High-Speed Amplifiers and Tuned Amplifiers

44 Pages·2005·0.32 MB·English

EECS 142
Lecture 3: High-Speed Amplifiers and Tuned Amplifiers
Prof. Ali M. Niknejad
University of California, Berkeley
Copyright (cid:13)c 2005 by Ali M. Niknejad
A.M.Niknejad UniversityofCalifornia,Berkeley EECS142Lecture3 p.1/44 –p.1/44
CS/CE Amplifier Bandwidth
R
L
C +
µ
v
o
R
S
−
+
C + C
v db L
s
−
Due to Miller multiplication, the input cap is usually the
dominant pole
1
ω R (C + A C )
0− s in v µ
≈ | |
1
ω = R C (1 + µ A ) R C µ A
0− s in v s in v
| | ≈ | |
A.M.Niknejad UniversityofCalifornia,Berkeley EECS142Lecture3 p.2/44 –p.2/44
CS/CE Amplifier Gain-Bandwidth
Assuming the voltage gain is given by the
low-frequency value of g R , we have
m L
C
1 in
ω = R C µg R = (g R )(g R ) µ
0− s in m L m s m L
g
m
R
1 2 s 1
ω = A ω µ
0− v −
T
| | R
L
2
The amplifier has a bandwidth reduction factor of A
v
R 1
2 L
ω A = ω
0 v T
× | | × R × µ
s
(cid:18) (cid:19)
A.M.Niknejad UniversityofCalifornia,Berkeley EECS142Lecture3 p.3/44 –p.3/44
Bandwidth Example
R
Say we need a gain of 60 dB (A = 1000) and L = 2.
v
R
s
The technology has a capacitance ratio of µ = 0.2:
2 6
ω A = 10 ω = ω 2 5
0 v 0 T
| | × ×
ω
T
ω =
0
5
10
Compare this to a current mirror amplifier. When we
follow the “normal” gain-bandwidth tradeoff, we have
ω ω
T T
ω = =
0
A 1000
i
A.M.Niknejad UniversityofCalifornia,Berkeley EECS142Lecture3 p.4/44 –p.4/44
Common Base Amplifier
C
µ
+
R
Cin vin gmvin ro L
R
S
−
+
vs RL
−
R
S
+
v
s
−
Write KCL at base node of circuit
v + v
s in
+g v +sC v = 0
v = v (1+g R +sC R )
m in in in
s in m s in s
R
s
A.M.Niknejad UniversityofCalifornia,Berkeley EECS142Lecture3 p.5/44 –p.5/44
Common Base Amp (cont)
And write KCL at the output node
1
(sC + )v + g v + sC v = 0
o o m in µ o
R
L
1
v + s(C + C ) = g v
o o µ m in
R
L
(cid:18) (cid:19)
The voltage gain is a product of two terms
v g R v
o m L x
A = =
v
v 1 + s(C + C )R v
s o µ L s
G R
m L
A =
v
C
(1 + s(C + C )R )(1 + sR in )
o µ L s 1+g R
m s
A.M.Niknejad UniversityofCalifornia,Berkeley EECS142Lecture3 p.6/44 –p.6/44
Common Base Bandwidth
Note the transconductance is degenerated,
G = g /(1 + g R ). Note that the input capacitance is
m m m s
also degenerated by the action of series feedback.
Unlike a CE/CS ampilifier, the poles do not interact (due
to absence of feedback capacitor)
First let’s take the limit of high loop gain, g R 1
m s
≫
R
L
R
A = s
v
(1 + s/ω )(1 + s/ω )
T L
1
where ω = ((C + C )R ) is the pole at the output.
L o µ L −
A.M.Niknejad UniversityofCalifornia,Berkeley EECS142Lecture3 p.7/44 –p.7/44
Matched Common Base Amp
The common-base amplifier has the nice property that
the input impedance is low (roughly 1/g ) and
m
broadband, thus easily providing a termination to the
driver (a filter, the antenna, or a previous stage). If we
assume that R = 1/g , we have
s m
1
g R
2 m L
A =
v
(1 + s/2ω )(1 + s/ω )
T L
The 3dB bandwidth is thus most likely set by the time
constant at the load.
A.M.Niknejad UniversityofCalifornia,Berkeley EECS142Lecture3 p.8/44 –p.8/44
Shunt Feedback Amp
V
CC
R
L
R
F
v
o
R C
S ∞
+
v
s
−
The shunt-feedback amplifier is a nice high-frequency
broadband amplifier building block. The action of the
shunt feedback is used to lower the input impedance
and to set the gain.
A.M.Niknejad UniversityofCalifornia,Berkeley EECS142Lecture3 p.9/44 –p.9/44
Shunt Feedback Gain / Input Resis
The in-band voltage gain and input impedance is given
by (see PS 1)
R
F
A =
v
R
s
R 1
F
R = (1 + )
in
R g
L m
R 1 R
For an input match, R = (1 + F ) , or g R = (1 + F )
s m s
R g R
L m L
Since the voltage gain sets R , the input impedance
F
match determines the required transconductance g
m
(and hence the power dissipation)
A bipolar version will dissipate much less power due to
the higher intrinsic g
m
A.M.Niknejad UniversityofCalifornia,Berkeley EECS142Lecture3 p.10/44 –p.10/44