TOPIC 3: ELECTRONICS | PHYSICS FORM 6
1. Conductors
Posses free electrons
Metals are all good conductors due to having low resistance to the flow of current.
2. Insulators
They do not have free electrons for conduction. They have high resistance to the flow of current.
All non metals are bad conductors. Eg. dry wood, paper and air.
3. Semiconductors
These are class of materials whose conductivity is between that of good conductors and insulators
Silicon and Germanium are examples of semiconductors elements widely used in electronic industry.
INTRINSIC SEMICONDUCTORS
These are pure semiconductors.
EXTRINSIC SEMICONDUCTORS
These are impure semi conductors material.
DOPING
Is a process of introducing a tiny amount of impurity into a semiconductors material to form extrinsic semiconductors shells.
N-SEMI CONDUCTORS
–Silicon and germanium atoms are tetravalent
-They have four electrons in their outermost shell.
-When a doner atom with fine electrons in its outer most shell (ie Arsenic) is added to a silicon crystal, the fifth electrons becomes a free change carriers since there is production of large number of negative charge carriers(electrons) the impure semiconductors is called N-Semiconductors
P-SEMICONDUCTORS
A-P- Semiconductor is made by adding a trivalent atom (an acceptor) such as B or on to pure semi conductor such as germanium. Since there is a production of large number of holes (positive charges) the impure semiconductor is called P- Semiconductor.
P-N=JUNCTION/DIODE:
This is formed when P and N semiconductors are melted to form a junction between them
The marrow region at the P-n junction which contains the negative and positive charge is called depletion layer.
A barrier dip is a p.d which oppose more diffusion of charges across the junction.
This is produced when the flow of +ve and -ve Charges ceases
P – N JUNCTION AS A RECTIFIER:
FORWARD BIAS.
Is said to be forward biased when its P- semiconductor is connected to the +ve terminals of the battery and its N- Semi conductors is connected to the -ve terminal at the battery.
In this case electrons and holes flow across the P-n junction. This happen because the +ve pole of the battery repel the +ve charge and –Ve pole rel the –ve charges.
REVERSE BIAS
A-P-N junction is said to be reverse biased when its P. Semiconductor is connected to the negative pole junction of a battery and N. Semiconductor is connected to the +ve [p;e pf the battery in this case only a very small a current flows.
P-N JUNCTION AS RECTIFIERS
The graph shows that P-N junction acts as a rectifier, it has low resistance in one direction of P.d (ie + v) and higher resistance in the opposite direction of P.d (-v)
RECTIFIER CIRCUITS:
HALF-WAVE RECTIFIER CIRCUIT
A rectifier is a circuit which allow the flow of current I P.d in one direction only:
FULL –WAVE RECTIFIERS CIRCUITS:
a) Using centre –tapped transformer.
b) Using bridge circuits
On one half of a cycle when P is +v relative to Q only diode D1 conducts
On one other half the same cycle only the diode D2 conducts.
In both cases the current gees through resistor RL in the same direction.
The large capacitor C is used for stabilizing the marying d.c voltage.
B) TRANSISTORS
Transistor is a component which amplifies current. It is made from three layers of P and n. Semiconductors. The layers are called the emitter (E) base (B) an collector (C)
There are two types of transistors.
I. n.p.n transistor
II. p.n.p transistor
Formation of a transistor
A transistor is formed by putting the doped semiconductors together in such a way that two junction are formed.
The pnp transistor (bipolar transistor).
-Bipolar means n p n and p n p transistor as they have two opposite polarity of doped semiconductors and voltages across terminals
P n p
n p n
Transistor configuration
There are 3 basic configurations
1. Common emitter configuration
2. Common base configuration
3. Common collector configuration
1. Common emitter configuration (n p n)
Under thus configuration the transistor has both voltage gain and current gain.
To get volt you need a resistance RL
= I
Ring is used for injecting only a small current for great amplification on E by C
Current gain
= very large 
But
Also
is the reflection of 
Common base configuration (PnP)
Under this configuration the transistor has voltage gain but no current gain
Earthing puts the common line at p.d=0
=
Common collector configuration
Under this configuration the transistor has current gain but no voltage gain
=Amplification factor
=
=
For common emitter
Vo = IC R2
Vi = IBRB
From, IE = IB + IC
Common emitter characteristic curve
The circuit above is for investigating the variation of current with voltage in the input and output circuits.
OUTPUT CHARACTERISTICS
IC-VCE with IB constant .
The results are plotted below.
The knee of the curves shown corresponds to a low P.d(0.2) the output for higher P.d the output IC varies linearly with VCE for a given value of base current IB.
The linear part of the characteristic is the one used in the audio frequency (a.f) amplifiercircuits so that the output is undistorted.
INPUT CHARACTERISTICS
IB-VBE with VCE constant
The results are as follows:-
The input characteristics is non-linear
TRANSFER CHARACTERISTICS
IC-IB with VCE constant
The results are as shown below:-
The output current IC varies linearly with the input current IB. The current transfer ratio or current gain is given by
In the figure below
Questions
1. An npn transistor has a current gain (Beta) value of 200. Calculate the base current 
required to switch a resister of 4µA.
2. An npn transistor has a dc base bias voltage 
of and an input base resistor 
of 100kΩ.What will be the of base current into the transistor
(The transistor is a silicon type)
Data
From Kirchhoff
law
For silicon
=10v
= 100kΩ
=0.6v (wasted voltage)
Solution
–
–=0
–=
Example
Given the circuit below, determine
The transistor has 
=150
The transistor of silicon type
Solution
Second Kirchhoff’s law in the input circuit
×
= 0
5V
A
A
=0.6v
10
= 3.4V
=5+ ( 
)
=5+ (10000×4.4×10-4) 
(6.6)
=2.8v
Example
1. A common emitter amplifier has 
= 1.2kÎ and supply Voltage of V=12v. Calculate the maximum collector current following throughout resistor when switched fully on (saturation assume
.Also find 
with a voltage drop of 1v across it, the transistor silicon.
Solution
A
Quiescent point:
It’s a point when the current flow is smooth i.e. not being clicked (excess) and transistor functions.
Saturation point:
If =0 transistor is in the cutoff region, there is a small current collector leakage, CEO
Normally 
is neglected so that
=
In cutoff both the base emitter and base collector junction are reverse based.
When base emitter becomes forward based. 
is increase, then IC also increases when 
decreases as a result.
When reaches its saturation value BC junction becomes forward based and 
can increase no further even with continued increase in
At the point of saturation (
) not longer valid)
VCE(Sat) for a transistor occurs somewhere below the knees of the collector curve.
The saturation value for (Sat) is usually a few tenth of volt for silicon transistors.
The DC load line, the cutoff and saturation can be illustrated by the load line.
Between the cutoff point and the saturation point is where the transistor is active and as most active at the quiescent point.
Self biasing /fixed bias
Outer loop
A
A
10 = 9.4×10-3×100 +
= 9.06V
Common emitter amplifier circuit
Faithful amplification- is the application or the output that is not distorted.
Question
a) Pd across base resistor
Consider loop (L), from Kirchhoff’s law
3V
=0 but =0.7
= 2.3V
b) From Ohms law
=
=
= 1.53
=+ 
= 
) + (
= (20- 3) + (2.3-1.836)
= (17+0.463)
=17.463v
d) Find
From Kirchhoff’s law
Given
––=0
β=20
=20v
So
20v-(1.5
×1.224
–=0
=18.164v
2. 
Q
= 
=
= +
= 25V
= 47mA = 4.7×10 -2A
= 
=
=0.3659Ω
Question
For the circuit above the transistor has a current gain 
=80 the collector supply voltage = 40 .
The required biased conditions are = 0.7V and = 1mA. Determine the suitable values for resistors
,
,
& ,
R2 = 10RE, VE = 1
VE = IV.
Given 
v
= 1×10 -3A
=80
=0.0000125
=
= 0.0010A
=1kΩ
=1
10
= 1.7 but
+ 
=10
Operational Amplifier (Op amp)
An operational amplifier (op amp) is an electronic device consist of a large number i.e. twenty and above.
It has 3 terminals two input terminals and one output terminal.
The op amp can perform electronically mathematically of such as additional, subtraction, multiplication, differentiation, integration
Properties of an op- amp
i) It has got a very high voltage gain called the open loop gain which typically is 105 for dc and low frequency but decrease with frequency.
ii) It has a very high input resistance typically 10, it draw a minute current from the signal source.
iii) It has a very low output resistanceR0, typically 100Î.
Description
It has one output and two inputs and one non inverting (
and one inverting (-).
Its operation must convenient from a dual balanced power supply giving its equal +ve and –ve voltage (+Vs, or,-Vs)
Inverting amplifier
=
(
–
)
=
Some of the output goes back to the input .This red called the amplification from A0 to A
But
I1, = I2
Example
(i)Find the closed loop gain of the inverting amplifier
From
= -10
ii) Supposed the voltage gain is to be increased to 40 and the current of 
remains the same .What are the values of the resistors required to gain this
NON INVERTING AMPLIFIER
The fraction β fed back via Rf
LOGIC GATES (Non inverting)
+ 
LOGIC GATES
SYMBOL
i. NOT GATE(INVERTER)
he
It has only one input and one output.
ii. OR GATE
This can have many number of inputs but only one input. It gives high output if either of the inputs is high or all inputs are high.
TRUTH TABLE FOR OR GATE
iii. AND GATE
It can have many number of inputs but only only one output. It gives high output when both input are high.
TRUTH TABLE FOR AND GATE
iv. NOR GATE
This is equivalent to OR gate followed by NOT gate.All outputs of OR gate are inverted
TRUTH TABLE FOR NOR GATE
v. NAND GATE
This is the AND gate followed by NOT gate . This is widely used gate . In this case the outputs of AND gate re inverted.
TRUTH TABLE FOR NAND GATE
All logic gates described can be connected together to form different function
(i)They are used to control traffic light
(ii)They are used in communication system
(iii)They are used in arithmetic and data processing
Questions
1. Find the expression for Y and form the truth table of the following diagram.
Solution
2. From the logic circuit below form the Boolean expression and draw the truth table
Solution
|
Truth Table |
|||
|
A |
B |
C |
Y |
|
1 |
1 |
1 |
1 |
|
1 |
1 |
0 |
1 |
|
1 |
0 |
1 |
0 |
|
1 |
0 |
0 |
0 |
|
0 |
0 |
0 |
0 |
Laws of Boolean algebra
T1: Commutative law
a) A+B=B+A
b) AB=BA
T2: Associative law
a) (A+B) +C=A+ (B+C)
b) (AB) C=A (BC)
T3: Distributive law
a) A (B+C) =AB+AC
b) A+BC= (A+B) (A+C)
T4: Identity law
a) A+A+=A
b) AA=A
T6: Redundancy law
a) A+AB=A
b) A (A+B) =A
T7: a) 0+A=A
b) 0A=0
T8: a) I+A=I
b) 1A=A
T11: De Morgan’s theorem
Example
1) Prove that
Algebraically
=A (I+B) +
B
Question
For a lift (L), these are the conditions:
i) The lift door must be closed giving d=1
ii) The appropriate floor button (B) must be pressed B
L= Bd but not L=B+ d
A boiler shut down solenoid (s) will operate if the temperature T reaches 50 and the circulating pump P ise turned off or if the pilot light L goes out.
That’s very good