TUNED AMPLIFIER

Rakesh Mandiyahttps://in.linkedin.com/in/rakeshmandiyarakesh.yadav1211@gmail.com

TUNED AMPLIFIER

IntroductionBand Pass AmplifiersSeries & Parallel Resonant Circuits & their

BandwidthAnalysis of Single Tuned AmplifiersAnalysis of Double Tuned AmplifiersPrimary & Secondary Tuned Amplifiers with

BJT & FETMerits and de-merits of Tuned Amplifiers

DEFINITION:-

An amplifier circuit in which the load circuit is a tank circuit such that it can be tuned to pass or amplify selection of a desired frequency or a narrow band of frequencies, is known as Tuned Circuit Amplifier.

CHARACTERISTICS OF TUNED AMPLIFIER Tuned amplifier selects and amplifies a single frequency

from a mixture of frequencies in any frequency range.

A Tuned amplifier employs a tuned circuit. It uses the phenomena of resonance, the tank circuit which is

capable of selecting a particular or relative narrow band of frequencies.

The centre of this frequency band is the resonant frequency of the tuned circuit .

Both types consist of an inductance L and capacitance C with two element connected in series and parallel.

RESONANCE CIRCUITS: When at particular frequency the inductive reactance

became equal to capacitive reactance and the circuit then behaves as purely resistive circuit. This phenomenon is called the resonance and the corresponding frequency is called the resonant frequency.

C L

Tuned circuit

Resonance circuits

Series Parallel

Classification of Tuned Circuits

Small signal amplifier, low power, radio

frequency

Class A

Single Tuned

circuit(one parallel circuit is

employed)

Double tuned

circuit(two tuned

circuit are employed)

Staggered

Tuned amplifi

er

Large signal amplifier, low power, radio

frequency

Class B&C

Shunt peaked tuned with higher

band width

CLASSIFICATION OF TUNED AMPLIFIER

Tuned amplifier

Small Signal Amplifier

Single Tuned

Amplifier

Double Tuned

Amplifier

Stagger Tuned

Amplifier

Large signal Amplifier

CLASSIFICATION OF TUNED AMPLIFIERS

Small Signal Tuned Amplifiers :- They are used to amplify the RF signals of small magnitude.

They are further classified as: (a) Single Tuned Amplifiers:- In this we use one

parallel tuned circuit in each stage. (b) Double Tuned Amplifiers:- In this we use two

mutually coupled tuned circuits for every stage both of tuned circuits are tuned at same freq.

(c) Stagger Tuned Amplifiers:- It is a multistage amplifier which has one parallel tuned circuit for every stage but tuned frequency for all stages are slightly different from each other.

(2) Large signal tuned amplifiers:- They are meant for amplifying large signals

in which large RF power is involved & distortion level is also higher. But tuned circuit itself eliminates most of the harmonic distortion.

BAND PASS AMPLIFIER:An amplifier designated to pass a definite band

of frequencies with uniform response.

The new band pass amplifier perform both function of low noise amplifier (LNA) & band pass filter is proposed for application of 900Mhz RF Front – end in wireless receivers .

BAND PASS AMPLIFIER:

It is having two differential stage comprising two transistor.

Main function of band pass filter to remove the band noise ,which also contributes to the rejection of image signals.

Finally a band pass amplifier amplifies only a band of frequency which lie in bandwidth of amplifier & thus named as band pass amplifier .

BAND PASS AMPLIFIER

BAND PASS AMPLIFIER

BAND PASS FILTER

SERIES RESONANT CIRCUITIt is the circuit in which all the resistive and reactive components are in series.

SERIES RESONANT LC

SERIES RESONANT CIRCUIT Impedance Of The Circuit: -

Z = R2 + (XL – Xc)21/2

Z = R2 + (ωL – 1/ ωC)21/2

For resonant frequency:-(XL = XC )

XL = ωL = 2 π frL XC = 1/ ωC = 1 / 2 π frC

SERIES RESONANT CIRCUITSince at resonance, XL = Xc

2 π frL = 1 / 2ПfrC fr = 1 / 2 π √LC ωr = 1 / √LC

RESONANCE CURVE OF SERIES RESONANT CIRCUIT :

QUALITY FACTOR It is voltage magnification that circuit produces at

resonance is called the Q factor.

Voltage Magnification = Imax XL / Imin R

= XL/ R

At Resonance XL/R = XC/R ωrL / R = 1 / ωrRC

THUS

Qr = ωrL / R = 1/ ωrC R = 2 π fr L / R = (2 π L / R) * (1 / 2 π √LC ) = √(L/C) / R = tanФ tan Ф = power factor of coil

IMPORTANT POINTS (1) Net reactance , X = 0. (2) Impedance Z = R . (3) Power factor is unity. (4) Power expended = 6 watt. Current is so large & will produce large voltage

across inductance & capacitance will be equal in magnitude but opposite in phase.

Series resonance is called an acceptor circuit because such a circuit accepts current at one particular frequency but rejects current at other frequencies these circuit are used in Radio – receivers .

REACTANCE CURVE SERIES RESONANT CIRCUIT

XL = 2ΠfL

X = XL - XC XC = 1

2ΠfC

R

fr

curr

ent

PARALLEL OR CURRENT RESONANCE

PARALLEL OR CURRENT RESONANCE

When an inductive reactance and a capacitance are connected in parallel as shown in figure condition may reach under which current resonance (also known as parallel or anti- resonance ) will take palace. In practice, some resistance R is always present with the inductor.

Such circuit is said to be in electrical resonance when reactive(watt less) components of line current becomes zero. The frequency at which this happened is known as resonant frequency.

Current will be in resonance if reactive component of R-L branch IR-L sinФ R-L = Reactive component of capacitive branch, neglecting leakage reactance of capacitor C

FREQUENCY V/S IMPEDANCE CURVE FOR LCR CIRCUIT

CURRENT AT RESONANCECapacitive Current Ic = 2 π fr CV

Coil Current IR-L = V/Z

= V / √ R2+ (ωrL)2 ФR-L = Cos-1(R/Z)

ФR-L = Sin-1(2ПfrL/Z)

( V / √ R2+ (ωrL)2 ) * (ωrL / √ R2+ (ωrL)2 ) = ωrCV

C = L / (R2+ (ωrL )2

ωr = (1 / √LC) * (( 1- CR2/L) )1/2

ωr = √(1/LC – R2/L2)

fr = ωr /2 π = (1/2 π ) * ((1/LC)- (R2 / L2))

RESONANCE CURVE OF PARALLEL RESONANT CIRCUIT :

With low resistance

With high resistance

current

Res

onan

t fr

eque

ncy

fR

Active component of coil IA = IR- L cosФR-L

= (V/Z) * (R/Z)

= VR/Z2

Reactive component of coil IR = IR-L sinФ = (V/Z ) * (2 π frL/Z )

Since at resonance Reactive component of coil current = Capacitive current (V / Z ) * (2 π frL / Z) = 2 π frCV Z = √(L/C) ………..(1) Line current IL = Active component of coil current = IA = IR-L cosФR-L

= VR/Z2 [using (1)] = VR(C/L) IL = [ V / (L/RC) ] (L/RC) = Effective or equivalent dynamic impedance of

parallel circuit at resonance.

IMPORTANT POINTS FOR CURRENT OR PARALLEL RESONANCE:

(1) Net susceptance is zero (1 / XC ) = ( XL / Z2 )(2) Admittance = Conductance(3) Power factor is unity as reactive ( wattles)

components of the current is zero(4) Impedance is purely resistive ZMax = (L / CR)(5) ILine(Min) = V / ( L/CR ) ( in phase with applied

voltage)

(6) f = (1/2П) * ( √(1/LC) – (R2/ L2)) Hz The frequency at which the net susceptance curve

crosses the frequency axis is called the resonant frequency .

At this point impedance is maximum or admittance is minimum & is equal to G , consequently (I) Line is minimum .

Band with of parallel resonant circuit B.W. = (f2 – f1) Quality Factor Q = XL / R = 2ПfrL / R Quality factor determines sharpness of resonance curve

and selectivity of circuit. Higher the value of quality factor more selective the tuned

circuit is.

CHARACTERSTICS OF PARALLEL OR CURRENT RESONANCE

Admittance is equal to conductance. Reactive or watt less component of line current is zero

hence circuit power factor is unity. Impedance is purely resistive , maximum in

magnitude and is equal to L/CR. Line current is minimum and is equal to

V / (L/CR) in magnitude and is in phase with the applied voltage.

REACTANCE CURVE PARALLEL RESONANT CIRCUIT

XL = 2ΠfL

X = XL - XC

curr

ent

1

XC =

2ΠfC

R

(1) SINGLE TUNED AMPLIFIER

+

Vs

Cin R1

R2

CeRe

Cc RL

LC

Vcc

(1) SINGLE TUNED AMPLIFIER• O/P of this amplifier may be taken either with the

help of Capacitive.

• A parallel tuned circuits is connected in the collector circuit.Tuned voltage amplifier are usually employed in RF stage of wireless communication , where such circuits are assigned the work of selecting the desired carrier frequency and of amplifying the permitted pass-band around the selected carrier frequency.

SINGLE TUNED AMPLIFIER Tuned amplifier are required to be

R1, R2, & Re = For biasing & stabilization circuit. Ce = By pass capacitor L-C = Tuned circuit connected in collector, the impedance of which depend upon frequency, act as a collector

load. If i/p signal has same frequency as resonant frequency

of L-C circuit . Large amplification will be obtain because of high impedance of L-C ckt.

SINGLE TUNED AMPLIFIER USING FET

SINGLE TUNED AMPLIFIER USING FET In the shown figure the single tuned amplifier

is depicted using a field effect transistor. The value of L and C is selected as per the

desired frequency level. One of the components either L or C is

variable so as to adjust the variable frequency.

THE HIGH FREQUENCY SIGNAL TO BE APPLIED BETWEEN BASE & EMITTER. THE RESONANT FREQUENCY OF CIRCUIT IS MADE EQUAL TO FREQUENCY OF I/P SIGNAL BY VARYING L OR C .

NOW TUNED CKT WILL OFFER VERY HIGH IMPEDANCE TO THE SIGNAL FREQ. & THUS LARGE O/P APPEAR ACROSS IT. AV = ( Β RAC )/ RIN RAC = TUNED CIRCUIT IMPEDANCE = Β(L/CR)/ RIN

AV = ΒL / ( CRRIN ) BANDWIDTH = (F2- F1 )

THE AMPLIFIER WILL AMPLIFY ANY FREQ. WELL WITHIN THIS RANGE.

CIRCUIT OPERATION

LIMITATION This tuned amplifier are required to be highly selective. But

high selectivity required a tuned circuit with a high Q- factor .

A high Q- factor circuit will give a high Av but at the same time , it will give much reduced band with because bandwidth is inversely proportional to the Q- factor .

It means that tuned amplifier with reduce bandwidth may not be able to amplify equally the complete band of signals & result is poor reproduction . This is called potential instability in tuned amplifier.

DOUBLE TUNED CIRCUIT :

DOUBLE TUNED CIRCUITThe problem of potential instability with a single tuned

amplifiers overcome in a double tuned amplifier which consists of independently coupled two tuned circuit :

(1) L 1C1 in collector circuit (2) L2 C2 in output circuit

A change in the coupling of two tuned circuit results in change in the shape of frequency response . By proper adjustment of coupling between two coils of two tuned circuits, the required results are :

High selectivity High voltage gain Required bandwidth

CIRCUIT OPERATION The resonant freq. of tuned circuit connected in collector circuit is

made equal to signal freq. by varying the value of C1. Tuned circuit (L 1C1) Offer very high impedance to

signal frequency & this large o/p is developed across it. The o/p of (L1C1) is transferred to (L2C2) through

mutual inductance. Thus the freq. response of double tuned circuit depends upon

magnetic coupling of L1 & L2. Most suitable curve is when optimum coefficient of coupling exists

between two tuned circuit .The circuit is then highly selective & also provides sufficient amount of gain for a particular band of frequency.

Volta

ge

gain

AV

Frequency fr

K=2

K=1.5K=1

fr

Critical coupling

Loose coupling

Resonance curve of Parallel Resonant circuit:

SHUNT PEAKED CIRCUITS FOR INCREASED BANDWIDTH

For expanding bandwidth we use various combinations of BJT & FET(MOS) in series or shunt so that we can use Stagger tuned amplifiers.

Shunt Peaking If a coil is placed in parallel (shunt) with the output signal path,

the technique is called SHUNT PEAKING. R1 is the input-signal-developing resistor. R2 is used for bias and temperature stability. C1 is the bypass capacitor for R2. R3 is the load resistor for Q1 and develops the output signal. C2 is the coupling capacitor which couples the output signal to the next stage.

SHUNT PEAKED CIRCUITS FOR INCREASED BANDWIDTH :

STAGGER TUNED AMPLIFIERS It is a multistage amplifier which has one parallel resonant

circuit for every stage, while resonant frequency of every stage is slightly different from previous stages. From circuit diagram it is clear that first stage of this

amplifiers has a resonant circuit formed by L1 & C1 that f1 = 1 / (2Π √L1 C1)

The o/p of stage is applied to second stage which is tuned to slightly higher frequency.

f2 = 1 / (2Π √L2 C2)

Second stage amplifiers the signals of frequency f2 by maximum amplitude while other frequency signal are amplified by less quantity . Thus frequency response

Curve of second stage has a peak of f2 which is

slightly higher than f1.

STAGGER TUNED AMPLIFIERS :

Over all response

Freq. response of first stage

Freq. response of second stage

f1 f0f2

Volta

ge

Frequency

STAGGER TUNED AMPLIFIER

STAGGER TUNED AMPLIFIERS

Over all response of these two stage is obtained by combining individual response & it exhibits a maximum flatness around the center frequency f0 . Thus overall bandwidth is better than individual stage.

Since two stages are in parallel (shunt) & overall bandwidth is increased thus, it behaves like shunt circuits for the increased bandwidth.

LARGE SIGNAL (NARROW BAND AMPLIFIER) TUNED AMPLIFIER

Single & double stage amplifier are not suitable for applications involving larger RF power , because of lower of efficiency of class A operation (single double) such as for excitation of transmitting antenna.

For such application larger signal tuned amplifier are employed because they are operation in class C operation that has high efficiency & capable of delivering more power in comparison to that of class A operation .

CIRCUIT DIAGRAM OF LARGE SIGNAL (NARROW BAND AMPLIFIER) TUNED AMPLIFIER :

+

Vcc

RBVs

C L

Cc

RL

Cs

Tuned class C amplifier

The resonant tuned circuit is tuned to freq. of i/p signal . When circuit has a high Q- factor , parallel resonance occur approximate freq. :

f = 1 / (2 π √LC)

At resonant freq. the impedance of parallel circuit is very large & purely resistive.

LARGE SIGNAL (NARROW BAND AMPLIFIER) TUNED AMPLIFIER

LARGE SIGNAL (NARROW BAND AMPLIFIER) TUNED AMPLIFIER

Higher the Q of circuit faster gain drops on either side of resonance freq.

A large Q leads to small bandwidth equal top sharp tuning this amplifier has Q>> 10,This means Bandwidth is less than 10% of fr & for this reason , it is called as narrow band amplifier.

COMPARISON BETWEEN TUNED AND AF AMPLIFIER

It has to amplify narrow band of frequencies defined by the tuned load at the collector

They are bulky and costlier

Used in radio transmitters and receivers, and television receivers circuits .

Works with a complete audio frequency range

More compact Amplifies sound

signals and act as drive for loud speakers

Tuned Amplifier AF Amplifier

APPLICATIONS OF TUNED AMPLIFIERTuned amplifiers serve the best for two

purposes:a)Selection of desired frequency.b)Amplifying the signal to a desired level. USED IN: Communication transmitters and receivers. In filter design :--Band Pass, low pass, High

pass and band reject filter design.

ADVANTAGES

It provides high selectivity. It has small collector voltage. Power loss is also less. Signal to noise ratio of O/P is good. They are well suited for radio transmitters and

receivers .

DISADVANTAGES

They are not suitable to amplify audio frequencies.

If the band of frequency is increase then design becomes complex.

Since they use inductors and capacitors as tuning elements, the circuit is bulky and costly.

Numerical

Numericals

Q-1) A single tuned amplifier consist of tuned circuits having R=5ohm,L=10mh,c=0.1mf. Determine

a)resonant frequency.b)quality factor of tank circuitc)band width of amplifier Ans - Given data-: R=50ohm;L=10mh;C = 0.1mf

We know: Resonant frequency = 1/ 2 π √[ (1/LC) – (R2 / L2 ) ]

= 1 / 2 π √[(1/10*10-3 - 25/100*10-6) ] = 5.034 KHz Q = 2 π * 5.034 * 10-3 * 10 * 10-3 / 5 = 63.227

BW = FR /Q = 5.034 / 63.227

= 79.62KHzRESULT:- Fr = 5.034 KHz

Q = 63.227BW = 79.62KHz

Q2.In a class c amplifier ckt C=300pf,L=50mH,R=40ohm, RL =4M ohm.

Determine:- a)Resonant frequency b)D.C load c)A.C load d)Quality factor

Ans :- Given: C=300pf L=50mH R=40ohm RL =4M ohm

Fr = 1/2 π √ LC

= 1/2 π √(50 * 10-6 * 300 * 10 -12 ) = 1.3 MHz

Rdc = 40 ohmXL = 2 π fr L = 2* 3.14 * 1.3 * 106 * 50 * 10-6

= 408.2

Qdc = 408.2/ 40 = 10.205

Rac = Rp ll RL Rp = Qdc * XL

= 10.205 * 408.2 = 4165.681

RL = 4* 106 Rac = 4161.34

Qac = [Rac / XL ] = 4161.34/408.2 = 10.194Result: -

Fr = 1.3 MHzQdc = 10.205Qac = 10.194

Q. A circuit is resonant resonant at 455 khz and has a 10khz bandwidth. The inductive reactance is 1255ohm. What is the parallel impedance of the circuit at resonance?

Solution:Given that: fr =455 khzFrequency BW=10khz and XL Let zp be the value of impedance at resonance We know that the value of bandwidth

(BW)=fr /QSo, 10*103 =455*103 /QQ=45.5Q=XL /RXL =1255

So,R=1255/45.5=27.6 ohm

1255=2∏fr L=2859*103 LL=1255/(2859*103)L=.439*103 H

Value of capacitance reactance at resonance:XC =XL

=1255Ω1/2∏fr = 1255Therefore, C=278.7*10-12 F

And value of circuit impedance at resonancezp =L/CR =0.439*103 / (278.7*10-12)*27.6 =57*103

=57 kΩRESULT: the parallel impedance at resonance is 57kΩ.

Q: A FET has gm=6 mA/v, has a tuned anode load consisting of a 400 microH inductance of 5 ohm in parallal with a capacitor of 2500pf. Find:-1. The resonant frequency2. Tuned circuit dynamic resistance3. Gain at resonance4. The signal bandwidth

Solution:-1. Fr => resonant frequency

= 1/(2π√(LC)) = 0.159/ √ (400*2500) = 1.59*105 Hz = 0.159 MHz

2. Rd => tuned circuit dynamic resistance = L/CR = 400/(2500*5)

= 106 * 80/2500 = 0.032 * 106

= 32 k ohm 3. Av = -gm rd = 6*32 = -192

4. BW= fr / Q

Q = Wr L/ R = (2 π * 0.159 * 10 -6 * 400 * 106) / 5 = 79.92BW = 0.159/ 79.92 = 1.98 KHz

RESULT:

Resonant frequency=0.159MHzDynamic Resistance=32kΩResonance gain=-192Bandwidth=1.98khz

Q: A tuned voltage amplifier, using FET with rd = 100 kohm and gm = 500 micro s has tuned circuit, consisting of L= 2.5 mH , C = 200 pF , as its load. At its resonant frequency , the circuit offers an equivalent shunt resistance of 100 kohm. For the amplifier determine:-1. Resonant gain2. The effective Q3. The Bandwidth Solution: -Given that: gm= 500Shunt resistance=100 kohm4. Resonant gain:-

Av = -gm (rd ll Rd )/(1 + jf/fr)Av = 500(100 ll 100)/(1 + j1)Av = 17.68

2. Effective Q: -Qeff = L/CR R = 100 ll 100Q = 2.5* 103 /(200*10-12 * 50 * 103) = 2.5 * 102

3. Bandwidth: -BW = fr /Qeff

fr = 1/2 π √ LC = 225 kHzBW = 225/ 2.5* 102

= 900 HzRESULT: Resonant gain=17.68

Qeff =2.5*102 BW = 900 Hz

Submitted By:Rakesh mandiya