy
In the component
Energy is “given” to the component.
Electrical energy  light + heat energy
Ohm’s law
In any circuit, providing the resistance of a component remains constant, if the
potential difference V across the component increases, the current through the
component will increase in direct proportion.
At a fixed temperature, for a given
conductor:
V  I,
ie V/I = constant
Voltage
This constant is defined as the
resistance,
ie
R
V
I
or
V  IR
Current
This is Ohm’s law.
The unit of resistance is ohm ()
The unit of current is ampere (A)
The unit of potential difference is volt (V).
Electric power
Given that E = QV and Q = It
We can conclude that E = ItV
And since P = E/t …. it follows that P = IV
Also, since
V = IR
P = I × IR (substituting for V)
P = I2R
or
P
V
 V (substitut ing for I )
R
P
V2
R
The expression I 2 R gives the energy transferred in one second due to resistive heating.
Apart from obvious uses in electric fires, cookers, toasters etc, consideration has to be
given to heating effects in resistors, transistors and integrated circuits.
The expression V 2 /R is particularly useful when the voltage of a power supply is fixed
and you are considering changes in resistance, eg different power of heating elements.
The potential divider
A potential divider provides a way of obtaining a variable voltage from a fixed voltage
supply.
Consider first two fixed resistors, R 1 = 10 Ω and R 2 = 20 Ω, connected in series across a
6 V supply:
6V
10 Ω
20 Ω
V1
V2
The current in this circuit can be calculated using Ohm’s law (total R = 10 + 20 = 30 Ω).
so,
I
VS
6

R 30
= 0.2 A
The voltage across R 1 : V 1 = IR 1 = 0.2 × 10
=2V
The voltage across R 2 : V 2 = IR 2 = 0.2 × 20
=4V
Notice that the voltages across the resistors are in the ratio of their resistances (1:2 in
this case),
ie,
V1 R1

V2 R2
The voltage across R 1 is given by:
V1 
R1
 VS
R1  R2
similarly, the voltage across R 2 is
V2 
R2
V
R1  R2
This is known as the potential divider rule.
Potential dividers can be used in a circuit called a Wheatstone Bridge. This type of
circuit can be used to find the value of an unknown resistor.
When the ratio of the resistors R 1 and R 2 is equal to the ratio of resistors R 3 and R 4 the
reading on the voltmeter is zero.
Worked potential divider example
20 kΩ
12 V
Y
10 kΩ
A
X
(a)
Calculate the voltage across XY.
R 1 = 10 kΩ
R 2 = 20 kΩ
V S = 12 V
V1 = ?
R1
VS
R1  R2
10
V1 
12
10  20
V1 
V1 = 4 V
(b)
The circuit attached to XY has a resistance of 10 kΩ. Calculate the effective
resistance between XY and thus the voltage across XY.
The resistance between XY comes from the two identical 10 kΩ resistors in
parallel. The combined resistance is therefore 5 kΩ. Use this resistance as R 1 in
the potential divider formula.
R 1 = 5 kΩ
R 2 = 20 kΩ
V S = 12 V
V1 = ?
R1
VS
R1  R2
5
V1 
12
5  20
V1 
V 1 = 2.4 V
Worked example 2
A 10 W model car motor operates on a
12 V supply. In order to slow the car
it is connected in series with a controller
(rheostat).
12 V
The controller is adjusted to reduce the
motor’s power to 4 W.
M
Assuming that the resistance of the motor
not change, calculate:
(a)
(b)
(a)
the resistance of the controller at this setting
the power wasted in the controller.
We cannot find the resistance of the controller directly. We need to find the
resistance of the motor and the total resistance of the circuit when the motor is
operating at 10 W.
For the motor at 10 W, P = 10 W, V = 12 V
P
V2
V 2 12 2
, so R 

R
P
10
R = 14.4 Ω
For the motor at 4 W, P = I 2 R, so I =
P
4

R
14  4
= 0.527 A
For the whole circuit: RT 
VT
12

 22.8 Ω
I 0.527
Resistance of controller = 22.8 – 14.4 = 8.4 Ω
(b)
does
Power in controller = I 2 R = 0.527 2 × 8.4
= 2.3 W
Electrical sources and internal resistance
In the National 5 course we assumed that a power supply is ideal, meaning that it’s
voltage remains constant whatever circuit is connected to it. In most cases this is a
good assumption, but taking a small battery and connecting lamps one after another in
parallel you will notice that eventually the brightness of the lamps becomes dimmer.
This is because it involves a significant current being supplied by the battery, which in
turn suggests a large current will flow through the supply.
If you are able to touch the battery you may notice it getting warm when in use. This
shows some energy is being wasted as heat. Like all conductors the inside of a cell has
some resistance and some of the chemical energy converted is dissipated as heat
within the cell and is not available to the circuit. We say that the supply has an internal
resistance, r.
In most cases this is so small (a few ohms) that it can be considered negligible.
However, when the current in the circuit is large, meaning the e xternal resistance is
small, the effect of the internal resistance can be significant.
The greater the current flowing through the supply the more energy will be “lost” in
the supply until eventually all the available energy is wasted and none is available
outside the power supply.
Energy will be wasted in getting the charge through the supply (this energy appears as
heat) and so the energy per unit charge available at the output (the terminal potential
difference or tpd) will fall. There will be ‘lost volts’. The lost volts = Ir.
We can represent a real supply as:
real cell = ideal cell + a resistance, r
Ideal cell
Internal
resistance
r
E
Real cell
To establish the key quantities in a circuit, it would be useful to know the voltage of
the ideal cell. This is called the emf of the cell , which can be defined as the energy
given to each coulomb of charge passing through the supply . It is found by
measuring the voltage across the cell on ‘open circuit’, ie whe n it is delivering no
current to a load. This can be done in practice by using a voltmeter or oscilloscope
connected across the cell.
r
I=0
E
Since no current is drawn from the supply, no voltage is dropped across the internal
resistance and a voltmeter across the real cell would register the voltage of the ideal
cell, the emf.
Under load
When a voltmeter is connected across a 1.5V cell, the reading on the voltmeter is 1.5V.
If a resistor is now connected across the terminals of the cell, the reading on the
voltmeter drops.
Explanation
When a real cell is delivering current in a circuit we say that it is under load and the
external resistance is referred to as the load. (the 11.4 resistor in the diagram below)
Consider the case of a cell with an internal resistance of 0.6 Ω delivering current to an
external resistance of 11.4 Ω:
Cell emf. = 6 V
current = 0.5 A
r
0.6 Ω
E
I = 0.5 A
V
11.4 Ω
The voltage measured across the terminals of the cell will be the voltage across the
11.4 Ω resistor, ie,
V = IR = 0.5 × 11.4
= 5.7 V
and the voltage across the internal resistance:
V = Ir = 0.5 × 0.6
= 0.3 V
So the voltage across the terminals is only 5.7 V (this is the terminal potential
difference, or tpd) and this happens because of the voltage dropped across the internal
resistance (0.3 V in this case). This is called the lost volts.
If Ohm’s law is applied to a circuit containing a battery of emf, E, and internal
resistance, r, with an external load resistance R L :
r
I
E
RL
and if I is the current in the circuit and V is the pd across R L (the tpd), then using
conservation of energy:
emf = tpd + lost volts
E = pd across R L + pd across r
and: E = V + Ir
This is the internal resistance equation, which can clearly be rearranged into different
forms, for example:
V = E – Ir
or, since V = IR,
E = IR L + Ir
ie E = I(R L + r)
Short circuit (R L = 0)
The maximum current is the short-circuit current. This is the current that will flow
when the terminals of the supply are joined with a short piece of thick wire (ie there is
no external resistance).
By substituting R = 0 in the above equation we get:
E = I(0 + r)
ie E = Ir
Measuring E and r by a graphical method
When we increase the current in a circuit the tpd will decrease. We can use a graph to
measure the emf and internal resistance. If we plot V on the y-axis and I on the x-axis,
we get a straight line of negative gradient. From above:
V = E – Ir
V = (–r) × I + E
Comparing with the equation of a straight line
y = mx + c
we can see that the gradient of the line (m) is equivalent to –r( or -  V/  I) and the
y-intercept (c) is E.
Second graphical method
When we change the external resistance (R L ) in a circuit, there will be a corresponding
change in the current flowing. If we plot R on the y-axis and 1/I on the x-axis, we get a
straight line of positive gradient and negative intercept. From above:
E = IR + Ir
Divide both sides by I:
E/I = R + r
R = (E × 1/I) – r
Comparing with the equation of a straight line:
y = mx + c
we can see that the gradient of the line (m) is equivalent to the emf, E and the yintercept (c) is –r.
R
Gradient (m)
= E (emf)
Intercept (c)
= –r (internal
resistance)
1
I
Maximum power transfer – load matching
The maximum power transferred between a source and an external circuit occurs when
the resistance of the external circuit is equal to the internal resistance of the source. The
maximum voltage transfer occurs when the external resistance is much higher than the
internal resistance of the source.
Worked example
A cell of emf 1.5 V is connected in series with a 28 Ω resistor. A voltmeter measures the
voltage across the cell as 1.4 V.
r
E
V
28 Ω
Calculate:
(a)
(b)
(c)
the internal resistance of the cell
the current if the cell terminals are short circuited
the lost volts if the external resistance R is increased to 58 Ω.
(a)
E = V + Ir
In this case we do not know the current, I, but we do know that the voltage across the
28 Ω resistor is 1.4 V, and I = V/R , so I = 0.05 A.
1.5 = 1.4 + 0.05r
r
(b)
0.1
 2Ω
0.05
Short circuit:
r
Total resistance = 2 Ω
1.5 V
I
(c)
E 1.5

 0.75 A
r
2
E = I(R + r)
1.5
= I(58 + 2)
I
= 1.5/60 = 0.025 A
lost volts = Ir
= 0.025 × 2 = 0.05 V
2Ω
I
Alternating current
The electricity supply from a cell or battery is d.c. This means that when the battery is
used it supplies a constant emf so the current is always in the same direction.
Alternating current is produced by rotating an electromagnet in a coil of wire.
This means that the induced voltage, which is constantly changing in the form of
a sine function, pushes the current one way and then the other.
Examples of each type of current when displayed on an oscilloscope are shown below.
a.c. waveform
d.c. waveform
Measuring frequency and peak voltage
On an oscilloscope the y-axis (voltage) scale is usually labelled ‘volts per division’ or
‘volts per centimetre’ and the x-axis (time) scale is called the ‘timebase’.
To calculate the frequency of an a.c. signal, we first have to find its period. This is
the time for one complete cycle of current, so we measure the horizontal distance on
the screen between crests.
An alternating voltage varies between a
maximum negative and positive value as the
voltage pushes first one way then the other.
This maximum is defined as the peak voltage.
It can be measured from an oscilloscope by
measuring the vertical distance on the screen
from the mid-point of the signal to the top
which is then multiplied by the setting of the
‘volts per division’ dial to calculate the peak
voltage.
Example
In the picture below each box on the oscilloscope screen has a side of 1 cm.
The distance between crests is 4cm. The distance from mid-point to top is 4cm.
The time base was set at 2 ms cm–1, the volts per division was set at 3V cm–1
The period of the wave is:
T = 4 × 2 ms = 8 × 0.001 = 0.008 s
f = 1/T = 1/0.008 = 125 Hz
The peak voltage is:
Vpeak = 4 x 3
Vpeak = 12 V
Alternating Supplies
a.c. is capable of transferring energy in the same way as d.c. The amount of power being
supplied can be calculated using the value of voltage and/or current at a particular point.
However, as we have already seen, in a.c. these values change constantly. To find the
effective value of alternating voltages or currents we need an average, but the average value
of the voltage during any complete cycle is zero!
Alternating voltage
A different ‘average’ must be found. The ‘average’ value of an alternating voltage that we
will find is called the root mean square (rms) voltage (Vrms).
The definition of the rms voltage is that value of alternating voltage that produces the same
power (eg heating or lighting) as the direct voltage. The rms value is that which is quoted on
a power supply, eg a 6 V battery will produce the same brightness of light bulb as a 6 V rms
a.c. supply.
Consider the following two circuits, which contain identical lamps.
The variable resistors are altered until the lamps are of equal brightness. As a result the d.c. has the
same value as the effective a.c. (ie the lamps have the same power output). Both voltages are
measured using an oscilloscope, giving the voltage equation below. Also, since V = IR applies to the
rms values and to the peak values, a similar equation for currents can be deduced.
V rms = 1 V peak
2
and
Irms = 1 Ipeak
2
Deriving the relationship between peak and rms values of alternating current.
The power produced by a current I in a resistor of resistance R is given by I2R.
A graph of I2 against t for an alternating current is shown below. A similar method can be
used for voltage.
The average value of I 2 is
2
I peak
2
and therefore the average power supplied is
2
I peak
2
R.
An identical heating effect (power output) for a d.c. supply is I2R
As we have seen the effective or rms value of a.c. is equivalent to the d.c. value and so
Setting both of these equal to each other gives:
Pac = Pdc
2
I peak
2
R  I rms
R
2
2
I peak
2
I peak
2
2
 I rms
 I rms
dc I = Irms.
Capacitors and Capacitance
The ability of a component to store charge is known as capacitance. A device that stores charge
is called a capacitor.
Typically capacitors are 2 conducting materials separated by an insulator. The simplest type
consists of two metal plates with an air gap between them. The symbol for a capacitor is based
on this:
Relationship between charge and pd
The capacitor is charged to a chosen voltage by setting the switch to X. The charge stored can be
measured directly by discharging through the coulombmeter with the switch set to Y. In this
way pairs of readings of voltage and charge are obtained.
It is found that the charge (Q) stored on a capacitor and the pd (V) across it are directly
proportional:
Q
Q/V is a constant
This constant is defined as the capacitance, C:
C
Q
V
The formal definition of capacitance is therefore the charge stored per unit voltage. The unit of
capacitance is the farad (F).
From the above formula: 1 F = 1 coulomb per volt.
The farad is too large a unit for practical purposes and the following are commonly used:
1 F (microfarad) = 1 x 10-6F
1 nF (nanofarad) = 1 x 10-9 F
Note: When a capacitor is charging, the current is not constant. This means the formula Q = It
should not be used to work out the charge stored.
Worked example
A capacitor stores 4 x 10-4 C of charge when the potential difference across it is 100 V. Calculate
the capacitance.
C = Q/V = 4 x 10 –4 /100 = 4 x 10 –6 F = 4 F.
Energy stored in a capacitor
Consider this circuit…
When the switch is closed a current flows. As the capacitor charges the negatively charged plate
will tend to repel the electrons approaching it. In order to overcome this repulsion, work has to
be done and so energy is supplied. This energy is supplied by the battery. Note that current does
not flow through the capacitor, electrons flow onto one plate and away from the other plate.
For a given capacitor the pd across the plates is directly proportional to the charge stored.
Consider a capacitor being charged to a pd of V and holding a charge Q.
Each charge moved onto the capacitor requires a different amount of work to move it towards the
plate, as it is working against the repulsive force of the charges already stored there. Work done in
charging the capacitor can therefore be found using the area under the Q-V graph, i.e. work = ½(QV).
This work is stored as electrical energy, so:
E = ½(QV)
(Contrast this with the work done moving a charge in an electric field where W = QV.)
Since Q = CV there are alternative forms of this relationship:
E = ½(CV2)
E = ½(Q2/C)
Worked example
A 40 F capacitor is fully charged using a 50 V supply. Calculate the energy stored in the
capacitor.
energy = ½(CV2)
= 0.5x 40 x 10-6 x 502
= 0.05 J
Charging a capacitor
Consider the following circuit:
The capacitor is initially completely discharged.
When the switch is closed, the voltage across the capacitor and the current flowing in the circuit
behave as shown in the graphs below.
When switch S is closed, the p.d. across the capacitor is zero, as it has no charge. The p.d.
across the resistor is V. The current flowing through the resistor is V/R
After a time, the capacitor has accumulated charge and the p.d. across it has increased.
The current through the resistor has fallen as the charges on the capacitor act against
the flow. When fully charged, the p.d. across the capacitor is equal to the voltage on the
battery. The current through the resistor is zero.
Discharging a capacitor
While the capacitor is discharging the voltage across the capacitor and the current in the circuit
behave as shown in the graphs below:
Although the current/time graph has the same shape as that during charging, the currents in
each case are flowing in opposite directions. The discharging current decreases because the pd
across the plates decreases as charge leaves them.
A capacitor stores charge, but unlike a cell it has no capability to supply more energy. When it
discharges, the energy stored will be used in the circuit, eg in the previous circuit it would be
dissipated as heat in the resistor.
Factors affecting the rate of charge and discharge
The time taken for a capacitor to charge is controlled by the resistance of the resistor R (because
it controls the size of the current) and the capacitance of the capacitor (since a larger capacitor
will take longer to fill and empty).
The values of R and C can be used together to form what is known as the time constant. The time
taken for the capacitor to charge or discharge is related to the time constant.
Large capacitance and large resistance both increase the charge or discharge time.
In the circuit above if the resistance and supply voltage are kept constant but the value of the
capacitance is varied then the I vs t graphs for capacitors of different value during charging are
shown below:
I
I
Large C
Small C
t
t
Both graphs show same initial charging current as the resistance in the circuits is the same.
However the time taken to reduce the current to zero ( ie fully charge the capacitor) is less with
the smaller capacitance.
If the supply voltage and the capacitance are kept constant and the resistance varied then then
the I vs t graphs for resistors of different value during the charging of the capacitor are shown
below:
I
I
Large R
Small R
t
t
The graphs show different initial charging current, as the resistance in the circuits is not the
same. The circuit with a large resistance has a smaller initial charging current.
The time taken to reduce the current to zero ( ie fully charge the capacitor) is greater with the
larger resistance, as the charging current will be smaller.
Worked example
The switch in the following circuit is closed at time t = 0.
(a)
Immediately after closing the switch what is
(i)
the charge on C?
(ii)
the pd across C?
(iii)
the pd across R?
(iv)
the current through R?
(b)
When the capacitor is fully charged what is
(i)
the pd across the capacitor?
(ii)
the charge stored?
(a)
(i)
The initial charge on the capacitor is zero.
(ii)
The initial pd across the capacitor is zero since there is no charge.
(iii)
pd across the resistor is 9 V
(VR = VS – VC = 9 – 0 = 9 V)
(iv)
Initial I = V/R = 9/550 = 0.016 A
(i)
Final pd across the capacitor equals the supply voltage, 9 V.
(ii)
Q = VC = 9 x 5 x 10-6 = 4.5 x 10-5 C .
(b)
Applications of capacitors
Blocking capacitor – a capacitor will stop the flow of a direct current but allow any alternating
component of a signal to flow.
Flashing indicators – a low value capacitor is charged through a resistor until it acquires sufficient
voltage to allow a discharge across a neon bulb.
Smoothing capacitor – in a simple rectifier circuit, a capacitor charges during the half cycle when a
diode conducts and discharges during the half cycle when the diode is reverse biased and so helps to
smooth out the wave form.
Timing circuit – a resistor and capacitor in series can delay the build-up of voltage across the
capacitor and hence delay the trigger of some response within the rest of the circuit.
And many more……
Conductors, Insulators and Semiconductors
By considering their electrical properties solids can be divided into 3 main groups.
Conductors: Materials that have many “free electrons” which can easily be made to flow
through the material e.g. metals.
Insulators: Materials that have very few free electrons, which cannot easily be made to
flow. e.g. plastics.
Semiconductors: Materials that are insulators when pure, but will conduct when an
impurity is added, or in response to heat, light etc.
Why do the different materials have such differing properties?
To answer this we have examine the positioning of electrons in solids.
In the previous unit we discussed the idea that in a single atom, electrons are arranged in
discrete orbits or energy levels. Electrons must exist in these orbits and cannot occupy the
spaces in between.
However in solids there are many atoms interacting leading to an interaction of the outer
energy levels, which results in the creation of bands of available energies instead of discrete
energy levels.
These bands are separated by gaps in which there are no electrons.
Band structure
When the atoms are all regularly arranged in the crystal lattice of a solid, the energy
levels become grouped together in a band. This is a continuous range of allowed
energies rather than a single level. There will also be groups of energies that are not
allowed, in what is known as a band gap. Similar to the energy levels of an individual
atom, the electrons will fill the lower bands first. The Fermi level gives a rough idea of
which levels electrons will generally fill up to, but there will always be some electrons
with individual energies above this.
In a conductor, the highest occupied band, known as the conduction band, is not
completely full. This allows the electrons to move in and out from neighbouring atoms
and therefore allowing conduction.
In an insulator the highest occupied band is full. This is called the valence band. The
first unfilled band above the valence band is the conduction band . For an insulator
the gap between the valence band and the conduction band is large and at room
temperature there is not enough energy available to move electrons from the valence
band into the conduction band, where they would have been able to contribute to
conduction.
Normally, there is almost no electrical conduction in an insulator however, if the
applied voltage is sufficiently high, then enough electrons can be lifted to the
conduction band to allow current to flow.
In a semiconductor the gap between the valence band and the conduction band is
smaller, and at room temperature there is sufficient energy available to move some
electrons from the valence band into the conduction band, allowing some conduction
to take place. An increase in temperature increases the conductivity of a
semiconductor as more electrons have enough energy to make the jump to the
conduction band.
Electron
energy
Conduction
band
Conduction
band
Conduction
band
overlap
Valence
band
Conductor
Fermi level
Valence
band
Semiconductor
Valence
band
Insulator
Band
gap
Semiconductors
The properties of semiconductors make them very important in electronic devices like
diodes and solar cells.
These electrical properties of semiconductor material can be changed dramatically by the
addition of very small quantities of impurity in a process known as doping.
The pure semiconductor (intrinsic) has a very high resistance and the addition of the impurity
reduces its resistance.
Commonly used semiconductor materials are silicon and germanium, both of which have 4
outer electrons available for bonding. In a pure crystal, each atom is covalently bonded to 4
other atoms. All of its outer electrons are bonded and so there are few free electrons available to
conduct.
(the few electrons which may become free are available due to thermal ionisation when the
material is heated).
n-type semiconductors
If an impurity such as the group 5 element arsenic (As), which has five outer electrons,
is grown into the crystal lattice, then four of its electrons will be used in bonding with
the silicon. The fifth will be free to move about and conduct. Since the ability of the
crystal to conduct is increased, the resistance of the semiconductor is therefore
reduced. Because of the extra electrons present, the Fermi level is closer to the
conduction band than in an intrinsic semiconductor.
This type of semiconductor is called n-type, since most conduction is by the movement
of free electrons, which are, of course, negatively charged.
It is important to realise that the semiconductor is electrically neutral as there are the
same number of positive and negative charges. The “n” refers only to the charge
carriers.
p-type semiconductors
The semiconductor may also be doped with a group 3 element like indium (In), which
has only three outer electrons. This produces a hole in the crystal lattice, where an
electron is ‘missing’. Because of this lack of electrons, the Fermi level is closer to the
valence band than in an intrinsic semiconductor.
An electron from the next atom can move into the hole created, as described
previously. Conduction can thus take place by the movement of positive holes. This is
called a p-type semiconductor, as most conduction takes place by the movement of
positively charged holes. The “p” refers to the charge carriers.
In terms of band structure we can represent the electrons as dots in the conduction
band, and holes as circles in the valence band. The majority of charge carriers are
electrons in n-type and holes in p-type, respectively.
Electron
energy
Conduction
band
Conduction
band
Fermi level
increased
Conduction
band
Fermi level
decreased
Valence
band
Valence
band
Valence
band
Intrinsic
semiconductor
n-type
semiconductor
p-type
semiconductor
Fermi
level
The p-n junction
One of the main functions of p-type and n-type semiconductors is their use in a p-n
junction.
When the 2 materials are brought together, and no pd is applied, some electrons in the
n-type combine with positive holes in the p-type.
These charge carriers near the boundary of the 2 materials are no longer free and a
region known as the depletion layer is formed.
This has the effect of giving the p-type side of the junction a small negative charge and
the n-type side an equal small positive charge. Once enough charge has built up, further
electron drift will be prevented. This depletion layer is a narrow region without holes or
electrons.
Forward and Reverse Biased Junctions (making the junction work for us)
In National 5 we found out about the diode,
p
n
The diode is a device that lets current flow in only one direction. The diode is a p-n
junction and works as follows:
Forward biased
The pd of the source overcomes the pd set up across the depletion layer and a current
flows in the circuit.
Reverse biased
The pd of the supply increases the pd across the junction and so increases the depletion
layer. The electrons cannot travel across the junction.
This effectively stops current flowing.
Current in a diode
Reverse biased
I/mA
Forward biased
Using the circuit shown above we can measure the current through a reverse and
forward biased diode for a range of voltages. The graph shows that the reverse biased
diode does not allow a current to flow. It also shows that for small voltages there is no
current in a forward biased diode. A voltage large enough to overcome the depletion
zone (about 0.7V) must be applied to force charge carriers across the zone.
Using the p-n junction
In addition to the “simple” diode there are a number of other electronic components
that will use a p-n junction.
An LED will use the p-n junction as follows :
In a forward-biased p–n junction diode, holes and electrons pass through the junction
in opposite directions. Sometimes holes and electrons will meet and recombine. The
electrons in the conduction band lose energy as they combine with the holes in the
lower valence band. This energy is emitted by the electron in the form of a photon.
Each recombination of electron and hole, results in one photon of radiation being
emitted. In most semiconductors this takes the form of heat, however in some
semiconductors such as gallium arsenic phosphide the energy is emitted as light. If the
junction is close to the surface of the material, this light may be able to escape. This
makes what we call a light emitting diode (LED). The colour of the emitted light (red,
yellow, green, blue) depends on the relative quantities of the three constituent
materials. The recombination energy can be calculated using
E = hf if the
frequency of the light emitted is measured.
A series resistor is always used to limit the current in the LED as a protection against
damage.
The Photodiode
A photodiode is a solid - state device used to detect light. Photodiodes are photo-sensors that
generate a current or a voltage when the p-n junction in the semiconductor is irradiated by
light. In one way of using such a device the p–n junction, in a transparent coating,
reacts to incident light in what is called the photovoltaic effect.
Each individual photon that is incident on the junction has its energy absorbed, assuming this
energy is larger than the band gap. The electrons within the crystal structure become
stimulated and are pulled up into the conduction band, leaving holes in their place in the
valence band.
The electron-hole pairs created occur throughout the p-layer, depletion layer and n-layer
materials.
In the depletion layer the electric field accelerates these electrons toward the n-layer and the
holes toward the p-layer. Of the electron-hole pairs generated in the n- layer, the electrons,
along with electrons that have arrived from the p-layer, are left in the n-layer conduction band.
The holes at this time are being diffused through the n-layer up to the depletion layer while
being accelerated, and collected in the p-layer valence band. In this manner, electron-hole pairs
which are generated in proportion to the amount of incident light are collected in the n- and players. This results in a positive charge in the p-layer and a negative charge in the n-layer. The
light has supplied an emf to the circuit. If an external circuit is connected between the p- and nlayers, electrons will flow away from the n-layer, and holes will flow away from the p-layer
toward the opposite respective electrodes.
The current produced is proportional to the light intensity.
Photodiodes are used as a source of electrical energy in the form of the solar cell.
The p–n junction can supply power to a load, eg a motor. Many photodiodes connected
together form a solar cell.
Photoconductive Mode
In this mode, a photodiode is connected to a supply voltage in reverse bias. In this mode we
would not expect the diode to conduct. This is true when it is kept in the dark.
However, when photons of light shine on the junction electrons are freed and create electronhole pairs similar to the situation in photovoltaic mode. This in turn creates a number of free
charge carriers in the depletion layer, decreasing the resistance and enabling current to flow.
As the irradiance of the light increases more free charge carriers are produced and therefore
there is less resistance in the junction and the current increases.
The photodiode connected to a supply voltage, in reverse bias, acts like a light dependant
resistor (LDR) and is said to be in photoconductive mode.
Acknowledgements
The compiler of these notes would like to thank the contributors to SPTR and
Education Scotland whose work has been used as a source of inspiration.