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Graphical solutions

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Graphical solutions. Fig. 5.18. Small signal equivalent circuits. Assuming operation in the saturation region. For small signal case is negligible. Where. The gate current for FET is negligible. g m = transconductance. Small signal equivalent circuit. Where,. - PowerPoint PPT Presentation
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Page 1: Graphical solutions

1

Page 2: Graphical solutions

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Graphical solutions

Page 3: Graphical solutions

3

Fig. 5.18

Page 4: Graphical solutions

4

Page 5: Graphical solutions

5

Assuming operation in the saturation region

For small signal case is negligible

The gate current for FET is negligible

Where,

Small signal equivalent circuit

gm = transconductance

Where

Small signal equivalent circuits

Page 6: Graphical solutions

6

μn- surface mobility of electronCox- capacitance of gate per unit area

We know that

and

K

IVV

DQtoGSQ

But from (5.3)

Therefore

Dependence of gm on Q-point and device parameters

Page 7: Graphical solutions

More complex equivalent circuits

7

• At higher frequencies small capacitance have to be added between device terminals

• Also in the saturation region iD versus VDS is considered to be constant. This is not actually the case. The drain current, iD increases slightly as VDS increases. In order to take care of that we must add a drain resistance rd in the small signal model.

Page 8: Graphical solutions

Example

8

Determine the values of gm and rd the MOSFET characteristics shown below.

From equation 5.34, we have, obtain iD = 6.7 mA at VDS = 4 V and iD = 8 mA at VDS = 14 V.

Thus, the reciprocal of rd is calculated as:

S

V

mA

v

i

r DS

D

d

31013.0414

7.681

Thus, rd = 7.7 KΩ

Page 9: Graphical solutions

Common source amplifier

9

C1, C2 – Coupling capacitors short circuit for AC signals and open circuit for DC bias calculation

CS – bypass capacitor small impedance for AC

Voltage Gain

Ldd

L

RRr

R111

1'

'0 )( Lgsm Rvgv

gsin vv

'0Lm

inv Rgv

vA

(5.38)

(5.39)

(5.40)

(5.41)

Page 10: Graphical solutions

Common source amplifier contd..

10

= 0 implies 1/rd = 0 or rd =

Input resistance

Output resistance

Example 5.6 (Modified)

Find the midband voltage gain, input resistance, and output resistance of the amp.

dd Rr

R11

10

21 RRRi

vR G

in

inin (5.42)

(5.43)

mW

mL

VV

VV

VAKP

t

DD

450

15

0

1

20

/40

0

2

VRR

RVV DDG 4

105

10206

6

21

2

Bias circuit

Thevenin Equivalent

Circuit

Page 11: Graphical solutions

Common source amplifier contd..

11

or

kRrR

rRR

MRRR

RgV

VA

rrask

RRr

R

mSIL

WKPg

VVmAIVVVV

KP

L

WK

KR

KVRKVRKVRV

Ddd

dD

in

Lmin

V

dd

Ldd

L

DQm

DSQDQGSQGSQ

s

tsGstsGSQ

411

5

4

4.5107

20109.1

]01

,,[;7

20

10101

1041

1111

1

3.1108.115

450104022

76.12,44.1,2.2,55.2

106.02

1040

15

450

2)(

2

14)12(

0

21

33'0

33

'

36

21

36

00

Page 12: Graphical solutions

The Source Follower

12

Section 5.6

Recall the similarity with Emitter Follower

Drain is grounded

Source is not grounded

R1 Gate to ground (drain)

R2 Gate to ground (drain)

RS Source to ground (drain)

gmvgs & rd drain to source

To draw the small-signal ac equivalent circuit, we have to first draw the following circuit like remembering that:

Thus,

Small-signal ac equivalent circuit

Page 13: Graphical solutions

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The Source Follower contd…

How to calculate the output resistance ?

Page 14: Graphical solutions

14

Current gain:

The current gain Ai = Av Rin/RL , like before.

Fig. 5.33

Page 15: Graphical solutions

15

Example contd….


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