Intro to Electronics P1

Mrinal K Mandal

[email protected]

Department of E & ECE

I.I.T. Kharagpur. 721302.

www.ecdept.iitkgp.ernet.in

Introduction to Electronics

1

People

Instructor:

• Mrinal Kanti Mandal ([email protected])

List of TAs:

• Nijwm Wary (RS),

• Indranil Som (RS),

• Navonil Chatterjee (RS),

• Priyajit Mukherjee (RS),

• Shashank Shekhar (MT),

• M. Vamsi Krishna (MT),

• Sebin Philip (MT),

• Baheti Priyanka Sushil (DD) .

Department of Electronics & Electrical Communication Engineering, I.I.T. Kharagpur [email protected]

2

Books

Electric Circuit Theory - Van Valkenburg, Prentice Hall.

Electronic Circuits Analysis and Design – Donald A. Neamean.

Digital Logic and Computer Design – M. Morris Mano, Prentice

Hall.

Reference books:

1. Introduction to Microelectronics, B. Razavi.

2. Electronic Devices and Circuit Theory – Robert L. Boylestad,

Prentice Hall.

3. Microelectronic Circuits, A.S. Sedra and K. C. Smith.

4. The Art of Electronics, Paul Horowitz and Winfield Hill.

5. Electronic Devices and Circuits, David A. Bell.


3

4

Passive Components

1. Resistor

Thin film carbon resistor Adjustable

wire wound

Rheostat

Resistors of different power

dissipation factors

Circuit symbol


5

Resistor Colour Coding

Thin film carbon resistor

•Band A - first significant figure (left side),

•band B - second significant figure,

•band C - the decimal multiplier,

•band D (if present) - tolerance of value in percent

(gold - ±5%, silver - ±10%, no band -20%).

• Inside material - A mixture of finely

powdered carbon and an insulating

material, usually ceramic. A resin holds the

mixture together. The resistance is

determined by the ratio of the powdered

ceramic to the carbon.

Black Brown Red Orange Yellow Green Blue Violet Gray White

0 1 2 3 4 5 6 7 8 9

Colour codes:

•Resistance – AB10^C ±D%

Example: yellow-violet-red-gold 4.7k Ω ±5%, between 4,465 Ω and 4,935 Ω.


6

2. Inductor

Passive Components

Different form of inductors

Magnetic field lines

• Inductor stores energy in magnetic field.

• Unit is Henry (H).


7

Circuit Components

3. Capacitor

Electrolytic capacitor Ceramic capacitor Polyester capacitor

Polarized capacitor

symbol

Non-polarized

capacitor symbol

• Capacitor stores energy in electric field.

• Unit is Farad (F).

•Ceramic capacitor marking – AB10^C pF ±10%.

•Example – 154 means 15×10000 pF±10%.

+


Active Components

Diode Bipolar transistor

Field effect transistor Operational amplifier


8

Signals

Signal generator

Different periodic waveforms

Effect of thermal noise

RMS value of a function:

rms values:


9

10

Cathode Ray Oscilloscope


11

Electronic Circuits

Mobile phone Computer motherboard


Concept of Ground

• Ground: a common reference point in any electrical circuit that may or may not

physically connected to the Earth.

• High power circuits: exposed metal parts are connected to ground to prevent

user contact with dangerous voltage if electrical insulation fails.

Connections to ground limit the build-up of static electricity when handling

flammable products or electrostatic-sensitive devices.

• In some power transmission circuits, the earth itself can be used as one

conductor of the circuit, saving the cost of installing a separate return conductor.

Signal ground Earth ground Chassis ground

• In portable electronic devices, a large conductor attached to one side of the

power supply acts as a "ground”.


12

Concept of Ground

• Planet earth is not a good conductor of dc voltage.

A B

1kΩ 1kΩ

RAB = ?

C D

1kΩ 1kΩ

Planet

earth RCD = ?

A typical earthing electrode. Printed circuit board (PCB)

PCB

ground


13

Thevenin’s Theorem

Any two-terminal linear, bilateral network containing impedances and energy

sources can be represented by an independent voltage source VTh and a single

impedance ZTh.

VTh is the open circuit output voltage, ZTh is the impedance viewed at the terminals

when all independent energy sources are replaced by their internal impedances.

VTh calculation:

Calculate the no load output voltage. It is equal to VTh.

RTh calculation:

Remove if any load.

Replace all sources by their input impedances.

Compute the total resistance between the load terminals.


+

- Vth

Rth

A

B

A

B

Black

box

14

Example

Obtain the Thevenin’s equivalence of the following circuit.

•VTh calculation:

•RTh calculation:


1 kΩ +

- 7.5 V

A

B

2 kΩ

15

Norton’s Theorem

Any two-terminal linear, bilateral network containing impedances and energy

sources can be represented by an independent current source IN in parallel with a

single impedance ZN (admittance YN).

IN is the short-circuit current between the terminals, ZN is the impedance viewed at

the terminals when all independent energy sources are replaced by their internal

impedances.

IN calculation:

Short the output terminals and calculate current through it.

RN calculation:

The same as RTh.


A

B

Black

box IN RN

A

B

16

Example

Obtain the Norton’s equivalence of the previous circuit.

•IN calculation:

•RN calculation:


17

Source Transformation


+

- Vth

Rth

A

B

A

B

Black

box IN

RN

A

B

Voltage source Current source

Th N

Th N N

ThN

Th

R R

V I R

VI

R

Calculations:

18

Voltage source Current source

Example

Obtain the Thevenin’s equivalent of the following circuit.

•RTh calculation:

RTh = (2||2) + 1 || 2 Ω

= 1Ω

•VTh calculation:


19

Capacitor Circuit

Current leads the phase of input voltage by 900.

Current and voltage wave forms.


0 0

0

0

cos [ sin ]

sin2

sin2

dv ti t

dt

dv tC

dt

C V t for v t V t

C V t

I t

Instantaneous current:

2

0

1sin 2

2p t v t i t CV t

Instantaneous power expended in charging:

Energy delivered in time interval t1: 1

2

1 0 1

0

11 cos 2

4

t

tW t p t dt CV t

Energy delivered in n half-cycles: 2

0

11 cos 2 0

4nW CV n

20

Capacitor Circuit

Capacitive reactance: 1 1

[where ].Cx s jj C sC

Joule loss due to an ideal capacitor is zero.

Representation of non-ideal capacitor.


Series and parallel connections:

1 2 ...eq nC C C C 1 2

1 1 1 1...

eq nC C C C

21

Inductor Circuit

0 0

0

cos [ sin ]

cos .

di tv t

dt

di tL

dt

L I t for i t I t

V t

Current lags the phase of input voltage by 900.

[where ].Lx j L sL s j Inductive reactance:


Instantaneous voltage:

2

0

1sin 2

2p t v t i t LI t

Instantaneous power expended in charging:

Energy delivered in time interval t1: 1

2

1 0 1

0

11 cos 2

4

t

tW t p t dt LI t

Energy delivered in n half-cycles: 2

0

11 cos 2 0

4nW LI n

22

Inductor Circuit

Joule loss due to an ideal inductor is zero.

Representation of non-ideal inductor.


Series and parallel connections:

1 2 ...eq nL L L L 1 2

1 1 1 1...

eq nL L L L

23

RC Circuit

Frequency domain analysis:

0, 1

, 0

f gain

f gain

RC circuit

(Lowpass filter)

Half-power points: 1 1

.2 2

out out

in in

P v

P v

2

2

33 3

1 1

21

1 2

1 1,

2 2

dd dB

CR

CR

fRC RC

B

B•f3dB: cut-off frequency (fc).

2

1

1

1 1

1 1

1| | (transfer function)

1

c in

c

in

c

in

j Cv v

R j C

v

v j CR sCR

v

v CR


24

LPF

Log100.5 = -0.301

CR Circuit

2

1

1 1

| | (transfer function)

1

R in

R

in

R

in

Rv v

R j C

v j CR sCR

v j CR sCR

v CR

v CR

0, 0

, 1

f gain

f gain

CR circuit

Half-power points:

2

3

33

1

21

1,

1.

2 2

d

ddB

CR

CR

RC

fRC

B

BThe same as RC circuit.

CR circuit

(High pass filter)


25

HPF

Comparison of CR & RC Frequency Responses

RC circuit CR circuit

Low pass filter High pass filter


26

θ

VR = RI

V = ZI VC = jxCI

Phasor diagram

I

LR & RL Circuits

0, 1

, 0

f gain

f gain

RL circuit

22


L in

L

in

j Lv v

R j L

v L

v R L

0, 0

, 1

f gain

f gain

22


R in

R

in

Rv v

R j L

v R

v R L

Output voltage across the inductor: Output voltage across the resistor:

HPF LPF

3 3/ , .2

d dB

R Rrad S f Hz

L L

B

Frequency domain analysis:

•Half-power points for the both cases:


27

Phasor diagram

θ

VR = RI

V = ZI VL = jxLI

I

Time Domain Analysis


28

Steady-state response: ( ) ( )Lts

t

f t f t

Transient response: response before the steady is achieved.

Laplace transform: 0

[ ( )] ( ) ( ) , ( ) , :realand positive, :constant

Re( ) > 0

st atf t F s f t e dt f t K e a K

s

L

Unit step function: ( ) 1, 0

0, 0

U t t

t

0

1[ ( )] stU t e dt

s

L

Inverse Laplace transform: -1 1

( ) [ ( )] ( )2

iT

st

iT

f t F s e F s dsi

L

U(t)

t

1

Time Domain Analysis


29

Differentiation: [ ] ( ) (0 ) , (0 ) is the initial value of ( )dy

sY s y y y tdt

L

Integration: 0

1[ ] ( )

t

y dt Y ss

L

1[ ]ate

s a

LExponential function:

For a sinusoidal wave in steady state:

1

sCsL

Capacitive reactance Inductive reactance

Time Domain Analysis of RC/CR Circuits

RC circuit

t = 0

V

0

Applying Kirchhoff’s voltage law

...(1)

Taking Laplace transform,

...(2)

Considering vc(0-) = 0,

...(3)


30

0

1(0 ) ( )

t

cv i dt Ri V U tC

(0 ) ( )( )cv I s V

R I ss Cs s

1( )

1 1

V C VI s

sCR R s CR

Taking inverse Laplace transform,

( ) ( )t CRVi t e U t

R

...(4)

Voltage across the resistor: t RC

Rv i R V e ...(5)

Time Domain Analysis of RC/CR Circuits

...(6)

• Time constant of the circuit:

Time taken to drop the voltage across the resistor to V/e.

...(7)

[Euler’s number e = 2.71828…]


31

Voltage across the capacitor: 1 t RC

C Rv V v V e

1 2 c

RC

f

• Rise time:

Time taken to reach the capacitor voltage from 10% to 90% of the final value.

Τr = 2.3 RC – 0.1RC

= 2.2 RC

~ 0.35 × 2πRC

= 0.35/ fc. ...(8)

1put t RCe e

Step response of a capacitor. Step response of the resistor.

Time Domain Response of RC/CR Circuits


32

vR / vin = 0.905 at t = 0.1τ

= 0.990 at t = 0.01τ

= 0.05 at t = 3τ

= 0.007 at t = 5τ

vC / vin = 0.95 at t = 3τ

= 0.993 at t = 5τ

1 t RC

Cv V e t RC

Rv V e

Example

Draw the output voltage waveform for the following circuit and calculate the rise

time.

Solutions:

At t = 0, the capacitor is shorted, so V0 = 0 and I0 = 10 mA.

Now,

Time constant = 1k × 1n = 1 μS.

•Rise time = 2.2 RC

= 2.2 1k × 1n

= 2.2 μS.

Output voltage waveform.


33

• Calculate the time when the output voltage is half

of that of the input. (Ans – 0.693 μS)

1 t RC

Cv V e

Time Domain Response of RC/CR Circuits

RC circuit

Low PRP (RC<<Ton) High PRP (RC>>Ton)

Charging phase:


34

1 t RC

Cv V e

t RC

Rv V e

Rectangular pulse:

onV U t U t T

Ton

Discharging phase:

tRC

Rv Ve

t RC

Cv V e

RC Integrator

Consider the output across the capacitor at high frequency i.e. f >>1/Ton.

Integrator circuit

Loop current is

The frequency condition, gives

...(1)

...(2)

Now, voltage across the capacitor is

...(3)

...(4)

At high frequency, the voltage across the capacitor is proportional to the time

integration of the input voltage.

Low pass filter at high frequency


35

1

invi

R j C

1 C R

invi

R

0

1 t

Cv i dtC

0

1 t

C inv v dtRC

RC Integrator Waveforms

Integrator circuit

vc

vc

at very high

frequency

at medium

frequency

Input:

Output:


36

RC Differentiator

Consider the output across the resistor at low frequency i.e. f <<1/ Ton.

Differentiator circuit

Loop current is

The frequency condition, gives

...(1)

...(2)

•The capacitor has enough time to charge up until vc

is nearly equal to the source voltage.

Now, voltage across the resistor is ...(3)

...(4)

At low frequency, the voltage across the resistor is proportional to the time

differentiation of the input voltage.

High pass filter at low frequency.


37

1

invi

R j C

1R C

1

inin c

v ii v v

j C j C

cR

dvv iR C R

dt

inR

dvv RC

dt

RC Differentiator Waveforms

Some other waveforms.


38

RL/LR Circuits


39

RL circuit

t = 0

V

0

Applying Kirchhoff’s voltage law

...(1)

Taking Laplace transform,

...(2)

Considering iL(0-) = 0,

...(3)

( )di

L Ri V U tdt

0V

L s I s i R I ss

1 1( )

V VI s

s R sL R s s R L

Taking inverse Laplace transform,

( ) 1 , 0Rt LVi t e t

R

...(4)

1 , .Rt L Rt L

R L Rv i R V e v v v Ve ...(5)

/

1 2 c

L R

f

Time constant:

Problem

A positive square wave of amplitude 10 V and PRP of 1 kHz is applied to the

following circuit. Draw the vR and vC waveforms for R = 1 kΩ, C = 10 nF.

Solution:

Time period of the input wave: 1 mS.

Ton = 0.5 mS

Time constant = 1 k ×10 n Sec

= 10 μS.


40

max 1

10V

ONT RC

Cv V e

0.51050t m

[email protected]

Ph. – +91-3222-283550 (o)

Department of E. & E.C.E.

I.I.T. Kharagpur, 721302.

Thank you

?

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Intro to Electronics P1

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