Examples Slides

8/3/2019 Examples Slides

1/28

Example : (a) Determine the a1 coefficient in the transfer

function of the amplifier stage whose equivalent circuit is

displayed in the figure. (b) Using the numerical valuesindicated, approximate the upper 3-dB frequency fH.

Solution :

(a) The circuit has three capacitors; hence we have

dsdsgdgdgsgs CRCRCRa000

1!


2/28

To evaluate the three time constants, and must

be found. By suppressing and open-circuiting the

capacitances, we can identify as depicted in the figure. Asthe same current exists in RD, rd, and QVgs, these elements

can be converted to their current source [QVgs/(rd + RD)]

parallel resistance (RD + rd) equivalent. This combination is in

parallel with Rs and can be reconverted to its Thevenin

equivalent as shown in the next figure. The equivalent voltagesource and series resistance Rs are

00, gdgs RR

0

dsR

svo

gsR

gsV'Q

)(||

'

dDSADd

A

rRRRRr

R!

!

Q

Q


3/28

Application of a test source Vt and computation of the current

It as indicated in the figure gives . With Vgs = Vt,

use of KVL gives

tt

o

gs IVR /!

As

t

t

RR

VI

!

)1( 'Q

from which

'1 Q

!! As

t

togs

RR

I

VR


4/28

The resistance can be compared form the circuit in the

figure. Application of KCL at D gives

o

gdR

21 III t !

)(21IIRRIRIRIV

tsstsstgs !!


5/28

Figure. The high-frequency equivalent circuit of the

common-source stage with source resistance.


6/28

Figure. (a) Circuit used to find . (b) Circuit in (a) with thedrain-source portion (Rs, QVgs, rd, and RD) replaced by a

Thevenin equivalent QVgs RA.

0gsR


7/28

Application of KVL for the right-hand loop gives

gsdsD VrRIRI Q! )(12

Combination of the above equations and solving for I2 yields

t

SdD

ssd

2 I)1(RrR

R)1(RrI Q

QQ!

The voltage Vt using KVL around the outside loop, is

D2sttRIRIV !


8/28

Substitution of the above equations and formation of Vt/Itgives

)1(RrR

]R)1(Rr[RR

I

VR

SdD

ssdD

s

t

t0

gdQ

QQ!!

The third resistance is obtained from the use of the figure.

Note that no current exists in Rs, so that Vgs appears as

indicated across Rs. Application of KCL requires

where

0

dsR

21 IIIt !

d

gst

1r

VVI

Q! and

D

gst

R

VVI

!

2


9/28

Figure. Circuit used to evaluate (a) and (b) for the

stage in previous figure.

0

gdR0

dsR


10/28

The control voltage Vgs = I2RS. Combining these equations

and solving for It,

we obtain

SdD

DSd

t

t

dsRrR

RRr

I

VR

)1(

)(0

Q

!!

(b) Numerical evaluation of a1 is as follows :

14.32080

96.116096.1)8020(||2)(||

160802

!

v!

!d

;!!!

!!!

Dd

A

dDSA

dm

Rr

RkrRRR

xrg

QQ

Q


11/28

and

;!

!

v!

;!

!

k17.42)1601(8020

)20

2(80R

)1601(28020

]2)1601(3.016080[203.0R

k546.0

14.31

96.13.0R

0ds

0

gd

0

gs

nsa

5.

29

5.117.416.

213

546.01

!vvv!

Thus

and

ZHMH

a

f 40.5105.292

1

2

1

9

1

!vv

!!

TT


12/28

Example : The parameter values used in the CE-CE cascade

in the figure are as follows: Rs = 600;, RC1 = 1.5k;, RC2 =

600;, rT1 = 1.2k;, gm1 = 0.1 , CT1=24.5pF, CT1= 24.5pF, CQ1= 0.5pF, rT2 = 2.4k;, gm2 = 0.05 , CT2 = 19.5Pf and CQ1=

0.5pF

(a) Determine the approximate value of fH and the

approximate location of the dominant pole. (b) Determine the

approximate location of the closest nondominant pole and

comment on the validity of the dominant-pole approximation.


13/28

Figure. (a) A CE-CE cascaded amplifier and (b) its

equivalent circuit at high frequencies.


14/28

Figure. Circuit used to compute the zero-frequencyresistances for CE-CE cascade in the previous figure.


15/28

Figure : Equivalent circuits used to calculate (a) , (b)

and (c)

1

22R 344R

2

44R


16/28

Solution :

(a) To obtain fH, we first evaluate the resistances

Thus

;!! kR 40.012006000

11

;!!! krRR CL 923.04.250.1211 Tand

1022

2.38923.0923.0100140.0 ;!v! kR

;!! kR 923.04.25.1033

;!v! kR 2.296.06.0501923.0044


17/28

nsa 5.612.295.05.19923.02.385.05.2440.01 !vvvv!

and

MHzfH

59.2105.612

19!

vv!

T

The dominant pole is located at p1, where

s/rad..a

p 79

1

1 1063110561

11v!

v!!

It is on interest to determine the values of fH1 and fH2 of the

amplifier. The first two terms in the calculation of a1 give

a11, the last pair of terms a12.


18/28

These are

nsa 9.

2811

! nsa 6.3212

!

from which

MHzafH 51.5109.282

1

2

19

11

1 !vv!!TT

and

MHza

fH

88.4106.322

1

2

19

12

2!

vv!!

TT


19/28

4

3

443

0

334

2

442

0

223

2

332

0

224

1

441

0

113

1

331

0

111

1

221

0

112CRCRCRCRCRCRCRCRCRCRCRCRa !

Correlation of the preceding equation above with the circuitin the figure indicates that only two terms, previously

unidentified, must be computed. In the figure we observe that

with C2 open-circuited, the portion of the circuit containing C1is decoupled from the second stage (C3 and C4). The

resistances seen by the C3 and C4 are thus independent ofwhether C1 is open-or short-circuited.

Hence , and the second and third

terms in the expression for a2 can be written as

0

44

1

44

0

33

1

33 RRandRR !!

121

0

114

0

443

0

331

0

11 aCR]CRCR[CR !


20/28

The resistance is computed from the circuit in the figure.

With C3 and C4 open-circuited, , is simply the value

of for the loaded first stage.

1

22R

1L

1

22RR !

1

22R

Similarly, short-circuiting C3 completely decouples the first

stage from the calculation of , which, as seen in the figure,

is RC2, and the value of for the second stage. Thus thepair of terms containing and are just the a2,

coefficients of the individual stages.

3

44R3

44R 122

R3

44R


21/28

The remaining pair of short-circuit resistance is obtained by

short-circuiting C2. It is noted in the figure that short-circuiting

C2 causes the voltage across the source gm1vT1 to become vT1.This v-I relationship represents a resistance vT1/ gm1vT1 = 1/gm1as shown in the circuit in the figure. As noted in the figure.

1/gm1 is much smaller than either RC1 or and hence makes

the parallel combination of these resistances approximateluy

1/gm1. The circuit in the figure is that used to compute and

except that the effective source resistance is 1/gm1 instead

of RC1. Thus

0

11R

033R

0

44R

0

11

1

12

2

33 ||1

|||| RgRrR mCT! and

2C2C2m

2

33

2

44 R)Rg1(RR !


22/28

Combination of all the terms in the three previous paragraphs

gives

)CRCR(CRCRCR)aCR(CRa4

2

443

2

332

0

2242C3

0

331221L1

0

112!

The resistances and are2

33R2

44R

;!! k01.04.0||01.0||5.1||4.2R233

;!v! k91.06.0)6.0501(01.0R244


23/28

Use of these and previously determined values in part a

yields

vvvvv! 6.05.19923.06.325.0923.05.2404.02a

S18103425.091.05.19010.05.02.385.0 v!vvv

The angular frequency of the pole is

s/rad1080.110342

105.61

a

ap

8

18

9

2

1

2 v!v

v!!


24/28

and

MHz6.282

1080.1

2

p

f

8

2

2 !T

v

!T!

,/1079.17

1sradp v!

sradp /101.16 72 v!

MHzf 85.21!

.6.252 MHzf !and


25/28

Example : The cascode amplifier has RC1 = RC2 = 1.5k; and

Rs = 300;. The transistors are identical and have rT = 2k;, gm= 0.05 , F

0= 100, C

T= 19.5 pF, C

Q= 0.5 pF, and [

T= 2.5 x 109 rad/s. (a) Determine fH for the circuit. (b)

Determine fH for a common-emitter stage having RC = 1.5k;,

driven from a source having Rs = 300; and using the

transistor whose parameter are given above. Compare the

result with that obtained in part a.

Solution : (a) For the common-emitter stage, we obtain

;!!T k..||.R

26100

2300

0

1


26/28

Using table for the common-base amplifier, we have

;!

! k..Ri 019801001

02

2

and

;!!! k..||.R||RR iCL 019500198051211

Hence

? A nsa 36.50195.00195.0501261.05.05.19261.011

!vv!

For the common-base stage gives

ns...

.

a 151515052

112 !v!


27/28

Thus

nsaaa 51.615.136.512111 !!!!

and

MHz..a

fH

424105162

1

2

19

1

!vvT

!T

!


28/28

(b) For the equivalent common-emitter stage,

? An

sa 8.155

.15.1

501

261.0

5.0

5.19

261.01 !vv!

and

MHz

..afH 110108152

1

2

1

91

!

vvT!

T!

Clearly, the cascode amplifier has higher value of fH than

does the common-emitter stage. If the load resistance RC2 is

made larger (say, 5k;), the improvement in fH is even moredramatic (18.7 MHz for the cascode and 3.82 MHz for the

common-emitter stage).

Date post:	06-Apr-2018
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