Sem3 all electronic asas lecturer note

Lecture 1

DC Circuits: Ohm’s Law

Series and parallel circuit

1

SSCP 2313 - Basic electronics2012/13 -1

Voltage

Review of V, I, and R

the amount of energy per charge available to

move electrons from one point to another in a

circuit and is measured in volts (V).

Current the rate of charge flow and is measured in

amperes (A)

Resistancethe opposition to current and is

measured in ohms ()

SSCP 2313- Basic electronics 2012/13-1 2

3

Ohm’s Law

Defines the relationship between voltage, current,

and resistance in an electric circuit

Ohm’s Law:

Current in a resistor is directly

proportional to the voltage across it and is

inversely proportional to the resistance.

Stated mathematically:

R

VI

V

I R

+ -

Where: I is the current (A)

V is the potential difference (V)

R is the resistance ()

V

I R

SSCP 2313- Basic electronics 2012/13-1

What is the voltage across a 680 resistor if the

current is 26.5 mA?

Example 1:

4SSCP 2313- Basic electronics 2012/13-1

5

Series Circuits

There is only a single path for current to flow.The amount of current is the same at all points.

R1

VSR2

R3

+

Total resistances in series:

Rtotal = R1 + R2 + R3...

V1 = IR1, V2 = IR2 and V3= IR3.

The voltage across each resistor:

Vs = V1 + V2 + V3

Total voltage:

= I (R1+ R2 + R3)

I

V1

SSCP 2313- Basic electronics 2012/13-1

SSCP 2313- Basic electronics 2012/13-1 6

Parallel Circuits

The voltage across every parallel component is equal.

The total current in the circuit is the sum of the currents in all the

branches.

I = I1 + I2 + ……. + In

nTotal RRRR

1......

111

21

RNR2R1Vs

IT

I3I2I1

Total resistances in parallel:V1 = V2 = Vs

Voltage drop in each resistor :

Special case for resistance of two

parallel resistors

R1 R2T

1 2

1

1 1R

R R

1 2

T

1 2

R RR

R R

or


PT = P1 + P2 + P3 +…..+ Pn

Or PT = Vs IT

thus

T

sT

R

VP

2

Power in series - Parallel circuits


House circuits contain parallel circuits

The parallel circuit will continue to operate even though one component may be open

Only the open or defective component will no longer continue to operate.

Advantages of parallel circuit


Example 2:

5 25

30 15

5 V 15 15

(Ans: Rs = 25 , Is = 0.2A)

Calculate: a) the total resistance.

b) curent flowing through the circuit.


Example 3: If Vs = 5 V, calculate: a) the total resistance.

b) curent flowing through the circuit.


Lecture 2

DC Circuits: Voltage Divider Rule

Current Divider Rule

1

SSP 2313- Basic electronics2012/13-1

The voltage drop across any given resistor in a series

circuit is equal to the ratio of that resistor to the total

resistance, multiplied by source voltage.

Voltage Divider Rule (VDR)

A potentiometer can act

as a variable voltage

divider, to control a

voltage.

I = Vin/ (R1 + R2)

V1 = I R1

21

1

RR

RVin


Example 1

The output voltage from the voltage divider is

a. 2.0 V

b. 4.0 V

c. 12 V

d. 20 V VS

VOUT

R2

R1

+

2.0 kW

10 kW

24 V


R1Vs

IT

I2I1

+

V1

-

+

V2

-

TR

R

21

21

RI T

R

R

21

12

RI

Current Divider Rule (CDR)

R2

2211 RIRIV

21 IIIT

21

21

RR

RRRT

21

21

RR

RRIRIV TTT

To determine how current entering a node is split between the various parallel resistors connected to the node.


Quiz

The current in R1 is

a. 6.7 mA

b. 13.3 mA

c. 20 mA

d. 26.7 mA

R1 R2

200 W100 WI = 20 mA

Example 2


Series voltage sources

Parallel current sources

Review


Voltage Sources in Parallel

Voltage sources with different potentials should never be connected in parallel

When two equal sources are connected in parallel

Each source supplies half the required current

If two unequal sources are connected

Large currents can occur and cause damage


Find: a) Voltage across R2

b) Current flows through R2.

Ans: VR2 = 3.3 V, IR2 = 0.335 mA

Example 3


If Vs = 5 V, calculate: a) voltage across AB

b) current flows through R6

Example 4


Ans: RT = 148.4 W IT = 0.034 A VAB= 1.63 V I6 = 0.02 A

Have a nice day….

Lecture 3

DC Circuits:

Kirchoff’s Law

1

SSP 2313- Basic electronics

2012/13-1

Kirchoff’s Law

Kirchhoff’s Voltage Law

(KVL)

Kirchhoff’s Current Law

(KCL)


Gustav Robert Kirchhoff

(12 March 1824 – 17 Oct

1887)

Kirchhoff’s Voltage Law (KVL)

VS = VR1 + VR2 + VR3

R1

I

VS

R3

VR1

VR3+-

VR2

+ - +-

R2

In any closed loop network, the total voltage source

around the loop is equal to the sum of all the voltage

drops within the same loop.


IRVs

The sum of the currents entering a junction is

equal to the sum of the currents leaving the

junction.

Kirchhoff’s Current Law (KCL)

I1

I2 I3

I1 = I2 + I3


4Ω

6Ω 12Ω

R1

R3

R2I1 I2

I3

V1 V2

36 V

72 V

Loop 1 Loop 2

Two loop and two-supply circuit

Referring to the circuit above, calculate:a) Current I1 , I2 , I3

b) Total power dissipated in the circuit.


KCL:

Current entering node A =

Current leaving node A

I1 + I2 = I3 (1)

KVL:

Loop 1: V1 = VR1 + VR3

36 = 6I1+ 4I3 (2)

Loop 2: V2 = VR2 + VR3

72 = 12I2 + 4I3 (3)

Substitute (1) into (2):

36 = 6 (I3 – I2) + 4I3

36 = 10 I3 - 6I2 (4)

(4) x -2 :

-72 = -20I3 + 12I2 (5)

(3)-(5):

144 = 24 I3

I3 = 6 ASubstitute I3 into (2) to get I1 and (2) to get I2.

4Ω

6Ω 12Ω

R1

R3

R2I1 I2

I3

V1 V2

36 V

72 V

A

Loop1 Loop 2

4Ω

6Ω 12Ω

R1

R3

R2I1 I2

I3

V1 V2

36 V

72 V

A

Loop 1 Loop 2

I1 = 2A and I2 = 4 A 66SSCP 2313- Basic electronics 2012/13-1

Power dissipated by each resistor

PR1 = I12 R1 = 24 W

PR2 = I22 R2 = 192 W

PR3 = I32R3 = 144 W

Total power dissipated

PT = PR1 + PR2 + PR3

= 360 W


Find the current flowing in the 40 Ω resistor, R3


I1= -1/7 A , I2 = 3/7 A, I3= 2/7 A

KCL:

)1(321 III

KVL:

IRVs

Loop 1:

)3(41

10)2(401010

31

31

II

II

Loop 2:

)5(31

20)4(6020

40)(20

402020

31

31

313

32

II

II

III

II

Solve eq (3) and (5)9SSCP 2313- Basic electronics 2012/13-1

Thank you…

Lecture 4

DC Circuits:

Thevenin’s Theorem

1

SSP 2313- Basic electronics

2012/13-1

M. Leon Thévenin (1857-1926),

published his famous theorem in 1883.


It is possible to simplify any linear circuit (network), no

matter how complex, to an equivalent circuit with just a

single voltage source (VTH) and a series resistance

(RTH) connected to a load.

Network •

•

A

B

VTH

RTH

A

B

+_

VTH: Thevenin’s voltage and

RTH : Thevenin’s resistance

Thevenin’s equivalent

circuit


Determination

of the

Thevenin’s

Voltage

VTH = Open circuit voltage

with load (RL) removed.

Determination

of the

Thevenin’s

Resistance

RTH = Total resistance in

open network with

sources were set to zero.

* If a voltage source - short circuit.

* If a current source - open circuit.


Find the voltage across

load resistor RL.

i) Disconnect RL to find VTH = VAB


V

RR

R

VV RAB

24

3621

2

2

ii) Short-circuit V to find RTH =RAB

iii) Draw the Thevenin’s equivalent circuit

2// 21 RRRTH


iv) Put back RL. Calculate the IL and VL.


ARR

VI

LTH

THL 6

Find the Thevenin’s equivalent circuit for:

i) VTH = Vab = 32 V

ii) RTH = Rab = 8


Lecture 5

DC Circuits:

Norton’s Theorem

1

SSCP 2313- Basic electronics2012/13-1

Edward Lawry Norton

1898-1983

Any two-terminal linear network can be replaced by

an equivalent circuit consisting of a current source (IN)

and a parallel resistor (RN).

Norton’s Theorem

IN: the output current when the output terminals or RL are

shorted.

RN: the total resistance between the two output terminals

when all sources have set to zero.

RNIN

Network •

•

A

B

Norton’s equivalent circuitRTH=RN and IN= VTH/RTH


3Ω

2Ω

10V

10Ω

RL

2Ω

Find Norton’s equivalent circuit and find the current that

passes through RL when RL = 1Ω

i) Short circuit RL to find IN = IAB

3Ω

2Ω

10V

10Ω

2Ω

IscIN

IN = 0.45 A


3Ω

2Ω 10Ω

2Ω

ii) Short-circuit V to find RN =RAB

RN = 13.2

RL13.20.45

iii) Draw the Norton’s equivalent circuit


iv) Put back RL. Calculate the IL and VL.

IL = 0.42 A


RL13.20.45 1Ω

Find the current in RL, if RL =2 k.


10 mA1 k

A

3 k

B

RL

5 k

IL= 0.833 mA

Maximum Power Transfer

The maximum amount of power will be dissipated by a

load resistance when that load resistance is equal to

the Thevenin’s/Norton’s resistance of the network

supplying the power.


RTH

VTH 12 V

10

10

WR

VP 6.3

10

622

The power delivered to

the matching load is

Lecture 6

DC Circuits:

Superposition’s Theorem

1


A circuit can be analyzed with only one source of

power at a time, the corresponding component

voltages and currents algebraically added to find

out what they'll do with all power sources in

effect.

Superposition Theorem

ii) Remove the rest:

Voltage – replace with short circuit

Current – replace with open circuit

Step to follow:

i) Pick one source at one time

iii) Sum the responses


R3V1 V2

100 W 20 W

10 W

15 V 13 V

R1 R2

Find the current flowing through R3 using

superposition theorem.

R1 R2

R3V1

100 W 20 W

10 W

15 VV2 shorted

REQ = 106.7 W, IT = 0.141 A and I’R3= 0.094 A


REQ = 29.09 W, IT = 0.447 A and I’’R3= 0.406 A

R1 R2

R3V2

100 W 20 W

10 W

13 V

V1 shorted


R1 R2

V1 V2

100 W 20 W15 V 13 V

Adding the currents gives IR3= 0.5 A

REQ ’ = 106.7 W, IT ’ = 0.141 A and I’R3

= 0.094 A

REQ’’ = 29.09 W, IT ’’ = 0.447 A and I’’

R3= 0.406 A

With V2 shorted

With V1 shorted

0.094 A 0.406 A

Therefore;


Using the superposition theorem, calculate the current in the R2.


I2’= 1 A , I2” = 1.2 A thus; I2= 0.2 A

With current source opened

With voltage source shorted


Thank you…

Lecture 7

•AC Current

•Phasor Diagram

•Capacitive Reactance

•Inductive Reactance

1


AC CURRENT

An alternating current has no direction in the sense that

direct current has. An “AC” circuit is one in which the

voltage and the current are sinusoidal in time.

Vp

Ip

time, t

V = Vp sin

=Vp sin (wt)

i = Ip sin (wt)

AC-voltage and current

Where : is angle in rad or degrees, w = 2π f (rad/s) and f =1/T (Hz).

V and i : the instantaneous voltage and current.

Vp and Ip : the peak current and voltage.

T


3

• The relationship between period and frequency is very useful to

know when displaying an AC voltage or current waveform on an

oscilloscope screen.

• By measuring the period of the wave on the horizontal axis of the

oscilloscope screen and reciprocating that time value (in

seconds), we can determine the frequency in Hertz.


4

• Other “waveforms” of AC produced within

electronic circuitry:


Peak-to-peak: is twice the peak value.Vpp = 2 Vp

5

AC voltage alternates in polarity and AC current alternates in

direction

One way to express magnitude (also called the amplitude), of

an AC quantity is to measure its peak height on a waveform

graph. This is known as the peak of an AC waveform.

• Another way is to measure the total height between opposite peaks.

This is known as the peak-to-peak (P-P) value of an AC waveform

Measurement of AC Magnitude


6

p

p

rms II

I 707.02

p

p

rms VV

V 707.02

Pave = Irms Vrms

RMS QUANTITIES IN AC CIRCUITS

The “RMS” stands for Root Mean Square, the

algorithm used to obtain the DC equivalent value from

points on a graph .

Sometimes the alternative terms equivalent or DC

equivalent are used instead of “RMS,” but the quantity

and principle are both the same.


7

Average Amplitude

The average value of a sine wave is defined over a half cycle rather than over a full cycle.

Vave = 0.637 Vp

Iave = 0.637 Ip

Average = 0 ; Peak = + / -

because all the positive points cancel out all the

negative points over a full cycle


8


9

PHASE SHIFT

Phase is always a relative measurement between two

waveforms


Phasor Diagrams

The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.Phasor diagrams are useful in solving complex AC circuits.

The sine wave can be represented as the projection of a

vector rotating at a constant rate. This rotating vector is

called a phasor.

450 900 1350

1800 2700 3600

V

Radius = Vp

v = Vp sin

10



I-V RELATIONSHIPS IN AC CIRCUITS: RESISTORS

V and I “In-phase”, there are

no phase difference, (f = 0).

V

wtp

I

2p

VR

~R

I

V

I

V

rms

rms

p

p

R is the Resistance, ()

VR12


I-V relationships in AC circuits: Inductor

L~Where XL is the Inductive Reactance, ()

V

wtp 2p

I

V and I “out of phase” by 90º.

The voltage leads the current by 90o.

(phase difference, f = +90).

L

rms

rms

p

pX

I

V

I

V

fLXL p2

where L is inductance (H).


V

~

C

I-V RELATIONSHIPS IN AC CIRCUITS: CAPACITOR

c

rms

rms

p

pX

I

V

I

V

fCXC

p2

1

V

wtp 2p

I

Where Xc is the Capacitive Reactance, ()

V and I “out of phase” by 90º.

The voltage lags the current by 90o

(phase difference, f = -90).

where C is capacitance (F).

IC

VC 14


The voltages across a resistor, an

inductor and a capacitor for the

same sinusoidal current.

15


Example:Consider a purely inductive circuit. The voltage across a 0.2 H

inductance is VL = 100 sin (400t + 70o)V. Determine iL and sketch it.

Solution:rad/s400w

Therefore,

8020400 .LXL w

AX

VI

L

mm 251

80

100.

The current lags voltage by 900, therefore

( )AtIL

020400251 sin.


17

Lecture 8


1

Complex Number Representation in Rectangular

Form and Trigonometry Form

A complex number W with magnitude M and direction

can be written as:

z = x + jy = r (cos + j sin ) Imaginary

real

r

z

0 x

jy

Complex Number

Rectangular

formTrigonometry

form

1j

j is the imaginary unit

The x-axis - real axis

The y-axis - imaginary axis


r

re

jyx)sinj(cosrz

j

The complex conjugate of a complex number, z = x + jy,

denoted by z* , is given by

z* = x – jy.

Polar

form


With siny r

x

y1tan

cos ,x r

r is called the absolute value or modulus or

magnitude of z and is denoted by |z|.

22 yxrz

)sin(cos jrz

Pz = x + iy

x

y

O

Im

Re

|z| =

r

θ


jyxz 22 yxr

)(tan 1

x

y

rej

x = r cos

z = rej atau r < x +jy

y = r sin

CONVERTING from Rectangular to Polar and Polar to Rectangular

Rectangular to Polar

Polar to Rectangular


Suppose we have 2 complex numbers, z1 and z2 given by :

2

1

2222

1111

j

j

erjyxz

erjyxz

2121

221121

yyjxx

jyxjyxzz

Easier with normal

form than polar form

Addition and Subtraction of Complex Numbers

2121

221121

yyjxx

jyxjyxzz


Example:

2. In figure below, if i2(t) = 2 cos( wt – 90) and I1 = 3 – j4,

calculate i3(t).

i1 i

2

i3

1. Calculate: (4 6i) + (3 + 7i)

= (4 + (-3) + i(-6 +7)

= 1 + i

Solution

From i2(t) = 2 cos( wt – 90), then

209022 jI

7.336.3

3

2tan23

23

)20()43(

122

213

j

jj

III

)7.33cos(6.3)(3w tti


)(

21

2121

21

21

j

jj

err

ererzz

magnitudes multiply! phases add!

Easier with polar form

than normal form

Multiplying


Example

45077

9010

.Z

and I

Solution:

45770

904507710

450779010

.

)().()(

).)((V

or

5050

55100

j

)j)(j(V

1. Find V, if given in a circuit


For a complex number z2 ≠ 0,

)(j

j

j

er

r

er

er

z

z21

2

1

2

1

2

1

2

1

magnitudes divide!

phases subtract!

Dividing

)( 21

2

1 r

r


Example

467130 .V

1535 .Z

)120(261.535

4.67130

Z

VI

5.12026

4.222.13

169

)200360()480150(

43

43

43

12050

j

j

j

j

j

j

Z

VI

Solution:

OR

12050 j

43 j


1. Convert i) z = 3 + j4 to polar form.

Exercise:

3)

j

ezii

to rectangular form.

2. If z1 = 3+ j5 dan z2 = 5 + j4 . Calculate:

i) z1 + z2 ii) z1z2 iii) z1z1*

)33

26()

65

43( :Calculate.3

j

j

j

j



RL AND RC SERIES CIRCUITS impedance,

Low-pass and high-pass filter.

Lecture 9


1

RC Series CircuitR

CVi

Total Impedance,= R –jXC

=

Magnitude impedance,Z =

Phase angle, =

R

XXR c

c

122 tan

22

cXR

R

XC1tan

R

XZ

V

Z

V

Ic

o

i

i

1tan

0

R

X

Z

V ci 1tan

Vi = Vi 0.

If input voltage, Vi has phase angle 0.

R

XCZ

The input voltage lags current by .

Impedance triangle:

VC

VR

Vi

Phasor diagram:


cOIZV

R

X

Z

V ci 1tan o

cX 90

R

X

Z

VXcoic 1tan90

Z

VX ic

R

Xtan90 c1o

To determine Vo:

I)If Vo is taken across capacitor, it is called low- pass filteror RC lag network.

The magnitude of the output voltage,Vo =

and the phase lag between the output & input voltage, =


Cut-off freq, fc- The freq at which Xc = R .fc=

RC2

1

Vo versus frequency for a low-pass R-C filter

As freq increases, Vout decreases low-pass filtering.


IRVO

ii) If Vo is taken across resistor, it is called high-pass filter

or RC lead network.

0tan 1 RR

X

Z

V ci

R

X

Z

RV ci 1tan

orZ

RVV i

o

R

Xtan c1

i

c

VXR

RV

220

where

and the phase lead, =

Vo increases with freq, because as Xc

becomes smaller, more of the Vi is

dropped across the resistor.5SSCP 2313- Basic electronics 2012/13-1

R

Vi

Example 1:Determine:i) the output voltage ii)the amount of phase lag from input to output in the lag network:iii)Draw the input and output waveforms showing the proper relationship.

Cf =1kHz

10 Vrms

680

0.1F

V0o

Z

VXVi ic

0)

Xc = 1/2fC = 1/2(1 kHz)(0.1 F) = 1.6 k

Z = R2 + Xc2 = 1.73 k

Vo = 9.2 Vrms

Solution:


iii) The waveforms:

ii) The phase lag,

2.23

tan90 1

R

Xco


Impedance,

= R +jXL

Magnitude impedance, Z =

Phase angle, =

R

XtanXR L

L122

2L

2 XR

R

Xtan L1

Vi

R

L

RL Series Circuit

Phasor diagram:

Impedance Triangle:

VR

VLVi

XL

Z

R

The input voltage leads the current by . 0ii VV

R

XtanZ

V

Z

V

IL

oi

i

1

0

R

Xtan

Z

V Li 1

If the input voltage Vi has phase angle 0.

=


IRVo

01 RR

Xtan

Z

V Li

R

Xtan

Z

RV Li 1

If Vo is taken across R is called low- pass filter

(phase lag)

R

VL

Vout

L

VoutVin

Vin

Vout

Vin


Lo IXV

901

LLi X

R

Xtan

Z

V

R

Xtan

Z

XV LoLi 190

If Vo is taken across L is called high- pass filter.

q

(phase lead)

R

VR

Vout

L VoutVin

Vin

Vout Vin


R

0.068 F

10 k

90 V

200 Hz

i) Determine I, Z, VC and VR

ii) Draw the waveforms for Vin, VR and Vc

(Ans: I= 5.84 mA, Z= 15.4 k,

VC = 68.3 V

40 V

50 kHz

R

30 mH

6.8 k

i)Determine I, Z, VL , VR.ii)Draw the waveforms for Vin, VR

and VL

(Ans: I= 3.44 mA, Z= 11.62 k,

VL = 32.4 V , VR = 23.4 V )

Problems



LECTURE 10

SERIES RCL•Impedance

•Series resonance•Quality factor1


THE SERIES RLC CIRCUIT

• This circuit acts as

both a low pass

and a high pass

filter at the same

time.

• It only allows

signal to pass from

a narrow range of

frequencies.

2 2SSCP 2313- Basic electronics 2012/13-1

R L C

VS

The total impedance for the RLC circuit is given

by

L CR jX jX Z

In polar form, this is written

22 1tan tot

L C

XR X X

R

Z

Impedance of series RLC circuits

X XL C

Z

R

“ Impedance Triangle”

3


Variation of XL and XC with frequency

In a series RLC circuit, the circuit can be capacitive or inductive,

depending on the frequency.

At the frequency where

XC=XL, the circuit is at series

resonance.

Below the resonant

frequency, the circuit is

predominantly capacitive.

Above the resonant

frequency, the circuit

is predominantly

inductive.

Rea

ctan

ce

f

XC XL

Series

resonance

XC=XL

XC>XL XL>XC


What is the total impedance and phase angle of the series

RLC circuit if R= 1.0 kW, XL = 2.0 kW, and XC = 5.0 kW?

VS

R L C

1.0 kW XC =

5.0 kW

XL =

2.0 kW

The circuit is capacitive, so I leads V by 71.6o.

The impedance is

WW

kjk

XXjRZ cL

)3(1

)(

3.16 kW

The total impedance is

2 2 2 21.0 k +3.0 ktot totZ R X W W

The phase angle is

1 1 3.0 ktan tan

1.0 k

totX

R W

W 71.6o

W 6.7116.3 kZ


What is the total impedance for the circuit below?

R L C

VS

330 mH

f = 400 kHz

470 W 2000 pF

786 53.3 W Z

The circuit is inductive.6


XL=XC

Notice that there is a frequency at which XC = XL.

This condition is called series resonance.

Series resonance

XL

f

XC

X

Series Resonance

1

2rf

LC

The resonant frequency:

At series resonance, XC and XL

cancel. VC and VL also cancel

because the voltages are equal

and opposite. The circuit is

purely resistive at resonance,

and Z = R..

- the total impedance is a

minimum.

Z = R


8

Resonance occurs at the frequency o where the current has its maximum value


What is the resonant frequency for the circuit?

R L C

VS

330 mH470 W 2000 pF

fr =196 kHz

9


• The sharpness of the resonance curve is usually described

by a dimensionless parameter known as the quality factor,

Q.

• For a series RLC circuit the quality factor is

– is the bandwidth of the curve, measured between the

two values of for which Pav has half its maximum value

• These points are called the half-power points

Quality Factor

10

C

L

R

R

Lf

R

LQ

s

s

1

2

0


The bandwidth (BW) of the

resonant circuit is the range of

frequencies for which the output

is 70.7% of the maximum current.

rfBWQ

RIPwherePPHPF

2

maxmaxmax2

1

f1 and f2 are referred to as the

critical frequencies, cutoff

frequencies or half-power

frequencies.



2

2:

2

1

12

fff

fffwhere

Q

ffffBW

r

r

r

• A high-Q circuit responds

only to a narrow range of

frequencies

– Narrow peak

• A low-Q circuit can detect

a much broader range of

frequencies

• Typical Q values in

electronics range from 10

to 100


a. For the series resonant circuit, find I, VR, VL, and VC at resonance.

b. What is the Q of the circuit?

c. If the resonant frequency is 5000Hz, find the bandwidth.

d. What is the power dissipated in the circuit at the half-power frequencies?


9050

9050

010

05.

VV

VV

VV

AIa

C

L

R

WPd

HzBWc

Qb

HPF

s

25.

1000.

5.

LECTURE 11

PARALLEL RCL•Conductance, Susceptance, Admittance

•Resonant frequency•Series-parallel RLC circuit1


1 1

Impedance of parallel RLC circuits

1 1 1 1

0 90 90L CR X X

Z

R L CVS

)11

(1

1111

LC

LC

XXj

R

jXjXRZ

Admittance, Y =

Conductance (G), Admittance (Y) and Susceptance (B)

1 1

0R

G

RConductance,

Capacitive Susceptance, 1

90C

CX

B

cjX

1

Inductive Susceptance, 1

90L

LX

B

LjX

1


R

10

Xc

5 Vi

10 V

XL

10

Determine the equivalent impedance for the circuit :

)XX

(jRZ

YLC

1111

)(j10

1

5

1

10

1

= 0.1 + j 0.1

= 0.141 45

45 0.141Z

1

= 7.1 - 45

IT = V Y = 10 (0.141 45)

= 1.41 A 45 ,

it lags the input voltage by 45.


Draw the admittance phasor diagram for the circuit.

1 11.0 mS

1.0 kG

R

1

0.629 mS2 10 kHz 25.3 mH

LB

2 22 2 + 1.0 mS + 0.629 mS 1.18 mSLY G B

VSL

25.3 mHf = 10 kHz

R

1.0 k

Y =

1.18 mS

G = 1.0 mS

BL =

0.629 mS


For parallel RLC circuits, the current phasors can be

obtained directly from Ohm’s law.

SR

VI

R

SC

C

V

IX

SL

L

V

IX

and that IR is plotted along the positive real

axis.

and that IC is plotted along the positive j axis.

and that IL is plotted along the negative j

axis.

+90o

90o

IC

IR

IL

The total current is given by:

22 1tan CL

tot R C L

R

II I I

I

I


Parallel resonance

Ideally, at parallel resonance, IC and IL cancel because the

currents are equal and opposite. The circuit is purely

resistive at resonance.

The resonant frequency in both parallel and

series circuits is the same

1

2rf

LC (ideal case)

• Capacitive and inductive susceptance

are equal.

• Total impedance is a maximum

(ideally infinite).

• The current is minimum.


f

Zmax

0.707Zmax

f1 fr f2

BW

Ztot

rfBWQ

Series-parallel RLC circuits

VS =R L

C

200 kHz

700 mH1.0 k

2700 pFCalculate the total

impedance of the circuit.

The impedance of the yellow box,


k2.0j43.0

49k66.0Z

L

L1

X

1j

R

1

jX

1

R

1

Z

1Y The total impedance , ZT

c1 jXZ

9.24k47.0

8

LECTURE 12

Power in AC Circuit•Real, Reactive and Apparent Power

•Power Factor•Maximum Power Transfer Theorem1



Power in Resistive Components

• Suppose a voltage v = Vp sin t is applied across a

resistance R. The resultant current i will be

• The power p :

• The average value of (1 - cos 2t) is 1, so

tIR

tV

R

vi P

P

sinsin

)2

2cos1()(sinsinsin 2 t

IVtIVtItVvip PPPPPP

rmsrmsPP

PP IVIV

IVP 222

1Power Average


Relationship between v, i and p in a resistor


Power in Capacitors

• In capacitors, the current leads the voltage by 90.

Therefore, if a voltage v = Vp sin t is applied across a

capacitance C, the current will be given by i = Ip cos t.

• Then

• The average power is zero

)2

2sin(

)cos(sin

cossin

tIV

ttIV

tItV

vip

PP

PP

PP


Relationship between v, i and p in a capacitor


Power in Inductors

• In inductors, the current lags the voltage by 90.

Therefore, if a voltage v = Vp sin t is applied across an

inductance L, the current will be given by i = -Ip cos t

• Therefore

• Again the average power is zero

)2

2sin(

)cos(sin

cossin

tIV

ttIV

tItV

vip

PP

PP

PP


Relationship between v, i and p in an inductor


Circuit with Resistance and Reactance

• When a sinusoidal voltage v = Vp sin t is applied

across a circuit with resistance and reactance, the

current i = Ip sin (t - ).

• Therefore, the instantaneous power, p is

)2cos(2

1cos

2

1

)2cos(cos2

1

)sin(sin

tIVIVp

tIV

tItV

vip

PPPP

PP

PP


• The expression for p has two components

• The second part oscillates at 2 and has an average

value of zero over a complete cycle

– this is the power that is stored in the reactive

elements and then returned to the circuit within each

cycle

• The first part represents the power dissipated in resistive

components. Average power dissipation is

)2cos(2

1cos

2

1 tIVIVp PPPP

cos)(cos22

)(cos2

1VI

IVIVP PPPP


• The average power dissipation

is termed the real power in units watts (W).

• The product of the r.m.s. voltage and current VI is termed

the apparent power, S in units volt amperes (VA).

cos)(cos2

1VIIVP PP

cos

cos

S

VIP

• This cosine is referred to as the power factor

cosfactor Power S

P


Active and Reactive Power

• When a circuit has resistive and reactive parts, the

resultant power has 2 parts:

– The first is dissipated in the resistive element. This is

the real power, P

– The second is stored and returned by the reactive

element. This is the reactive power, Q , which has

units of volt amperes reactive (VAR).

Reactive power (inductive)

QL = VI = I 2 XL (VAR)

Reactive power (capacitive)

QC = VI = I 2 XC (VAR)


Power triangle

S = VI

S = VI

S = P – jQC

S = P + jQL


Real Power P = I2R = VI cos watts

Reactive Power Q = VI sin var

Apparent Power S = VI VA

S 2 = P 2 + Q 2

Note: V and I are in rms value

)(2CL XXIQ

Summary


Maximum power transfer theorem

• When the output of a circuit has a reactive element

maximum power transfer is achieved when the load

impedance is equal to the complex conjugate of the

output impedance.


Determine the total PT

and QT for the circuit.

Sketch the power triangle.


16

1

Analysis of Series RCL Circuit :

Kirchhoff’s Laws

Lecture 13

1


2

Kirchhoff’s Laws in Complex FormKirchhoff’s Current Law (KCL):

At any junction in a circuit with impedances, the total complex

current is zero. 0I

Kirchhoff’s Voltage Law (KVL):

For any closed loop with impedances, the total complex

e.m.f. is equal to the total potential differences across the

circuit components in the loop.

IZV

Based on Kirchhoff’s law:

kVVVV ....0 21


3

Z1

Z2

Z3

V1

V2

V3

V• From KVL:

eqZI

)ZZZ(I

VVVV

321

321

where Zeq is the equivalent impedance. Thus

321 ZZZZ eq

Analysis of Series RCL Circuit Using Kirchhoff’s Laws


Complex Power

*IVS, PowerComplex

jQP


5

For the circuit shown below, calculate:

(a) the impedances Z1, Z2 dan Z3

(b) the total impedance of the circuit

(c) the current I

(d) the voltage V across Z1

(e) the power dissipated in Z1

Z1 Z2

Z3

0.1 H 10 5 10-3

F

0.2 H

5

V = 28 V

f = 50/ Hz

Example 1:


6

(a) The impedances Z1, Z2 dan Z3 are

4514

1010

)1.0)(100(10

111

j

j

LjRZ

4.6311

105

10100

15

1

3

22

j

j

CjRZ

7621

205

)2.0)(100(5

333

j

j

LjRZ

Solution:


(b) The total impedance is

453.28

2020

)205()105()1010(

ZZZZ 321

j

jjj

eq

(c) The current I is

451453.28

028

Z

VI

eq

A

The current I is 1A rms and lagged behind the voltage by 45.

(d) The voltage across Z1 is

0V41)4514)(451(ZIV 111

The voltage across Z1 is 14V in phase with the source voltage.


8

(e) The dissipated power on Z1 is

1010

4514

)451014(

11

j

)(

IVS

The apparent power is 14 VA,

The real power is 10 W and

The reactive power is 10 VAR


Analysis of Parallel RCL Circuit Using Kirchhoff’s Laws

Z1 Z2 Z3

I1 I2 I3

IT

V

321 IIII • From KCL:

321

111

ZZZV=

eqZ

V=

321

1111

ZZZZeq

where Zeq is the equivalent impedance. Thus

I

200 V5F

400 f= 1000/ Hz

I1 I2

Example 2:

For the circuit shown below, calculate:

(a) the total impedance of the circuit

(b) the current I1, I2 dan I.

(c) the total power dissipated in the circuit

0.15 H

Cj

CjZ

111

Solution:

= -j100 = 100 -90o

LjRZ 2 = 400 + j300 = 500 37o

21

111

ZZZeq

(a)

300400

1

100

1

jj

Zeq = 112 -79.6o

o

o

o

Z

V

I 90290100

0200

11

A,(b)

AZ

V

I o

o

o

374.037500

0200

22

AZ

V

I o

o

o

6.7979.16.79112

0200

*IVS

= 200 0o (1.79 -79.6o )= 358 -79.6o

= 64.6 –j352

(c)

S =358 VA, P = 64.6 W andQ = 352 VAR (capacitive).


Lecture 14

Analysis of RCL Circuit :

Mesh Current


2

RCL circuit analysis using: Mesh Current

1Z

2Z

3Z

aI cI

bI

1V 2V

• We can use Kirchhoff Laws to obtain three equations

and the values of Ia, Ib and Ic can be solved.

• This problem can be simplified by using mesh current

analysis method


3

Loop 1:

1Z

2Z

3Z

1I 2I1V 2V

22211

221111

)(

)(

ZIZZI

ZIIZIV

Loop 2:

21322

221322

)(

)(

ZIZZI

ZIIZIV

• Only two equations required to solve the values of I1 and I2 , to obtain the values of Ia, Ib and Ic .


Cramer’s Rule:

a11 x1 + a12 x2 = k1

a21 x1 + a22 x2 = k2

Can be written in matrix forms:

a11 a12 x1 = k1a21 a22 x2 = k2

x1 , x2 and x3 can be found by :

x1 = k1 a12k2 a22

a


x2 = a11 k1a21 k2

a

where a = a11 a12

a21 a22


Referring to the figure below, calculate the values of Ia, Iband Ic

2 - j5 j6

4 010 305

Ia

Ib

Ic


Example 1

Solution

The circuit can be redrawn as shown below:

621 jZ += 53 jZ -=

42 =Z

o

010

o305

1I 2I

Loop 1:

21

21

211

4)66(

4)642(

4)()62(010

II

II

III

j

j

j


Loop 2:

12

212

4)54(

4)()5(305

II

III

j

j

The two simultaneous equations can be solved using

Cramer rules to find I1 and I2.

305

010

544

466

2

1

I

I

j

j

95.38

638

(4)(4)-j5)-j6)(4-(6

544

466

j

j

j


9

Thus,

3.60A67.1

95.38

60j7.22

95.38

)305)(4()5j4)(010(

95.38

5j4305

4010

I1


10

Therefore:

45A3.1

95.38

7.533.50

95.38

)010)(4()305)(6j6(

95.38

3054

0106j6

I2

1.53A95.2III

45A3.1II

3.60A67.1II

21b

2c

1a



Thank you…

Lecture 15

RCL circuit analysis using:



Any circuit having two output terminals A and B, can be

replaced by an equivalent Thevenin’s circuit having a

Thevenin’s voltage source VTH in series with a Thevenin’s

impedance, ZTH.

• Thevenin’s output voltage, VTH is the open circuit

voltage measured at terminals A and B.

• Thevenin’s equivalent impedance, ZTH is the impedance

measured at the terminals A and B where the voltage

source is shorted and the current source is opened.

A

B

CIRCUIT

A

B

VTH

zTH

RCL circuit analysis : Thevenin’s Theorem


• If a load resistor, RL is connected at the output terminals A

and B, the current in RL :

LT

TL

RZ

VII

A

B

RL

ZT IL

VT

I


Example 1:

Determine the Thevenin’s equivalent voltage VTH and the

Thevenin’s equivalent impedance ZTH for the circuit below:

010

16j 10

10

9010

4j

5jA'

B'

A

B

I


Solution:

VT = VAB = VA’B’

Loop 1,

ZIE

A 5.0

20j20

10j10

20j20

9010010I

)4j1016j10(I9010010

V 4.632.11

10j5

j10-(10)0.5

9010)10(IVT


The value of ZTH can be calculated by shorting all the

voltage sources:

16j 10

10

4j

5jA'

B'

A

B

6.188.7

5.2j4.7

5j1020j10

)20j10(10

5j]10||)20j10[(ZTH


If the terminals A and B are connected to an impedance ,

ZL = 10 + j10, what is the current flowing in ZL ?

.

A 7.8659.0

5.7j4.17

4.632.11

)10j10()5.2j4.7(

4.632.11

RZ

VII

LTH

THL

A

B

VTH

zTH

ZL= 10 +j10

Draw the Thevenin’s equivalent circuit:

By using Kirchhoff’s voltage law,


-j2

3

5 0o

j4

ZL= 5

Find current in ZL using Thevenin Theorem.

Example 2:

Ans:( IL = 1.07 43.5o )



Lecture 16

RCL circuit analysis using:

Norton’s Theorem


• Any circuit having two output terminals A and B, can be

replaced by an equivalent Norton’s circuit having a

Norton’s current source IN in parallel with a Norton’s

impedance, ZN, :

A

B

CIRCUIT

A

B

ZN Z

LIN

RCL circuit analysis : Norton’s Theorem

NLN

NL I

ZZ

ZI


Example 1 :

For the circuit shown below, calculate IN , ZN and IL

9050

5

3

4j

5jA

B

050


• The Norton’s current source, IN is the current which flows

through the short circuit at the terminals A and B.

• The Norton’s impedance, ZN is the equivalent impedance

between terminals A and B when the current source is

opened and the voltage source is shorted.

To calculate ZN, terminals A and B are opened and

all sources are shorted:5 5jA

B

ZN = ((5 + j0) // (0 + j5)

=3.53 45


Solution:

Loop 1:

9050

5

050

5jA

B

Loop 1 Loop 2

I1 I2

IN

90A10

5

9050I

)5)(I(9050

1

1


To calculate IN, terminals A and B are shorted.

Loop 2 :

90A10

905

050

5j

050I

)5j)(I(050

2

2

90A20

20j

10j10j

90109010

III 21N


To calculate IL, draw the Norton equivalent circuit:

A

B

IN

9020

ZN

4553.3 Z

L

Using the current divider rule, the load current IL

2.58A3.8

)4j3(5453.3

5453.39020

ZZ

ZII

LN

NNL


-j2

3

5 0o

j4

ZL= 5

Find current in ZL using Norton’s Theorem.

Example 2:

Ans:( IL = 1.07 A 43.5o )


Semiconductor

LECTURE 17


• Atom is the smallest particle for an element which

maintain the characteristics of the element.

• Each atom has an orbital type structure consisting of

a nucleus surrounded by orbiting electrons.

• The nucleus consists of protons and neutrons.

• Electrons are negatively charged particles and

protons are positively charged particles

• Neutrons are particles with no charge.

• The number of electrons in an electrically neutral

atom of a material gives the atomic number for that

material

Atomic Structure


• The goal of electronic materials is to generate andcontrol the flow of an electrical current.

• Electronic materials include:

1. Conductors: have low resistance which allowselectrical current flow

2. Insulators: have high resistance whichsuppresses electrical current flow

3. Semiconductors: can allow or suppresselectrical current flow

Electronic Materials


• Good conductors have low resistance so

electrons flow through them with ease.

• Best conductors include:

– Copper, silver, gold, aluminum, & nickel

• Alloys are also good conductors:

– Brass & steel

• Good conductors can also be liquid:

– Salt water

Conductors


• The atomic structure of good

conductors usually includes

only one electron in their

outer shell.

– It is called a valence

electron.

– It is easily striped from

the atom, producing

current flow.

Conductor Atomic Structure

Copper Atom


• Insulators have a high resistance so current

does not flow in them.

• Good insulators include:

– Glass, ceramic, plastics, & wood

• Most insulators are compounds of several

elements.

• The atoms are tightly bound to one another so

electrons are difficult to strip away for current

flow.

Insulators


• Semiconductors are materials that essentially

can be conditioned to act as good conductors, or

good insulators, or any thing in between.

• Common elements : carbon (C), silicon (Si),

and germanium (Ge)

• Silicon is the best and most widely used

semiconductor.

Semiconductors


The name “semiconductor” implies that it conducts

somewhere between the two cases (conductors or

insulators)

Conductivity :

σmetals ~1010 /Ω-cm

σinsulators ~ 10-22 /Ω-cm

The conductivity (σ) of asemiconductor (S/C) lies between these two extreme cases.

S/C


Semiconductor Valence Orbit

• The main characteristic of a semiconductor element is

that it has four electrons in its outer or valence orbit.


• The unique capability of semiconductor atoms is theirability to link together to form a physical structure calleda crystal lattice.

• The atoms link together with one another sharing theirouter electrons.

• These links are called covalent bonds.

Crystal Lattice Structure


Intrinsic Semiconductor

Pure (good insulating) semiconductor material, i.e.

undopped semiconductor


Since the outer valence electrons of each atom aretightly bound together with one another, theelectrons are difficult to dislodge for current flow.

Silicon is a great insulator.

Types of Semiconductor Materials

Doping forms two types of semiconductor:

i. N type :-The Si doped with extra e (donor).

-“N” is for -ve, which is the charge of

an electron.

ii. P type” :- Si doped with material missing e that

produce holes (acceptors).

- “P” is for +ve, which is the charge

of a hole

Extrinsic Semiconductor Dopped semiconductor materials

To make the semiconductor conduct electricity, otheratoms called impurities must be added.This processis called doping.


An impurity, like arsenic or phosphorous, has 5 valenceelectrons.

Adding arsenic (doping) will allow four of the As valenceelectrons to bond with the neighboring Si atoms.

The one electron left over for each As atom becomesavailable to conduct current flow.

The free electrons are the majority carriers.

N type


P type

An impurity, like Boron, has 3 valence electrons.

The 3 e in the outer orbit form covalent bonds with its

neighboring semiconductor atoms. But one e is missing from

the bond.

This place where a fourth electron, is referred to as a hole.

The hole assumes a + charge so it can attract e from some

other source. Thus, holes are the majority carriers to

support current flow.


http://en.wikipedia.org/wiki/Majority_carrier




• If you use lots of arsenic atoms for doping, there

will be lots of extra electrons so the resistance of

the material will be low and current will flow freely.

• If you use only a few boron atoms, there will be

fewer free electrons so the resistance will be high

and less current will flow.

• By controlling the doping amount, any resistance

can be achieved.

Resistance Effects of Doping


Current Flow in N-type

Semiconductors

• The battery has a + terminal

that attracts the free electrons in

the semiconductor and pulls

them away from their atoms

leaving the atoms charged

positively.

• Electrons from the - terminal of

the supply enter the

semiconductor material and are

attracted by the + charge of the

atoms missing one of their

electrons.

• Current (electrons) flows from

the + terminal to the – terminal.


Current Flow in P-type

Semiconductors

• Electrons from the - battery

are attracted to the positive

holes and fill them.

• The + terminal of the

supply pulls the electrons

from the holes leaving the

holes to attract more

electrons.

• Current (electrons) flows

from the - terminal to the +

terminal.


Thank You…

LECTURE 18PN JUNCTION- Diode


• A diode is made by joining p and n-type semiconductor

materials

• Diodes are unidirectional devices that allow current to

flow in one direction.

Structure of PN junction diode

Symbol of PN junction diode


The PN Junction Diode

The Open-Circuited p-n JunctionA p-n junction is the junction between an n-type

semiconductor and a p-type semiconductor.

Near the junction, electrons diffuse across to combine

with holes, creating a "depletion region".


http://hyperphysics.phy-astr.gsu.edu/hbase/solids/pnjun.html

In the p-type region there are holes from the

acceptor impurities and in the n-type region there

are extra electrons.

When a p-n junction is formed, some of the

electrons from the n-region which have reached

the conduction band are free to diffuse across the

junction and combine with holes.

Filling a hole makes a - ion and leaves behind a +

ion on the n-side. A space charge builds up,

creating a depletion region which inhibits any

further electron transfer unless it is helped by

putting a forward bias on the junction.

Depletion Region


The Biased p-n Junction: A. Forward Bias

Because + charge is repelled by a + voltage and - charge is

repelled by a - voltage, both the free electrons in the n-type and the

holes in the p-type are forced toward the junction.

This causes the width of the depletion region to decrease.

The potential across the junction which opposes diffusion is

decreased by the applied bias to the value VB − VS, where VB is the

built-in potential and Vs is the biased voltage.

This caused the diffusion current

increases rapidly as VS is increased.

Because a large current flows, the

junction is said to be forward biased.

current is permitted


B. Reverse Bias

Because - charge is attracted by a + voltage and + charge is

attracted by a - voltage, both the free electrons in the n-type

and the holes in the p-type are pulled away from the junction.

This causes the width of the depletion region to increase.

The potential across the junction which opposes diffusion is

increased by the applied bias to the value VB + VS.

This causes a very small

current to flow, thus the

junction is said to be reverse

biased.

so current is prevented


Forward Biased diode

The diode behaves like a ‘ON’ switch in this mode. A

conduction diode has approximately a constant voltage

drop across it. It’s called turn-on voltage.

Resistance R and diode’s body resistance limits the

current through the diode

Vs has to overcome VB in order for the diode to

conduct.

VV

VV

onD

onD

25.0

7.0

)(

)(

For silicon

For germanium


Reverse Biased diode

The diode behaves like a ‘OFF’ switch in this

mode.

If we continue to increase reverse voltage Vs

breakdown voltage of the diode is reached.

Once breakdown voltage is reached, diode

conducts heavily causing its destruction.


Breakdown diode

Diode breakdown is caused by thermally generated

electrons in the depletion region.

When the reverse voltage across diode reaches

breakdown voltage, these electrons will get sufficient

energy to collide and dislodge other electrons.

The number of high energy electrons increases

rapidly, leading to an avalanche effect causing heavy

current and ultimately destruction of diode.


In the forward bias

region, current increases

rapidly after the barrier

potential (0.7 V for Si) is

reached.

The voltage across the

diode remains equal to the

barrier potential.

The reverse-biased diode

effectively acts as an

insulator until breakdown is

reached.

I-V Characteristics Curve

VR VF

IF

IRReverse voltage

Forward

voltage

0.7 V

Barrier potential

VBR (breakdown voltage)


Leakage current

Avalanche current

Diode models

The ideal model assumes the diode is either an open or closed switch.

VR

VF

IF

IR

Reverse

bias

Forward

bias

The complete model includes the forward resistance of the diode.

The practical model includesthe barrier voltage in theapproximation.

0.7 V

The diode is designed to allow current to flow in only one

direction. The ideal diode would be a perfect conductor in one

direction (forward bias) and a perfect insulator in the other

direction (reverse bias). In many situations, using the ideal

diode approximation is acceptable.


Thank you….

Chapter 19

i) Rectifier Circuits


• Power supply is a group of circuits that convert ac

energy to dc energy.

• Two types:

– linear power supply: provides constant current path

between its input and its load.

- switching power supply: provides intermittent current

path between its input and its output.

Introduction


Linear Power Supply

Basic components:1. Rectifier – diode circuit that converts the ac to what is called

a pulsating dc.2. Filter – circuit that reduces the variations in the output of

the rectifier.3. Voltage regulator – maintain a constant power supply output

voltage.


2 types:

Half-wave rectifier

Full-wave rectifier

Rectifier


Half-wave Rectifier


The effect of the barrier potential on the half-wave rectified output voltage


Average and RMS value of HWR

π

VV P

DC 2

VV P

rms

VDC

Vrms

t0

t1

t2

t3

Vp

VO

t



A center-tapped full-wave rectifier.


Basic operation of a center-tapped full-wave rectifier.


Center-tapped full-wave rectifier with a

transformer turns ratio of 1.

Vp(pri) is the peak value of the primary voltage.


Center-tapped full-wave rectifier with a

transformer turns ratio of 2.


Full - wave Bridge Rectifier

can buy pre-packaged diode bridges


Full-wave Rectifier - operation

π

V2V P

DC

2

VV P

rms



The End


Lecture 20

Capacitive filter in power supply


Transformer Rectifier FilterInput Voltage

240 V 50 Hz

DC Output

Voltage

(Unregulated)

Unregulated Power Supply

A simplest power supply consists of a transformer,

a rectifier and a filter.

In a power supply, a transformer is required to step-up or

step-down mains voltage source of 240 Vrms at 50 Hz to a

suitable voltage.

The function of a rectifier is to convert the two way current

from the transformer to a one way current


A

C

B

D

AC source

load

capacitor

An RC circuit has a time constant = RC

The function of a filter is to remove ripples on the rectifier

output and create a power supply with a constant DC

value.

The simplest filter consists of a large capacitor which is

connected in parallel with a load resistor RL .


Filter

Vr = ripple voltage


How the Filter Works?

When the diode conducts, thecapacitor charges up to thepeak of the sine wave.

When the sine voltage drops,the charge on the capacitorremains. Since the capacitor islarge it forms a long timeconstant with the load resistor.The capacitor slowlydischarges into the loadmaintaining a more constantoutput.

The next positive pulse comesalong, recharging the capacitorand the process continues.


Large capacitor capable of storing charge in longer time, and will not

discharge too much during the time between t1 and t2.

If RL is small, the capacitor will discharge faster because the

discharge current through RL is large.

- the more charge will be removed from the capacitor and the larger

will be the ripple

Effect of RL and CF on the ripple


Filter Output Voltage

Vpk : peak rectifier output voltage

Vdc : average (or dc) value

Vr : ripple

Example:

Assuming the line frequency is 60 Hz, the time between charging peak for half-wave rectifier,

t = 1/f = 1/60 = 16.7 ms

For full-wave rectifier, the frequency is 2x60 = 120 Hz. Therefore,

t = 1/120 = 8.33 ms


Ripple Voltage

)fw(RCf2

V

C

tIV

L

LL)pp(r

where IL = the dc load current

t = the time between charging peaks

C = the capacitance (in Farads)


DC Output Voltage with Capacitor Filter

For Full-wave:

Cf4

IV

VVV

Lp

)p(rpDC

For Half-wave:

Cf2

IVV L

pDC

Vr and VDC determine the ripple factor,r

Vp(rect) : unfiltered peak rectified voltage

Vr(pp) : peak to peak ripple voltage

VDC : average value of ripple voltage

rhwr Vr( pp)

VDC

1

2 3 fRLC

rfwr Vr(pp)

VDC

1

4 3 fRLC


The full-wave rectifier has approximately half the ripple output

produced by the half-wave rectifier. This is due to the shortened

time period between capacitor charging pulses. Therefore, full-wave

rectifier are typically used in power supplies.11SSCP 2313- Basic electronics 2012/13-1

Regulated Power Supply

Unregulated power supply gives an output voltage which

is not constant at the value that has been specified.

The variation of the output voltage can be caused by

several factors.

1. Variation of Load Current

The current flows in the load varies when the load

resistance is varied.

2. Variation of Input Voltage

The primary voltage for the transformer is normally

taken from mains supply.

-This voltage varies between 220 V to 260 V

depending on the usage of electric power in an

area.


Although the variation of the output voltage is small, it can

cause problem to some electrical equipments:

- test instruments and computer circuits.

To overcome this problem a voltage regulator circuit is

added to the output of the unregulated power supply.

- to maintain the output voltage of the power supply at

the required value despite of the variation of the input

voltage or the variation of load current.

Transformer Rectifier Filter

input voltage

240 V 50 Hz

unregulated output

voltage

regulated output

voltage

Voltage

Regulator


Parameters for a Regulated Power Supply:

Load Regulation

100% V

VVRegulation Load %

load) (full o

load) (full oload)(without o

Line Regulation ( Source Regulation)

ILILIH

OLOLOH

V/)VV(

V/)VV(

variationageinput volt %

variation voltageload %Regulation Line


Determine the peak to peak ripple voltage and the ripple

factor for a bridge rectifier using capacitive filter of 100 F.

The load resistance is 2 k and the voltage across load is

12 V. Assume supply freq to be 50 Hz and ideal diodes.

Answer:

%44.1r,V6.0V )pp(r



LECTURE 21

Application of diodes:

ii) Voltage Regulator

1


Many of the fixed voltage regulator ICs have 3 leads and look

like power transistors. Ex:

7805 (+5V) , 7905 (-5 V) 7812 (+12 V), 7912 (-12V).

(a) Voltage Regulator IC


For low current power supplies, a

simple voltage regulator can be made

with a resistor and a zener diode

connected in reverse.

Zener diodes are rated by their

breakdown voltage Vz and maximum

power Pz (typically 400mW or 1.3W).

The resistor limits the current. The

current through the resistor is constant,

so when there is no output current all

the current flows through the zener

diode and its power rating Pz must be

large enough to withstand this.

(b) Zener diode regulator


4

Circuit Symbol :

• When forward-biased, zener diode behave as standard

rectifying diodes: it has a forward voltage drop about 0.7 V.

• In reverse-bias mode, it does not conduct until the applied

voltage reaches the zener voltage- at which point the diode is

able to conduct substantial current.

• As the power dissipated by this reverse current does not

exceed the diode's thermal limits, the diode will not be

harmed.

Zener Diode


Choosing a zener diode

1.The zener voltage Vz is the output voltage required.

2.The input voltage Vs must be a few volts greater than Vz

(this is to allow for small fluctuations in Vs due to ripple) .

3.The maximum current Imax is the output current required

plus 10% .

4.The zener power Pz is determined by the maximum

current: Pz > Vz × Imax or P = I2 Z


Example:

6

A 1N754A Zener diode has a dc power dissipation rating of 500 mW and a nominal Zener voltage of 6.8 V. What is the value of IZM for the device?

(max) 500mW73.5mA

6.8V

D

ZM

Z

PI

V

Solution:


Zener Diodes – Operating Range

A zener diode is much

like a normal diode, the

exception being is that it

is placed in the circuit in

reverse bias and

operates in reverse

breakdown.

Note:

Its forward characteristics

are just like a normal

diode.Operating range


Zener Diodes – Breakdown Characteristics

Note very small reverse current (before “knee”).

Breakdown occurs @ knee.Breakdown Characteristics:• VZ remains near constant• VZ provides:

-Reference voltage-Voltage regulation

• IZ escalates rapidly• IZMAX is achieved quickly• Exceeding IZMAX is fatal


Zener Diodes – Voltage Regulation

Regulation occurs between:

VZK - knee voltage

to

VZM - Imax


Zener Diodes – Equivalent Circuit

• Ideal Zener exhibits a

constant voltage,

• Ideal Zener exhibits no

resistance characteristics.


Cathode

Anode

Reverse

biasing

Zener Diodes – Equivalent Circuit

Zener exhibits a near

constant voltage, varied by

current draw through the

series resistance ZZ.

As Iz increases, Vz also

increases.


Zener Diodes – Characteristic Curve

Vz results from Iz.

Range at which voltage is maintained while current is

changing

Maximum current the diode can handle

Minimum current required to operate in breakdown

region


Zener Diode - Applications

Regulation with varying input voltage

The zener diode will “adjust” its impedance based on

varying input voltages. Zener current will increase or

decrease directly with voltage input changes. The zener

current, Iz, will vary to maintain a constant Vz.

Note: The zener has a finite range of current operation.

Vo remains constant


Zener Diode - Applications

Regulation with varying load

The zener diode will “adjust” its impedance based on varying

input voltages and loads (RL) to be able to maintain its

designated Vz. The zener current will increase or decrease

inversely with varying loads. The zener has a finite range of

operation.

VZ

remains constant


Example 2:

Vinmin= VRmin + VZ = 2.26V + 10V = 12.26V.

Vin(max) = VRmax + VZ = 24.2V + 10V = 34.2V

VReg is ≈12.3 V to 34.2 V.

1N4740

PDMAX = 1W, VZ = 10V, RL= 1 k

IZmin = 0.25mA to

IZmax = 100mA.

Calculate the range of input voltage,

VReg for which output will remain

constant.

IL= Vz/RL=10 mA

VRmin = IR= (Izmin+ IL) x 220

= 2.26 V

VRmax = I R= (Izmax+ IL) x 220

= 24.2V.

Solution:


Thank you….


Lecture 22

Diode applications:

iii) Clippers


Clipper Circuit

• Clipper circuits have the ability to „clip‟ off a portion of

the input signal without distorting the remaining part

of the alternating waveform.


There are several types of clippers:

Series clippers

Parallel clippers

Biased clippers

Zener clippers

Two stages clippers

(i) Series Clipper

The diode in a series clipper “clips” any voltage that

does not forward bias.

The diode is in series with load.

RL

Vi

Vo

Negative Series Clipper

RL

Vi

Vo

Positive Series Clipper

Forward biased: Vo = Vi – 0.7

Reverse biased: Vo = 0


(ii) Parallel Clippers (Shunt)

The diode in a parallel clipper circuit “clips” any voltage that forward bias it.

The diode is in parallel with load. The series resistor RS acts as a

current limiter to avoid any damage to the diode or any component in

the circuit.

RS

Vi

RL

Vo

Positive Shunt Clipper

RS

Vi

RL

Vo

Negative Shunt Clipper

iSL

Lo V

RR

RV

Reverse biased:

Forward biased: Vo = 0 or 0.7 V


(iii) Biased Series Clippers

Adding a DC

source in series

with the clipping

diode changes the

effective forward

bias of the diode.


Biased Parallel Clipping Circuit

RL

Vi

RS

D1

VB

VB + 0.7 V

i) Positive biased clipperD1

VB

RL

RS

- (VB + 0.7 V)

ii) Negative biased clipper


iv) Positive and Negative biased clipper

Two Batteries


Batteries arereplaced byZener diodes.


v) Zener Clipper

The End


Lecture 23

Diode applications:

iv) Clamper


Clamper

A clamping circuit is a circuit to shift a waveform either

above or below a different dc level without distorting

the waveform.

The network has a capacitor, a diode and a resistor.

The magnitude of R and C must be chosen such that thetime constant t = RC is large enough to ensure that thevoltage across the capacitor does not dischargesignificantly during the interval the diode is non-conducting.


Two Types of Clamper:

Simple Clamper

Biased Clamper

Negative Clampers

Negative clamper shifts the input waveform so that the

positive peak voltage of the output waveform will about

the same as the DC reference voltage of the clamper.


Operation of negative clamper+ ve region

+

R Vo

-

+

-

Vi

C • 0 - T/2: Diode is ON (shorted)• Assume RC time is small and capacitor charge to V volts very quickly• Vo=0 V (ideal diode)

- ve region • T/2 -T: Diode is OFF (opened)• Both for the stored voltage across capacitor and applied signal current through cathode to anode

•KVL: Vo = -2V

+

R Vo

-

+

-

V

C

Vo

-

+V

Vi

t0 T/2 T

V

-V


Positive clamper shifts the input waveform so that the

negative peak voltage of the output waveform will about

the same as the DC reference voltage of the clamper.

Positive Clampers


Operation of a positive clamper circuit:

RL

D1

ON

C1

+ 2.5 V

- 2.5 V

0 V

(a) square wave input waveform

(b) Capacitor charging path

D1

OFF

RL

I

C1

+ 5.0 V

0 V

(d) Output of clamper circuit

(c) Small discharge through RL


KVL: Vo = 2V

• Biased clamper enable us to shift a waveform so

that it occurs above or below a DC reference voltage

other than 0 V.

Biased Clamping Circuits

(a) Positive biased clamper


Vs

VB

Vo

Vm

-Vm

VB

2Vm+VB

C

The End


LECTURE 24

Diode Applications:

iv) Voltage multiplier


Voltage-Multiplier Circuits

• Voltage Doubler

• Voltage Tripler

• Voltage Quadrupler

Voltage multiplier circuits use a combination of diodes and

capacitors to step up the output voltage of rectifier circuits.

VOLTAGE MULTIPLIERS are used primarily to develop high

voltages where low current is required, such as picture tube

in TV receivers, oscilloscopes, etc.


Two types:

Half wave voltage doubler

Full wave voltage doubler

Voltage Doubler

A voltage doubler is an electronic circuit which charges

capacitors from the input voltage and switches these

charges - voltage is twice at the output (in the ideal case)

The switching elements are diodes. The output is taken

across the second capacitor.


Half wave voltage doubler

• +ve Half-Cycle

o D1 conducts

o D2 is switched off

o Capacitor C1

charges to Vm

• -ve Half-Cycle

o D1 is switched off

o D2 conducts

o Capacitor C2

charges to 2Vm

Vout = VC2 = 2Vm


Full wave voltage doubler

+ve half cycle:

D1 conducts, charging C1

to the peak voltage Vm.

-ve half cycle:

D2 conducts, charging C2

to Vm,

Since both capacitors

C1 and C2 are in

series, the output

voltage is

approximately 2Vm.


Voltage Tripler

First +ve half cycle:

D1 conducts, charging the

C1 to Vm.

First -ve half cycle:

D2 conducts charging the

C2 to 2Vm.

second +ve half cycle:

D3 and D1 will conduct,

charging C1 and the voltage

across C2 charges C3 to

the same value 2Vm

Thus the voltage across C1

and C3 is 3Vm.

D1D2


Voltage Quadrupler

D1D2

Second -ve half cycle:

D2 and D4 will conduct, leading to C3 charging C4 to the

same peak value 2Vm.

Thus the voltage across C2 and C4 is 4Vm.


The End…


LECTURE 25

BIPOLAR JUNCTION TRANSISTOR (BJT)


Introduction

• The famous and commonly use of semiconductor device isBJTs (Bipolar Junction Transistors).

• It was invented by Shockley in 1951.

• It can be used as amplifier ( radio and TV signals) and logicswitches.

• BJT consists of three terminal:

collector : C

base : B

emitter : E

• A BJT consists of a three-layer "sandwich" of doped (extrinsic) semiconductor materials:

P-N-P or N-P-N.


The Two Types of BJT Transistors:


BJTs – Practical Aspects

Heat sink


• An NPN transistor can be considered as two diodes with a shared

anode.

• In typical operation, the base-emitter junction JE of the NPN

transistor is forward biased and the base-collector junction JC is

reverse biased.

Operation of NPN BJT

• Due to f-b, e in N-type emitter

flow towards the base and

constitues the IE. These e flow

through P-type base where

they tend to combine with holes

(less than 5%) to constitutes

base current, IB. Remaining e

crossover into the collector

region which constitutes

collector current, Ic.


Transistor Currents

• Emitter current (IE) is the sum of the collector

current (IC) and the base current (IB) . • Kirchhoff’s current law;

BCE III


Collector Current (IC)

• Collector current (IC) comprises two components;

majority carriers (electrons) from the emitter

minority carriers (holes) from reverse-biased BC junction

→ leakage current, ICBO

• Total collector current (IC);

• Since leakage current ICBO is usually so small, it can be

ignored.

EC IImajority

CBOC II minority

CBOEC III


…Collector Current (IC)

• Then;

E

C

I

I

• The value of DC represents the quality of a transistor.

• Usually its value lies between 0.95 to 0.99, which means

95% to 99% of the emitter current reach the collector.


Base Current (IB)

• IB is very small compared to IC;

• The ratio of IC to IB is the dc current gain of a transistor,

called beta (β)

B

C

I

I

• Typical value of : 50 – 400, and usually designated

as an equivalent hybrid (h) parameter, hfe on transistor

datasheets.

• Beta can vary with temperature and also varies from

transistor to transistor.


Circuit Configuration

Common Base (CB)

Common Emitter (CE)

Common Collector (CC)


Common Base (CB)

The base is common to both

the input signal and output

signal.

The collector current is less

than the emitter current –

current gain less than 1, it

attenuates the input signal.

is not very common due to its

high voltage characteristics.

It has high voltage gain but

small current gain.


Common Emitter (CE)

is generally used for current

amplifiers - highest voltage

gain and power gain of all

three configurations.

is an inverting amplifier –

output signal being 180 out

of phase with input signal.


Common Collector (CC) is known as a voltage followeror emitter follower circuit - itsoutput is taken from the emitterresistor.

is very useful for impedance

matching because of its high

input impedance and low output

impedance.

is a non-inverting amplifier-

input signal is in phase with the

output signal.

It has low voltage gain (less

than 1) but high current gain.13SSCP 2313- Basic electronics 2012/13-1

Thank you….


LECTURE 26

Static Characteristics of BJT


• When a transistor is operating as an amplifier, a signal given at the input side and a signal gain is obtained at the output side

• The static characteristics of a BJT is the characteristics of the device when it is given bias voltages at the junctions without any signal at the input side of a circuit containing the transistor.

• There are 3 important static characteristics:

1. Input Characteristics – this characteristic relates the input current to the input voltage

2. Output Characteristics – this characteristic relates the output current to the output voltage

3. Transfer Characteristics – this characteristic relates the output current to the input current.

Introduction


• The input characteristic

curve is almost similar

to the forward bias

operation curve of a

PN junction diode.

Input Characteristics

Input characteristic: input current (IB) against input voltage (VBE) for several output voltage (VCE)

From the graph :IB = 0 A VBE < 0.7V (Si)IB = value VBE > 0.7V (Si)


• Output characteristic:

Output current (IC) against output voltage (VCE) for

several input current (IB).

• In this circuit, VBB is used to set IB, while VCC is used to varyVCE.

• For each value of IB, the value of VCE is varied and IC isrecorded.

mA

IB

IC

VCC

VBB

VCE

u A

Circuit used to

obtain the output

characteristics of

a BJT.

Output Characteristics


For other values of IB, the

current IC saturated to a value

which is independent of VCE and

only controlled by IB .

As IB become larger the slope of

the characteristic curve

increases.

This slope is due to the base

width modulation effect which is

known as Early effect.

In theory, CEOBC III

For IB = 0, CEOC II

That is the reverse saturation

current.

ICEO

IB = 0

IB1

IB2

IB3

IB4

IB5

VCE

IC

IB

increases

0

The output characteristics of a BJT

transistor


Operating Regions of BJT Transistor

IB = 0

IB1

IB2

IB3

IB4

IB5

VCE

IC

0

cut-off region

saturation region

power limit region

active region

Four operating

regions:

1. Cutoff region

2. Active region

3. Saturation region

4. Power limit region


• The active region of operation is the region where thecollector curves are almost horizontal.

• Active region – in which the transistor can act as alinear amplifier, where the BE junction is forward-biased and BC junction is reverse-biased.

• The collector current, IC varies with base current IB as

IC = IB .

Active Region


• Saturation region – in which both junctions areforward-biased and IC increase linearly with VCE

• In this region the relationship IC = IB does not occur.

– When the IB is high enough, the entire Vcc isdropped across Rc, with no voltage between CE.This condition is called saturation.

– The saturation current,

Saturation Region

CR

CCV

CsatI

V 0CEV


Cut-Off Region

In this region the transistor is said to be in a non-operatingmode.

Cut-off region – when IB = 0, no IC flows, IC is essentiallyzero with increasing VCE, therefore no voltage drop acrossRc. As a result, the VCE is nearly equal to Vcc.

In this condition the collector-emitter circuit is open circuit.

The saturation mode and cut-off mode are useful in digitalelectronic where the transistor is acting as a switch.

In saturation mode the transistor is operating as a closedswitch, while in cut-off mode it is operating as an openswitch.


When the collector-base voltage is too large, the

collector-base junction will breakdown, giving rise to a

large and unwanted collector current to flow.

This large current generates large heat which may

damage the transistor.

This region of operation should be avoided in transistor

circuits.

Power Limit Region


Note: VCE is at maximum and IC is at minimum (ICmax=ICEO) in the cutoff

region. IC is at maximum and VCE is at minimum (VCE max = VCEsat = VCEO) in

the saturation region. The transistor operates in the active region between

saturation and cutoff.11SSCP 2313- Basic electronics 2012/13-1

Because a transistor's collector current is

proportionally limited by its base current, it can be

used as a sort of current-controlled switch.

Transistor as Switch

A transistor when used as a switch is simply being biased so that it

is in cutoff (switched off) or saturation (switched on). Remember

that the VCE in cutoff is VCC and 0 V in saturation.


• The BJT can be used as a switch by driving it back and

forth between saturation (closed) and cutoff (open).

The BJT as a switch


Thank you…

Lecture 27

Transistor Bias Circuit

•The DC Operating Point

•Fixed Base Bias Configuration


The DC Operating point

The analysis or design of any electronic amplifier has two

components: DC and AC.

A transistor must be properly biased with a dc voltage in

order to operate as a linear amplifier.

Bias establishes the dc operating point.

A dc operating point must be set so that signal variations at

a terminal input are amplified and accurately reproduced at

the output terminal.

If not, it can go into saturation or cutoff region.

The dc operating point is often referred to as Q-point

(quiescent point).


DC Load Line

The straight line is known as

the DC load line.

The dc load line is a graph that

represents all the possible

combinations of IC and VCE

for a given amplifier. VCE

IC

C

CCC

R

VI (sat)

CCCE VV (off)

• This load line is

obtained from the

output equation when

IC = 0 and VCE = 0 V.


Selection of Q-Point

• The intersection of the dc bias

value of IB with the dc load line

determines the Q-point.

• It is desirable to have the Q-point

centered on the load line. Why?

• When a circuit is designed to have

a centered Q-point, the amplifier

is said to be midpoint biased.

• Midpoint biasing allows optimum

ac operation of the amplifier.

• This includes the AC voltage gain

and the maximum AC output

voltage swing allowed by the

amplifier.


When the Q-point is selected in the middle of a load line,

the voltage VCEQ will have a value half of VCC and ICQ will

be half of IC(sat) .


VCE

V

IC

mA

IB = 0 A

IB = 10 A

IB = 20 A

IB = 30 A

IB = 40 A

Q

0

2

4

6

8

10

2 4 6 8 10

(sat)21

CI

(sat)CI

0

CCV210 (off)CEV

• If the Q-point is at the upper side of the load line center, theinput can cause the transistor to saturate.

• When this occurs, part of the output sinusoidal waveform will beclamped.

VCE V

IB = 0 A

IB = 10 A

IB = 20 A

IB = 30 A

IB = 40 A

Q

0

2

4

6

8

10

2 4 6 8 10

(sat) CI

0

0aCEV Q

Qa

(off) CEV

aCQI

IC mA


If the Q-point is at the lower side of the load line center, theinput can cause the transistor to be in cut-off.

When this occurs, part of the output sinusoidal waveform willalso be clamped.

VCE V

IC mA

IB = 0 A

IB = 10 A

IB = 20 A

IB = 30 A

IB = 40 A

Q

0

2

4

6

8

10

2 4 6 8 10

n)(saturatio CI

0

0bCEV Q

Qb

(off) CEV

bCQI


The selection of IBQ, ICQ and VCEQ will influence

the performance of the transistor when it is used

to amplify an AC signal.

The external circuit required to be connected to a

transistor to produce the values of IBQ, ICQ and

VCEQ is called the biasing circuit.


• The purpose of biasing is to fix the operation of a transistorin a linear condition.

• Although the operating point has been fixed, there aresome transistor parameters which can vary and causingthe selected operating point to shift.

• There are two factors affecting an operating point of atransistor:

- transistor replacement (the change in value)

- temperature.

Stability of Biasing Circuit


• The value of for a transistor differs from other

transistors, although they are from the same type.

• For example, transistors 2N2219 have values between

50 and 200.

• The collector current :

COBC III )1(

• So, if the transistor that has been biased in a circuit isdamaged and replaced by other transistor from thesame type, the value of is different from the originaltransistor.

• This causes the operating point of the circuit to change.

1. Transistor Replacement


• Parameters vBE, and ICBO of a transistor amplifier circuit are very sensitive to temperature.

• vBE reduces by 2.5 mV/C when temperature increases

• increases linearly with the increase in temperature

• ICBO doubles for every 10C temperature increase for temperature above 25C.

• ICBO is the most sensitive to temperature change.

• The effect of IC against temperature variation is given by the stability factor S.

• Thus, to enable the stability factor of a bias circuit to be evaluated, all equations of IC should include the leakage current (ICBO) term.

2. Temperature


• Solve the circuit using

KVL to find Q-point:

• 1st step: Replace

capacitors with an open

circuit

• 2nd step: Locate 2 main

loops which;

BE loop

CE loop

FIXED BASE BIAS CIRCUIT



• 1st step: Replace capacitors with an open circuit


• 2nd step: Locate 2 main loops.

12

1

2

BE Loop CE Loop




• BE Loop Analysis

1 From KVL;

IB

B

BECCBQ

BEBBCC

R

VVI

0VRIV



• CE Loop Analysis

From KVL;

As we known;

BQCQ II

2

ICCCQCCCEQ

CECCCC

RIVV

0VRIV


Load Line Analysis

ICsat = VCC / RC

VCEcutoff = VCC

The dc load line are:



• Example 1 Find the Q-point for the

circuit below and draw

the dc load line:

Answers;

ICQ = 2.35 mA

IBQ = 47.08 μA

VCEQ = 6.83V


Thank you…..

LECTURE 28

BJT BIASING CIRCUIT

ii) Fixed Bias with Emitter Resistor Circuit



• DISADVANTAGE

Unstable – because it is too dependent on β and

produce width change of Q-point

For improved bias stability , add emitter resistor to

dc bias.


FIXED BIAS WITH EMITTER RESISTOR

• An emitter resistor, RE is

added to improve stability

• Solve the circuit using KVL

• 1st step: Replace capacitors

and with an open circuit

• 2nd step: Locate 2 main loops

which;

BE loop

CE loop

Resistor, RE added







2

BE Loop CE Loop

1

12


• BE Loop Analysis From KVL;

Since

Substitute for IE

0 EEBEBBCC RIVRIV

EB

BECCBQ

EBBEBBCC

R)1(R

VVI

0RI)1(VRIV

BE II )1(

1





From KVL;

Assume;

Therefore:

Also:

ECQ II

0 EECECCCC RIVRIV

2)RR(IVV ECCCCCEQ

EBEBBCCB

CCCCECEC

EEE

V V RI– V V

RI - V V V V

RI V


Saturation Level

VCEcutoff: ICsat:

The endpoints can be determined from the load line.

mA 0 I

V V

C

CCCE

ERCR

CCV

CI

CE V 0V



• Example 2 Find the Q-point of

the circuit below and

draw the dc load

line:

Answers;

ICQ = 2.01 mA

IBQ = 40.1 μA

VCEQ = 13.97V


Thank you…..

Lecture 29

BJT Biasing Circuit:

iii) Voltage Divider Bias Circuit


• Provides good Q-point stabilitywith a single polarity supply voltageand independent of .

• Most widely used.

• Solve the circuit using KVL:

• 1st step: Replace capacitors with anopen circuit

• 2nd step: Simplified circuit usingThevenin Theorem

• 3rd step: Locate 2 main loopswhich;

BE loop

CE loop

VOLTAGE DIVIDER BIAS CIRCUIT





Simplified Circuit

Thevenin Theorem;

2nd step: : Simplified circuit using Thevenin Theorem

21

2121 //

RR

RRRRRTH

CCTH VRR

RV

21

2

From Thevenin Theorem;




1

2

BE Loop CE Loop

1

2



• BE Loop Analysis From KVL;

Recall;

Subtitute for IE

0 EEBETHBTH RIVRIV

ERTH

BETHBQ

EBBETHBTH

R)1(R

VVI

0RI)1(VRIV

BE II )1(

1




From KVL;

Assume;

Therefore;

ECQ II

0 EECECCCC RIVRIV

)RR(IVV ECCQCCCEQ

2




Example 3: Find the Q-point of the

circuit below and draw

the dc load line:

Answers;

RTH = 3.55 kΩ

VTH = 2V

IBQ = 6.05 μA

ICQ = 0.85 mA

VCEQ = 12.22V


Thank you….


Lecture 30

BJT Biasing Circuit:

vi) Collector feedback bias

v) Emitter bias


Another way to

improve the stability

of a bias circuit is to

add a feedback path

from collector to

base.

In this bias circuit

the Q-point is only

slightly dependent

on the transistor

beta, .

iv) Collector to Base Bias Circuit (Collector Feedback)


BE Loop

)R(RR

VVI

ECB

BECCBQ

From KVL:

0RI–V–RI–RI– V EEBEBBCCCC

Where IB << IC:

CI

BI

CI

CI'

Knowing IC = IB and IE IC, the

loop equation becomes:

0RIVRIRI– V EBBEBBCBCC

Solving for IB:


CE Loop

IE RE+ VCE + I’CRC = VCC

Since IC IC and IC = IB:

IC(RC + RE) + VCE – VCC =0

Solving for VCE:

VCEQ = VCC – ICQ(RC + RE)

From KVL:


Example 4:

5

Determine the values of ICQ and VCEQ for the amplifier shown.

RB

RC

1.5 k

+10 V

180 k

hFE = 100

1

10V 0.7V28.05μA

180kΩ 101 1.5kΩ

CC BEB

B FE C

V VI

R h R

( 1)

10V 101 28.05μA 1.5kΩ

5.75V

CEQ CC FE B CV V h I R

Answer:

100 28.05μA

2.805mA

CQ FE BI h I


v) Emitter bias.

6

RC

RE

RB

IC

IE

IB

Q1

Input

Output

+VCC

-VEE

0.7V

1

EEB

B FE E

VI

R h R

C FE BI h I

CE CC C C E E EEV V I R I R V

CE CC C C E EEV V I R R V

• Emitter bias circuit uses both +ve and

-ve power supply.

• Provides excellent bias stability.

Loop BE: KVL

EEBEBBEE RIVRIV

Loop CE: KVL

ECQ II:Assume


Example 5:

RC

750

RE

1.5k

RB

100

IC

IE

IB

Q1

Input

Output

+12 V

-12 V

hFE

= 200

Determine the values of ICQ and VCEQ for the amplifier shown.

12V 0.7V

( 1)

11.3V37.47μA

100Ω+201 1.5kΩ

B

B FE E

IR h R

200 37.47μA

7.49mA

CQ FE BI h I

( )

24V 7.49mA 750Ω 1.5kΩ

7.14V

CEQ CC C C E EEV V I R R V

CE CC C C E EEV V I R R V


Load Line forEmitter-Bias Circuit

8

(sat)

( )CC EE CC EEC

C E C E

V V V VI

R R R R

( )CE off CC EE CC EEV V V V V

VCE

IC

IC(sat)

VCE(off)


The end…


Lecture 31


BJT AC Analysis: Circuit 1 and 2










Example 1


Answer




Example 2

circuit in Fig. E

Answer


Lecture 32

BJT AC Analysis: Circuit 3 and 4



Example 2


Answer






Example 3


Answer





Junction Field Effect Transistor

(JFET)

Lecture 33


Introduction

1. Field effect transistors control current by voltage

applied to the gate.

2. The FET’s major advantage over the BJT is high input

resistance.

3. Overall, the purpose of the FET is the same as that of

the BJT.


N-channel Junction field-effect transistor (JFET)


BJT vs JFETBipolarJunctionTransistor

• Current-based device

• IBase controls ICollectorEmitter

JunctionFieldEffectTransistor

•voltage controlled devices

•VGate controls IDrainSource

ID


The JFET – Primary Characteristics

• The JFET uses voltage to control the current flow.

• You will recall, the transistor uses current flow through the

base-emitter junction to control current.

• JFETs can be used as an amplifier just like the BJT.

• VGG voltage level controls current flow in the VDD, RD circuit.


There are 2 types of JFET

n-channel JFET

p-channel JFET

Three Terminal

Drain – D

Gate -G

Source – S


– Major structure is n-type material (channel) betweenembedded p-type material to form 2 p-n junction.

– In the normal operation of an n-channel device, the Drain(D) is positive with respect to the Source (S). Currentflows into the Drain (D), through the channel, and out ofthe Source (S)

– Because the resistance of the channel depends on thegate-to-source voltage (VGS), the drain current (ID) iscontrolled by that voltage.

N-channel JFET


The JFET - Biasing

• The “source-drain” current is

controlled by a voltage field at

the “gate”.

• That field is developed by the

reverse biased gate-source

junction.

• With more VGG (reverse bias)

the field grows larger.

• This field or resistance limits

the amount of current flow

through RD.

• With low or no VGG current

flow is at maximum.8SSCP 2313- Basic electronics 2012/13-1

JFET Characteristics and Parameters Ohmic

Let’s first take a look at the effects with a VGS of 0V. ID increases proportionally with increases of VDD (VDS increases as VDD is increased). This is called the ohmic region (point A to B).

• The N-channel JFET can be explained by the output

characteristic curve.

• The output characteristics curve shows the relationship

between iD and vDS for different values of vGS .

• In this characteristics curve it can be identified that there are

three regions of operation:

• cut-off region,

• ohmic region and

• saturation region


JFET Characteristic Curve

• To start, suppose VGS=0

• Then, when VDS is increased, ID increases. Therefore,ID is proportional to VDS for small values of VDS

• For larger value of VDS, as VDS increases, the depletionlayer become wider, causing the resistance of channelincreases.

• After the pinch-off voltage (Vp) is reached, the IDbecomes nearly constant (called as ID maximum, IDSS-Drain to Source current with Gate Shorted)


Pinch-off (VGS = 0 V, VDS = VP).


JFET Characteristics and Parameters Pinch-Off

The point when ID stop to increase regardless of VDD increases (constant

current source) is called the pinch-off voltage (point B) (Note: VGS = 0).

This current is called maximum drain current (IDSS).Breakdown (point C) is reached when too much voltage is applied. This is

undesirable, so JFETs operation is always well below this value.

• In saturation condition,

one end (at the drain

end) the JFET will

become pinch-off due

to the depletion region.


JFET Characteristics and Parameters Drain CurvesFrom this set of curves you can see increased negative voltage applied to

the gate (-VGS) produces no change in ID. ID is limited and the pinch-off

voltage (VP) is reduced. Note: VGS controls IDSS

p(off) GSGS VVV

If VGS is less than pinch-off voltage, the resistance becomes an

open-circuit ;therefore the device is in cutoff (VGS=VGS(off))


Pinch-Off “Vp” vs Cutoff “VGS(off)”

• VGS(off) and VP are always = and opposite is sign

• VP(pinch off) - the value of VDS where ID becomes constant

with VGS = 0.

• VP(pinch off) also occurs for VDS<VP if VGS ≠ 0.

Note: Although VP is constant, VDSmin where ID is constant, varies.

VGS(off)= -5VVp=+5V


JFET Characteristics and Parameters

The transfer characteristic curve illustrates the control VGS has on

ID from cutoff (VGS(off) ) to pinch-off (VP). Note the parabolic shape. The

formula below can be used to determine drain current.

Square-law device: Parabolic curve of the JFET Transfer Characteristic

Curve.

VGS(OFF)

Vp

2

(off)

1

GS

GSDSSD

V

VIi

Note:(VGS = 0 to VGS(off)

controls ID)

JFET Characteristics and Parameters

gm2

gm1

The transfer curve is also known as the transconductance

curve.The value of drain current when vGS = 0 is marked as

IDSS .

JFET Transfer Characteristic Curve JFET Characteristic Curve


• Defined by Shockley’s equation:

• Relationship between ID and VGS.

• Obtaining transfer characteristic curve axis point from

Shockley:

– When VGS = 0 V, ID = IDSS

– When VGS = VGS(off) or Vp, ID = 0 mA

)(

2

)(

1offGSP

offGS

GSDSSD VV

V

VII

Transfer Characteristics…


Exercise 1

Sketch the transfer defined by IDSS = 12 mA dan VGS(off) = Vp= - 6V

Answer

2GS

D DSS

P

V I = I 1 -

V

DGS P

DSS

I V = V 1 -

I

VGS ID

0 IDSS

0.3Vp IDSS/2

0.5Vp IDSS/4

Vp 0 mA

VGS =0.3VP

VGS =0.5VP

IDSS/2

IDSS/4


Exercise 2

DGS P

DSS

I V = V 1 -

I

Sketch the transfer defined by IDSS = 4 mA and

VGS(off) = 3V

2GS

D DSS

P

V I = I 1 -

V

VGS =0.5VP

VGS =0.3VP

VP

IDSS/2

IDSS/4

Answer


Thank you…

Lecture 34

JFET Biasing Circuit:• Fixed – Bias

• Self-Bias


• The general relationship of all FET amplifiers for dc Biasing analysis:

• For the JFET, the relationship between input and output quantities is nonlinear due to the squared term in Shockley’s equation:

• Nonlinear functions results in curves as obtained for transfer characteristic of a JFET.

• JFETs differ from BJTs:

• Nonlinear relationship between input (VGS) and output (ID)

• JFETs are voltage controlled devices, whereas BJTs are current controlled

Introduction to FET Biasing

2

(off)

1

GS

GSDSSD

V

VIi

SDG IIandA0I


Common FET Biasing Circuits:

i) Fixed – Bias or Gate Biasedii) Self-Biasiii) Voltage-Divider Bias

Remember: the purpose of biasing is to set a point of

operation (Q-point).


• The simplest biasing arrangement.

• The configuration includes the ac levels Vi and Vo

and the coupling capacitors.

Fixed-Bias Configuration


• For the DC analysis,

• Capacitors are open circuits

• The zero-volt drop across RG permits

replacing RG by a short-circuit

Applying KVL:VGS= -VGG


2 k

1 M

2 V

IDSS = 10 mA

VP = -8 V

Determine the following of the network and plot the transfer

characteristic and load line :

i) VGSQ

ii) IDQ

iii)VDS

iv)VG

Answer:

V2VV)i GGGSQ

mA625.5

V

V1Ii)ii

2

(off) GS

GSDSSDQ

V75.4

RIVV)iii DDDDDS

V2VV)iv GSG

Example 1


• The self-bias configuration eliminates the need for

two dc supplies.

• The controlling VGS is now determined by the

voltage across the resistor RS

Self- Bias Configuration


Self-bias is the most common type of biasing method for

JFETs. No voltage is applied to the gate. The voltage to

ground from here will always be 0V.

ID = IS for all JFET circuits.

VGS

Self bias is more stable than Fixed bias.

SD

SGGS

RI

VVV

The drain voltage, VD = VDD - IDRD

Since VS= IDRS, the drain to source voltage,

VDS is

)RR(IV

VVV

SDDDD

SDDS


Example:

3.3 k

1 M1 k

Determine the following of the network and plot the transfer characteristic

and load line :

i) VGSQ

ii) IDQ

iii) VDS

iv) VS

v) VD

Answer:

V8V

,mA8IChoosing

V4V

,mA4IChoosing

RIV)i

GS

D

GS

D

SDGS

ii) At the Q- point:IDQ = 2.6 mA

mA82.8

)RR(IVV)iii SDDDDDS

iv) VS = IDRS= 2.6 V

v) VD = 11.42 V

IDSS = 8 mAVP = -6 V

Thus; At the Q- point: VGSQ = -2.6 V



Lecture 35

JFET Biasing Circuit:

Voltage Divider Bias


• The arrangement is the same as BJT but the DC

analysis is different

• In BJT, IB provide link to input and output circuit, in

FET VGS does the same

Voltage-Divider Bias


IG = 0A ,Kirchoff’s current law requires that IR1= IR2 and

the series equivalent circuit appearing to the left of

the figure can be used to find the level of VG.

SDGGS RIVV

DDDDD RIVV

)RR(IVV SDDDDDS

DC equivalent circuit3SSCP 2313- Basic electronics 2012/13-1

Example:

Determine the following of the network and plot the transfer

characteristic and load line :

i) VGSQ and IDQ

ii) VS

iii)VD

iv)VDS

16 V

2.4 k

2.1 M

270 k

1.5 k

IDSS = 8 mA

VP = -6 V


Answer:

V82.1

RR

VRV)i

21

DD2G

and

)k5.1(IV82.1

RIVV

D

SDGGS

When ID=0 mA, VGS= +1.82 V

VGS = 0V, ID=1.21 mA

Thus VGSQ= -1.8 V and IDQ = 2.4 mA

ii) VS= 3.6 V

iii) VD= 10.24 V

iv) VDS= VD - VS = 6.64 V

Plot the

transfer

curve



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