Introduction to Electronics

Mrinal K Mandal [email protected] Department of E & ECE I.I.T. Kharagpur. 721302. www.ecdept.iitkgp.ernet.in

Introduction to Electronics

1

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

University Press.

2. Introduction to Microelectronics, B. Razavi.

3. Electronic Devices and Circuit Theory – Robert L. Boylestad, Prentice Hall.

4. The Art of Electronics, Paul Horowitz and Winfield Hill. 5. Electronic Devices and Circuits, David A. Bell.

Department of Electronics & Electrical Communication Engineering, I.I.T. Kharagpur [email protected] 2

Some Important Points

•Attendance.

•Risk of deregistration: if attendance is lower than a certain percentage.

•Total marks distribution = surprise class test + attendance + mid-sem + end-sem evaluation results.

•n+1 surprise class tests: best n results will be considered.

•Register in Intinno paathsaala.


Signal

Generation of discrete time signal by sampling a continuous signal.


Signal is a function that conveys information about the behavior or attributes of some phenomenon. • In the physical world, any quantity exhibiting variation in time or variation in space

(such as an image) is potentially a signal that might provide information on the status of a physical system, or convey a message.

• Popular forms: audio, video, speech, image, communication, geophysical, sonar, radar, medical and musical signals.

• In electronics engineering: time varying voltage/ current/ electromagnetic waves. • Analog and digital signals.

Digital approximation of the analog signal.

Signal

RMS value of a function:


( )2

0

1limT

rms Tf f t dt

T→∞= ∫ ( )

2

1

2

2 1

1lim .T

rms TT

f f t dtT T→∞

=− ∫

signal

transducer

Electronic signal

Signal processing

Transmitter

Electromagnetic wave

Electromagnetic wave

Receiver

Signal processing

Electronic signal

transducer

signal

Transmitter.

Receiver.

Function Generator

Signal generator

Different periodic waveforms

rms values:


Effect of thermal noise

V(t)

t

7

Cathode Ray Oscilloscope

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

8

Electronic Circuits

Electronics in daily life.

Nokia phone circuit board.


9

Electronic Circuits

Electronic warfare and communication systems.


10

Passive Components

1. Resistor

Thin film carbon resistor Adjustable wire wound

Rheostat

Resistors of different power dissipation factors

Circuit symbol


11

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

12

2. Inductor

Passive Components

Different form of inductors

Magnetic field lines

• Inductor stores energy in magnetic field. • Voltage leads the current. • Unit is Henry (H).


13

Circuit Components 3. Capacitor

Electrolytic capacitor Ceramic capacitor Polyester capacitor

Polarized capacitor symbol

Non-polarized capacitor symbol

• Capacitor stores energy in electric field. • Current leads the voltage. • 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


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”.


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


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

17

Example Obtain the Thevenin’s equivalence of the following circuit.

•VTh calculation:

•RTh calculation:


1 kΩ + -

7.5 V

A

B

2 kΩ

18

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

19

Example

Obtain the Norton’s equivalence of the previous circuit.

•IN calculation:

•RN calculation:


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 RV I R

VIR

==

=

Calculations:

21

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:


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

dtdv t

Cdt

C V t for v t V t

C V t

I t

ω ω ω

πω ω

πω

∞

=

= =

= +

= +

Instantaneous current:

( ) ( ) ( ) 20

1 sin 22

p t v t i t CV tω ω= =

Instantaneous power expended in charging:

Energy delivered in time interval t1: ( ) ( ) ( )1

21 0 1

0

1 1 cos 24

t

tW t p t dt CV tω= = −∫Energy delivered in n half-cycles: ( )2

01 1 cos 2 04nW CV nπ= − =

23

Capacitor Circuit

Capacitive reactance: 1 1 [where ].Cx s j

j C sCσ ω

ω= = = +

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= + + +

24

Inductor Circuit

( ) ( )

( )

( )0 0

0

cos [ sin ]cos .

di tv t

dtdi t

Ldt

L I t for i t I tV tω ω ω

ω

∞

=

= =

=

Current lags the phase of input voltage by 900.

[where ].Lx j L sL s jω σ ω= = = +Inductive reactance:


Instantaneous voltage:

( ) ( ) ( ) 20

1 sin 22

p t v t i t LI tω ω= =

Instantaneous power expended in charging:

Energy delivered in time interval t1: ( ) ( ) ( )1

21 0 1

0

1 1 cos 24

t

tW t p t dt LI tω= = −∫Energy delivered in n half-cycles: ( )2

01 1 cos 2 04nW LI nπ= − =

25

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= + + +

26

RC Circuit Frequency domain analysis:

0, 1, 0

f gainf gain

∴ → =→∞ =

RC circuit

(Lowpass filter)

Half-power points: 1 1 .2 2

out out

in in

P vP v

= ⇒ =

( )( )

2

2

33 3

1 121

1 21 1,

2 2d

d dB

CR

CR

fRC RC

ω

ωω

ω ωπ π

=+

⇒ + =

⇒ = = ∴ = =BB •f3dB: cut-off frequency (fc).

( )2

11

1 11 1

1| | (transfer function)1

c in

c

in

c

in

j Cv vR j C

vv j CR sCR

vv CR

ωω

ω

ω

=+

∴ = =+ +

∴ =+


LPF

Log100.5 = -0.301

CR Circuit

( )2

1

1 1

| | (transfer function)1

R in

R

in

R

in

Rv vR j C

v j CR sCRv j CR sCR

v CRv CR

ωωωω

ω

=+

∴ = =+ +

∴ =+

0, 0, 1

f gainf gain

∴ → =→∞ =

CR circuit

Half-power points:

( )2

3

33

121

1 ,

1 .2 2

d

ddB

CR

CR

RC

fRC

ω

ω

ω ω

ωπ π

=+

⇒ = =

⇒ = =

B

B The same as RC circuit.

CR circuit (High pass filter)


HPF

Comparison of CR & RC Frequency Responses

RC circuit CR circuit

Low pass filter High pass filter


θ

VR = RI

V = ZI VC = jxCI

Phasor diagram

I

LR & RL Circuits

0, 1, 0

f gainf gain

∴ → =→∞ =

RL circuit

( )22| | (transfer function)

L in

L

in

j Lv vR j L

v Lv R L

ωω

ω

ω

=+

∴ =+

0, 0, 1

f gainf gain

∴ → =→∞ =

( )22| | (transfer function)

R in

R

in

Rv vR j L

v Rv R L

ω

ω

=+

∴ =+

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

HPF LPF

3 3/ , .2d dB

R Rrad S f HzL L

ωπ

= =B

Frequency domain analysis:

•Half-power points for the both cases:


Phasor diagram

θ

VR = RI

V = ZI VL = jxLI I

Time Domain Analysis


Steady-state response: ( ) ( )Ltst

f t f t→∞

=

Transient response: response before the steady is achieved.

Laplace transform: 0

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

Re( ) >

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

s a−

∞−= = <∫L

Unit step function: ( ) 1, 00, 0

U t tt

= ≥= ≤

0

1[ ( )] stU t e dts

−

∞−= =∫L

Inverse Laplace transform: -1 1( ) [ ( )] ( )2

iTst

iT

f t F s e F s dsi

γ

γπ

+

−

= = ∫L

U(t)

t

1

Time Domain Analysis


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[ ]ates a

=−

LExponential function:

For a sinusoidal wave in steady state:

1sC

sL

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)


0

1(0 ) ( )t

cv i dt Ri V U tC− + + =∫

(0 ) ( ) ( )cv I s VR I ss Cs s− + + =

1( )1 1

V C VI ssCR R s CR

∴ = = + +

Taking inverse Laplace transform,

( ) ( )t CRVi t e U tR

−∴ = ...(4)

Voltage across the resistor: t RCRv 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…]


Voltage across the capacitor: ( )1 t RCC Rv V v V e−= − = −

1 2 c

RCf

τπ

==

• 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


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 RCCv V e−= −

t RCRv 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.


• Calculate the time when the output voltage is half of that of the input. (Ans – 0.693 μS)

( )1 t RCCv V e−= −

Time Domain Response of RC/CR Circuits

RC circuit

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

Charging phase:


( )1 t RCCv V e−= −

t RCRv V e−=

Rectangular pulse:

( ) ( )onV U t U t T− −

Ton

Discharging phase:

t RCRv V e−=−

t RCCv 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


1invi

R j Cω=

+

1 C Rω <<

inviR

≈

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:


1:5 approximation: τ ≥ 5/PRF (=5T)

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.


1invi

R j Cω=

+

1R Cω<<

1in

in cv ii v vj C j Cω ω

≈ ⇒ ≈ =

cR

dvv iR C Rdt

= =

inR

dvv RCdt

∴ ≈

RC Differentiator Waveforms Some other waveforms.


[Differentiation]

1:5 approximation: 1/PRF (=T) ≥5τ

RL/LR Circuits


RL circuit

t = 0

V

0

Applying Kirchhoff’s voltage law

...(1)

Taking Laplace transform,

...(2)

Considering iL(0-) = 0,

...(3)

( )diL Ri V U tdt

+ =

( ) ( ) ( )0 VL 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 tR

−∴ = − ≥ ...(4)

( )1 , .Rt L Rt LR L Rv i R V e v v v Ve− −∴ = = − = − = ...(5)

/1 2 c

L Rf

τπ

==

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

Check for 1:5 approximation: Time constant = 1 k ×10 n Sec = 10 μS.


43

( )max 1

10V

ONT RCCv V e−= −

=

0.5 10 50t mτ µ∴ = =

Semiconductor A semiconductor is a material which has electrical conductivity between that of a conductor such as (copper, 5.96x107 S/m) and an insulator (glass, ~10-13 S/m). In semiconductors •The conductivity increases with increasing temperature. •Current conduction occurs via free “electrons” and "holes", charge carriers. •Doping greatly increases the number of charge carriers. •Contains excess holes p-type, and excess free electrons n-type.

Simplified band-diagrams (a graphic representation of the energy levels)


Intrinsic and Extrinsic Semiconductors Intrinsic semiconductor:

When the number of impurities is extremely low so that it can be considered as a pure semiconductor material.

Example: Semiconductor carriers/cc at room temp Ga As 1.7× 106

Si 1.5× 1010

Ge 2.5× 1013

Ge atom Si atom

Electron configuration: 2, 8, 18, 4. Electron configuration: 2, 8, 4.


Intrinsic Semiconductor

Covalent bond of the Si atoms at t = 0 K

Creation of electron-hole pair because of thermal agitation at t >0K.

electron

hole

•For intrinsic semiconductor, electron density ni = hole density pi.

•Si about 1 free electron for every 1012 atoms at 300K)


Extrinsic Semiconductors A semiconductor with doping material is called extrinsic semiconductor.

1. n-type material: Created by introducing impurity elements that have five valance electrons ( eg. antimony, arsenic, phosphorus etc).

•Impurities with five valance electrons are called donor atoms.

2. p-type material: Created by introducing impurity elements that have three valance electrons ( eg. boron, gallium, indium etc). •Impurities with three valance electrons are called acceptor atoms.


Extrinsic Semiconductors

Majority carriers and minority carriers: Excess carriers are majority carriers.

•In n-type: electrons are majority carriers and holes are minority carriers. •In p-type: holes are majority carriers and electrons are minority carriers.

• For extrinsic semiconductors, np = ni2, n: electron concentration, p: hole

concentration in the extrinsic material, ni: intrinsic free electron density.


•The resulting material is electrically neutral.

Drift and Diffusion Currents


•Charge carriers moves under the influence of electric field. •They cannot travel in a straight line. •Drift current depends on carrier concentration, carrier mobility and the applied electric field.

E

Drift current

atom electron hole

Electrical conductivity ( )n pe n pσ µ µ= + n: electron concentration/ cc p: hole concentration/ cc μn: electron mobility cm2/V-S μp: hole mobility cm2/V-S

(Ω-cm)-1

Drift and Diffusion Currents


•Charge carriers (one type) when injected from an external source produces concentration gradient. •Current density is proportional to the concentration gradient.

Diffusion current Diffusion

where

p-n Junction: Semiconductor Diode


p-n diode

Formation of depletion region

• Depletion region contains an equal number of immobile ionized atoms on both sides of the pn junction.

• The barrier voltage opposes the flow of majority carriers across the junction but assists the flow of minority carriers across the junction.

p-n Junction: Semiconductor Diode


Reverse-Bias Condition

pn junction under reverse-bias condition

•The holes in the p-side and the electrons in the n-side are attracted by the electrodes the depletion region widens barrier voltage increases. •Minority carriers are generated on each side causes a reverse saturation current. •Reverse saturation current is almost independent of applied voltage but increases with increasing temperature.


Forward-Bias Condition

pn junction under forward-bias condition

•The holes in the p-side region and the electrons in the n-side region are driven toward the junction width of the depletion region is reduced barrier voltage decreases. •With increasing forward-bias voltage, the barrier voltage finally disappears current increases suddenly.


Diode current can be given by Shockley equation: ( )0 1D TV nVDI I e= −

n = 1 for Ge and 2 for Si, VT = kT/e = 26 mV at room temp, T – temperature (K), k = Boltzman’s const (1.38x10-23 J/K), e = electronic charge (1.6x10-9 C)

I-V Characteristics

I-V characteristics of a resistor. I-V characteristics of a diode.


Temperature coefficient for a pn junction: -1.8 mV/0C for Si and -2.02 mV/0C for Ge.

Diode Breakdown Under Reverse Bias

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

1. Avalanche breakdown:

• Thermo-generated electron-hole pair accelerates under the applied electric field. •Under the right circumstances, they gain sufficient energy to knock other electrons free (by collision), creating more electron-hole pairs. •The procedures is regenerative avalanche of carriers conductivity increases suddenly.

Ein + Eext

Avalanche breakdown Avalanche of carriers

Etotal

Diode Breakdown Under Reverse Bias


2. Zener breakdown: • High electric field (~107 V/m) enables tunnelling of electrons from the valence to the conduction band of a semiconductor (tear out a covalent bond) a large number of free minority carriers suddenly increases the reverse current. • High doping concentration is required. • Occurs at lower reverse bias voltage (~ 6 V).

Diode characteristics Zener breakdown

•Usually, Rbreakdown <RFB

•Applications: voltage regulator, clipper circuit, voltage shifter etc

ID

VD

Zener breakdown

Avalanche breakdown

p n

Symbol

Ein + Eext

Iz

Diode Modelling

Ideal diode response. Ideal diode as a switch. More accurate: use to solve problems.

58

ID

VD

ID

VD


Si-diode Ge-diode Zener-diode Tunnel-diode Power-diode

dc and ac resistances:

VD

Diode characteristics

ID

ΔID

ΔVD dc resistance, , ac resitance, .D D

D dD D

V VR rI I

∆= =

∆

A good approximation for Vi >12 V (error <5%)

Diode Load Line


Vi

R

VD ID

Load line

( ) = − 0 exp 1 ...(1)D D TI I V nV

Diode with a resistance

Diode current:

VD

ID

Vi /R

Vi

IDQ

VDQ

Q-point

Load line Shockley diode model:

• Difficult to solve for VD from (1) and (2). • Plot them in a graph and obtain the solution.

Applying KVL,

[ ]

= +

= − = + form ...(2)

i D D

i DD

V I R VV VI y mx CR R

So, from (1) and (2),

( ) − = − 0 exp 1 . ...(3)i DD T

V VI V nVR R

Diode Analysis


Diode characteristics

vD

iD

IDQ

VDQ

t

t

Diode circuit

ac comp.,dc comp.,total instantaneous comp.

d

D

D

iIi

→→→

( )

( )( )( )( ) ( ) ( )

( ) ( )

( )

0

0

0 02

0

0

exp 1

exp 1

exp exp exp

exp

linear approximati

1 [assume ], 1 ...2!

[ exp , ]on

D D T

DQ d T

DQ d T DQ T d T

xDQ T d T d T

DQ d DQ DQ T d DQ d T

i I v nV

I V v nV

I V v nV I V nV v nVxI V nV v nV v V e x

I i I I V nV i I v nV

= − = + −

≈ + =

= × + << = + + +

= + = =

vD

iD

IDQ

iD

[ ]d T DQ d d dr V I v i r= ∴ =Diffusion resistance

Vi

R

VD ID

vi

Diode Modelling: Small Signal


Small signal approximation: ( )

( )exp

1 [for 1 ].D DQ d T

xDQ d T d T

i I v nV

I v nV v V e x

=

≈ + << ⇒ ≈ +

[ ]

( ) ( )

2 2 3 3

2 3

sin .

sin sin1 sin ... where2! 3!

1 1Now, sin 1 cos 2 , sin 3sin sin 3 ...2 4

d

D DQ T

v a t

b t b ti I b t b a nV

t t t t t

ω

ω ωω

ω ω ω ω ω

=

∴ = + + + + =

= − = −

• There are many frequency components those cannot be neglected. • The output signal waveform is not an exact replica of that of the input non-linear device.

Problem with large signal:

Mathematical models used to approximate the actual behaviour of real diodes to enable calculations and circuit analysis.

Diode Modelling: Large Signal


Vi

R

VD ID


Iterative Solution:

( )

( )

= −

⇒ + =

⇒ = +

− ⇒ = + = − −

⇒ = +

0

0

0

0

0

exp 1

1 exp

1

1 put

1 ...(3)

D D T

DD T

DD T

i D i DD T D

i DD T

I I V nV

I V nVI

IV nV lnI

V V V VV nV ln IRI R R

V VV nV lnRI

Guess a starting value of VDQ on R.H.S. and compute the function. Guess a second value and continue unless it converges.

Load line

VD

ID

Vi /R

Vi

IDQ

VDQ

Q-point

Load line

Graphical solution: Plot the two equations on a graph paper (computer) and note down the value.

Diode Modelling: Large Signal


Vi

R

VD ID


Piecewise linear model:

Piecewise linear approximation.

VD

ID

Vi /R

Vi

IDQ

VDQ

Load line

A function is broken down into several linear segments.

Equivalent model.

ID

VD

Diode characteristics.

(Forward bias) (Reverse bias)

Vγ rD

• rD represents the 1/slope at Q-point.

26 mV( at room temp).TD

DQ DQ

VrI I

≈ ≈

Diode Circuit


vi RL

vD

id

+ - +

- vL

Half-wave rectifier:

Half-wave rectifier circuit.

vi RL id

+

- vL

Vγ

vi RL id

+

- vL

Equivalent circuits. (in forward bias) (in reverse bias)

A rectifier converts alternating current to direct current.

Response to solve problems. (Consider the ideal diode model if nothing is specified)

ID

VD

Ideal diode

ID

VD

Half-Wave Rectifier


Peak Inverse Voltage (PIV): peak voltage in reverse bias condition that a rectifier can block. It is limited by diode break down voltage.

vi RL id

+

- vL

vi RL id

+

- vL

(in forward bias)

(in reverse bias)

Voltage waveforms considering ideal diode.

vi (V) t (mS) T/2 T 3T/2

vL (V) t (mS) T/2 T 3T/2

vD (V) t (mS) T/2 T 3T/2

Vm

Vm

-Vm

Half-Wave Rectifier


sini mv V tω=

vi (V) t (mS) T/2 T 3T/2

Vγ

vL (V) t (mS) T/2 T 3T/2

Vγ

vD (V) t (mS) T/2 T 3T/2

Vγ

Vm

mV Vγ−

Voltage waveforms.

vi RL id

+ - +

- vL

dv

vi RL id

+

- vL

(in forward bias)

(in reverse bias)

( ) ( )( )

| |d peak i peak L D

m L

i v V R R

V V Rγ

γ

= − +

≈ −

-Vm


( ) ( )01

cos sinn nn

v t V a n t b n tω ω∞

=

= + +∑

Fourier series: Any periodic function can be represented as a summation of sinusoidal functions.

Representation of a square wave according to Fourier series.

Half-Wave Rectifier: Analysis

• Rectified output voltage can be represented by a Fourier series.

• It contains a dc term + many sinusoidal (time varying) components.

• Sinusoidal signal is represented by its rms value.

t (mS) f (t)

1

2

6



[ ]2

0

0

1 sin where2

1 sin 02

dc av m

m

m

V V V d t

V d

V

π

π

θ θ θ ωπ

θ θππ

= = =

= +

=

∫

∫

dc part:

ac part:

22 2

0

2

0

1 sin2

1 cos 22 2

2

rms m

m

m

v V d

V d

V

π

π

θ θπ

θ θπ

=

−=

=

∫

∫

vL (V) t (mS) T/2 T 3T/2

Vm

2T 5T/2

Output voltage waveforms considering an ideal diode.

• rms value of a 12 V DC voltage is 12 V. • DC contribution could not be avoided.



Ripple voltage and ripple factor:

( ) 2 2

2 2

24

0.14870.385

r rms dc

m m

m

m

v rms v V

V V

VV

π

= −

= −

==

Vdc

t (mS)

(V)

vr|p-p

vt = Vdc + vr (ripple voltage) (V)

Ripple voltage

ripple voltage (rms)Ripple factor = 100%dc voltage

100%

0.385 100% 121%

r

dc

m

m

vV

VV π

×

= ×

= × =

2

2

2

20.406

dc Ldc ac

rms L

m

m

V RP Pv R

VV

η

π

= =

=

=

Efficiency:

Half-Wave Rectifier


• In the circuit, both of the diodes have a cut in voltage of 0.7 V. Calculate the current components.

10 0.7330

28.182 14.09

R R

D R

I V R

mAI I mA

−= =

== =

Solution:

vi RL

vD

id

+ - +

- vL

• A sinusoidal source vi = 12sin100πt V is used in the half-wave rectifier circuit with RL = 1 kΩ. The diode has a cut-in voltage of 0.7 V. Calculate the PRV and the power rating of the diode.

Answer: PRV = 12 V. Power rating = Imax x Vdmax = 7.91 mW.

10 V ID

IR

330 Ω D1 ID D2

N.B.: If the diodes have different cut-in voltages, the diode with lower cut-in voltage will be switched on.

Half-Wave Rectifier : Experiment


Transformer: • Controllable voltage ratio. • Impedance matching.

2

1

s

i

v Nv N=

Half-wave rectifier.

RL

vD + - +

-

vL Line voltage (220 V, 50 Hz)

Center tapped transformer (12-0-12 V)

Primary

Secondary

Center tapped transformer

Voltage regulation:

Voltage regulation 100%NL FL

FL

V VV−

= ×

VNL: open circuited output voltage VFL: output voltage with minimum load resistance.

Full-Wave Rectifier


Full-wave rectifier.

Voltage waveforms considering ideal diodes.

vi (V) (Diode 1)

t (mS) T/2 T 3T/2

iD2 (mA) t (mS) T/2 T 3T/2

vL (V) t (mS) T/2 T 3T/2

iD1 (mA) t (mS) T/2 T 3T/2

vi (V) (Diode D2)

t (mS) T/2 T 3T/2

1 1 2m sV v=

2 2sv

1 2s Lv R

2 2sv

2 2s Lv R

1 2sv

D1 D1

D2

RL

vD1 + - +

-

vL vs1

vs2

+

+ -

- vD2 + -

D1

D2

1 12sv Vγ−For non-ideal diode

Consecutive peaks have different values if • cut-in voltages are different. • vs1 ≠ vs2


Full-Wave Rectifier: Analysis

2

0

0

1 sin2

2 sin 022 2

dc av m

m

m dc half wave

V V V d

V d

V V

π

π

θ θπ

θ θπ

π−

= =

= +

= =

∫

∫

dc part:

ac part:

22 2

0

2

0

1 sin2

1 cos 222 2

2 2

rms m

m

m rms half wave

v V d

Vd

V v

π

π

θ θπ

θ θπ

−

=

−= ×

= =

∫

∫

vL (V) t (mS) T/2 T 3T/2

Vm

2T 5T/2

2

2

2

8

0.812 2

dc Lfull wave dc ac

rms L

half wave

V RP Pv R

η

πη

−

−

= =

=

= =

Efficiency:

Output of a full-wave rectifier.


Full-Wave Rectifier: Analysis

vL (V) t (mS) T/2 T 3T/2

Vm

2T 5T/2

Ripple voltage and ripple factor:

( ) 2 2

2 2

2

42

0.09470.308

r rms dc

m m

m

m

v rms v V

V V

VV

π

= −

= −

==

Ripple voltage

Ripple factor 100%

0.308 100% 48%2

r

dc

m

m

vV

VV π

= ×

= × =

Output of a full-wave rectifier.

C

vD1 + - +

- v= ? vs1

vs2

+

+ -

- vD2 + -

D1

D2

• Vm should not exceed PIV.

Full-Wave Rectifier



RL

vD1 + - +

-

vL vs1

vs2

+

+ -

- vD2 + -

D1

D2


vi (V) t (mS) T/2 T 3T/2

iD2 (mA) t (mS) T/2 T 3T/2

vL (V) t (mS) T/2 T 3T/2

iD1 (mA) t (mS) T/2 T 3T/2

vi (V) t (mS) T/2 T 3T/2

1 1 2m sV v=

2 2sv

1 2s Lv R

2 2sv

2 2s Lv R

1 2sv

D1

D2

Bridge Rectifier


Equivalent circuit in negative cycle. Voltage waveforms considering ideal diodes.

vL D1

D2

D3

D4

RL Line voltage (220 V, 50 Hz)

Bridge rectifier circuit.

Equivalent circuit in positive cycle.

vL D1

D2

D3

D4

RL +

-

vL D1

D2

D3

D4

RL

+

-

vi (V) t (mS) T/2 T 3T/2

iD2 (mA)

vL (V)

iD1 (mA)

mV

D1, D2

D3, D4

m LV R

m LV R

m LV R−

mV

D1, D2

1 12 2m sV v Vγ= −For non-ideal diode:

Rectifier With Filter


• Rectified output contains a dc + many sinusoidal terms (amplitude of the sinusoidal components decreases as the frequency increases).

• A series inductor attenuates and a shunt capacitor bypasses high frequency signals.

Shunt capacitor as a filter.

C L

R

C C

L

C C

Rectifier.

RL vs1

vs2

+

+ -

-

D1

D2

Low Pass Filter

ac source

Series inductor as a filter.

LC π filter. RC π filter.

Different types of filters. Rectifier with a filter.

Rectifier With Filter: Half-Wave


Half-wave rectifier.

RL

vD + - +

-


C

Voltage waveforms considering ideal diode.

vi (V) t (mS) 2T

vL (V) t (mS) T 2T

Vm

Vm

T

Vmexp(-t/RLC)

T1 T2

Analysis:

• Charging (T1) time constant ≈ (RD + Rind ) x C ≈ 0

• Discharging (T2) time constant = RL xC

( )( )( )( )

( ) ( )( )

min

When the capacitor discharges:exp /

exp /

1 1 /

/ /

/

L c m L

L m L

r m Lp p

m L L

m L

v v V t R C

v V T R C

v V T R C

V T R C I T C

V f R C

−

= = −

≈ −

≈ − −

= =

=

• Time period T = T1 + T2

Output voltage:

Rectifier With Filter: Half-Wave


Half-wave rectifier with filter.

RL

vD + - +

-


C

• A repetitive surge current IS (during T1) flows to recharge the capacitor.

• IS is spike shaped. For simplified analysis, it is modeled by a rectangular pulse.

• Average input current to rectifier is equal to the average load current:

Surge current:

1

2

1

1

L Sav

S L av

I T I T

TI IT

× = ×

⇒ = +

Surge limiting resistor.

Rs

Surge current through the diode.

vL (V)

t (mS) T

Vm

T1 T2

t (mS)

Id (mA) Surge current

Triangular wave approximation: ( ) | 2 3r r p pv rms v −=

Problem


Half-wave rectifier with filter.

RL

vD + - +

-

vL Line voltage

C

As shown in the figure, a half-wave rectifier dc power supply is to provide 10 V (dc) to a 1 kΩ load. Calculate the capacitance required so that the peak-to-peak ripple voltage does not exceed ±10% of the average output voltage. Source frequency is 50 Hz. What should be the amplitude of the input voltage?

Solutions:

( )

2

1 20

2010% of 10V 9V - 11V

11 V

= / 110 .

r p p

m

m L r p p

T mSf

T T mSv

V

C V f R v Fµ

−

−

= =

≈ =

= ± ≈

∴ =

=

vL (V)

t (mS) T

Vm

T1 T2

Triangular wave approximation: ( ) | 2 3r r p pv rms v −=

10 V

Rectifier With Filter: Full-Wave



Analysis:

• Charging (T1) time constant = RD xC = 0

• Discharging (T2) time constant = RL xC

( )( )( )

( )( ) ( )

minexp / 2

1 1 / 2

/ 2

/ 2 2

L m L

r m Lp p

m L

m L L

v V T R C

v V T R C

V T R C

V f R C I f C

−

≈ −

≈ − −

=

= =

• Time period T/2 = T1 + T2

Output voltage: vi (V) t (mS)

2T

vL (V) t (mS) T 2T

Vm

Vm

T

Vmexp(-t/RLC)

T1 T2

T/2 3T/2 5T/2


RL

vD1 + - +

-

vL vs1

vs2

+

+ -

- vD2 + -

D1

D2

C

Problem


Calculate the peak-to-peak ripple voltages for a half-wave and a full-wave rectifiers with a capacitor filter. (RL = 1 kΩ, C = 470 μF, vi = 12 V, 50 Hz) Solutions:

T = 1/50 = 20 mS Discharging time constant = RLC = 470 mS. T1 = 0 approximation is valid.

• Full-wave rectifier:

( )( )

/

12 2 50 1 4700.722 ( 0.707 using exp. function)

r m Lp pv V f R C

kV V

µ−

=

= × ×

= =

( )( )

/ 2

12 2 50 1 4700.361 ( 0.357 using exp. function)

r m Lp pv V f R C

kV V

µ−

=

= × ×

= =

• Half-wave rectifier:

Becomes a complex problem for RL = 1 kΩ, C = 47 μF.

( )

( )

2

2

3 4

4

Half-wave: sin

Full-wave: sin

L

L

T T R Cm

T T R Cm

t V e

t V e

ω

ω

− +

− +

=

=

Solve for T2 first:

Triangular wave approximation: 2 3r rp p rmsv v

−=

Voltage Regulator


• Automatically maintains a constant voltage level. • Output voltage is independent of load or source voltage variation.

A dc voltage regulator using Zener diode.

1. The diode must be fired (in break down mode). 2. Diode current must be limited to avoid burn out.

Zener breakdown

ID (mA)

-VD (V) VZ

RZ = 0

Iz|allowed

Iz|allowed /10 Operating

region

Vi

Rs

Vz

I RL

+

-

VL IZ

IL

Conditions for proper operations:

( )

(1)(2)

z L

i s z

i s L

I I IV I R VV I R R

= += +

= +

When the diode operates in breakdown region.

When the diode is not in breakdown region. ×

Voltage Regulator


A dc voltage regulator using Zener diode.

( )

minmin

min

minmin

without the diodeCalculate for the voltage divider formedby and

1

L Z

L

s L

Li Z

L s

is L

z

V VV

R RRV V

R R

VR R

V

>

⇒ × >+

⇒ < −

max

max

max

wherez z z z zallowed allowed

i zz allowed

s

i zs

z z

I I I P V

V VI

RV V

RP V

≤ = −

⇒ ≤

−⇒ ≥

Condition 1: diode in breakdown Condition 2: avoid diode burn out

Vi

Rs

Vz

I RL

+

-

VL IZ

IL

Both Vi and RL vary:

min max

min max

max min

tototo

i i i

L L L

L L

V V VR R R

I I

⇒⇒

Take RLmax as infinite if unspecified.


Voltage Regulator • In the following circuit, Vi can vary between 9

and 12 V, and RL between 1 kΩ and infinity. A Zener diode with Vz = 6 V and PD = 400 mW is used to design the voltage regulator. Choose a suitable value of Rs to avoid diode burn out.

Vi

Rs

Vz

I RL

+

-

VL IZ

IL

Solutions: The diode must be fired:

minmin

min max

max

max

||| |19 6

1 || 0.5

Li z

L s

s

s

RV VR R

kk R

R k

× ≥+

⇒ × ≥+

⇒ ≤ Ω

Maximum allowed Zener current:

/ 66.67z allowed D zI P V mA= =

maxmin

min min

||

12 6| | 90 .66.67

i zs

z allowed

s s

V VRI

R Rm

−≥

−⇒ ≥ ⇒ ≥ Ω

The range of Rs is 90 Ω <Rs <500 Ω.


Voltage Regulator

100%, No load output voltage

Full load output voltage.

NL FLNL

FL

FL

V VVoltage regulation VV

V

−= × →

→

100%.L

i

VSource regulationV

∆= ×∆

ΔVL change in output voltage for a change of ΔVi input voltage.

Ripple rejection ratio = output ripple voltage/ input ripple voltage.

Vi

Rs

Vz IZ

I Vi

Rs

Vz IZ

I Vi

Rs

Vz

IZ

I Rz

Voltage regulator without the load.

Equivalent circuit (ideal Zener).

Equivalent circuit


Clipping Circuits

Series clipper:

• A clipper (limiter) clips off an unwanted portion of a waveform. • Often used to protect a circuit from a large amplitude signal. • Example: half-wave rectifier.

RL v0

+

-

vin

+

- RL

v0

+

-

vin

+

-

RL v0

+

-

vin

+

- 0 V

5.3 V

0 V

6 V

Negative series clipper. Output voltage. Input voltage.

Equivalent circuit when the diode is in forward bias.

Equivalent circuit when the diode is in reverse bias.

ID

VD

Diode characteristics.

OFF ON


Series Clippers Positive clipper:

RL v0

+

-

vin

+

-

0 V

-5.3 V

0 V

6 V

Positive series clipper. Output voltage. Input voltage.

Series noise clipper:

Series noise clipper. Output voltage. Input voltage.

RL v0

+

-

vin

+

- 0 V

6 V 0.7 V

-0.7 V

Dead zone

0 V

5.3 V

• Unwanted lower level noise can be eliminated by a series clipper circuit.


Shunt Clippers

0 V

V1 V

Input voltage.

Positive shunt clipper:

0 V

V1 V

Input voltage.

Negative shunt clipper:

Positive shunt clipper.

RL v0

+

-

vin

+

-

R1 IL

Vγ

Negative shunt clipper.

RL v0

+

-

vin

+

-

R1 IL

Vγ

Output voltage.

(V1 – ILR1) V

0 V Vγ−

Output voltage.

-(V1 – ILR1) V

0 V

Vγ

• Diode is in reverse bias condition: shunt branch has no effect on the output voltage.


Biased Clipping Circuits Series clipper:

Input voltage.

RL v0

+

-

vin

+

-

Series clipper with negative shift.

VγBV

0 V

V1 V

Output voltage.

0 V

V1 V ( )1 BV V Vγ− +

( )in BV V Vγ= +

• Check the diode biasing condition for Vin = 0 V. • Replace the diode by its equivalent model. • Determine the value of Vin required to change the above biasing condition. • Calculate the output voltages for max./min. values of Vin. • Check for zero crossing (calculate the value of Vin for which the output is zero).

Solution steps:

No voltage values are specified: consider 1 .BV V Vγ> >


Biased Clipping Circuits

Input voltage.

RL v0

+

-

vin

+

-

Series clipper with positive shift.

VγBV

0 V

V1 V

Input voltage. Positive shunt clipper.

0 V

V1 V

• Consider ideal diode if the type of the diode (Si, Ge etc) or the cut-in voltage is unspecified.

• If unspecified consider R1 = 0.

R1

vin RL v0

+

-

+

-

IL

vB

Output voltage.

0 V

V1 V ( )in BV V Vγ= +

F.B.

R.B.

1 1LV I R−

Output voltage.

0 V

V1 V ( )1 BV V Vγ+ −

( )in BV V Vγ= −



RL v0

+

-

+

-

R1 IL

vB vin

Input voltage.

0 V

V1 V

Output voltage.

0 V

V1 V ( )in BV V Vγ= −

R.B. F.B.

Negative shunt clipper with positive shift.

RL v0

+

-

+

-

R1 IL

vB vin

Input voltage.

0 V

V1 V

Output voltage.

0 V

V1 V

in BV V Vγ= − +

R.B. F.B.

Positive shunt clipper with negative shift.



RL v0

+

-

+

-

R1 IL

vB vin

Input voltage.

0 V

V1 V

Output voltage. Negative shunt clipper.

Input voltage.

0 V

V1 V

Output voltage. Negative shunt clipper.

RL v0

+

-

+

-

R1

vB1 vin vB2

D1 D2 0 V

V1 V

( )2 2in BV V Vγ= − +

D1 in F.B.

D2 in F.B.

( )1 1γ= +in BV V V

0 V

V1 V

( )in BV V Vγ= − +

R.B.

F.B.


Clampers • A clamping circuit (dc restorer) shifts the entire signal by a dc value. • Change the dc voltage level but does not affect its shape. • Always use a capacitor.

Negative voltage clamping circuit.

0 V

V1 V

Input voltage.

v0

0 V

V1 V

-2V1 V

vin

v0

Output voltage (ideal diode).

Some important points: • Identify the charging and discharging path of the capacitor. • Diode in forward bias: input voltage appears across the capacitor (charging). • Diode in reverse bias: capacitor holds the voltage (discharging). • Charging time constant >>discharging time constant. • Total input swing = total output swing.

RL

+

-

vin

+

-

C

IL

Vγ

vC + -


Positive Voltage Clamper

Positive voltage clamping circuit. Input voltage. Output voltage.

v0 RL

+

-

vin

+

-

C

IL

Vγ

vC + -

0 V

12 V

-12 V

-0.7 V RL

+

-

11.3 V + -

-12 V 0.7 V

23.3 V

-0.7 V

During the negative cycle. During the positive cycle.

23.3 V RL

+

-

11.3 V + -

12 V

5Tτ ≥

• Consider capacitor charging step first.


Biased Clamping Circuits

v0 = 4.3 V

+

- 5 V

0.7 V -12 V

16.3 V - +

v0 = 28.3 V

+

- 5 V

0.7 V 12 V

16.3 V - +

Positive shunt clamper. Input voltage. Output voltage.

v0 RL = 10 kΩ

+

-

vin

+

-

C = 4.7 μF IL

Vγ

5 V

0 V

12 V

2 mS

-12 V

28.3 V

2 mS 4.3 V

Equivalent circuit in negative half-cycle.

Equivalent circuit in positive half-cycle.

• Charging time const. = 0, discharging time const. = 47 mS<< time period = 2 mS.


Biased Clamping Circuits

v0 = -4.3 V

+

- -5 V

0.7 V 12 V

16.3 V + -

v0 = -28.3 V

+

- -5 V -12 V

16.3 V + -

Negative shunt clamper. Input voltage. Output voltage.

v0 RL = 10 kΩ

+

-

vin

+

-

C = 4.7 μF IL

Vγ

-5 V

0 V

12 V

2 mS

-12 V

Equivalent circuit in positive half-cycle.

Equivalent circuit in negative half-cycle.

-28.3 V

2 mS

-4.3 V

Problems


Q1. (i) What type of filter is it? Calculate the cutoff frequency of the filter. (ii) The switch is closed at t = 0, calculate the output voltage at t = 0 and at t = 10 mS.

10 kΩ

+

-

vo 1 μF

12 V 1 kΩ

(i) Lowpass filter. Cutoff frequency = 175.1 Hz. (ii) At t = 0, the capacitor is shorted. Vo = 0 V. Now, time constant = Req×C = 0.909 mS. t ≥ 5τ Vo = 1.09 V.

909 Ω

+

-

vo 1 μF

10 kΩ

+

-

vo 1 μF 12 V

10 kΩ

• Calculate the time constant of the following circuit.

The output is open-circuited. time constant = R×C = 10 mS.

Problems


R

+

-

vc C C vi

+

-

Q2. In the circuit, the capacitors are fully charged at t = 0- so that vi = 12 V (C = 47 μF, and R = 1 kΩ). (i) The switch is closed at t = 0, calculate the time

when vc = 6 V. (ii) Calculate the minimum power rating of the resistor.

(i) Ceq = 94 μF.

ln

65.2 .

eqt RCC o

Ceq

o

v V ev

t RCV

t mS

−=

⇒ = −

⇒ =

(ii). ( )2

2

Power rating

12 11

144 .

peaki R

kk

mW

=

=

=

Problems


Q3. In the following circuit, a current source ii = sin(2πft) mA with internal resistance Ri = 10 kΩ is connected to a RC circuit. Calculate the output voltages (magnitudes) at f = 10 kHz and 100 kHz. Given that C = 2.2 nF, and RL = 10 kΩ.

Transform the current source into a voltage source vi. ( )

( )

( )

( )

( )

22 2

22 2

3

10

4

100

10sin 2 V.

Now,1

10 , 1.382, 7.643 0.47

4.7sin 20 10 V.

13.82100 , 0.4996 5sin 20 10 V.1 764.3

L

L

L

L

L

i i i

R L

i i L

RL i L

i

R kHz

RR kHz

i

v i R ft

v CRv C R R

vat kHz CR C R R

v

v t

vat kHz v t

v

π

ω

ω

ω ω

π

π

∴ = =

=+ +

∴ = + = ⇒ =

∴ = ×

∴ = = ⇒ = ×+

Ri

+

-

v0

C

RL

ii

Problems


Q4. In the following circuit, a pulse of height V and width a is applied at t = 0. Find an expression for the current.

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( )

0

.

1Now, applying KVL, 0

Taking Laplace transform,0

1

1 1Assuming 0 0,1 1 1

Taking inverse Laplace transfo

in

t

c

c as

as as

c

v t V U t U t a

v i dt Ri V U t U t aC

v I s VR I s es Cs s

e V ev I s VCCRs R s CR s CR

−

−

− −

− −

−

∴ = − −

+ + = − −

∴

+ + = −

−= = = − + + +

∴

∫

( ) ( ) ( ) ( )

rm,

.t a CRt CRVi t e U t e U t aR

− −− = − −

R

+

-

vc C V

Problems


Q5. The circuit is in steady state. The switch is closed at t = 0. Find an expression for the vc.

( )

( )

( ) ( ) ( )

( ) ( )

( ) ( )

0

20 .3

But, for >0, looking from the capacitor terminal,

the Thevenin's voltage = .2

1Now, applying KVL, 02 2

Taking Laplace transform,0

,2 2

2 ,2 3 2

32

c

t

c

c

v V

tV

R Vi v i dtC

v I sR VI ss Cs s

I sR V VI ss Cs sV R

I ss C

−

−

−

−

∴ =

+ + =

+ + =

⇒ + + =

⇒ = −+

∫

( ).

R

R +

-

vc C

V

R

R

R/2 +

-

vc C

V/2

Equivalent circuit for t >0.

Problems


( ) ( ) ( ) ( )( )

( ) ( )( ) ( )

( ) ( )( )

( ) ( )

( )( )

0

2

Capacitor voltage for >0, 0 32 ,

3 2

2 3.

22

Now, expanding into partial fractions,

2 322

2 3.

6

.2 6 2

Taking in

cc

s

s CR

c

tv I s V RCVV s

s Cs s s s CR

Vs V RC A Bs s CRs s CR

Vs V RC VAs CR

Vs V RC VBs

V VV ss s CR

−

=

=−

∴

= + = −+

+= = +

++

+= =

+

+= =

∴ = ++

∴

( ) 2

verse Laplace transform,

, (for 0).2 6

t CRc

V Vv t e t−= + >

R +

-

vc C

V

R

R

Problems


( ) ( ) ( ) ( )( )

( ) ( )( ) ( )

( ) ( )( )

( ) ( )

( )( )

0

2

Capacitor voltage for >0, 0 32 ,

3 2

2 3.

22

Now, expanding into partial fractions,

2 322

2 3.

6

.2 6 2

Taking in

cc

s

s CR

c

tv I s V RCVV s

s Cs s s s CR

Vs V RC A Bs s CRs s CR

Vs V RC VAs CR

Vs V RC VBs

V VV ss s CR

−

=

=−

∴

= + = −+

+= = +

++

+= =

+

+= =

∴ = ++

∴

( ) 2

verse Laplace transform,

, (for 0).2 6

t CRc

V Vv t e t−= + >

R +

-

vc C

V

R

R

Additional Questions


Q1. In the following circuit, charge on the capacitor is zero for t <0. R1 = 10 kΩ, R2 = 10 kΩ R3 = 1 kΩ C = 10 μF. (i) At t = 0 Sec, the switch is closed. Find I1 and I2 at t = 0 and at t = 1 Sec. (ii) The switch is reopened at t = 2 Sec. Find I1 and I2 at t = 2 Sec.

Answer:

( ) ( )( ) ( )1 1 2 3

2 3 2 3

(i) At 0, the capacitor isshorted.

0 9 || 0.825 mA. and

0 0.825 0.075 mA.

t

I R R R

I R R R

=

∴ = + =

= × + =

R1 +

- vc C

9 V R2

R3 I2

I1

( ) ( ) ( )1 2 1 2

= 60 1 Sec the capacitor is fully charged.

1Sec 1Sec 9 0.45 mA.eqR C mS

I I R R

τ = << ⇒

∴ = = + =

( ) ( ) ( )1 2 2 3

(ii) At 2 Sec, left-hand part is open.

2 Sec 0 and 2 Sec 4.5 0.409 mA.

t

I I R R

=

∴ = = + =



Q2. A Series RC circuit is excited by a voltage source of variable frequency. The output is taken across the R. Sketch the variation of the steady state transfer function with angular frequency ω.

Hints: Obtain H(jω) and represent in magnitude and phase form.

ω

0.707 1.0

M(ω)

1/RC

Magnitude response

ω

900

θ(ω)

1/RC

450

Phase response

Q3. Initially the switch is connected to a and the circuit is in steady state. The switch is moved from a to b at t = 0. Find the current in the inductor. What is the power dissipated in R at t = 0 and t = ∞?

R

iL L 6 V

a

b

Answer: 6 A, 06 A, 0

Rt Li R e tR t

−= − ≥= <

36 W,0 W.

RP R==



Q4. In the following circuit, the switch is closed at t = CR. Assuming that all currents and voltages are zero at t = 0-, determine the output voltage for 0≤ t ≤∞.

Answer:

Q5. In the following circuit, the switch is connected to a. At t = 0, it is moved to position b. Find an expression for the voltage v0 for t >0. Given that R1 = 2 kΩ, R2 = 1 kΩ, L1 = 2 mH, L2 = 1 mH.

Hints: ( )0 3 mA,i − =

( ) 6 3 20 3 10 , 0.tv t e t− −= − × >

v(t)

R +

- C V0 U(t)

C

( )

( ) ( )00

0 , 0 ,

1 exp 2 V,2

v t t CRV

V e t CR CR CR te

= ≤ <

= − + − − ≤ < ∞

R1

6 V

a b

R2 L1 L2 vo

+

- ( ) 00 0

1 2 20 0

Applying KCL at ,

1 10 0 0.t t

b

vi v dt v dt

L R L− −

−

+ + + + =

∫ ∫

Answer:



Q6. In the following circuit, obtain the complex transfer function and express them in magnitude and phase forms. Obtain the 3 dB cutoff frequency and the frequency at which the phase difference between the output and input voltage is 450. (5+5)

R

+

-

vo C

vin R

Q7. In the following circuit, the switch is connected to a and the circuit is in steady state. At t = 0, it is moved to position b. Find an expression for the voltage v0 for t >0. Given that R1 = 2 kΩ, R2 = 1 kΩ, L1 = 2 mH, L2 = 1 mH. (10)

R1

6 V

a b

R2 L1 L2 vo

+

-

0

+1

-1

t T/2 T 3T/2

f(t)

Q8. Express the following periodic function in terms of step functions and determine the corresponding Laplace transform. (5)

[email protected] Ph. – +91-3222-283550 (o) Department of E. & E.C.E. I.I.T. Kharagpur, 721302.

Thank you

?

109

Date post:	12-Apr-2017
Category:	Engineering
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