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FORM SIX: PHYSICS STUDY NOTES-TOPIC 3: ELECTRONICS


 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


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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.

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P-N=JUNCTION/DIODE:

This is formed when P and N semiconductors are melted to form a junction between them

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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.

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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.

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                                      P-N JUNCTION AS RECTIFIERS

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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.

 

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).

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.

 

TOPIC 3: ELECTRONICS

  1. Conductors

Posses free electrons

Metals are all good conductors due to having low resistance to the flow of current.

  1. 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.

  1. 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:

  1. a)     Using centre –tapped transformer.
  2. 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.

  1. 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.

  1. n.p.n transistor
  2. 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.
  1. 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 = 1.2kΏ 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

  1. a) Pd across base resistor

Consider loop (L), from Kirchhoff’s law

3V =0 but =0.7

= 2.3V

  1. b) From Ohms law

=

=

= 1.53

=+

= ) + (

= (20- 3) + (2.3-1.836)

= (17+0.463)

=17.463v

  1. d) Find

From Kirchhoff’s law

Given

–=0

β=20

=20v

So

20v-(1.5 ×1.224 – =0

=18.164v

  1. 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

=110

= 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

  1. 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.
  2. 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 100Ώ.

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

  1. 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

  1. NOT GATE(INVERTER)

he
It has only one input and one output.

  1. 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

  1. NOR GATE
    This is equivalent to OR gate followed by NOT gate.All outputs of OR gate are inverted

TRUTH TABLE FOR NOR GATE

  1. 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

  1. 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

  1. a) A+B=B+A
  2. b) AB=BA

T2: Associative law

  1. a) (A+B) +C=A+ (B+C)
  2. b) (AB) C=A (BC)

T3: Distributive law

  1. a) A (B+C) =AB+AC
  2. b) A+BC= (A+B) (A+C)

T4: Identity law

  1. a) A+A+=A
  2. b) AA=A

T6: Redundancy law

  1.    a) A+AB=A
  2.    b) A (A+B) =A

T7:  a) 0+A=A

  1. b) 0A=0

T8: a) I+A=I

  1. 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:

  1. i) The lift door must be closed giving d=1
  2. 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.

READ TOPIC 4

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