REPASO EXAMEN 1INEL4202 Electrnica II

60dB/dec

40dB/dec

20dB/dec

midband range

(3 zeros at origin)

low frequencies high freqs.

fp,1fp,2

fp,3fp,4

fp,5

FREQUENCY RESPONCE

Low-frequency poles and zeros High-frequency poles

Midband gain

A(s) = AM

s

s+ p,1

s

s+ p,2

s

s+ p,3

p,4

s+ p,4

p,5

s+ p,5

BAJAS FRECUENCIAS

Mtodo complejo: reemplazar cada condensador externo por impedancia y analizar

Mtodo simplificado: desde los terminales de cada condensador

buscar resistencia equivalente REQ

calcular frecuencias de los polos

!

para condensador de bypass, calcular frecuencia del cero

!

frecuencia polo mas alta es la dominante y determina el ancho de banda si esta a mas de una decada de las demas. Si no sumar todos los polos para obtener una aproximacin de la frecuencia de 3dB.

fL =1

2REQC

fZ =1

2REC

FRECUENCIAS ALTAS

Efecto del condensador: reduce A segn aumenta f

representar circuito por equivalente ac

rep. condensadores parasticos como externos y ver si puede determinar la REQ por inspeccin si posible

analizar el circuito para determinar REQ despus de sustituir modelo del transistor si no puede hacerlo por inspeccin

iIN = Y (vIN vOUT )= sC(1AM )vIN

Entrada

Acta como un condensador C(1-AM)Salida

iOUT = Y (vOUT vIN )= sC

1 1

AM

vOUT

Acta como un condensador C(1-1/AM)

Efecto de Miller

FRECUENCIAS ALTAS - CE/CS

Usar Miller para reemplazar C con una capacitancia equvalente que aparece en paralelo con C

CM = C+ (1 - AM)C

AM es usualmente -gmRL para CE/CS

RL = RL||RC = carga equivalente en ac

Determine polo de alta frecuencia usando

REQ = res. equivalente desde terminales de CM

fH =1

2CMREQ

R1

R2 RE

RC

RLCE

CC2CC1RTH

vS

VCC

RBRLL=RC||RL

RTH

vS

C

C/gm v/

E

B

r/ v/

C

RB=R1||R2

Common-base/-gate

Av = vcve C emisor a tierra C colector a tierra dos polos

fH1 =1

2CREQ,1fH2 =

1

2CREQ,2

-VEE +VCC

vS

RTHRE RC

RL

vOUT

VCC

vOUT

C1

RTH

vS

RE

vSRTH

RE r/ C/v/+-

gmv/

C RC||RLRLL

vOUT

Common-collector/common-drain

RTH

vS

R1

R2 RLvOUT

C1

VCC

RTH

vS Rb=R1||R2

r/

C/C gmv/

RL

vOUT+ -v/

RTH

vS RB

RLvOUTC1

FRECUENCIA DE GANANCIA UNITARIA ft Unity-gain frequency frecuencia a la cual (s) = iC/iB = 1 concepto se usa tambien en los amplificadores operacionales mas facil de medir que C problemas pueden especificar ft y C. C se puede obtener como sigue:

ft =

2r (C + C)=

gm2 (C + C)

C =

2rft C = gm

2ft C

Para el mosfet, useft =

gm2 (Cgs + Cgd)

+ out, y por lo tanto in debe estar asociado af3. Consecuentemente,

30MHz =1

2CgdRD! Cgd = 1

2 30 106 35460 = 0.15pF

y

1MHz =1

2 10k(Cgs + 101 0.15pF )

Cgs + 101 0.15pF = 12 8k 106 = 19.9pF = Cgs + 15.15pF

Cgs = 4.7pF

EJEMPLO

Para el siguiente amplificador, R1 = 10k, R2 = 30k, RC = 5k, RE =14.3k, hfe = 100, C = 3pF y C = 0.5pF . Determine

1. la ganancia de frecuencias intermedias (midband gain AMB),

2. las frecuencias de los polos de baja frecuencia, y

3. las frecuencias de los polos de alta frecuencia.

C1=100F

C2=1F vout

-15V

+15V

RS=501

vS

R1

R2

RC

RE

RS=10k1Q2

Q1

RESPUESTA

voutRS=501

vS RC||RSRE

Q2Q1

la ganancia de frecuencias intermedias (midband gain AMB),

Av =14.3kk r1hfe+1

14.3kk r1hfe+1 + 50(gm1 r2

hfe + 1)(gm2 3.33k)

ICQ1 =14.3V

14.3k= 1mA ' ICQ2

gm1 ' gm2 = 40mA/Vr1 = r2 = 2.5k

Av =14.3kk 2.5k101

14.3kk 2.5k101 + 50(40mA/V 2.5k

101)(40mA/V 3.33k)

' 24.7574.75

(1)(133.2) = 44.1V/V

frecuencias de los polos de baja frecuencia, yRespuesta:

fL1 =1

2 100F 74.75 = 21.3Hz

fL2 =1

2 1F 15k = 10.6Hz

voutRS=501

vS 14.3k1

Q2Q1

3.33k1C/ C+C/ C

frecuencias de los polos de alta frecuencia:Podemos representar el circuito de alta frecuencia usando el circuito equiv-

alente de frecuencias intermedias y condensadores C y C externos,

fH1 =1

2C 50k24.75 = 3.2GHz

fH2 =1

2(C + C) 24.75 = 1.8GHzy

fH3 =1

2C 3333 = 95.5MHz

vOUT

vIN

CG

CC1

CC2

Cb

-10V

+10V

R1

R2

R3

RD

RLRS

vOUTvIN

RP

RLRS vOUTvIN

RPRL

RS

Cgs gmvgs+

-vgs

Cgd

RS

R1 = 179.5kR2 = 179kR3 = 145.5kRS = 2kRD = 3kRL = 10kRS = 10k

KN = 1.2mA/V 2

VTN = 2V = 0

Cgs = 5pFCgd = 0.8pF

Ejemplo

Los siguientes problemas provienen de otro libro de texto escrito por Naeman

( )( ) ( )

3

3

1 1 1 15.92 ms2 2 2 1015.9 10 1.89 F4.41 4 10

L LL L

LL D L C C C

D L

f rr frr R R C C CR R

= = = =

= + = = =

+ +

7.16 a.

( )( )( )( )

( ) ( )( )( )

( ) ( )( )

2

2

2

2

9

9 0.5 12 4 46 23 15 0

23 23 4 6 15 3 V2 62 2 0.5 3 2 1 mA/V1 1 12 0.923 k

SGD P SG TP

S

SG SG SG

SG SG

SG SG

m P SG TP m

o S om

V I K V VRV V V

V V

V V

g K V V g

R R Rg

= = +

= +

+ =

= =

= + = =

= = =

b. ( )0 L Cr R R C= + c. ( )

( )3

3

1 1 1 7.96 ms2 2 2 207.96 10 0.729 F0.923 10 10

LL

C Co L

f r frC CR R

= = = =

= = =

+ +

7.17 a.

( ) ( ) ( )( )( )( )

( ) ( )( ) ( ) ( )( )( )

( )( )

1 2

1

1

1

22

2

11 mA, 0.00833 mA120|| 0.1 1

0.1 121 4 48.4 kon 1

1 0.00833 48.4 0.7 121 0.00833 4

1 48.4 12 5.135

113 k113 48.4 84.7113

CQ BQ

E

TH BQ TH BE BQ E

TH CC

I I

R R R

V I R V I R

R VR

RR

R R kR

= = =

= +

= = = + + +

= + +

=

=

= = +

b.

( )( )0 0

0

0 0

1120 0.026 3.12 k1

80 80 k13.12 4 80 0.02579 4 80 25.6121

ErR R r

r

r

R R

= += =

= =

= = =

c.

( )( )

( )

0 23 6 3

3

0.0256 4 10 2 10 8.05 10 s1 1 19.8 Hz2 2 8.05 10

L Cr R R Cr

f fr

= +

= + =

= = =

7.18 (a)

( ) ( ) ( )( )

( )( )

5 0.7 1.075 mA 1.064 mA410 1.064 2 1.075 43.57 V

1.064 40.92 mA/V0.026100 0.026 2.44 K1.064

EQ CQ

CEQ

CEQ

CQm

T

T

CQ

I I

VV

Ig VVr I

= = =

=

=

= = =

= = =

(b)

1 1

1 1 1 1

2 2

2 2 2

2440For , 200 40001 101224.0 , 1.053 ms

For , 2 47 49 K49 ms

C eq S E

eq eq C

C eq C L

eq c

rC R R R

R r R CC R R R

R C

= + = ++

= = =

= + = + =

= =

(c) ( )1 1311 1 151 Hz2 2 1.053 10

f f

= = =

7.19 (a)

( ) ( )

( )

3 12

8

8

2 47 10 10 101.918 10 s

1 1 8.30 MHz2 2 1.918 10

H C L L

H HH

R R C

f f

= =

=

= = =

(b)

( )( )

( )

2

22

8

1 0.11 .21 100 1 .20.1

99 992 2 1.918 1082.6 MHz

H

H

H

f

f

f

f

=

+

= = +

= =

=

7.20 (a)

( )( )

( )

0 23 6 3

3

0.0256 4 10 2 10 8.05 10 s1 1 19.8 Hz2 2 8.05 10

L Cr R R Cr

f fr

= +

= + =

= = =

7.18 (a)

( ) ( ) ( )( )

( )( )

5 0.7 1.075 mA 1.064 mA410 1.064 2 1.075 43.57 V

1.064 40.92 mA/V0.026100 0.026 2.44 K1.064

EQ CQ

CEQ

CEQ

CQm

T

T

CQ

I I

VV

Ig VVr I

= = =

=

=

= = =

= = =

(b)

1 1

1 1 1 1

2 2

2 2 2

2440For , 200 40001 101224.0 , 1.053 ms

For , 2 47 49 K49 ms

C eq S E

eq eq C

C eq C L

eq c

rC R R R

R r R CC R R R

R C

= + = ++

= = =

= + = + =

= =

(c) ( )1 1311 1 151 Hz2 2 1.053 10

f f

= = =

7.19 (a)

( ) ( )

( )

3 12

8

8

2 47 10 10 101.918 10 s

1 1 8.30 MHz2 2 1.918 10

H C L L

H HH

R R C

f f

= =

=

= = =

(b)

( )( )

( )

2

22

8

1 0.11 .21 100 1 .20.1

99 992 2 1.918 1082.6 MHz

H

H

H

f

f

f

f

=

+

= = +

= =

=

7.20 (a)

( )( )( )( ) ( )( )

( )( )( )( )

2

12 2

2

5

5 1 1.2 1.5 1.2 3 2.251.2 2.6 2.3 0 2.84

1.8 10 1.8 1.2 1.2 5.68

2 2 1 1.8 2.683 /

SGP SG TP

SG SG SG SG

SG SG SG

DQ

SDQ SDQ

m P DQ

o

V K V VRV V V VV V V V

I mAV V V

g K I mA Vr

= +

= = +

= ==

= + =

= = =

=

(b) 1 1 0.3727 2.68

1.2 0.373 0.284 is

m

i

R kgR k

= = =

= =

For ( )( )61 1, 284 200 4.7 10 2.27C sC ms = + = For ( ) ( )3 3 62 2, 1.2 10 50 10 10 51.2C sC x ms = + = (c) CC2 dominates,

( )3 321 1 3.12 2 51.2 10dB s

f Hz

= = =

7.21 Assume 21 , 80 / , 0TN nV V k A V = = = Neglecting 200SiR = , Midband gain is:

v m DA g R= Let 0.2 , 5DQ DSQI mA V V= =

Then 9 5 200.2D DR R k

= =

We need 210 0.5 /20v

mD

Ag mA VR= = = and 2 2 2n

m n DQ DQk Wg K I IL

= =

or ( )0.0800.5 2 0.2 7.812W WL L

= =

Let

( ) ( )( )( )( ) ( ) ( )

( ) ( ) ( )

1 2

2 2 2

1 2

2 1

1 24

41 3

1

9 9 2250.2 0.2 0.20.0800.2 7.81 1 1.80 9 92 225

45 , 180 180 45 36

1 1 7.96 107.958 10 or 2 2 200 200 36 10

DQ

DQ GS GS

TH

Si TH C C

C

R R kIR RI V V R R

R k R kR R R k

s R R C CfC

+ = = =

= = = = = +

= = = = =

= = = = + =

+

=

( )5

52 33

2

0.022 1 1 5.31 105.305 10 or 2.65 2 20 102 3 10 D L L L

F

s R C C C nFf x

= = = = = =

7.22 a.

f =1

2r (C + C)

( )

( )3

12

21 38.46 mA/V0.026

38.46 102 10 2 10510 MHz

510 4.25 MHz120

mT

CQm

T

T

T

T

gfC C

Ig V

f

fff f

=

+

= = =

=

+

=

= = =

7.39

( )

( )

( )( )

39

12

3

12 9

5000 MHz 33.3 MHz150

20.5 19.23 mA/V0.026

19.2 105 10 2 0.15 1019.2 100.15 0.612 pF

2 10 5 100.462 pF

T

mT

m

ff f

gfC C

g

C

C

C

= = =

=

+

= =

=

+

+ = =

=

7.40 a. 2000 MHz 13.3 MHz150

Tff f = = = = b.

( )

( )

( )

2

2 2

1501 /

150 101 /

1501 22510224 13.33 224 199.6 MHz

fe

fe

hj f f

hf f

ff

f f f

=

+

= =

+

+ = = = = =

7.41 (a)

( )

0

1 1

11

1 1

where1

11

1

11

m L

i i

bb

i ib b b b

V g V Rrr sC sr CV V Vr rr r sr CsC

r rV Vr r sr r C r r s r r C

=

+= =

+++

= = + + + +

( )

( ) ( )

1 2

3 12

8

4.11 40 5 0.50.405 k0.405 10 15.6 149 10

6.67 10 s1 2.39 MHz2

H eq M

eq S

eq

H

H HH

r R C CR r R R RRr

f fr

= +

= =

=

= +

=

= =

For lower frequency:

( )( )

1

1 2

3 6 2

0.5 40 5 4.112.64 k2.64 10 4.7 10 1.24 10 s1 12.8 Hz2

L eq C

eq S

eq

L

L LL

r R CR R R R rRr

f fr

=

= + = +

= = =

= =

b.

39.5

fHfL

!A!!

f ( )

( )( )( )

0

1 2

1 2

2.135 0.81022.135 0.529.23 0.8102 5 2.5

39.5

m C L

iS

i i

V

V

V g V R RR R r

V VR R r R

V V V

AA

=

= +

= = + =

=

7.57

( )( )( )( )

( ) ( )( )( )

( ) ( )( )

( )( )

( )( )

2

2

2

2

2

0 0

9

2 1.2 4 4 92.4 8.6 0.6 0

8.6 8.6 4 2.4 0.62 2.4

3.512 V2 2 2 3.512 26.049 /2 3.512 2 4.572 mA1 1 21.9 k0.01 4.56

SGD P SG TP

S

SG SG SG

SG SG

SG

SG

m P SG TP

m

D

o

VI K V V RV V V

V V

V

Vg K V Vg mA VI

r rI

= + =

+ =

+ =

=

=

= + =

=

= =

= = =

a. ( )( )( ) ( )( )

11 1 6.04 21.9 1 6.785 pF

M gdT m o D

M M

C C g r RC C

= +

= + =

b. ( )( )( ) ( )

( )

( ) ( )

3 12

8

0

2 100 10 10 6.78 103.29 10 s

1 4.84 MHz2

100102

1006.04 21.9 11025.67

H i G gsT M

H

H

H HH

m o D gs

Ggs i i

G i

v

v

r R R C Crr

f f

V g r R VRV V VR R

A

A

= +

= +

=

= =

=

= = +

= =

7.58

( ) ( )

( )( )( )( )

( )( )( )

( ) ( )( )

2

1 2

2

2

2

2220 10 20 1022 84.67 V10 4.67

5.33 1 0.5 4 40.5 3.33 0

1 1 4 0.5 3.33 3.77 V2 0.52 2 1 3.77 23.54 /

G

G

SGD P SG TP

S

SG SG SG

SG SG

SG SG

m p SG TP

m

RV R RV

VI K V VRV V V

V V

V V

g K V Vg mA V

= = + + =

= = +

= +

=

+= =

= + =

=

b. ( )( )

( ) ( )( )1

3 1 3.54 2 5 18.2 pFM gdT m D L

M M

C C g R RC C

= +

= + =

a. ( )

( )( )1 2

3 12

8

0.5 8 22 0.461 k0.461 10 15 18.2 10

1.53 10 s1 10.4 MHz2

eq gsT M

eq i

H H

r R C C

R R R R

r

f fr

= +

= = =

= +

=

= =

c. ( )

( )( )( )( )

0

1 2

1 2

5.87 0.92155.87 0.53.54 0.9215 2 5 4.66

m gs D L

gs i i gs ii

v v

V g V R RR RV V V V V

R R RA A

=

= = = + = =

CgsT CgdT

!

"

gmVgsRi

RS Vgs RD RL

13 frequency due to :gsT eq S im

dB C R R Rg =

( ) ( )12

12

1 4 0.5 0.246 k1.811 162 MHz

2 246 4 10

Aeq gsT

eq

A

f R C

R

f

=

= =

= =

3 frequency due to gdTdB C

( )( ) 3 12

12

12 2 4 10 10

119 MHz

BD L gdT

fR R C

f

=

=

=

Midband gain

!

"

gmVgsRiV0

Vi Vgs RS RD RL!"

( )( )( )( )

0

1 1 41.811 1 4 0.51.810.492

0.492 1.81 4 2 1.19

Sm

gs i i

S im

i

m gs D L

v v

RgV V VR RgV

V g V R RA A

= =

+ +

=

=

= =

7.63

( ) ( )120 0.026 3.059 k1.0239.23 mA/Vm

r

g = =

=

a.

( )

( ) ( )( )

( )

( )( )

( )

2 3

12 9

9

3 12

9

9

1Input: 22

0.1 20.5 28.3 3.06 0.096 k96 12 2 2 10 1.537 10 s

1 103.6 MHz2 1.536 10

1Output: 2

15 10 10 2 106.67 10

1 23.9 MHz2 6.67 10

H

s

eq

H

H

C L

H

f rr R R R r C C

Rr

f

f rr R R C

f

=

= + = =

= + =

= =

=

=

=

=

= =

b.

( )

( )( )

2 3

2 3

2 3 20.5 28.3 3.059 2.433 k2.43339.23 5 10 125.62.433 0.1

m C LS

R R rA g R R R R r RR R r

A A

=

+ = =

= = +

& &&& &

c. 15 pFL LC C C= > dominates frequency response.

( )

( ) ( )( )

( )

( )( )

( )

2 3

12 9

9

3 12

9

9

1Input: 22

0.1 20.5 28.3 3.06 0.096 k96 12 2 2 10 1.537 10 s

1 103.6 MHz2 1.536 10

1Output: 2

15 10 10 2 106.67 10

1 23.9 MHz2 6.67 10

H

s

eq

H

H

C L

H

f rr R R R r C C

Rr

f

f rr R R C

f

=

= + = =

= + =

= =

=

=

=

=

= =

b.

( )

( )( )

2 3

2 3

2 3 20.5 28.3 3.059 2.433 k2.43339.23 5 10 125.62.433 0.1

m C LS

R R rA g R R R R r RR R r

A A

=

+ = =

= = +

& &&& &

c. 15 pFL LC C C= > dominates frequency response.