Chapter 5 small signal midfrequency JFET

Val de Loire Program p.67 CHAPTER 5: SMALL-SIGNAL MIDFREQUENCY FET Table of Contents 5.1.. INTRODUCTION In this chapter, all voltage and current signals are considered to be in the

Val de Loire Program p.67

CHAPTER 5:

SMALL-SIGNAL MIDFREQUENCY FET

Table of Contents

5.1 INTRODUCTION 68

5.2 SMALL-SIGNAL EQUIVALENT CIRCUITS FOR THE FET 68

5.3 CS AMPLIFIER ANALYSIS 70

5.4 CD AMPLIFIER ANALYSIS 72

5.5 CG AMPLIFIER ANALYSIS 74

Table of Figures Fig 5.1 Drain charactersistics 69

Fig 5.2 Small-signal models for the CS FET 70

Fig 5.3 CS Amplifier 71

Fig 5.4 CD Amplifier 73

Fig 5.5 CG Amplifier 75

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CHAPTER 5:

SMALL-SIGNAL MIDFREQUENCY FET

5.1 INTRODUCTION

In this chapter, all voltage and current signals are considered to be in the midfrequency range, where all capacitors appear as short circuits

There are three basic FET amplifier configurations: the source (CS), drain (CD) or source-follower (SF), and common-gate (CG) configurations The CS amplifier, which provides good voltage

amplification, is most frequently used The CD and CG amplifiers are applied as buffer amplifiers (with high input impedance and near-unity voltage gain) and high-frequency amplifiers, respectively

5.2 SMALL-SIGNAL EQUIVALENT CIRCUITS FOR THE FET

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Fig 5-1 Drain charactersistics

From the FET drain characteristics, it is seen that if i is taken as D

the dependent variable, then

D GS DS

i f v v

For small excursions (ac signals) about the Q point, i D i ; thus, d

application of the chain rule:

d D D m gs ds

ds

r

Where g and m r are defined as follows: ds

Transconductance:    

m

GS Q GS Q

g

Source-drain resistance:    

ds DS Q DS Q

or  

ds

D Q D Q

r

As long as the JFET is operated in the pinchoff region, i G i g  0, so that the gate acts as an open circuit This leads to the current-source equivalent circuit Either of these models may be used in analyzing an amplifier, but one may be more efficient than the other in a particular circuit

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Fig 5-2 Small-signal models for the CS FET 5.3 CS AMPLIFIER ANALYSIS

A simple common-source amplifier is shown in Fig 5-3(a) and its associated small-signal equivalent circuit is displayed in Fig 5-3(b) Source resistor R is used to set the Q point but is bypassed by s C for s

midfrequency operation

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Fig 5-3 CS Amplifier Example 5.1 In the CS amplifier, let R D3k ,   60 ,

30 

ds

r k

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

i

v A

v

(b) Evaluate A using the given typical values v

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Solution

(a) By voltage division,



 



D

D ds

R

Substitution of v gs v and rearrangement give : i



  



v

i D ds

A

(b) The given values lead to

  5.45

v A

Where the minus sign indicates a 180 phase shift between 0 v and i o

v

5.4 CD AMPLIFIER ANALYSIS

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Fig 5-4 CD Amplifier

A simple common-drain (or source-follower) amplifier is shown in Fig 5-4(a); its associated small-signal equivalent circuit is given in Fig 5-4(b), where the voltage-source equivalent of Fig 5-2(b) is used to model the FET

Example 5.2 In the CD amplifier, let R S 5k ,   60 ,

30 

ds

r k

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

i

v A

v

(b) Evaluate A using the given typical values v

Solution

(a) By voltage division,





S gd S

R v R

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Replacement of v by gd v and rearrangement give i





v

A

(b) Substitution of the given values leads to

 0.895

v A

Note that the gain is less than unity; its positive value indicates that

o

v and v are in phase i

5.5 CG AMPLIFIER ANALYSIS

Its small-signal equivalent circuit, incorporating the current-source model of Fig 5-2(a), is given:

(a) CG amplifier

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(b) Small-signal equivalent circuit

Fig 5-5 CG Amplifier Example 5.3 In the CG amplifier, let R D 1k ,  

 2 10 3

m

30 

ds

r k

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

i

v A

v

(b) Evaluate A using the given typical values v

Solution

(a) By KCL, i r i d g v Applying KVL around the outer loop m gs

gives:

o d m gs ds gs

v i g v r v

But v gs  v and   i o

d

D

v i

R ; thus

o

D

v

R

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

1

m ds D o

v

i D ds

v A

(b) Substitution of the given values yields

 1.97

v A