Chapter 6 frequency effects in amplifiers

LOW-FREQUENCY EFFECT OF BYPASS AND COUPLING CAPACITORS .... 6-7 Bode plot of low frequency effect of coupling capacitor ..... When the frequency is low enough, the coupling and bypass ca

Val de Loire Program p.77

CHAPTER 6:

FREQUENCY EFFECTS IN AMPLIFIERS

Table of Contents

6.1 INTRODUCTION 79

6.2 BODE PLOTS AND FREQUENCY RESPONSE 80

6.3 LOW-FREQUENCY EFFECT OF BYPASS AND COUPLING CAPACITORS 83

6.3.1 Low-frequency effect of bypass capacitor 84

6.3.2 Low-frequency effect of coupling capacitor 87

6.4 HIGH FREQUENCY HYBRID- BJT MODEL 90

6.5 HIGH-FREQUENCY FET MODELS 92

Table of Figures Fig 6-1 Frequency response and transfer function 80

Fig 6-2 Bode plots of Ex 6.1 82

Fig 6-3 CE amplifier 84

Fig 6-4 Small-signal equivalent circuit 85

Fig 6-5 Low- and mid-frequency asymptotic Bode plot 87

Fig 6-6 Low frequency effect of coupling capacitor 88

Fig 6-7 Bode plot of low frequency effect of coupling capacitor 90

Val de Loire Program p.78

Fig 6-8 High frequency hybrid- bjt model 91 Fig 6-9(a) Mid-frequency small-signal current-source FET model 92 Fig 6-9(b) High-frequency small-signal current-source FET model 92

Val de Loire Program p.79

CHAPTER 6:

FREQUENCY EFFECTS IN AMPLIFIERS

6.1 INTRODUCTION

In the previous chapters on amplifiers, the coupling and bypass capacitors were considered to be ideal shorts and the internal transistor capacitances were considered to be ideal opens This treatment is valid when the frequency is in an amplifier’s midrange

As you know, capacitive reactance decreases with increasing frequency and vice versa When the frequency is low enough, the coupling and bypass capacitors can no longer be considered as shorts because their reactances are large enough to have a 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 a significant effect on the amplifier operation

In this chapter, you will study the frequency effects on amplifier gain and phase shift

Val de Loire Program p.80

6.2 BODE PLOTS AND FREQUENCY RESPONSE

Fig 6-1 Frequency response and transfer function

   /  

T s N s D s is the Laplace-domain transfer function

In amplifier analysis, transfer functions are the current-gain ratio

  i 

T s A s and voltage-gain ratio T s A s v  

For convenience, with s j , we make the following definitions:  

1 Call A j , the frequency transfer function

2 Define M  A j  , the gain ratio

Val de Loire Program p.81

3 Define M db 20 logM 20 logA j  , the amplitude ratio, measured in decibels (db)

The graph of M (simultaneously with db  if desired) versus the logarithm of the input signal frequency (positives values only) is called a

Bode plot

Example 6.1 A simple first-order network has Laplace-domain

transfer function and frequency transfer function







1 ( )

1

A s

s and   

1 ( )

1

A j

j

Where  is the system time constant

(a) Determine the network phase angle  and the amplitude ratio

db

M

(b) Construct the Bode plot for the network

Solution

(a) In polar form, the given frequency transfer function is













1 2

1 2

tan

1 1

1

A j

Hence,  tan1





2 2

1

1

db

(b) If values of M and db  are calculated and plotted for various values of , then a Bode plot is generated This is done in Fig 6-2,

Val de Loire Program p.82

where  is given in terms of time constants  rather than, say, hertz

This particular system is called a lag network because its phase angle  is negative for all 

Fig 6-2 Bode plots of Ex 6.1 Example 6.2 A simple first-order network has Laplace-domain

transfer function and frequency transfer function

  1 

A s s and A j( ) 1  j 

Determine the network phase angle  and the amplitude ratio M , db

and discuss the nature of the Bode plot

Solution

After A j  is converted to polar form, it becomes apparent that

tan1 

Val de Loire Program p.83

And M db 20 logA j  20 log 12 10 log 1 2

Thus, the complete Bode plot consists of the mirror images about zero of M and db  of Fig 6-2 Since here the phase angle  is

everywhere positive, this network is called a lead network

A break frequency, cutoff frequency or corner frequency is the

frequency 1/ For a simple lag or lead network, it is the frequency at 

which M2 A j 2 has changed by 50 percent from its value at  0 Corner frequencies serve as key points in the construction of Bode plots

6.3 LOW-FREQUENCY EFFECT OF BYPASS AND COUPLING CAPACITORS

As the frequency of the input signal to an amplifier decreases below the midfrequency range, the voltage (or current) gain ratio decreases in magnitude

The low-frequency cutoff point  L is the frequency at which the gain

ratio equals 1 / 2 ( 0.707) times its midfrequency value, or at which

db

M has decreased by exactly 3 db from its midfrequency value

Low-frequency amplifier performance (attenuation, really) is a

consequence of the use of bypass and coupling capacitors to fashion the

dc bias characteristics

When viewed from the low-frequency region, such amplifier

response is analogous to that of a high-pass filter

Val de Loire Program p.84

6.3.1 Low-frequency effect of bypass capacitor

Example 6.3 For the amplifier of Fig.6-3 , assume that C   C

but that the bypass capacitor C cannot be neglected Also, let E

0

re oe

h h  and R  i 0

(a) Find an expression that is valid for small signals and that gives the voltage-gain ratio A s at any frequency v  

(b) find the voltage-gain ratio at low frequencies

(c) the voltage-gain ratio at higher frequencies

(d) the low-frequency cutoff point

(e) Sketch the asymptotic Bode plot for the amplifier (amplitude ratio only)

Fig 6-3 CE amplifier

Val de Loire Program p.85

Solution

(a) The small-signal low-frequency equivalent circuit (with the approximation implemented) is displayed in Fig 6-4 In the Laplace domain, we have

 1 / 

1

||

Fig 6-4 Small-signal equivalent circuit

Note that

i  h  i

Using KVL: v i h i ie b Z i E e  h ie h fe 1Z Ei b

By Ohm’s law:    ||  fe C L

h R R



 

1 1

fe C L

v

h R R

A s



  

Val de Loire Program p.86

(b) The low-frequency voltage-gain ratio is obtained by letting

 0

s :

 





0

0 lim

1

fe C L L

h R R v

A

(c) The higher-frequency (midfrequency) voltage-gain ratio is obtained by letting s   :

 

 

 



1/

fe C L

fe C L

h R R

A

h R R

(d) A s can be rearranged to give: v 

 



 

1

1

v

E E ie

Which clearly is of the form









1 2

1 1

s

s

With 



1 1

E E

C R





1 2

1

E E ie

R C h

Typically, h fe  1 and h R fe E h , so a reasonable approximation ie

of 2 is

Val de Loire Program p.87

2  1

/

E ie fe

With R E h ie/h , fe 2 is an order of magnitude greater than 1 (e) The low- and midfrequency asymptotic Bode plot is depicted in Fig 6-5

Fig 6-5 Low- and mid-frequency asymptotic Bode plot 6.3.2 Low-frequency effect of coupling capacitor

Example 6.4 In the circuit of Fig 6-6(a), the impedance of the coupling

capacitor is not negligibly small

(a) Find an expression for the voltage-gain ratio

 

 v  o/ S

(b) Determine the midfrequency gain of this amplifier

(c) Determine the low-frequency cutoff point  L, and sketch an asymptotic Bode plot

Val de Loire Program p.88

(a)

(b)

Fig 6-6 Low frequency effect of coupling capacitor

Solution

(a) The small-signal low-frequency equivalent circuit is shown in Fig 6-6(b) By Ohm’s law:



 || 1/

S S

V I

Val de Loire Program p.89

Then current division gives:

But Ohm’s law requires that

 

V h R I

Thus,  



   || 1

fe C B o

h R R Cs V

A s

Now, with s  j , it magnitude is 







fe C B

h R R C

(b) The midfrequency gain follows from letting s  j    We have:





fe C B mid

h R R A

(c) Cutoff frequency:

1 1/

||

L

The asymptotic Bode plot is sketched in Fig 6-7

Val de Loire Program p.90

Fig 6-7 Bode plot of low frequency effect of coupling capacitor

Because of capacitance that is inherent within the transistor, amplifier current- and voltage-gain ratios decrease in magnitude as the frequency of the input signal increases beyond the midfrequency range

The high-frequency cutoff point  H is the frequency at which the gain ratio equals 1 / 2 times its midfrequency value, or at which M db

has decreased by 3 db from its midfrequency value The range of frequencies above  H is called the high-frequency region Like  L,  H

is a break frequency

The most useful high-frequency model for the BJT is called the

hybrid-  equivalent circuit In this model, the reverse voltage ratio h re

Val de Loire Program p.91

and output admittance h are assumed negligible The base ohmic oe resistance r bb, assumde to be located between terminal B and the base

junction, has a constant value (typically 10 to 50 ) that depends directly

on the base width The base-emitter-junction resistance r b e is usually much larger than r bb and can be calculated as

b e

r

Capacitance C  is the depletion capacitance associated with the reverse-biased collector-base junction; its value is a function of V BCQ

Capacitance C  C  is the diffusion capaciatnce associated with the forward-biased base-emitter junction; its value is a function of I EQ

Fig 6-8 High frequency hybrid- bjt model

Val de Loire Program p.92

6.5 HIGH-FREQUENCY FET MODELS

The small-signal high-frequency model for the FET is an extension

of the midfrequency model Three capacitors are added: C between gate gs

and source, C between gate and drain, and gd C between drain and ds

source They are all of the same order of magnitude - typically 1 to

10pF

Fig 6-9(a) Mid-frequency small-signal current-source FET

model

Fig 6-9(b) High-frequency small-signal current-source FET

model