1

FETFREQUENCY RESPONSE

LOW FREQUENCY

2

LOW FREQUENCY – COMMON SOURCE

R Si C 1

R D C 2

+V DD

v i R G

R S

C 3

R L

v o

3

vi gmVgs

S

DG

Vgs

+

-

RS

RDRLRG

RSi

C3

C1C2

Input RC circuit Output RC

circuit

Bypass RC circuit

Low-frequency equivalent circuit

4

The cutoff frequencies defined by the input , output and bypass circuits can be obtained by the following formulas.

112

1

CRf

Cc where

RC1=RSi+RG

222

1

CRf

Cc where

RC2=RD+RL

332

1

CRf

Cc where

RC3=RS||1/gm

Input RC circuit

Output RC circuit

Bypass RC circuit

5

EXAMPLE

Determine the lower cutoff frequency for the FET amplifier. Given K = 0.4mA/V2, VTN= 1V, = 0

10K

v i

R Si C 1

0.01 mFR G

1 MWR S

1 kW

C 2

0.5 mF

R D

4.7 kW

R L

2.2 kW

v o

+V DD

20 V

C 3

2 mF

6

HzMKCR

fC

c 8.15)01.0)(110(2

1

2

1

11

m

HzmKCR

fC

c 73.238)2)(211(2

1

2

1

33

m

HzKKCR

fC

c 13.46)5.0)(2.27.4(2

1

2

1

22

m

Input RC circuit

Output RC circuit

Bypass RC circuit

7

Since fc in bypass RC circuit, is the largest of the three cutoff frequencies and is separated from the next highest frequency by more than two octaves, the dominant pole approximation is applicable. Hence the low cutoff frequency fH for the amplifier is

fH = fc = 238.73Hz

8

LOW FREQUENCY – COMMON GATE

Analyse for mid-band gain and lower cut-off frequency

V 1TNV 2mA/V 3NK 0 pF 15gsC pF 4gdC

9

.

.

R D

R L

R i C C 1

C C 2

R S

V DD = +5 V

V SS = -5 V

v o

v i

Q

2 kW

10 kW

5 kW

4 kW

1 mF

2 mF

10

R S

R D

V DD = +5 V

V SS = -5 V

5 kW

10 kW

I G = 0

I D

I D

V S

V D

DC ANALYSIS

V 349.1GSV

mA/V 093.2mg

11

AC ANALYSIS – MID-BAND

. .

.

R i

2 kW

R S

R D R Lg mv gs

v i

v o

G D

S

10 kW

5 kW 4 kW

+v s

-

+

v gs

-i i

i s

RRR DL //'

12

sgs vv

'Lgsmo Rvgv

'Lsm Rvg

'Lms

o Rgv

v

Ssiii RiRiv

gsmis vgii

smi

si

S

s vgR

vv

R

v

smi

s

S

s

i

i vgR

v

R

v

R

v

13

iSmSi

S

i

s

RRgRR

R

v

v

i

s

s

o

i

ov v

v

v

v

v

vA

iSmSi

SLm RRgRR

RRg

'

14

210093.2102

105//4093.2

vA

V/V 864.0

15

AC ANALYSIS – LOW FREQUENCY

. .

.

R i

2 kW

R S

v i

S

10 kW

C C 1

C C 2

R D

5 kW 4 kWR L

v o

1 mF

2 mF

g mv gs

DG

+

v gs

-

R in

16

W

4561010093.21

10

1 43

4

Sm

Sin Rg

RR

Effective resistance across CC1 is;

W k 456.2456.021 iniC RRR

rad/s 40710110456.2

1163

111

CCL CR

Hz 8.642

407

21

1

LLf

17

rad/s 56102109

1163

222

CCL CR

Effective resistance across CC2 is;

W k 9542 LDC RRR

Hz 8.82

56

22

2

LLf

Since the two frequencies are separated by more than two octaves, the dominant pole approximation can be applied. The lower cut-off frequency is the higher of the two frequencies i.e. 64.8 Hz

18

LOW FREQUENCY – COMMON DRAIN

.

.

R 1

R 2

R i

R S R L

C 1

C 2

V DD = 10 V

v o

v i 1 kW

17.1 MW

50 MW

10 kW8 kW

10 µF

1 µF

mA 12.3DSSI

V 25.1tV

pF 4.3gdC pF 6.3gsC

0

Analyse for AM & fL

19

R 1

R 2

R S

V DD = 10 V

17.1 MW

50 MW

8 kW

I D

I D

V G

I G = 0

V S

DC ANALYSIS

V 548.0GSV

mA/V 8.2mg

20

.

.

.

+

v gs

_

g mv gsR G

R i

v i

R L '

G D

Sv o

+

v g

_

1 kW

AC ANALYSIS – MID-BAND

21 // RRRG

50//1.17

W M 74.12

LSL RRR //'

10//8W k 44.4

21

ogsg vvv oggs vvv

'' LogmLgsmo RvvgRvgv

'1

'

Lm

Lm

g

o

Rg

Rg

v

v

iisG

Gg vv

RR

Rv

1i

g

v

v

22

'1

'

Lm

Lm

i

g

g

o

i

oM Rg

Rg

v

v

v

v

v

vA

Substituting values;

V/V 93.0

10//812.31

10//812.3

1

Om

OmM Rg

RgA

23

AC ANALYSIS – LOW FREQUENCY

.

.

.

g mv gsR G

R i

R S R L

C 1 G D

S

v i

v o

+

v g

_

+

v gs

_ C 2

24

The effective resistance across C1 is;

WWW M 74.12M1.17//50k 11 GsC RRR

mHz 2.110101074.122

1

2

166

11

CR

fCs

L

25

To find the effective resistance across C2, we apply a voltage source vx as follows;

..

R i C 1

R G

G D

+

v gs

_

v x

i x

R S

g mv gs

i S

S

26

x

xi i

vR

S

xS R

vi

xgs vv

At node S;

xgsmS ivgi or xxmS

x ivgR

v

xS

Smx v

R

Rgi

1

Sm

S

x

xi Rg

R

i

vR

1

27

The output circuit is simplified to;

R i R L

C 2

The effective resistance across C2 is;

iLC RRR 2

Sm

SL Rg

RR

1

W

k 342.1080000028.01

800010000

28

222 2

1

CRf

CL

Hz 4.151010342.102

163

By dominant pole approximation, the lower cut-off frequency is;

Hz 4.152 LL ff

29

• The cut-off frequency is also known as 3-dB frequency or half-power frequency or corner frequency.

• The lower cut-off frequency is determined by the components (R and C) external to the transistor.