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The Common Base Amplifier - Microelectronic Circuits II - Slides | EGR 392, Study notes of Engineering

Material Type: Notes; Professor: Hu; Class: Microelectronic Circuits II; Subject: Engineering; University: Central Michigan University; Term: Fall 2009;

Typology: Study notes

Pre 2010

Uploaded on 11/12/2009

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81
6.7 The Common-Gate and Common-base Amplifiers with active Loads
6.7.2 The Common-Base Amplifier
(a) Active-loaded common-base amplifier.
82
1. Analysis with R
L
Input Resistance R
in
:
i
i
in
i
v
R=
(b) Small-signal analysis
performed directly on the
circuit diagram with the BJT T
model used implicitly.
roeii
irvi += /
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rvvrv /)(/ +=
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rRivrv /)(/ +=
oLiiiei
rRrvivrv /])/([/
π
+=
i
o
L
o
iL
o
i
e
i
v
rr
R
r
iR
r
v
r
v
π
++=
i
o
L
oe
i
o
L
v
rr
R
rr
i
r
R)
11
()1(
π
++=+
π
r
R
r
r
Rr
L
e
o
Lo
++
+
=
1
e
L
e
o
Lo
r
R
r
r
Rr
)1(
1
β
+
++
+
=
pf3
pf4
pf5
pf8
pf9
pfa

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81

6.7 The Common-Gate and Common-base Amplifiers with active Loads

6.7.2 The Common-Base Amplifier

(a) Active-loaded common-base amplifier.

82

  1. Analysis with RL Input Resistance R in :

i

i in

i

v

R =

(b) Small-signal analysis

performed directly on the

circuit diagram with the BJT T

model used implicitly.

ii = vi/ re+i ro =vi / re+(vi−vo)/ro

=vi / re+(vi−ioRL)/r o

=vi /re+[vi−(ii−vi/r π)RL]/r o

i o

L

o

Li

o

i

e

i v rr

R

r

Ri

r

v

r

v

π

i o

L

e o

i o

L v rr

R

r r

i r

R

π

r π

R

r

r

r R

L

e

o

o L

e

L

e

o

o L

r

R

r

r

r R

  • β

out x x 83 R =v /i

Output Resistance Rout:

i x = v

Rr

R r

r

v

R

v

e

e

e

π

π

π

R e′ =Re||r π

i x Re′= v

ix R′e =

mo

x xo

g r

v ir

v (^) x −v=(i (^) x+gmv)r o

m o

x xo

g r

v ir

v

= ro +( 1 +gmro)R e′ o vo e

=r +A R′

  1. Take RL out; 2. Take voltage source out; 3. add vx

84

(c) Small-signal analysis with the

output open-circuited.

Input Resistance (open circuit) Ri:

i

i i

i

v

R π

π

r

v r

v

i

i

Ro =r o

Open Circuit Voltage Gain Avo:

i

o vo v

v A = (^) mo i

m io i g r v

g vr v = +

= 1

Output Resistance (open circuit) Ro:

87

Summary

Common-Gate, Common-Base

Common-Source, Common-Emitter

Open Circuit

Voltage Gain Avo

Input Resistance <<

Output Resistance (^) >>

High-frequency

response

Better (no Miller effect)

Avo = 1 +gmro Avo =−gmR L′

e

L e

o

o L in

r

R r

r

r R R

( 1 )

1

  • β

= Rin=∞

Rout = ro+AvoR e′ out o

R =r

88

6.8 The Cascode Amplifier

6.8.1 The MOS Cascode

Cascode configuration: by placing a common-gate (common-base) amplifier stage in cascade with a common-source (common-emitter) amplifier stage.

Common-source

Common-gate

89

Small Signal Analysis:

Q 1 : Avo 1 =−gm 1 ro 1

Rin = ∞( ii = 0 )

Rout 1 =ro 1

Q 2 :

2 2 2

2

vo

L

m mb

in A

R

g g

R +

Avo 2 = 1 +(gm 2 +gmb 2 )ro 2

Rout =ro 2 +Avo 2 ro 1

≅( gm 2 ro 2 )ro 1

Rd 1 =Rout 1 || Rin 2

90

vo 1 =−gm 1 ro 1 v i

The cascode with the output open-circuited (or RL→∞).

=−A 01 v i

vo=

vo 01 vo 2

A =−A A

Avo 2 vo 1 =Avo 2 ( −Avo 1 vi)

01 02

≅−A A

For the usual case of equal intrinsic gains,

2 2 Avo =−A 0 =−( gmro)

93

( for CS, on Slide 55.)

Rgs 1 = R sig

Rgd 1 =( 1 +gm 1 Rd 1 )Rsig +Rd 1

Rgd Rsig gmRL R L

Cgs 1 sees

Cgd 1 sees

( C (^) db 1 +Cgs 2 ) sees

( C (^) L +Cgd 2 ) sees

Rd 1

RL ||R out

94

[( 1 ) ]

1 2 1 2

1 1 1 1 1

db gs d L gd L out

H gs sig gd m d sig d

C C R C C R R

C R C g R R R

H

fH

In the case when Rsig is very small,

τH ≅( C L+Cgd 2 )(RL||Rout )

L gd 2 L out

H C C R R

f

95

A 0 times larger

A 0 times smaller

96

Example 6.

This example illustrates the advantages of cascoding

by comparing the performance of a cascode

amplifier with that of a common-source amplifier in

two cases:

(a) The resistance of the signal source is significant, Rsig = 10kΩ

(b) Rsig is negligibly small.

Assume all MOSFETs have W/L of 7.2μm/0.36μm and are operating at ID = 100μA, gm = 1.25mA/V, =0.2, ro = 20kΩ, Cgs = 20fF, Cgd = 5fF, Cdb = 5fF, and CL=5fF. For case (a), let RL = ro = 20kΩ for the CS amplifier and RL =

Rout for the cascode amplifier. For all cases, determine Av, fH, and ft.

99

[( 1 ) ]

1 2 1 2 2

1 1 1 1 1

db gs d L db gd L out

H gs sig gd m d sig d

C C R C C C R R

C R C g R R R

( 5 5 5 ) ( 20 || 640 ) 653 ps

20 10 5 [( 1 1. 5 ) 1. 22 ] ( 5 20 ) 1. 22

+ + + × =

= × + + + + + ×

244 MHz

× ×

fH

f t =| Av| f H = 23. 5 × 244 = 5. 73 GHz