ELECTRONICS - I

1

I – term 2012-2013 ELECTRONICS – I (EE311)

Prepared by : Dr. Lotfi LADHAR

EE-311

ELECTRONICS - I

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CHAPTER 9

AMPLIFIER

FREQUENCY

RESPONSE

3



Outlines

Introduction to Frequency Response of an

Amplifier

Gain of an Amplifier in Decibels (dB)

Frequency Response of a BJT Amplifier

Frequency Response of an FET Amplifier

Frequency Response of a Multistage

Amplifier

4



Amplifier Frequency Response -Introduction

Most amplifiers have a finite range of

frequencies in which it will operate. We will

discuss what determines the frequency

response of an amplifier circuit and how it is

measured.

5



Basic Concepts

When frequency is low enough, the coupling and bypass

capacitors can no longer be considered as shorts because their

reactances are large enough to have significant effect. Also,

when the frequency is high enough, the internal transistor

capacitances can no longer be considered as opens because

their reactances become small enough to have significant

effect on the amplifier operation. We will discuss how the

capacitor limits the passage of certain frequencies. This is

called the frequency response of an amplifier.

6



Basic Concepts – Effect of Coupling Capacitors

Coupling capacitors C1 and C3 limit the passage of very low

frequencies. Emitter bypass C2 capacitor will have high

reactance to low frequencies as well, limiting the gain. Also

the capacitance causes a phase shift of the signal.

fCX

C

2

1

7



When the frequency is sufficiently high

XC ≈ 0 Ω and the voltage gain of the CE

amplifier is

At lower frequencies, XC >> 0 Ω and

the voltage gain is

Basic Concepts – Effect of Bypass Capacitors

e

C

v

r

RA

'

ee

C

v

Zr

RA

'

At lower frequencies, the reactance of the emitter bypass

capacitor, C2 in previous Figure, becomes significant and

emitter is no longer at ac ground. The capacitive reactance XC2

in parallel with RE creates an impedance that reduces the gain

as shown in Figure.

Nonzero reactance of the bypass capacitor in

parallel with RE creates an emitter impedance,

(Ze), which reduces the voltage gain.

8



Basic Concepts – Effect of Internal Transistor Capacitances

At high frequencies, the coupling and bypass capacitors become

effective ac shorts and do not affect an amplifier’s response. Internal

transistor junction capacitances, however, do come into play,

reducing an amplifier’s gain and introducing phase shift as the signal

frequency increases.

Cbe is the base-emitter junction capacitance and

Cbc is the base-collector junction capacitance.

9



Basic Concepts – Effect of Internal Transistor Capacitances

When the reactance of Cbe becomes small enough, a significant

amount of the signal voltage is lost due to a voltage-divider effect of

the signal source resistance and the reactance of Cbe as illustrated in

Figure (a). When the resistance of Cbc becomes small enough, a

significant amount of output signal voltage is fed back out of phase

with input (negative feedback), thus effectively reducing the voltage

gain as shown in Figure (b).

10



Basic Concepts – Miller’s Theorem

At high frequencies, the coupling and bypass capacitors become

effective ac shorts and do not affect an amplifier’s response. Internal

transistor junction capacitances, however, do come into play,

reducing an amplifier’s gain and introducing phase shift as the signal

frequency increases.

Miller’s theorem states that C effectively appears as a capacitance

from input and output to ground, as shown in Figure (b).

Cin(Miller) = C(Av + 1)

v

v

MilleroutA

ACC

1)(

ET212 Electronics – Amplifier Frequency Response Floyd 9

11



Basic Concepts – Miller’s Theorem

Millers theorem allows us to view the internal capacitances

as external capacitors for better understanding of the effect

they have on the frequency response.

12



The Decibel

The decibel is a common unit of measurement of voltage

gain and frequency response. It is a logarithmic

measurement of the ratio of one power to another or one

voltage to another. The formulas below are used for

calculation of decibels for power gain and voltage gain.

Ap(db) = 10 log Ap

Av(db) = 20 log Av

13



Table shows how doubling or having voltage gains translates into dB values. Notice in the table that every time the voltage gain is doubled, the dB value increases by 6 dB, and every time the gain is halved, the dB value decreases by 6 dB.

VOLTAGE GAIN (Av) dB (WITH RESPECT TO ZERO REFERENCE)

32 20 log(32) = 30 dB

16 20 log(16) = 24 dB

8 20 log(8) = 18 dB

4 20 log(4) = 12 dB

2 20 log(2) = 6 dB

1 20 log(1) = 0 dB

0.707 20 log(0.707) = - 3 dB

0.5 20 log(0.5) = - 6 dB

0.25 20 log(0.25) = - 12 dB

0.125 20 log(0.125) = - 18 dB

0.0625 20 log(0.0625) = - 24 dB

0.03125 20 log(0.03125) = - 30 dB

14



Ex 10-1 Express each of the following ratios in dB: (a) (b) (c) Av = 10 (d) Ap = 0.5 (e)

250in

out

P

P100

in

out

P

P

707.0in

out

V

V

(a) Ap(dB) = 10 log(250) = 24 dB (b) Ap(dB) = 10 log(100) = 20

dB

(c) Av(dB) = 20 log(10) = 20 dB (d) Ap(dB) = 10 log(0.5)

= - 3 dB

(e) Av(dB) = 20 log(0.707) = - 3 dB

15



The Decibel – The Critical Frequency

The critical frequency also known as the cutoff

frequency is the frequency at which the output power

drops by 3 dB, which represents one-half of it’s midrange

value. An output voltage drop of 3 dB represents about a

70.7% drop from the midrange value.

Power is often measured in units of dBm. This is

decibels with reference to 1mW of power. This means

that 0 dBm = 1mW.

16



Ex 10-2 A certain amplifier has a midrange rms output voltage of 10 V. What is the rms output voltage for each of the following dB gain reductions with a constant rms input voltage? (a) – 3 dB (b) – 6 dB (c) – 12 dB (d) – 24 dB

Multiply the midrange output voltage by the voltage gain

corresponding

to the specified dB value in Table.

(a) At – 3 dB, Vout = 0.707(10 V) = 7.07 V

(b) At – 6 dB, Vout = 0.5(10 V) = 5 V

(c) At – 12 dB, Vout = 0.25(10 V) = 2.5 V

(d) At – 24 dB, Vout = 0.0625(10 V) = 0.625 V

17



Low-Frequency Amplifier Response

In looking at the low frequency ac equivalent circuit of a

capacitor coupled amplifier we can see there are three RC

circuits which will limit low frequency response. The input at

the base, the output at the collector, and the emitter.

A capacitively coupled amplifier.

The low-frequency ac equivalent circuit of the

amplifier in Figure (left) consists of three high-pass

RC circuits.

e

c

midv

r

RA

')(

18



Low-Frequency Amplifier Response – The Input RC Circuit

in

Cin

inbase V

XR

RV

22

1

in

c

CR

CfX

1

1

2

1

12

1

CRf

in

c

The input RC circuit for the BJT

amplifier is formed by C1 and the

amplifier’s input resistance and is

shown in Figure.

The input circuits effects on the

signal at a given frequency can

be more easily understood by

looking at this simplified input

circuit. The frequency at which

the gain is down by 3dB is called

the lower critical frequency (fc).

This frequency can be

determined by the formula below.

1)(2

1

CRRf

ins

c

Input RC circuit formed by the input coupling

capacitor and the amplifier’s input resistance.

19



Ex 10-3 For an input RC circuit in a certain amplifier, Rin = 1.0 kΩ and C1 = 1 μF. Neglect the source resistance. (a) Determine the lower critical frequency. (b) What is the attenuation of the input RC circuit at the lower critical frequency? (c) If the midrange voltage gain of the amplifier is 100, what is the gain at the lower critical frequency?

HzFkCR

fin

c159

)1)(0.1(2

1

2

1

1

(b) At fc, Xc1 = Rin. Therefore

707.0in

base

V

VnAttenuatio

(c) Av = 0.707 Av(mid) = 0.707(100) = 70.7

(a)

20



Low-Frequency Amplifier Response – Voltage gain roll-off at low frequency

1.010

1

101

1

101

)1001(100

)10(

222

222

1

2

in

in

in

in

inin

in

inin

in

Cin

in

in

base

R

R

R

R

RR

R

RR

R

XR

R

V

V

The decrease in voltage gain with frequency is called roll-off.

Let’s take a frequency that is one-tenth of the critical frequency (f =0.1fc).

Since Xc1 = Rin at fc, then Xc1 = 10 Rin at 0.1fc because of the inverse

relationship of XC1 and fc. The attenuation of the input RC circuit is,

therefore,

dBV

V

in

base 20)1.0log(20log20

Attenuation

The dB attenuation is

21



Low-Frequency Amplifier Response – dB/decade

The decrease in voltage

gain with frequency is

called the roll-off. A ten

times change in frequency

is called a decade. The

attenuation measured in dB

at each decade is is the

dB/decade. This typical dB

Av vs frequency illustrates

the relationship. Sometimes

roll-off is expressed in

dB/octave, which is a

doubling or halving of a the

frequency.

dB voltage gain versus frequency

for the input RC circuit.

22



Ex 10-4 The midrange voltage gain if a certain amplifier is 100. The input RC circuit has a lower critical frequency of 1 kHz. Determine the actual voltage gain at f = 1 kHz, f = 100 Hz, and f = 10 Hz.

When f = 1 kHz, the voltage gain is 3 dB less than at midrange. At – 3 dB,

the voltage gain is reduced by a factor of 0.707.

Av = (0.707)(100) = 70.7

When f = 100 Hz = 0.1fc, the voltage gain is 20 dB less than at fc. The

voltage gain at – 20 dB is one-tenth of that at the midrange frequencies.

Av = (0.1)(100) = 10

When f = 10 Hz = 0.01fc, the voltage gain is 20 dB less than at f = 0.1fc or

– 40 dB. The voltage gain at – 40 dB is one-tenth of that at – 20 dB or one-

-hundredth that at the midrange frequencies.

Av = (0.01)(100) = 1

23



Low-Frequency Amplifier Response – Phase shift in the input RC circuit

in

C

R

X11tan

o

inR

0)0(tan0

tan 11

o

in

in

R

R3.84)10(tan

10tan 11

In addition to reducing the voltage gain, the input RC circuit also

causes an increasing phase shift through an amplifier as the

frequency decreases.

o

in

in

R

R45)1(tantan 11

A decade below the critical frequency,

Xc1 = 10Rin, so

At the critical frequency, Xc1 = Rin, so

For midrange frequencies, Xc1 ≈ 0 Ω, so

Phase angle versus frequency for the input RC circuit.

Input RC circuit causes the base voltage to lead the input voltage

below midrange by an amount equal to the circuit phase angle.

24



Low-Frequency Amplifier Response – The Output RC Circuit

The output RC circuit affects the response similarly to the

input RC circuit. The formula below is used to determine

the cutoff frequency of the output circuit.

fc = 1/2(RC + RL)C3

Development of the equivalent low-frequency output RC circuit.

25



Low-Frequency Amplifier Response- The Bypass RC Circuit

The bypass RC circuit is no different in it’s effect on the gain at low

frequencies. For midrange frequencies it is assumed that XC2 ≈ 0 Ω,

effectively shorting the emitter to ground so that the amplifier gain is

Rc/r’e. As frequency is reduced, XC2 increases. The impedance from

emitter to ground increases, gain decreases. Av = Rc / (r’e + Re)

26



Low-Frequency Amplifier Response- The Bode Plot

An RC circuit and its low-frequency response. (Blue is ideal; red is actual.)

A plot of dB voltage gain versus frequency on semilog paper

(logarithmic horizontal axis scale and a linear vertical axis

scale) is called a Bode plot. A generalized Bode plot for an

RC circuit like that shown in Figure (a) appears in part (b) of

the figure.

27



High-Frequency Amplifier Response

A high-frequency ac equivalent circuit for the BJT amplifier in

Figure. Notice that the coupling and bypass capacitors are treated as

effective shorts and do not appear in the equivalent circuit. The

internal capacitances, Cbe and Cbc, which are significant only at high

frequencies, do appear in the diagram.

Capacitively coupled amplifier and its high-frequency equivalent circuit.

28



High-Frequency Amplifier Response – Miller’s Theorem in High-Frequency Analysis

Looking in from the signal source, the capacitance Cbc appears in

the Miller input capacitance from base to ground.

Cin(Miller) = Cbc(Av + 1)

Cbe simply appears as a capacitance to ac ground, as shown in

Figure, in parallel with Cin(Miller). Looking in at collector, Cbc appears

in the Miller output capacitance from collector to ground. As shown

in Figure.

v

v

bcMilleroutputA

ACC

1)(

High-frequency equivalent circuit after applying Miller’s theorem.

29



Low-Frequency Amplifier Response- The Input RC Circuit

XCtot = Rs||R1||R2||βac r’e

1/(2πfc Ctot) = Rs||R1||R2||βacr’e

fc = 1/(2π(Rs||R1||R2||βacr’e )Ctot

and

Therefore,

At high frequencies, the input circuit is as shown in Figure (a), where

βac r’e is the input resistance. By combining Cbe and Cin(Miller) in

parallel and repositioning shown in Figure (b). By thevenizing the

circuit to left of capacitor, as indicated, the input RC circuit is

reduced to the equivalent form shown in Figure (c).

Where Rs is the resistance of the signal source and Ctot = Cbe + Cin(miller).

30



Ex 10-10 Derive the input RC circuit for the BJT amplifier in Figure. Also determine the critical frequency. The transistor’s data sheet provides the following: βac = 125, Cbe = 20 pF, and Cbc = 2.4 pF.

First, find r’e as follows:

1.1125

'

26.2470

06.1

06.17.0

76.1107.26

7.4

21

2

E

e

E

E

E

BE

CCB

I

mVr

mAV

R

VI

VVVV

VVk

kV

RR

RV

31



The total resistance of the input circuit is

Rin(tot) = Rs||R1||R2||βac r’e = 600 Ω||22 kΩ||4.7 kΩ||125(11.1 Ω) = 378 Ω

Next, in order to determine the capacitance, you must calculate the

midrange gain of the amplifier so that you can apply Miller’s theorem.

991.11

1.1

'

//

')(

k

r

RR

r

RA

e

LC

e

c

midv

MHz

pF

CRtotintotin

62.1

)260)(378(2

1

))((2

1

)()(

Apply Miller’s theorem.

Cin(Miller) = Cbc(Av(mid) + 1) = (2.4 pF)(100) = 240 pF

The total input capacitance is Cin(Miller) in parallel with Cbe.

Cin(tot) = Cin(Miller) + Cbe = 240 pF + 20 pF = 260 pF

fc

32



Total Amplifier Frequency Response

Figure (b) shows a generalized ideal response curve (Bode plot) for

the BJT amplifier shown in Figure (a). The three break points at the

lower critical frequencies (fc1, fc2, and fc3) are produced by the three

low-frequency RC circuits formed by the coupling and bypass

capacitors. The break points at the upper critical frequencies, fc4 and

fc5, are produced by the two high-frequency RC circuit formed by the

transistor’s internal capacitances.

33



Total Amplifier Frequency Response - Bandwidth

An amplifier normally operates with signal frequencies between fcl

and fcu. The range (band) of frequencies lying between fcl and fcu is

defined as the bandwidth of the amplifier, as illustrated in Figure.

The amplifier’s bandwidth is expressed in units of hertz as

BW = fcu – fcl

34



General Frequency Considerations

• At frequencies above and below the midrange, capacitance and any

inductance will affect the gain of the amplifier.

• At low frequencies the coupling and bypass capacitors lower the gain.

• At high frequencies stray capacitances associated with the active device lower

the gain.

• Also, cascading amplifiers limits the gain at high and low frequencies.

The frequency response of an amplifier refers to the frequency range in which the

amplifier will operate with negligible effects from capacitors and device internal

capacitance. This range of frequencies can be called the mid-range.

34

35



Bode Plot

A Bode plot indicates the

frequency response of an

amplifier.

The horizontal scale

indicates the frequency (in

Hz) and the vertical scale

indicates the gain (in dB).

35

36



Cutoff Frequencies

The mid-range frequency

range of an amplifier is

called the bandwidth of

the amplifier.

The bandwidth is defined

by the lower and upper

cutoff frequencies.

Cutoff – any frequency at

which the gain has

dropped by 3 dB.

36

37



BJT Amplifier Low-Frequency Response

At low frequencies, coupling

capacitor (CS, CC) and bypass

capacitor (CE) reactances

affect the circuit impedances.

37

38



Coupling Capacitor (CS)

The cutoff frequency due to CS can be calculated by

sisLs

)CR(R2

1f

e21i βr||R||RR

where

38

39



cLoLC

)CRR(π2

1f

oCo r||RR

Coupling Capacitor (CC)

The cutoff frequency due to CC can be calculated with

where

39

40



Bypass Capacitor (CE)

EeLE

CRπ2

1f

)rβ

R(||RR e

sEe

21ss R||R||RR

The cutoff frequency due to CE can be calculated with

where

and

40

41



BJT Amplifier Low-Frequency Response

The Bode plot indicates

that each capacitor may

have a different cutoff

frequency.

It is the device that has

the highest lower cutoff

frequency (fL) that

dominates the overall

frequency response of the

amplifier.

41

42



Roll-Off of Gain in the Bode Plot

The Bode plot not only

indicates the cutoff

frequencies of the various

capacitors it also indicates

the amount of attenuation

(loss in gain) at these

frequencies.

The amount of attenuation

is sometimes referred to as

roll-off.

The roll-off is described as

dB loss-per-octave or dB

loss-per-decade.

42

43



Roll-off Rate (-dB/Decade)

-dB/decade refers to the

attenuation for every 10-fold

change in frequency.

For attenuations at the low-

frequency end, it refers to

the loss in gain from the

lower cutoff frequency to a

frequency that is one-tenth

the cutoff value.

In this example:

fLS = 9kHz gain is 0dB

fLS/10 = .9kHz gain is –20dB

Thus the roll-off is 20dB/decade

The gain decreases by –20dB/decade

43

44



Roll-Off Rate (–dB/Octave)

-dB/octave refers to the

attenuation for every 2-fold

change in frequency.

For attenuations at the low-

frequency end, it refers to

the loss in gain from the

lower cutoff frequency to a

frequency one-half the cutoff

value.

In this example:

fLS = 9kHz gain is 0dB

fLS / 2 = 4.5kHz gain is –6dB

Therefore the roll-off is 6dB/octave.

This is a little difficult to see on this graph because

the horizontal scale is a logarithmic scale.

44

45



FET Amplifier Low-Frequency Response

At low frequencies,

coupling capacitor (CG,

CC) and bypass capacitor

(CS) reactances affect the

circuit impedances.

45

46



Coupling Capacitor (CG)

GisigLC

)CR(Rπ2

1f

Gi RR

The cutoff frequency due to

CG can be calculated with

where

46

47



Coupling Capacitor (CC)

CLoLC

)CR(Rπ2

1f

dDo r||RR


CC can be calculated with

where

47

48



Bypass Capacitor (CS)

SeqLS

CRπ2

1f

Ωrm

Seq dg

1||RR


CS can be calculated with

where

48

49



FET Amplifier Low-Frequency Response

The Bode plot indicates that

each capacitor may have a

different cutoff frequency.

The capacitor that has the

highest lower cutoff

frequency (fL) is closest to the

actual cutoff frequency of the

amplifier.

49

50



Miller Capacitance

Any p-n junction can develop capacitance. In a BJT amplifier,

this capacitance becomes noticeable across:

• The base-collector junction at high frequencies in

common-emitter BJT amplifier configurations

• The gate-drain junction at high frequencies in common-

source FET amplifier configurations.

These capacitances are represented as separate input and output

capacitances, called the Miller Capacitances.

50

51



Miller Input Capacitance (CMi)

Note that the amount of

Miller capacitance is

dependent on inter-

electrode capacitance

from input to output (Cf)

and the gain (Av).

fvMi )CA(1C

51

52



Miller Output Capacitance (CMo)

If the gain (Av) is

considerably greater

than 1, then

fMo C C

52

53



BJT Amplifier High-Frequency Response

Capacitances that affect the

high-frequency response are

• Junction capacitances

Cbe, Cbc, Cce

• Wiring capacitances

Cwi, Cwo

• Coupling capacitors

CS, CC

• Bypass capacitor

CE

53

54



Input Network (fHi) High-Frequency Cutoff

iThiHi

CRπ2

1f

i21sThi R||R||R||RR

bcvbeWi

MibeWii

)CA(1CC

CCCC

where

and

54

55



Output Network (fHo) High-Frequency Cutoff

MoceWoo CCCC

oLCTho r||R||RR

oThoHo

CRπ2

1f

where

and

55

56



hfe (or ) Variation

The hfe parameter (or ) of a

transistor varies with

frequency

)C(Crβπ2

1f

bcbeemidβ

56

57



BJT Amplifier Frequency Response

Note the highest lower cutoff frequency (fL) and the lowest upper cutoff

frequency (fH) are closest to the actual response of the amplifier.

57

58



FET Amplifier High-Frequency Response

Capacitances that affect the

high-frequency response are

• Junction capacitances

Cgs, Cgd, Cds

• Wiring capacitances

Cwi, Cwo

• Coupling capacitors

CG, CC

• Bypass capacitor

CS

58

59



Input Network (fHi) High-Frequency Cutoff

iThiHi

CRπ2

1f

GsigThi R||RR

MigsWii CCCC

gdvMi )CA(1C

59

60



Output Network (fHo) High-Frequency Cutoff

oThoHo

CRπ2

1f

dLDTho r||R||RR

ModsWoo CCCC

gdv

Mo CA

11C

60

61



Multistage Frequency Effects

Each stage will have its own frequency response,

but the output of one stage will be affected by

capacitances in the subsequent stage. This is

especially so when determining the high frequency

response. For example, the output capacitance (Co)

will be affected by the input Miller Capacitance

(CMi) of the next stage.

61

62



Multistage Amplifier Frequency Response

Once the cutoff frequencies have been determined for each stage (taking into

account the shared capacitances), they can be plotted.

Note the highest lower cutoff frequency (fL) and the lowest upper cutoff

frequency (fH) are closest to the actual response of the amplifier.

62

63



Square Wave Testing

In order to determine the frequency

response of an amplifier by

experimentation, you must apply a wide

range of frequencies to the amplifier.

One way to accomplish this is to apply a

square wave. A square wave consists of

multiple frequencies (by Fourier

analysis: it consists of odd harmonics).

63

64



Square Wave Response Waveforms

If the output of the

amplifier is not a perfect

square wave then the

amplifier is ‘cutting’ off

certain frequency

components of the square

wave.

64

65



Questions

66



Questions

TOGETHER WE CAN MAKE THE DIFFERENCE

KNOWLEGE IS A POWER

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