EEE 241

ANALOG ELECTRONICS

CLASS 15&16

DR NORLAILI MOHD NOH

2

GM can also be obtained by inspection of the

circuit. Since for Q2, collector current is the

output current and emitter current is the input

current (CB configuration), then current gain of

Q2 is ≈ 1.

Hence, current gain of the cascode is equal to

the current gain of the CE Q1.

For the CE, ais = β.

From the previous slide, we know that

We know that

Hence, is 1

M m1i π1

a βG = = = g

R r

o oiMis i

ii i

Gvi ia = = =R

vi i

iR = r

r

+

1v m1 1g v

2v

1 2,C ER

m2 2g vo2r

2C

r

2B

+

ov

1E

+iv

o1r

OR

1B

iR

3

At node C2 :

To determine RO :

When vi=0, then v1=0.

Hence, gm1v1=0

t m2 2 ro2i =g v +i

o2t ro2 2

t m2 2 o2t 2

t t o2 m2 2 o2 2

m2 o2t t o2 2

-v +i r -v = 0

-v + i -g v r -v = 0

-v +i r -g v r -v = 0

v =i r - g r +1 v

+2v

m2 2g v o2r

2C

r

2B

+tv

o1r

+

2v ti

1Cro2i

t

ot

v 0i

vR =

i

KVL at the output loop,

r

+

1v m1 1g v

2v

1 2,C ER

m2 2g vo2r

2C

r

2B

+

ov

1E

+iv

o1r

OR

1B

iR

4

+2v

m2 2g v o2r

2C

r

2B

+tv

o1r

+

2v ti

1C ti

m2 o2t t o2 2

2t

o1 π2

m2 o2 t o1 π2t t o2

t m2 o2

O o2t

o1 π2

v =i r - g r +1 v

-vi =

r //r

v = i r - g r +1 -i r //r

v g r +1R = = r +

1 1i+

r r

m2 o2

O o2m2

o1 o2

m2

o2O o2

m2

o1 o2

m2

o2O o2

m2 o1

o1 o2

g r +1R = r +

1 g+

r β

1g +

rR = r 1+

1 g+

r β

1g +

rR = r 1+

1 g r1+

r β

where o

πm

βr =

g

← enough

5

If ro1 = ro2 , m2 o1

O o2m2 o1

o2

g r +1R = r 1+

g r1+

β

If gm2 ro1 >> βo2 ,

m2 o1

O o2m2 o1

o2

o2 m2 o1

O o2m2 o1

g r +1R = r 1+

g r

β

β g r +1R = r 1+

g r

o1m2 o1 m2

o2 o2O o2 o2

m2 o1 m2 o1

o1 o2 o2

1 rg + r g +

r rR = r 1+ = r 1+

1 g r g r1+ 1+

r β β

If βo2 >>1, then gm2 ro1 >> 1,

o2 m2 o1

O o2m2 o1

o2O o2

O o2 o2

β g rR = r 1+

g r

R = r 1+β

R r β

6

Hence, the CE-CB connection displays an output resistance that is

larger by a factor of about βo2 then the CE stage alone (RO of CE is ro

if RC is neglected)

o o ovo O M

i o i

v v ia R G

v i v

Open circuit voltage gain :

Hence, vo o2 o2 m1a = r β g

O o2 o2R r β

For a CE amplifier, if RC is not

included. Thus, the magnitude of the

maximum available voltage gain is higher

by a factor βo2 than for the case of a single

transistor.

v m oa = g r

7

3.4.2.2 The MOS Cascode

s2 ds1

gs1 i

gs2 s2

i

v = v

v = v

v = v

R =

1M

BIASV

2M

R

DDV

OR

iV+

m1 gs1g v

R

o2r

2D2G

+ov

1S

+iv

o1r

oi

1G 1 2,D S

ds1v+

m2 gs2

m2 s2

g vg v

gs1v

+

gs2v

+

8

oM

i v 0o

iG =

v

KCL at node D2,

ds1o m2 ds1

o2

o ds1 m2o2

vi = -g v -

r

1i = v -g -

r

KCL at node D1,

o m1 i

o1 m2

o2

o m1M

i

o1 m2

o2

1i 1+ = g v

1r g +

r

i gG = =

1v1+

1r g +

r

ds1o m1 i

o1

oo m1 i

o1m2

o2

vi = g v +

r

1ii = g v +

1r -g -r

m1 gs1g v

R

o2r

2D2G

+ov

1S

+iv

o1r

oi

1G 1 2,D S

ds1v+

m2 gs2

m2 s2

g vg v

gs1v

+

gs2v

+

m1 gs1g v

o2r

2D2G+o 0v

1S

+iv

o1r

oi

1G 1 2,D S

ds1v+

m2 gs2

m2 s2

g vg v

gs1v

+

gs2v

+

← enough

9

o m1M

i

o1 m2

o2

m1 o1 m2

o2M

o1 m2

o2

o1m1 m2 o1 m1

o2M

o1m2 o1

o2

M m1o1

m2 o1

o2

i gG = =

1v1+

1r g +

r

1g r g +

rG =

1r g + +1

r

rg g r + +1 -g

rG =

rg r + +1

r

1G = g 1-

rg r + +1

r

If ro1 =ro2, then

M m1m2 o1

1G = g 1-

g r +2

If gm2ro1 >>1, then M m1G g

This equation shows that the

transconductance of the cascode is

always < gm1. If gm2ro1>>1, the

difference is small. The main point here

is that the cascode configuration has

little effect on the transconductance

since the overall GM for the CS is gm.

10

Another way of deriving to this

conclusion is by inspecting the circuit.

For CG,

Ri =1/gm.

The current gm1vi will mostly flow

through this Ri rather than ro1 as

1/ gm << ro1 .

For a CG,

ai = 1

Hence, the output current will then be

the same as the input current to the

CG, which is gm1vi.

GM = io /vi = gm1vi /vi = gm . m1 gs1g v

R

o2r

2D2G

+ov

1S

+iv

o1r

oi

1G 1 2,D S

ds1v+

m2 gs2

m2 s2

g vg v

gs1v

+

gs2v

+

11

t

Ot

v 0i

vR =

i

When vi=0, gm1vi=0.

ro2t m2 ds1

o2

r t ds1o2

t ds1m2t ds1

o2

tm2t ds1

o2 o2

vi = -g v +

rv = v -v

v -vi = v -g +

r

1 vi = v -g - +

r r

Determining Ro

KCL at node D2, m1 gs1g v

R

o2r

2D2G

+ov

1S

+iv

o1r

oi

1G 1 2,D S

ds1v+

m2 gs2

m2 s2

g vg v

gs1v

+

gs2v

+

o2r

2D2G

+tv

o1r

ti

1 2,D S

ds1v+

m2 gs2

m2 s2

g vg v

gs2v

+

oR

12

ds1 t o1

tm2t t o1

o2 o2

o1o1 m2t o2 t

o2

O o2 o1 o2 m2 o1

O m2 o1 o2

v =i r

1 vi = i r -g - +

r r

ri 1+r g + r = v

r

R = r +r r g +r

R g r r

KCL at node D2,

o2r

2D2G

+tv

o1r

ti

1 2,D S

ds1v+

m2 gs2

m2 s2

g vg v

gs2v

+

oR

tm2t ds1

o2 o2

1 vi = v -g - +

r r

13

The output resistance can further be increased by using more

than one level of cascoding. However, number of levels of

cascoding is limited by the power supply voltage and signal

swing requirements.

This equation shows that the MOS cascode increases the output

resistance by a factor of about gm2ro as the RO for CS is ro if RD is

not included.

vo M O

vo m1 m2 o1 o2

2m ovo m1 m2 o1 o2

a = -G R

a = -g g r r

a = - g r if g g and r r

O m2 o1 o2R g r r

14

Two-stage Amplifier (Cascade)

BJT Cascade

B1 11 12

B2 21 22

i

C2

C1

π1,2

o1,2

m1,2

R = R //R =5kΩ

R = R //R =5kΩ

R =1kΩ

R = 20kΩ

R =10kΩ

r =1500Ω

r =33.333Ω

g = 26.67mS

Given :

Find : (a)

(b)

(c)

(d)

(e)

ov2

o1

va

v

o1v1

in

va

v

ov

i

va

v

in2Z

in1Z

1T2T

+inv EC

o1v

+

+iv

iR iC

12R C1R

11R ER

2C

22RC2R

ov

+

o1Z

in1Z

in2Z

21R

CCV

15

+

inv

o1v

+

12R

C1R

11R

22R C2R

ov

+

o1Z in2Zin1Z

1r

m1 ing v

o1r 21R 2r

m2 o1g v

o2r

iR1B 1 2,C B o1i

+iv

1E 2E

2C

1T2T

+inv EC

o1v

+

+iv

iR iC

12R C1R

11R ER

2C

22RC2R

ov

+

o1Z

in1Z

in2Z

21R

CCV

16

(a) Final stage voltage gain, v2a

ov2

o1

m2 o1 o2 C2

v2o1

v2

v2

va =

v

-g v r //Ra =

v

a = -26.67m 33.333k//20k

a = -333.37

+

inv

o1v

+

12R

C1R

11R

22R C2R

ov

+

o1Z in2Zin1Z

1r

m1 ing v

o1r 21R 2r

m2 o1g v

o2r

iR1B 1 2,C B o1i

+iv

1E 2E

2C

17

(b) Final stage input impedance, in2Z

// //

// 1.154

o1in2

o1

in2 21 22 2

in2

vZ

i

Z R R r

Z 5k 1500 k

(c) Initial stage voltage gain, v1a

o1v1

in

m1 in o1 C1 21 22 π2

v1in

v1

va =

v

-g v r //R //R //R //ra =

va = -26.76

+

inv

o1v

+

12R

C1R

11R

22R C2R

ov

+

o1Z in2Zin1Z

1r

m1 ing v

o1r 21R 2r

m2 o1g v

o2r

iR1B 1 2,C B o1i

+iv

1E 2E

2C

18

(d) Amplifier input impedance, Zin1

// //

inin1

i

in1 11 12 1

in1

vZ

iZ R R r

Z 5k//1500=1.154k

(e) // //

// //

11 12 1in i

11 12 1 i

in

i

R R rv vR R r R

v 5k//1500=0.5357

v 5k//1500 1k

+

inv

o1v

+

12R

C1R

11R

22R C2R

ov

+

o1Z in2Zin1Z

1r

m1 ing v

o1r 21R 2r

m2 o1g v

o2r

iR1B 1 2,C B o1i

+iv

1E 2E

2C

19

Overall voltage gain of the cascaded amplifier, av

0.5357 4778.97

o in o1 ov

i i in o1

v v1 v2

v v v vav v v v

a a a

The total gain of the cascaded amplifier is the multiplication of the

gain at each stage of the cascaded network.

+

inv

o1v

+

12R

C1R

11R

22R C2R

ov

+

o1Z in2Zin1Z

1r

m1 ing v

o1r 21R 2r

m2 o1g v

o2r

iR1B 1 2,C B o1i

+iv

1E 2E

2C

ii

20

Find (a) Voltage gain ratio

(b) The current gain ratio

Lv

i

vav

Lii

iai

+inv

+iv

12R

D1R11R

CC

22R

D2R

Lv

+

oRinR

21R

1M2M

LR

CC

CCii Li

DDV

MOSFET Cascade

21

+inv

+iv

12R

D1R11R

CC

22R

D2R

Lv

+

oRinR

21R

1M2M

LR

CC

CCii Li

DDV

+gs1v

gs2v

+12R11R 22R

+

oRinR

m1 gs1g v

o1r 21R

m2 gs2g v

o2r

1G o1i

+iv

D1R

Lv

LR

ii Li2G 2D1D

1S

D2R

2S

22

(a) Voltage gain ratio

L m2 gs2 o2 D2 Lv2

gs2 gs2

v2 m2 o2 D2 L

gs2 m1 gs1 o1 D1 21 22v1

gs1 gs1

v1 m1 o1 D1 21 22

v -g v r //R //Ra = =

v v

a = -g r //R //R

v -g v r //R //R //Ra = =

v v

a = -g r //R //R //R

+gs1v

gs2v

+12R11R 22R

+

oRinR

m1 gs1g v

o1r 21R

m2 gs2g v

o2r

1G o1i

+iv

D1R

Lv

LR

ii Li2G 2D1D

1S

D2R

2S

gs1 i

L gs2 L gs1v

i gs1 gs2 i

m1 m2 o1 o2 D1 21 22 D2 L

vo1 D1 21 22 o2 D2 L

v = v

v v v va = = × ×

v v v v

g g r r R //R //R R //Ra =

r + R //R //R r + R //R

23

L 11 12 v 11 12Li

i L i L

m1 m2 o1 o2 D1 21 22 D2 L

vo1 D1 21 22 o2 D2 L

m1 m2 o1 o2 D1 21 22 D2 L 11 12

io1 D1 21 22 o2 D2 L L

v R //R a R //Ria = = =

i R v R

g g r r R //R //R R //Ra =

r + R //R //R r + R //R

g g r r R //R //R R //R R //Ra =

r + R //R //R r + R //R R

(b) The current gain ratio,

+gs1v

gs2v

+12R11R 22R

+

oRinR

m1 gs1g v

o1r 21R

m2 gs2g v

o2r

1G o1i

+iv

D1R

Lv

LR

ii Li2G 2D1D

1S

D2R

2S