Exercises for Cascode Amplifiersaries.ucsd.edu/NAJMABADI/CLASS/ECE102/12-F/NOTES/... · Cascode Amp...

Exercises for Cascode Amplifiers

ECE 102, Fall 2012, F. Najmabadi

F. Najmabadi, ECE102, Fall 2012 (2/26)

Exercise 1: Compute R (assume ro1 = ro2 = ro3 and gm1 = gm2 = gm3)

Every Cascode stage increases R by gmro

oomoom

oomo

rrgrrgrrgr

)(

)1(2

1122

≈+≈

++

oomomom

omommo

rrgrgrgrgrggr

2232

2233

)(

)1(

≈+≈

++

2omrg

Double Cascode


Exercise 2: Compute all indicated R’s, v’s, and i’s . Assume transistors are identical. (S&S Problem 7.34)


1 orR =

)(

)1( 2

2

oomoom

oomo

rrgrrgrrgrR

≈+≈

++=

1

)(1

23 o

om

oomo

om

o rrg

rrgrr g

RrR =+

+=

++

=


oom

oomo

om

o rrg

rrgrr g

RrR 1

)(1

23 =

++

≈+

+=

From previous slide:

CS amplifier:

)||()||( 3 oomomv rrgRrgA −=−=

5.0 1 iom vrgv −=

5.05.0 3

13 im

o

iom vgr

vrgRvi ==−=



A simpler method:

)(5.0)(5.0 22242 iomomim vrgrgvgRiv −=−=−=

5.0 34 imvgii ==

)( 2 oom rrgR ≈

Cascode amplifier (we can use cascode gain to find v2)



5.05.0 153 iomoim vrgrvgRiv −=−=−=

5.0 45 imvgii ==

1 orR =

Q3 is NOT an amp configuration It is a part of the active load!


1 orR =

543 iii == Currents are the same!

Looking into drain of Q4

Looking into drain of Q3, R1 is increased by 1+gmro

Looking into source of Q2, R2 is decreased by 1+gmro

)( 2 oom rrgR ≈

3 orR ≈


)(5.0 22 iom vrgv −= 5.0 1 iom vrgv −=

CS Amp with a load of ro Q3 increased R1 and Q2 decreased R2 by the same amount leading to R3 = ro (Bad circuit)

Cascode Amp with a practical load (Good circuit)


Output is NOT taken across the load (BAD circuit)

CS Amp (input at the gate Output at the drain)

CG Amp (input at the source Output at the drain)

Q3 is NOT an Amp (input at the drain)

5.0)1(

)(5.0

23

22

iomomo

o

iom

vrgvrgr

rv

vrgv

−≈+

=

−=

ro

ro (1+gm ro )

v2 )1( 2 oomo rrgrR ++=


Exercise 3: Due to manufacturing error, the input terminal of a cascode amplifier is mis-configured as is shown. Compute vo. Assume identical transistors are identical.


From previous Problem:

)( 2 oom rrgR =

oS rR =

This is a CG configuration )||( 2 omomv rgRrgA ≈=


)( 2 oom rrgR =

oS rR =

Q1 does not affect the gain, but it changes the input resistance to:

orR 3 =From previous problem:

5.0|||| 3 oooSi rrrRRR ===

Exercise: Compute Ro (answer: Ro ≈ ro )


Exercise 4: Due to manufacturing error, a parasitic resistance has appeared between drain and source of Q1. Compute vo. Assume identical transistors are identical.


From previous Problem:

3 orR ≈

)||( 22 omomv rgRrgA ≈= )||||()||||( 31 poompomv RrrgRRrgA −=−=Q2: CG configuration: Q1: CS configuration:

)||||( 221 pooomvvv RrrrgAAA −==

)( 2 oom rrgR ≈


When a resistor is placed between Drain and Source of a transistor, there is a simpler and more elegant way to solve the signal circuit

Rrr oo ||=′


)||||( 221 pooomvvv RrrrgAAA −≈=

Rrr oo ||=′


Exercise 5: Due to manufacturing error, a parasitic resistance has appeared on the circuit. Compute vo. Assume identical transistors are identical.


)|| ()|||| ()||( 222 pompomomv RrgRRrgRrgA ≈=′≈ )||( 31 RrgA omv −=

Q2: CG configuration: Q1: CS configuration:

)||)(|||| ( 322

21 RrRRrgAAA opomvvv −==

)()1( 2 oomoomo rrgrrgrR ≈++=

poo Rrr ||=′

)(11

23

o

oo

om

oom

mom

o

rrr

rgrrg

gr g RrR

′×≈

′+≈

′++′

=


Exercise 6: Find the open-loop gain, overall gain and output resistance of the circuit if VOV1 = 0.3 V and RL = 100 k (µnCox = 200 µA/V2, µpCox = 50 µA/V2, Vtp = − 0.6 V, λn = 0.1 /V, λp = 0.2/ V and (W/L) = 20/0.18 for all transistors).

11 final –P5

2621

11 )3.0(

18.020102005.05.0 ××××=

= −

OVoxnD VL

WCI µ

mA 11234 ==== DDDD IIII

V 6.03 =OVV

V 3.0 1212 ==→= OVOVDD VVII

23

623

3

33 18.0

2010505.05.010 OVOVoxpD VVL

WCI ××××=

== −− µ

V 6.0 3434 ==→= OVOVDD VVII


A/V 1067.63.0

1022 33

1

112

−−

×=×

===OV

Dmm V

Igg k 10101.0

1 13

112 =

×=== −

Dnoo I

rrλ

mA 11234 ==== DDDD IIII V 6.0 & V 3.0 3412 ==== OVOVOVOV VVVV

A/V 1033.36.0

1022 33

3

334

−−

×=×

===OV

Dmm V

Igg k 5102.0

1 13

134 =

×=== −

Dpoo I

rrλ

Previous page:

k 93.3105)105103.33(1105

)1( 333-3

1

44331

=×+×××+×=

++=

RrrgrR oomo

From Elementary R forms:

k 687 )1( 11222 =++= oomo rrgrR

We need only to compute two of Avo, Av, and Ro set as they are related: o We calculate all three below to

show the method. o Note: Cascode Amp formula gives Avo

oL

Lvov RR

RAA+

⋅=


k 5 A/V, 1033.3 k, 10 A/V, 1067.6 343

34123

12 ==×====×== −−oommoomm rrggrrgg

From Previous page: k 100 k, 687 k, 93.3 21 === LRRR

Av Avo (RL→ ∞ )

Q2 (CG):

3.55 )||(k29.8k4.48||k 10 ||

k4.48k3.93||k 100||

222

2

1

=′≈==′

===′

Lomv

Lo

LL

RrgARr

RRR

83.8)||( 2111 −=−= iomv RrgA

2.60 )||(k03.9k3.93||k 10 ||

k3.93

222

2

1

=′≈==′

==′

Lomv

Lo

L

RrgARrRR

k 82.2k 687||k 93.3 || 21 === RRRo

R’L1 = Ri2:

k 862.0 1

22

22 =

++

=om

oi r g

RrR

k 48.3k 100||k 93.3 || 1 === LRRR k 93.3 1 == RR

k 53.1 1

22

22 =

++

=om

oi r g

RrR

Q1 (CS):

29.5)||( 2111 −=−= iomv RrgA

293 21 −=×= vvv AAA 532 21 −=×= vvvo AAA

Resistance in the drain of Q2

Exercise: Show

oL

Lvov RR

RAA+

⋅=


Exercise 7: In the folded cascode circuit below, all transistors have the same µCox (W/L), the same λ and ID2 = 2ID1 . Find the gain and the output resistance of the amplifier (in terms of gm1 and ro1 only).


Bias

Problem: ID2 = 2ID1 , the same µCox (W/L), and the same λ

2 & 2 :KCL

121345

133112

DDDDDD

DDDDDD

IIIIIIIIIIII

=====→+==

121345 2 & OVOVOVOVOVOV VVVVVV ====

→

= 5.0 2

OVoxD VL

WCI µ

→= 2

OV

Dm V

Ig

11

1

1

1

2

22

1345

22 2 2222

mOV

D

OV

D

OV

Dm

mmmm

gV

IV

IV

Ig

gggg

=×=×

==

===

→= 1

Do I

rλ

1112

2

1345

5.01 5.0211

oDDD

o

oooo

rIII

r

rrrr

=×=×

==

===

λλλ


1112

55442

)()1(

oom

oomo

rrgRrrgrR

≈++=

CS Amp

CG Amp

111

1111

33

233

11111112333

1)(

1

])(||[)||(/

oom

oomo

om

oi

omoomomomxov

rr g

rrgrr g

RrR

rgrrgrgRrgvvA

=+

+=

++

=

≈=≈=

Q1 (CS):

111111111

111321

25.0)]33.0(||[)||(/33.0||)5.0(||

omoomLomixv

oooioL

rgrrgRrgvvArrrRrR

−=−=′−=====′

211111131 )(25.0)( 25.0/ omomomvviov rgrgrgAAvvA −=×−=×==

Q3(CG):

121345

121345

5.0 , 2 ,

oooooo

mmmmmm

rrrrrrgggggg

========

Signal


121345

121345

5.0 , 2 ,

oooooo

mmmmmm

rrrrrrgggggg

========

21 || RRRo =

21155442 )1( omoomo rgrrgrR ≈++=

21111111

11331

111421

1

33.033.0)33.01(

)1(33.0||)5.0(||

:

omomomo

oomo

ooooo

rgrgrgrRRgrR

rrrRrRR

≈++=

++====

211

211

211

21

25.0||)33.0(

||

omomomo

o

rgrgrgRRRR

==

= 1112

55442

)()1(

oom

oomo

rrgRrrgrR

≈++=

(Assumed 0.33 gm1 ro1 >> 1)

Exercises for Cascode Amplifiersaries.ucsd.edu/NAJMABADI/CLASS/ECE102/12-F/NOTES/... · Cascode Amp...

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