Advanced Mixers

Transcription

Advanced Mixers

Berkeley
Advanced Mixers
Prof. Ali M. Niknejad
U.C. Berkeley
c 2014 by Ali M. Niknejad
Copyright Niknejad
Advanced IC’s for Comm
Sampling Versus Mixing
Note that the essential difference between a “mixer” and a
“sampler” is the hold operation. Most samplers have a
zero-order hold, so the output is held high by the capacitor.
In a “mixer”, the output returns to zero due to the lack of
energy storage elements. The conversion gain is simply the
fundamental Fourier coefficient of the LO waveform. For a
50% square wave
gc =
2
2π
Z
Pi
sin(x)dx =
0
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1
≈ 0.318 (-10 dB)
π
Sampling
For a ZOH sampler, we can model it as an ideal sampler
followed by an a LPF with bandwidth 1/TD . Note that TD is
the duty cycle over which the output is held. Assume TD is
small so we can model the system as follows:
yout (t) = (p(t) · x(t)) ∗ h(t)
Yout (f ) = (P(f ) ∗ X (f ))H(f )
For a sinusoidal input, we have
1
X (f ) = (δ(f − fx ) + δ(f + fx ))
2
and
∞
1 X
P(f ) =
δ(f − nfLO )
TLO n=−∞
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Sampling
Taking the convolution of X and P, we note that things don’t
overlap unless the input is shifted to the peaks of P
∞
X
1
(δ(f − nfLO − fx ) + δ(f − nfLO + fx ))
X (f )∗P(f ) =
2TLO n=−∞
The hold action can be modeled by a filter with constant
finite impulse response of duration TD . Taking the Laplace
Transform, we get a sinc frequency response.
H(f ) =
sin(πfTD ) −jπfTD
e
= TD sinc(πfTD )e −jπfTD
πf
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Sampling Conversion Gain
1.0
0.8
0.6
0.4
0.2
-3
-2
-1
1
2
3
Finally, multiplying by the sinc with our impulse train, we get
the overall transfer function. For a IF frequency, reducing the
hold time improves the gain of the system. This is kind of a
non-intuitive result !
There’s actually a trade-off between the bandwidth of the IF
and the hold time, so there’s no free lunch.
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Tayloe’s Sampling Receiver
Product detector and method therefor, US Patent 6230000 B1
Note that the mixer is realized by four samplers, each
sampling with 25% duty cycle.
Such a sampler has I/Q quadrature down-conversion and is
balanced.
As we have seen, this bilateral mixer also creates a
narrowband RF filtering effect which is beneficial for rejection
of blocker signals.
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Harmonic Reject (HR) Mixers
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The Need for Harmonic Rejection
If the RF bandwidth is wider than an octave, then harmonics
of the LO are downconverter to the same IF and corrupt the
desired signal.
Imagine operating a receiver at 800 MHz. A PA operating at
3X this frequency or 2400 MHz will couple into the receive
chain and get downconverter by the third harmonic of the LO.
Even with an Rx filter, the coupled signal may be very strong
(30 dB of isolation implies a 0 dBm 3rd harmonic blocker)
Cable receivers are among the systems that process very wide
swaths of bandwidth. Reconfigurable LTE transceivers operate
from 500 MHz to 3600 MHz and have HR requirements.
Wideband Software Defined Radios (SDR) also need it for
similar reasons.
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Why Mixers Like Harmonics
Even if we drive the LO with a sine wave, as long as the
amplitude exceeds the threshold voltage of the switching, then
the output is steered more like a square wave, which is rich in
harmonics.
∞
4 X sin(2π(2n − 1)fLO t)
s(t) =
π
2n − 1
n=1
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…
F
…
…
…
How to Synthesize
a Sine Wave?
-T/2
T/2
t
T
-7/T
-5/T
-3/T
-1/T
0
1/T
3/T
5/T
f
7/T
-1
(b)
|So(f)|
so(t)
1
f
…
…
-T/2
T/2
T
…
t
…
F
-9/T
-7/T
-5/T
-3/T
-1/T
0
1/T
3/T
5/T
7/T
9/T
-1
(c)
Notice that sampling a sine wave produces a square wave. If
Figure 4.10 Sample and hold operation on sinusoid in time and frequency domains for three sampling
we oversample the
sine (a)
wave
cycle, we
frequencies.
fs=2fi. eight
(b) fs=4fi.times
(c) fs=8fiper
.
produce a staircase approximation to the sine wave.
As the
sampling
rate is pushed
even higher,
the resulting is
SHS
waveform
The
key
observation
is that
this waveform
free
from will
thehave
3rd
harmonic!
Can
somehow
use thisthewaveform
even and
fewer 5th
harmonics.
Intuitively
thiswe
makes
sense by inspecting
time-domain?signal
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which is an SHS waveform, is the key element to the rejection of unwanted harmonics in
Summing
Four Square
Waves
the harmonic-rejection
mixer presented
in the following section.
T/16
ss(t)
…
…
-T/2
t=0
T/2
T
T/2
T
T/2
T
T/2
T
t
p0(t)
…
…
2
-T/2
t=0
t
p1(t)
…
…
1
-T/2
t=0
t
p2(t)
…
-T/2
1
t=0
…t
Figure 4.14 SHS waveform resulting from shifting of sampling position and the square waves which
compose the SHS waveform.
The staircase approximation
to sine can be synthesized by
summing
three square waves of the right magnitude.
The square waves illustrated in Figure 4.14 need to be scaled in amplitude and
phase shifted appropriately to reject
the third and
fifth harmonics. Signal p0(t) is scaled
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Fourier Series of Square Waves
The wave p0 (t) is a 50% duty cycle square wave
√
1
1
1
4 2
(cos(ωt)− cos(3ωt)+ cos(5ωt)− cos(7ωt)+· · · )
p0 (t) =
π
3
5
7
The waveform p1 (t) is just a time and amplitude shifted
version of p0 (t)
√
2 2
1
1
1
p1 (t) =
(cos(ωt) + cos(3ωt) − cos(5ωt) − cos(7ωt) + · · · )+
π
3
5
7
√
2 2
1
1
1
(sin(ωt) − sin(3ωt) − sin(5ωt) + sin(7ωt) + · · · )
π
3
5
7
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Synthesizing a Sine from Squares (II)
Similarly, p2 (t) is also a time shifted copy
√
2 2
1
1
1
p2 (t) =
(cos(ωt) + cos(3ωt) − cos(5ωt) − cos(7ωt) + · · · )+
π
3
5
7
√
2 2
1
1
1
(− sin(ωt) + sin(3ωt) + sin(5ωt) − sin(7ωt) + · · · )
π
3
5
7
The desired waveform, a sum of the three, is free from 3ω and
5ω
√
8 2
1
p(t) = p0 (t) + p1 (t) + p2 (t) =
(cos(ωt) − cos(7ωt) + · · · )
π
7
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99
4.5 Harmonic-Rejection Mixer
Weldon/Berkeley Architecture
RL
RL
Vout
(W/L)sw
I
(W/L)sw
2 (W/L)sw
LO-45
Vin
(W/L)in
I
2 (W/L)sw
(W/L)sw
I
LO0
(W/L)in
2 (W/L)in
Ibias
(W/L)sw
LO45
2 (W/L)in
(W/L)in
I
2 Ibias
(W/L)in
Ibias
Figure 4.16 Simplified circuit diagram of the harmonic-rejection mixer.
Three parallel mixers are used, each driven by p0 (t), p1 (t),
and
p2 (t).potentially
We must
scale
the achieved
devicesbyappropriately
totherealize
The scaling
could
have √
been
scaling the load of
three
the right amplitude (need 2) [Weldon]
mixers. However, in this case the current from each mixer is summed and an output
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is multiplied by
the SHS waveform.
Matching
Requirements
Therefore comparing the output of the HRM to
square wave represents the relative improvement in the suppression of the harmonics.
T
2
(1+ )p0(t+ T/2 )
2 (1+ )
t
square
wave
HD3
1
t
p1(t)
HD5
HR3
HR5
t
p2(t)
1
fLO
3fLO
5fLO
7fLO
f
Figure 4.17 Graphical representation of HR3 and HR5.
Time mismatch and/or amplitude mismatch leads to finite 3rd
and 5th.
Based on the harmonics of a square wave from Equation (4.5), the relationship
between and HD3 in dB is given by
HDNiknejad
HR3
3
1
20Advanced
log IC’s for Comm
Matching Requirements (3rd)
40
35
=0.01
30
HR3 (dB)
=0.05
25
20
15
10
=0.1
0
0.5
1
1.5
2
2.5
3
3.5
4
(degrees)
GoodFigure
HR requires
good
matching
! of gain and
4.18 Thirdvery
harmonic
rejection
as a function
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phase error.
113
4.6 Matching in HRM
Matching Requirements (5th)
40
35
=0.01
30
=0.05
25
HR5 (dB)
20
15
=0.1
10
5
0
0.5
1
1.5
2
2.5
3
3.5
4
(degrees)
Figure 4.19 Fifth harmonic rejection as a function of gain and phase error.
Niknejad
labeled for the sakeSynthesis
of simplicity.
LO Waveform
D
D1M
__
D
Q
__
clk
Q
Q1M
D
D1S
__
D
Q
__
clk
Q
D1
Q1
D
D2M
__
D
Q
__
clk
Q
Q2M
D
D2S
__
D
Q
Q2
__
clk
Q
D2
LO
LO
Q1M
Q1
Q2M
Q2
Figure 4.22 Expanded view of divide-by-four and timing diagram.
Requires an LO to be provided at 4 times higher frequency
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Harmonic Reject Transmitter
135
5.2 Harmonic Rejection Transmitter Architecture
IBB
IIF
DAC
fRF
0
90
`
QBB
PA
RF Filter
QIF
DAC
= HRM
LO2
8 phase
generator
LO1
Figure 5.1 Block diagram of a Harmonic Rejection Transmitter.
Similar concept applies for the receiver.
The basic function of the HRT is as follows. The baseband is identical to the
homodyne and heterodyne transmitter which were discussed in Chapter 3 as digital inphase and quadrature baseband signals
pass through
a DAC
then a low-pass filter.
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Advanced
IC’s and
for Comm
Forming Gains at Baseband
Prof. Al Molnar’s MS thesis had this 8-phase passive mixer
with HR suppression [Molnar].
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Sub-Harmonic Mixers
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Sub-Harmonic Mixer Design
Sub-Harmonic Mixer
fRF
fLO = 0.5fRF
The subharmonic mixer is driven by an LO signal that is an
integer fraction, or subharmonic, of the desired LO frequency.
For example, if the RF signal is 2GHz, and the desired LO is
2GHz for direct conversion, a subharmonic mixer will be
driven by a 1GHz LO signal. The advantages are:
Lower LO re-radiation through the antenna (LO leakage)
Lower LO self mixing (lower DC offset at IF)
Relaxed requirement on the device switching speed.
Lower LO buffer current
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25% LO Waveform
The idea is to generate an LO signal that is half the desired
frequency, yet is rich with a second harmonic, then somehow
use the second harmonic, along with the RF, to get desired IF.
A 50% duty cycle LO has no second harmonic. However a
25% duty cycle LO has the desired harmonic.
VLO
TLO /4
TLO
The Fourier series expansion of the above square wave is:
VΦ1 =
"
#
√
√
1 √
2
2
cos(3ω0 t) −
cos(5ω0 t) + ...
2 cos(ω0 t) + cos(2ω0 t) +
π
3
5
Where ω0 is the subharmonic LO frequency, which in this case
is half the desired LO frequency.
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Sub-Harmonic Mixer
LO1
+
RF
IF
LO2
The subharmonic mixer topology uses two identical mixers
excited by two phases of the 25% duty cycles. The RF signal
is multiplied by these two delayed 25% LO signals and the IF
is added in phase at the output, as shown in the above figure.
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Sub-Harmonic Mixer Waveforms
The two LO signals, VF 1 and VF 2 , are T /2 delayed relative
to each other.
VLO
TLO /4
TLO
TLO /2
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Sub-Harmonic Waveforms
VΦ2
"
#
√
√
√
1
2
2
=
− 2 cos(ω0 t) + cos(2ω0 t) −
cos(3ω0 t) +
cos(5ω0 t) + · · ·
π
3
5
Looking at the combined IF output of the subharmonic mixer
we can write:
2
VIF = VRF VΦ1 + VRF VΦ2 = VRF
cos(2ω0 t)
π
As seen, the resulting IF output is the product of the RF
signal and an effective LO that has twice the subharmonic LO
frequency, which is in fact the desired LO. The above analysis
was for only one mixer. The question then, how can one build
quadrature subharmonic mixer for both I and Q channels?
The answer comes in the way to generate a sin(2ω0 t) LO
signal using similar 25% duty cycle signals as follows.
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I/Q Sub-Harmonic Mixer
VLO
TLO /8
TLO
5 TLO /8
The LO signal LOF 3 is delayed T /8 relative to LOF 1 . LOF 4 is
delayed by T /2 relative to LOF 3 .
The Fourier series expansion of the above square waves is:
VΦ3 =
VΦ4 =
1
"
√
π
1
π
2 cos(ω0 t −
π
2
√
) + sin(2ω0 t) +
2
3
sin(3ω0 t −
√
2
"
π
2
√
)+
2
5
cos(5ω0 t −
√
2
π
2
#
) + ...
√
π
π
π
− 2 cos(ω0 t − ) + sin(2ω0 t) −
sin(3ω0 t − ) −
cos(5ω0 t − ) + ...
2
3
2
5
2
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#
Generation of LO
Looking at the combined IF output of the subharmonic mixer
we can write:
2
sin(2ω0 t)
VIF Q = VRF VΦ3 + VRF VΦ4 = VRF
π
In order to generate all 4 phases of the subharmonic LO
signals, a divide by 4 prescalar needs to be used. This means
the VCO needs to run at 4X the subharmonic LO, 2X the RF
signal for direct conversion.
IFI
LO1
LO2
RF
2f0
LO3
V CO
÷4
4f0
LO4
IFQ
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Sub-Harmonic Realization
IF (+)
IF (−)
LO(+)
LO(+)
LO(−)
LO(−)
RF(+)
RF(−)
The above is a simplified implementation of one subharmonic
mixer. Since a subharmonic mixer is two mixers in one driven
by the same RF input but has two different LO phases, a
single common Gm stage is used.
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Mixer First Architecture
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Why Not Mixer First
Getting rid of the LNA seems like a really bad idea. Isn’t an
LNA a must ?
The LNA input impedance is matched to the antenna (or
filter) for optimal power transfer into the receiver.
Furthermore, the impedance match simplified board design
since transmission lines can be used to bring in the signal from
the antenna without worrying about any potential impedance
transformation (except length should be minimized to reduce
losses).
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Why Not (Noise and LO Leakage)
Recall that the LNA is there to minimize the noise
contribution of sub-sequent stages, so it should be designed
with low noise and as high of a gain as possible.
Also the LO signal leakage to the antenna port is minimized,
an issue if we remove the LNA.
Niknejad
Maybe ?
In practice the gain is limited by linearity considerations, so a
trade-off is made. The first potential benefit of a mixer first
architecture is therefore to realize much higher linearity.
If the linearity can be made good enough to reduce or
eliminate off-chip filters (LC, SAW, or other bulky
components), then there’s a potential win. But this means
mixer noise cannot be too high.
Since the filters introduce about 1-2dB of loss at the input of
the Rx, perhaps a mixer first Rx can be about 1-2 dB noisier
than an LNA first counterpart.
Niknejad
Cook’s Mixer First Receiver
Fig. 1. Transceiver block diagram.
Fig. 2. Simplified transceiver front-end schematic.
Ben Cook, a Berkeley grad working with Prof. Pister,
proposed a mixer first receiver and was among the first of
recent IC designers to realize the benefits of n-path filtering
offered by passive bilateral mixers. [Cook]
Fig. 3.
Ben Cook and Axel Berny started a company based on this
concept (Passif) that was acquired by Apple.
Quadrature VCO utilizing back gate coupling.
and DC level of the VCO signal driving the switches. The
mixer outputs differentially drive a bandpass filterNiknejad
comprising
Fig. 4. Circuit model for tapped-capacitor resonator.
Advancedcoupled
IC’s forVCO
Comm
back-gate
architecture was used here because it
Mixer First Motivation and Benefits
Use a passive mixer to achieve very good linearity. Take
advantage of the bilateral nature to realize a bandpass
response to improve out-of-band linearity.
Make the system tunable by controlling the LO frequency,
rather than tuning capacitors or inductors, potentially
realizing a very broadband receiver that is also linear and
robust to out-of-band interference.
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Andrew/Molnar Mixer First Proposal
Fig. 9. Equivalent baseband amplifier noise models.
Fig. 11. Simulated and analyt
Fig. 10. Proposed passive-mixer first receiver with resistive feedback
amplifiers.
the squared noise by ) a
), yield
Use a multi-phase voltage mode passive mixer driventiplying
by
by
sistive feedback. We find the new effective
present on each
delayed copies
of
the
LO.
As
we
shall
see,
we
can
combine
the
branch by applying the Miller effect to the feedback resistor
:
output phases to form an I/Q signal with desirable properties.
This multi-phase idea was inspired by the Weldon
(31)
sub-harmonic
mixer
and
a
similar
structure
using
8-phases was
Substituting the new
into the impedance matching LTI
modelM.S.
shows thesis.
that we can perform impedance matching using
in Molnar’s
the amplifier feedback resistors.
Once we have added the feedback amplifiers to implement
, the noise performance
changes
as well. Whereas most of
Niknejad
We have verified this
lation using periodic ste
,
Equivalent Model
REWS AND MOLNAR: IMPLICATIONS OF PASSIVE MIXER TRANSPARENCY
3093
1. (a) Simplified circuit model of 4-phase passive mixer. (b) LO driving waveforms. (c) Equivalent model to (a), with
ng nature of the waveforms in the middle.
lumped with
based on nonoverAssume non-overlapping 25% clock drive to the mixers.
Assuming the switches turn on instantly, and have an
ng range. These advantages imply that passive mixer-first with fundamental frequency
and time varying
on-resistance
Rsw .the flex- phase and amplitude , which capture both modulation
ivers will likely provide
the next step in of
improving
ty and performance of highly integrated wireless receivers.
and offset frequency of the received signal. If the amplitude and
Note that at any given time,phase
oneoffset
switch
is always
by can
thebe approxthey
change slowly
relative seen
to
PASSIVE MIXER TRANSPARENCY: FIRST-ORDER ANALYSIS imated as constant over a given LO period, and the input can be
source, so we can move switch resistance and lump it with the
as
he passive mixer analyzed here contains four switches (tran- approximated
antenna
resistance
Ra to form Ra0 = Ra + Rsw .
ors) which are successively
turned
on in four nonoverlap-
(2)
g, 25% duty-cycle phases over the course of one local ostor (LO) period [1], [10], [15]–[17]. These nonoverlapping
To compute the input impedance presented by the mixer to
es are necessary for preventing the I-Q crosstalk described the antenna, we start by computing the voltage across each of
12]. The input port of the mixer is connected directlyNiknejad
to the
Charge Balance
Assume that the baseband acts like a sampler and holds the
voltage over the LO cycle. In other words, the time constant
RB CL TLO .
Using this assumption, we can invoke charge balance and
derive the magnitude of the baseband voltage on stage VC ,m
(m = 0, · · · , 3)
Qm =
VC ,m TLO
= charge flowing into RB
RB
Z
=
(m+1)TLO
4
mTLO
4
+
T
− LO
8
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TLO
8
VRF − VC ,m
dt
Ra0
Charge Balance for Sinusoidal Drive
If the input VRF is assumed to be a tone at ω0 , we have
VRF = A cos(ωLO t + φ)
VC ,m TLO
A sin(ωLO t + φ) T2 VC ,m TLO
=
− R0
RB
ωLO Ra0
4
a
T1
1
1
A
(sin(ωLO T2 + φ) − sin(ωLO T1 + φ))
VC ,m TLO
+
=
0
RB
4Ra
ωLO Ra0
√
2 2
RB
mπ
VC ,m =
A cos(φ +
)
π RB + 4Ra0
2
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Up-Convert Baseband Voltage
The baseband voltage will induce an RF current to flow from
the antenna
VRF (t) − Vx (t)
Ia (t) =
Ra0
To find Vx (t), we note that it’s a staircase function due to
the periodic rotation of the switches with a fundamental
period of TLO .
Four successive outputs in one cycle are given by
{cos(φ), − sin(φ), − cos(φ), sin(φ)} = {i, −q, −i, q}
If we find the fundamental coefficient of the Fourier series, we
can find Vx,ωLO :
Vx,ωLO = A
8
RB
8
RB
cos(ωLO t + φ) = VRF 2
2
0
π RB + 4Ra
π RB + 4Ra0
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RF Current
We can now calculate the RF current flowing from the
antenna, and hence the impedance seen by the antenna
Ia,ωLO =
VRF − Vx,ωLO
Ra0
4Ra0 + 1−8
R
π2 B
= VRF (t) 0
Ra RB + 4Ra02
Note that as RB → 0, Ia = VRF /Ra0 . RB is in series with Ra0 .
Note that as RB → ∞, Ia = VRF (1 − 8/π 2 )/Ra0 . This means
that there’s an additional equivalent shunt impedance in
parallel to RB !
Niknejad
S ON CIRCUITS AND SYSTEMS—I: REGULAR PAPERS, VOL. 57, NO. 12, DECEMBER 2010
Mixer LTI Equivalent Circuit
ause
ed to
step
ding
hase
y the
o the
Fig. 3. LTI equivalent circuit for passive mixer with
due to harmonics and
The first Ra0 term is. obvious, but the next two terms are not
impedance-transformed
at all obvious.
The term γRB is the baseband equivalent resistance seen by
the antenna.
The Rsh term is in shunt and represents additional power loss.
It will be shown that this is actually the power loss due to the
harmonics of the harmonics of the LO. Up to now it was
derived using charge balance but if we go through each
harmonic and calculate it’s loss as seen from the input port,
we arrive at the same result.
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Equivalent Circuit Elements
From the model we can calculate the equivalent impedance
Ia (ωLO ) = VRF (ωLO )
Ra0 γRB
γRB + Rsh
+ Ra0 Rsh + Rsh γRB
Equate this to the expression derived from charge balance to
get
2
γ = 2 ≈ 0.203
π
4γ
Rsh = Ra0
≈ 4.3Ra0
1 − 4γ
Note that Rsh was derived assuming Ra0 is constant over all
frequencies, which is not a good assumption for most
antennas or filters that may precede the antenna.
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Impedance Matching
Given that the input impedance has a real part that depends
on the switch resistance (naturally) but also on the baseband
impedance, we can turn the baseband impedance to realize a
match
Rin = Rsw + γRB ||Rsh
Rsw < Rin < Rsw + Rsh
We’d like to make Rsw Ra but we’ll return to this point
later. For now, solve for RB
RB =
1 Rsw Ra − Rsw Rsh
γ Rsw + Rsh − Ra
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Antenna Impedance Variation
In the above charge balance derivation, it was assumed that
the antenna impedance was constant, even at LO harmonics.
Since the voltage Vx has harmonics of the LO, the current
draw from the antenna at these harmonics will differ. If we
write Vx as a Fourier series and account for the antenna
impedance variation by defining
Ra0 (nωLO ) = Ra (nωLO ) + Rsw
Then the current through the switches is given by
∞
X
Vx,n (t)
VRF (t)
Ia (t) = 0
−
Ra (ωLO )
Ra0 (nωLO )
n=1,3,···
Invoking charge balance, we arrive at

−1
∞
X
1

Rsh = 
n2 Ra0 (nωLO )
n=3,···
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Rsh Harmonic Interpretation
We now have an alternative expression for Rsh

Rsh = 
∞
X
n=3,5,···
−1
1

n2 Ra0 (nωLO )
Let’s verify that this agrees with our earlier calculation when
Ra is constant
−1

∞
X
1

Rsh = 
n2 (Ra + Rsw )
n=3,5,···
Rsh =
π2
−1
8
1
(Ra + Rsw )
Niknejad
−1
= Ra0
4γ
1 − 4γ
Non-Zero IF
Excite the circuit with a frequency above or below the LO (to
produce a non-zero IF)
VRF (t) = A cos((ωLO + ωIF )t)
Assume baseband impedance is an RC but excited at non-zero
ZB (ωIF ) = RB ||
1
RB
=
jωIF CL
1 + jωIF CL RB
Repeating the math and assuming ωIF ωLO , an essentially
identical expression is derived (now with ZB ):
Vx (ωRF ) = VRF (ωRF )
Niknejad
ZB
8
π 2 ZB + 4Ra0
ANDREWS AND MOLNAR: IMPLICATIONS OF PASSIVE MIXER TRANSPAREN
General LTI Model
Given
above result,
it’sfor
now
easy to show that the .
Fig. 6. New
LTIthe
equivalent
circuit
frequency-dependent
antenna port is loaded by an impedance given by
to (6)
γRB ||Rsh
, we findZinan
(ωIFexpression
) = Rsw + becomes essentially identic
1 + jωIF CL γRB ||Rsh
From the RF port, this is transformed into a bandpass
characteristic similar to what we have seen before (due to
mixer transparency, a baseband low-pass is converted into a
bandpass response).
This then produces the same
derivation
of the current
Niknejad
(28
a
Noise Model
IEEE TRANSACTIONS ON CIRCUITS AND SYSTEMS—I: REGULAR PAPERS, VOL. 57, NO. 12, DECEMBER
. Equivalent noise models for LTI model from Fig. 3.
.
Given that this is an LTV system, we must account for all the
noise folding that occurs from every harmonic of the LO. It is
fortunate that this can be simply modeled by an equivalent
Rsh . In fact, we have already derived that Rsh models the
power loss due to the harmonics, so this is not completely
Equivalent baseband amplifier noise models.
surprising
With this insight, we have a very simple LTI model.
Niknejad
Noise Analysis
Each noise term is summed to calculate the total output
noise. This expression is then converted to a NF by
normalizing to the source noise
Rsh
Rsw
+
F =1+
Ra
Ra
Ra + Rsw
Rsh
2
RB
+
Ra
Ra + Rsw
γZB
2
In practice, the baseband resistance is realized as a shunt
feedback resistance with an amplifier. In this situation, the
baseband amplifier contributes noise
Rsh Ra + Rsw 2 γRF Ra + Rsw 2
Rsw
+
+
+
F =1 +
Ra
Ra
Rsh
Ra
γRF
va2
Ra + Rsw + Rsh 2
Ra + Rsw
γ
+
4kTRa
γRF
Rsh
Niknejad
Noise Optimization
Zsh should be maximized to minimize contributions of
harmonics to noise and loss.
Zin = Rsh + γZB ||Zsh
Also, this will maximize the tuning range of the system (by
varying RB ).
But note that maximizing Zsh requires one to maximize the
antenna impedance at the harmonics of the LO, opposite to
an LC tank or a resonant antenna.
Niknejad
Optimum Switch Size
Since the switch adds parasitic capacitance in parallel to Zsh ,
we cannot arbitrarily make the switch large.
Za (ω) → 0 at ω → ∞
Za0 (ω) → Rsw for ω → ∞
Reducing Rsw reduces Zsh at higher frequencies, which
increases the noise! There’s an intermediate width for optimal
performance.
If we have a very high Q antenna such that the antenna
impedance is zero at harmonics, then Rsh = 4.3Rsw and so
F =1+
Rsw
(Ra + Rsw )2 (1 − 4γ)
+
Ra
4γRa Rsw
For Fmin ≥ 4.05 dB, choose Rsw
p
Rsw = Ra 1 − 4γ
Niknejad
d
8-Phase Architecture
for varying
showing shift in center frequency
ase passive mixer.
e repeat this analysis for the case where
Noise figure can be improved if we can minimize the
harmonic contributions to Rsh . One way to do this
is to build a harmonic reject mixer.
By using an 8-phase mixer with 12.5% duty cycle,
the number of harmonics is cut in half compared to
a 4-phase mixer (eliminate 3rd, 5th, 11th, 13th, and
so on)!
This boosts the value of Zsh and also changes the
value of γ
√
2
γ8ph = 2 (2 − 2)
π
8γ8ph
(Rsw + Za ) ≈ 18.9(Rsw + Za )
Zsh,8ph =
1 − 8γ8ph
The optimum switch size is also modified and the
minimum achievable noise figure is close to 2 dB !
Niknejad
References
Weldon Jeffrey Arthur Weldon, High Performance CMOS Transmitters for Wireless Communications, Ph.D.
Dissertation, 2005.
Cook B.W. Cook, A. D. Berny, A. Molnar, S. Lanzisera, K.S.J. Pister, “An Ultra-Low Power 2.4GHz RF
Transceiver for Wireless Sensor Networks in 0.13/spl mu/m CMOS with 400mV Supply and an Integrated
Passive RX Front-End,” IEEE Internationa Solid-State Circuits Conference(ISSCC), Digest of Technical
Papers, pp.1460,1469, 6-9 Feb. 2006.
Molnar A. Molnar, B. Lu, S. Lanzisera, B.W. Cook, K.S.J. Pister, “An ultra-low power 900 MHz RF transceiver for
wireless sensor networks,” Custom Integrated Circuits Conference, 2004. Proceedings of the IEEE 2004 ,
vol., no., pp.401,404, 3-6 Oct. 2004.
AndrewsTCAS C. Andrews and A. Molnar, “Implications of Passive Mixer Transparency for Impedance Matching and
Noise Figure in Passive Mixer-First Receivers,” IEEE Trans. on. Circuits and Systems-I: Regular Papers,
vol. 57, no. 12, Dec. 2010.
AndrewsJSSC C. Andrews and A. Molnar, “A Passive Mixer-First Receiver With Digitally Controlled and Widely Tunable
RF Interface,” IEEE J. of Solid-State Circuits, vol. 45, no. 12, Dec. 2010.
Niknejad

Advanced Mixers

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