Berkeley
Broadband Amplifiers
Prof. Ali M. Niknejad and Dr. Ehsan Adabi
U.C. Berkeley
c 2014 by Ali M. Niknejad
Copyright Niknejad
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Outline
Broadband Amplifiers
Shunt-Peaking
Distributed Amplifiers
Multi-section Matching (Bode-Fano Limits)
Transformer Matching Networks
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Cascade Amplifier Bandwidth Shrinkage
Consider an amplifier consisting of a cascade of identical
single-pole stages
G0
G (s) =
1 + sτ
The bandwidth of n stages can be derived as follows
Gn (s) = G (s)n =
|Gn (jω0 )| =
G0n
(1 + sτ )n
G0n
G0n
G0n
√
=
=
|1 + jω0 τ |n
|1 + ω02 τ 2 |n
2
21/n − 1 = ω02 τ 2
p
ω0 τ = 21/n − 1
Bandwidth shrinks rapidly compared to the single stage.
Three stages =⇒ bandwidth drops by half
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Common Source Amplifier Bandwidth
RL
Cgd
Rs
CL
Cgs
Classic amplifier has several poles. The poles can be
calculated as
τgs = Cgs Rs
τgd = Cgd (Rs + RL + gm Rs RL )
τL = CL RL ||ro
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Minimizing the effect of Cin
RL
RL
M3
M2
CL
CL
Rs
Rs
M1
The effect of Cgd can be minimize with a cascode
configuration.
The load can be isolated with a buffer M3 (τL can be reduced)
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More Buffers
RL
Rs
CL
Similarly, the input capacitance can be isolated with a buffer
(τgs reduced)
We see that we can trade speed for power consumption.
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Feedback Amplifiers
G0
0 dB
ω0
ω0 G0
f
For low order systems (with one dominant pole), product of
gain and bandwidth is constant
G=
G0
1 + sτ
G
G
GCL
0
0
G
G0
G0 f +1
1+sτ
=
=
=
=
G0
1 + Gf
1 + sτ + G0 f
1 + sτ G0 f1+1
1 + 1+sτ
f
=
G0
1+T
1+
τ
s 1+T
Gain×BW =
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G0
1+T
G0
×
=
1+T
τ
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Shunt-Series Amplifier
RF
RL
Cgd
Rs
CL
Cgs
R
By using feedback, Gain ↓
Ri and Ro ↓, matching acquired
BW ↑
By using feedback, we reduce the gain, reduce Ri and Ro
(desired for output matching), and increase the bandwidth
−RL RF − RE
RE RF + RE
gm
RE =
1 + gm R1
Av =
BW × Av =
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Rin =
RE (RF + RL )
RE + RL
Rout =
RE (RF + RS )
RE + RS
1
Cgs
gm
+
RL Cgd
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Taking Advantage of a Zero
Rs
vout
RL
vs
CL
Consider the step function of a low pass circuit. The output
tracks the input with a time constant of τ :
τ = CL (Rs ||RL )
vout
t
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Feedforward with a Capacitor
Cs
Rs
vout
RL
vs
CL
Insight: Add a feedthrough capacitance CS so that the edge
of our signal propagates to the output immediately :
at t = 0+ →
− vout =
Cs
vs
CL + Cs
at t = ∞ →
− vout =
RL
vs
RL + Rs
vout
RL
R L + Rs
Cs
Cs + CL
t
τ = Rs ||RL (CL + Cs )
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Transfer Function with Zero
To see this, we can derive the full transfer function:
vout
=
vs
=
RL
1+RL CL s
RL
RS
1+RL CL s 1+RS CL s
−1
Rs Cs
−1
pole p =
(RL ||Rs )(CL + Cs )
zero z =
1 + Rs Cs s
RL
RL + Rs 1 + (RL ||Rs )(CL + Cs )s
If we equalize the pole / zero, the pole zero cancel and we
have a perfect step !
if p = z →
− Rs Cs = RL CL
vout
t
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Application of a Zero
RL
CL
vs
Rs
Cs
If Rs Cs = RL CL , then the gain is approximately constant over
a fraction of the device fT
vout
Zout
≈
→
− constant
vs
Zin
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Shunt Peaking Amplifier
RL
|Z|
CL
RL network
L
vout
RC network
f
vin
Use an inductor at the drain to produce a zero. The zero
“peaking” location should occur at a high frequency to
compensate for the gain roll-off due to the pole(s)
Note that inductor does not need to be a high Q component
since it’s in series with a large resistor. Can build it using
multiple layers in series to make the inductor compact.
R s RL + 1
1
ZL = (R + Ls )||
= 2
sC
s LC + sRC + 1
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Shunt Peaking Design Equations
|ZL |
=
R
Max BW
|Z | = R at ω = 1/RC
Maximally flat
Best group delay
No shunt peaking
s
1 + (ωτ )2
(1 − ω 2 τ 2 m2 ) + (ωτ m)2
m
√
2
2√
1+ 2
3.1
∞
Normalized BW
1.85
1.8
1.72
1.6
1
Normalized Peak
1.19
1.03
1
1
1
Can trade off between bandwidth (85% increase) versus group
delay variation (60% increase in bandwidth).
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More Shunt Peaking
Shunt and series peaking
Cc
Shunt and double series
peaking. T-coil bandwidth
enhancement.
k
Basically the order of the
matching network is increasing
and it’s resembling a
synthesized transmission line.
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Why does this help?
In the above structures, parasitic capacitors are charged and
discharged serially so that the current available to charge a
capacitor is more and hence the rise time is shorter at the
expense of delay
Can we take this idea to the limit?
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Distributed Amplifier
d
γd , Zd
Zd
Zd
M1
M2
M3
M4
+
vo
−
Zg
+
vs
−
γg , Zg
Zg
g
The goal is to convert the lumped amplifier into a distributed
structure. The idea is to take a fixed gm (transistor width
W ), and split it into parallel fingers that are embedded into a
transmission line at the gate and drain.
Both transmission lines need to be properly terminated to see
flat impedance with frequency. The propagation constant on
the gate and rain line need to be matched so that the waves
add constructively.
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Distributed Amplifier Gain
Zd
gm v1
ro
Co
gm v2
ro
Co
gm v3
ro
Co
gm v4
ro
Co
Zd
+
vo
−
Zg
+
vs
−
Cπ
Ri
+
v1
−
Cπ
Ri
+
v2
−
vgs,i =
Cπ
Ri
+
v3
−
Cπ
Ri
+
v4
−
Zg
vs −j(i−1)βg `g
e
2
βg is the propagation constant on the gate line. The load
current is also a summation of N currents each coming from
the input transistors
N
1X
Id =
id,i e −(N−i)jβd `d
2
i=1
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Gain (cont)
id,i = −gm vgs,i resulting in
N
Id = −
gm X −(i−1)jβg `g −(N−i)jβd `d
vs
e
e
4
i=1
N
X
gm
−Njβd `d jβg `g
e
e −ij(βg `g −βd `d )
= − vs e
4
i=1
The above equation applies for any arbitrary line, but
obviously we’d like to synchronize the delay on the gate and
drain line
βg `g = βd `d = θ
gm
Id = − vs e −(N−1)jθ · N
4
1
|Id |2 Zd
g 2 N 2 Zd Zg
Pout
G=
= 12 2
= m
Pin
4
8 |vs | /Zg
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Artificial Distributed Amplifiers
d
Zd
Zd
M1
M2
M3
M4
+
vo
−
Zg
+
vs
−
Zg
g
Additive gain versus multiplicative gain obtained in cascade.
Bandwidth is extremely high, up to fT /2. In practice the gain
will vary due to the properties of the artificial transmission
line, particularly the cut-off frequency.
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The Right Terminations
m-derived Sections
4 Stage DA Block
2
3
1
4
m-derived Sections
To improve the performance of a DA, it’s important to take
into account the frequency variation of the impedance of the
line due to the fact that it’s actually an artificial lumped line
rather than a truly distributed line.
The m-derived sections are loads that terminate the line in
order to to provide a match over a frequency range
approaching the line cut-off. Otherwise spurious reflections
would occur and cause the gain to roll-off faster.
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A New Twist on Distributed Amplifiers
LOAD
M
3
2
M
Output DA
ZY
M
4
1
Zx
M
2
3
Input DA
M
1
4
M
Internal Feedback
Core DA
2
3 Filter
1
4
M
M-Derived Sections
Notice that a drain line wave can be fed back into the DA and
it can travel back through the gate line in the opposite
direction, thereby generating a cascade gain from the same
DA!
Input and output DA’s used to provide broadband match.
A. Arbabian and A. M. Niknejad “A broadband distributed amplifier with internal feedback providing 660 GHz
GBW in 90 nm CMOS,” Int. Solid-State Circuits Conf. Tech. Dig., pp.196 -197 2008
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Tapered-Line Amplifier
Tapering the Line Tapered1/ 3 ID2
2 / 3 ID2
1/ 3 ID1
4 / 3 ID1
o
ID1
2ID1
o
ID2
o
Two section (n=2) example
In a DA, half theif power
wasted
on the second drain
ID1 ID2isand
properly delayed:
termination.Forward
In a PA,
that’s
a
lot
of
power to throw away.
traveling currents add constructively
Reverse
currents cancel
By tapering the line,
wetraveling
can eliminate
reverse-wave
propagation and hence termination.
In this two-section example, the reflection and transmission
coefficient are given by
© 2009 IEEE International Solid-State Circuits Conference
ρ1 =
(Z0 /2) − Z0
(Z0 /2) + Z0
=
−1
© 2009 IEEE
τ1 =
3
2Z0
(Z0 /2) + Z0
=
4
3
Note that if the currents are properly delayed, the reverse
currents can cancel
J. Roderick and H. Hashemi “A 0.13 µm CMOS power amplifier with ultra-wide instantaneous bandwidth for
imaging applications”, IEEE Int. Solid-State Circuits Conf. Tech. Dig., vol. 1, pp.374 -376 2009
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Broadband Matching Networks
Yin
YS
+
vs
−
Yout
Input
Match
Output
Match
∗
Yin
YL
∗
Yout
Consider that many core amplifiers are broadband but to
obtain the optimal gain requires matching, and the LC
matching networks introduce bandwidth limitations.
Can we make broadband matching networks? Bode-Fano
provides a clue ...
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Bode-Fano Criterion
What’s the best we can do with a matching network in terms
of the quality of the match Γ ∼ 0 and bandwidth?
Surprisingly, there is a theoretical answer to this question and
the answer depends on the load (reactance versus resistance)
[Bode][Fano]. In other words, if we’re trying to match to a
device that has capacitance input impedance with some real
part, there is a fundamental limit to the bandwidth
achievable. For an RC shunt load
Z ∞
π
1
dω ≤
ln
|Γ(ω)|
RC
0
and for an RC series load
Z ∞
ln
0
1
dω ≤ πRC
|Γ(ω)|
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Bode-Fano Criterion
For example, imagine a Brickwall match shown over a
bandwidth B. This result implies that
Z ∞
1
1
dω = B ln
≤ πRC
ln
|Γ(ω)|
|Γ0 |
0
So we cannot in particular go to zero reflection (perfect
match) over an interval of frequencies, only at a finite number
of frequencies. Moreover, there’s a trade-off between the
obtainable bandwidth and the quality of the match. They are
not independent quantities.
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Active Load Filter Matching
RL
LL
+
vo
−
C1
+
vs
−
Lg
L1
C2
L2
Cp
Ls
active filter
Note that the input impedance of an inductively degenerated
amplifier looks like an LCR network. Make that the
termination of a ladder section (broadband) filter !
The core amplifier can be made broadband by using inductive
peaking.
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Balanced Amplifiers
Z1
o
0
90
o
Zin
90o
Z0
Z0
0o
Z0
If a core amplifier is broadband but poorly matched, we can
also use a coupler to drive two amplifiers in parallel as shown.
Note that the reflected signal from the top amplifier is 180◦
out of phase with the reflected signal from the bottom
amplifier ! The reflected signals cancel out.
The bandwidth limitation now comes from the design of a
broadband quadrature coupler.
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Stagger Tuning
1
0
-5
0.8
-10
0.6
-15
-20
0.4
-25
0.2
-30
0.25
0.5
0.75
1
1.25
1.5
1.75
2
0
0.25
0.5
0.75
1
1.25
1.5
A multi-stage amplifier can be made broadband by stagger
tuning the various stages
There’s a trade-off in bandwidth versus gain and gain flatness.
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Transformer Matching
Writing the mesh equations for the transfer function of an
ideal transformer
vout
RL Ms
= 2
2
vs
s (L1 L2 − M ) + s(Rs L2 + RL L1 ) + Rs RL
RL M
s
vout
=
vs
L1 L2 − M 2 (s − p1 )(s − p2 )
If p2 p1 , we can simplify the transfer function
p1 =
−Rs RL
Rs L2 + RL L1
p1 + p2 ≈ p2 =
Mid band gain
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−Rs L2 + RL L1
L1 L2 − M 2
vout
RL M
=
vs
Rs L2 + RL L1
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Low k Coupling Transformers
If we include the capacitance in the transformer, it’s actually a
fourth order circuit. Intuitively, there are two modes due to
resonance and anti-resonance.
In resonance the mutual coupling adds to the effective
inductance and the coupling capacitance is neutralized.
In anti-resonance, the mutual coupling subtracts from the
effective inductance and the coupling is excited. We can see
that the anti-resonance mode is at a higher frequency than
the resonance mode.
If the coupling is strong, these modes are very far apart in
frequency. If the coupling is weak, these modes can be moved
close together to give two peaks in the transfer function.
Similar to stagger tuning, we can broadband the response by
moving peaks close together in an optimal fashion.
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Coupled Resonator Matching
Cc
R1
C1
L1
L2
C2
R2
The transfer function is a 4th order circuit. We can build it as
is or convert it into a transformer coupled circuit by using
Duality (and Y-∆ transformation):
L1 − M
R1
C1
M
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R2
Capacitively Coupled Resonator Stages
Vecchi et. al., “A Wideband Receiver for Multi-Gbit/s Communications in 65 nm CMOS”, JSSC 2011
The transfer function of two coupled resonators can be
approximated by the product of two second-order transfer
functions.
Depending on the strength of the coupling, the poles of the
system can be staggered optimally to provide a flat passband.
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Transfer Function Equations
40f
50
30f
20f
40
30
10f
20
10
30
40
50
60
70
80
90
√
100
s 3 kQω0 R1 R2
(Q(1 +
+ sω0 + Qω 2 )(Q(1 − k)s 2 + sω0 + Qω 2 )
p
p
ω0 = 1/ L1 (C1 + Cc ) = 1/ L2 (C2 + Cc )
GCR (s) = −gm
k)s 2
Q = R1 /ω0 L1 = R2 /ω0 L2
p
k = Cc / (C1 + Cc )(C2 + Cc )
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Matching Network “Filter” Design
Matching network design is very similar to filter design.
Given a transfer function, you can trade-off flatness for group
delay variation, or try to maximize the attenuation and
particular frequencies.
For example, for maximally flat, make as many derivatives
zero near ω = 1
s
H(s) = 4
s + a3 s 3 + a2 s 2 + a1 s + a0
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Optimal Fourth Order Transfer Function
R1
is
vo
io
= −RL
C1
L1 − M
L2 − M
R2
vout
C2
M
RL
sM
s 3 M 2 C1 (1 + sRL C2 ) − RL (1 + sR1 C1 + s 2 L1 C 1) − (R2 + sL2 )(1 + sRL C2 )(1 + sR1 C 1 + s 2 L1 C 1)
For M 1, (low coupling factor k)
vout
sM
=
2
is
(1 + sR1 C1 + s L1 C1 )(1 + RRL2 + s( RL2L + R2 C2 ) + s 2 L2 C2 )
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Maximizing Gain
vout
sM
=
2
is
(1 + sR1 C1 + s L1 C1 )(1 + RRL2 + s( RL2L + R2 C2 ) + s 2 L2 C2 )
L1 C1 ≈ L2 C2 ≈
|Z (jω0 )| =
1
Q1
√
1
ω0
√
k L1
L2
RL
+
√ 1
L 2 ω0 Q 2
Increasing L1 and L2 increases the gain. Limited by quality
factor and resonance.
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References
1
H. W. Bode, Network Analysis and Feedback Amplifier
Design, Van Nostrand, N.Y., 1945.
2
R. M. Fano, “Theoretical Limitations on the Broad-Band
Matching of Arbitrary Impedances,” Journal of the Franklin
Institute, vol. 249, Jan. 1950.
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