Lecture 18: Balanced Mixers/PNoise and...
Transcript of Lecture 18: Balanced Mixers/PNoise and...
EECS 142
Lecture 18: Balanced Mixers/PNoise andPSS/Transformers
Prof. Ali M. Niknejad
University of California, Berkeley
Copyright c© 2005 by Ali M. Niknejad
A. M. Niknejad University of California, Berkeley EECS 142 Lecture 8 p. 1/28 – p. 1/28
Double Balanced Mixer
IEE
+LO
−LO
VCC
Q1 Q2 Q3 Q4
Q5 Q6
+LO
+RF −RF
RE
IF
RIF
RE
The single balancedGilbert cell withdifferential outputrejects the RF at theIF port but the LOfeed-through remains.
A double balancedmixer has a dif-ferential RF and adifferential LO (doublebalanced)
A. M. Niknejad University of California, Berkeley EECS 142 Lecture 8 p. 2/28 – p. 2/28
Double Balanced Mixer LO Rejection
As before, the RF stage is a transconductance stage.Degeneration can be used to linearize this stage.
Because the bias current at the output is constantIEE/2 regardless of the LO voltage, the LO signal isrejected. This relies on good matching betweentransistor Q2 and Q4.
The differential operation also rejects even orderdistortion. Viewed as two parallel Gilbert cells, thismixer is also more linear as it processes only half of thesignal.
The noise of this mixer, though, is higher since thenoise in each transistor is independent.
A. M. Niknejad University of California, Berkeley EECS 142 Lecture 8 p. 3/28 – p. 3/28
Mixer Operation
IEE
+LO
−LO
VCC
Q1 Q2 Q3 Q4
Q5 Q6
+LO
+RF −RF
RE
IF
RIF
RE
+is
−is
IEE
+LO
−LO
VCC
Q1 Q2 Q3 Q4
Q5 Q6
+LO
+RF −RF
RE
IF
RIF
RE
+is−is
The AC operation nearly identical to a single balancedGilbert cell.
Even a single ended output, though, effectivelymultiplies the signal by ±1, thus rejecting the RF signal.
A. M. Niknejad University of California, Berkeley EECS 142 Lecture 8 p. 4/28 – p. 4/28
Mixer Noise DefinitionBy definition we have F = SNRi
SNRo
. If we apply this to areceiving mixer, the input signal is at the “RF” and theoutput signal is at “IF”. There is some ambiguity to thisdefinition because we have to specify if the RF signal isa single or double sideband modulated waveform.
For a double sideband modulated waveform, there issignal energy in both sidebands and so for a perfectmultiplying mixer, the F = 1 since the IF signal is twiceas large since energy from both sidebands fall onto theIF. For a single-sideband modulated waveform, though,the noise from the image band adds doubling the IFnoise relative to RF. Thus the F = 2.
If an image reject filter is used, the noise in the imageband can be suppressed and thus F = 1 for a cascadeof a sharp image reject filter followed by a multiplier.
A. M. Niknejad University of California, Berkeley EECS 142 Lecture 8 p. 5/28 – p. 5/28
Mixer Noise FoldingSince the mixer will downconvert any energy at adistance of IF from the LO and it’s harmonics, all thenoise from these image bands will be downconverted tothe same IF.
For a Gilbert cell type mixer, the current of the Gm stageproduces white noise which is downconverted.Summing over all the harmonics, we have
∞∑
k=−∞
|ck|2 =1
T
∫ T
0
s(t)2dt
where the last equality follows from Parseval’s Theorem.
For a square waveform s(t), the harmonic powers falllike 1/k2. Thus the third harmonic contributes about 10%to the switching pair noise.
A. M. Niknejad University of California, Berkeley EECS 142 Lecture 8 p. 6/28 – p. 6/28
Mixer Noise Due to Switching Pair
Q1 Q2
In addition to the noise fold-ing discussed above, theswitching pair itself will con-tribute noise at IF. If Q1/Q2are both on, they act as adifferential amplifier and in-troduce noise.
When only Q1 or Q2 is on, though, the noise is rejecteddue to the degeneration provided by thetransconductance device.
Thus we generally use a large LO signal to minimizethe time when both devices are conducting. It’stherefore not surprising that the noise figure of themixer improves with increasing LO amplitude.
A. M. Niknejad University of California, Berkeley EECS 142 Lecture 8 p. 7/28 – p. 7/28
Simulating Mixers with SPICE
Let’s take a typical example of RF = 1GHz, IF = 1MHz.This requires an LO = RF ± IF , close to the RF.
To resolve the IF frequency components, we need tosimulate several cycles of the IF, say 10, so Tsim = 10TIF
But that means that we must simulate 10, 000 RF cycles
Tsim
TRF
=10TIF
TRF
=10fRF
fIF
=10 · 1000
1= 10, 000
If we are conservative, we may insist that we simulateat least 10 points of the RF cycle, that implies 100, 000points per simulation.
This is a long and memory intensive simulation. Weencountered a similar problem when simulating IM3
A. M. Niknejad University of California, Berkeley EECS 142 Lecture 8 p. 8/28 – p. 8/28
Periodic Steady-State (PSS) Simulation
Transient simulation is slow and costly because wehave to do a tight tolerance simulation of several IFcycles with a weak RF.
The SpectreRF PSS analysis is a tool for finding theperiodic steady-state solution to a circuit. In essence, ittries to find the initial condition or state for the circuit(capacitor voltages, inductor currents) such that thecircuit is in periodic steady state.
It can usually find the periodic solution within 4-5iterations.
In the mixer, if we ignore the RF signal, then theperiodic operating point is determined by the LO signalalone.
A. M. Niknejad University of California, Berkeley EECS 142 Lecture 8 p. 9/28 – p. 9/28
PSS Iteration
Since typical PSS run converges in 4-5 cycles of theLO, or a simulation time of about 5TLO, the overallsimulation converges several orders of magnitude fasterthan transient at the IF frequency.
PSS requires that the circuit is not chaotic (periodicinput leads to a periodic output).
High Q circuits do not pose a problem to PSSsimulation since we are finding the steady-statesolution. The high Q natural response takes roughly Qcycles to die down, thus saving much simulation time.
A. M. Niknejad University of California, Berkeley EECS 142 Lecture 8 p. 10/28 – p. 10/28
PSS OptionsWe can perform PSS analysis on driven or autonomouscircuits. An autonomous circuit has no periodic inputsbut produces a periodic output (e.g. an oscillator).
For PSS analysis we need to specify a list of “large”signals in the circuit. In a mixer, the only large tone isthe LO, so there is only one signal to list.
We also specify the “beat frequency” or the frequencyof the resulting periodic operating point. For instance ifwe drive a circuit with two large tones at f1 and f2, thebeat frequency is |f1 f2|. Spectre can auto-calculatethis frequency.
An additional time for stabilization tstab can be specifiedto help with the convergence. For a mixer this is notneeded.
A. M. Niknejad University of California, Berkeley EECS 142 Lecture 8 p. 11/28 – p. 11/28
Periodic AC (PAC)
Once a PSS analysis is performed at the “beatfrequency”, the circuit can be linearized about this timevarying operating point.
Note that for a given AC input, there are as manytransfer functions as there are harmonics in the LO.
We specify the frequency range of the AC input signalas either an absolute or relative range. A relative rangeis a frequency offset with respect to the beat frequency.
We also specify the maximum sidebands to keep for thesimulation. Note that this does not affect the accuracyof the simulation but simply the amount of saved data.
A. M. Niknejad University of California, Berkeley EECS 142 Lecture 8 p. 12/28 – p. 12/28
PAC Picture1.0
0.9
0.1
2.0
3.01.9
2.91.1 2.1
3.1
fo 2fo 3fofifo − fi
2fo − fi 3fo − fi 4fo fi−fo + fi 2fo + fi
The input frequency fi is translated to frequenciesfi + kfo, where fo is the beat (LO) frequency. The k = 0sideband corresponds to the DC component of the LOsignal (e.g. time invariant behavior). The non-zerocomponents, though, correspond to mixing. E.g. k = 1correspond to frequency down-conversion. k = +1 isthe normal up-conversion. k = 2 is the 2nd harmonicmixing fi 2fo.
A. M. Niknejad University of California, Berkeley EECS 142 Lecture 8 p. 13/28 – p. 13/28
PNoise
fo 2fo 3fofifo − fi
2fo − fi 3fo − fi 4fo fi−fo + fi 2fo + fi
PNoise is a noise analysis that takes the frequencytranslation effects into account. The simulationparameters are similar to PAC with the exception thatwe must identify the input and output ports (for noisefigure) and the reference side-band, or the desiredoutput frequency. For a mixer, this is k = 1.
A. M. Niknejad University of California, Berkeley EECS 142 Lecture 8 p. 14/28 – p. 14/28
Coupled Inductors
V1
I1
V2
I2
+
−
+
−
L1 L2
M
Consider a set of coupled inductors. In particular, fortwo coupled inductors, we have
V1 = jωL1I1 + jωMI2
V2 = jωMI1 + jωL2I2
We usually express M as a fraction of the geometricmean:
M = k√
L1L2
A. M. Niknejad University of California, Berkeley EECS 142 Lecture 8 p. 15/28 – p. 15/28
Magnetic Flux
+ V1 − + V2 − + V3 −
By imposing physical arguments, we can show that1 ≤ k ≤ 1. When M > 0, a positive current in one
winding induces a positive voltage across the secondwinding. Likewise, M < 0 induces a negative voltage.
Recall that the magnetic flux can couple in a positive ornegative orientation depending on the physicalorientation of the windings.
A. M. Niknejad University of California, Berkeley EECS 142 Lecture 8 p. 16/28 – p. 16/28
Magnetic Flux (cont)
Magnetic coupling is due to magnetic flux linkage. Acurrent leads to flux which couples to other nearbycircuits. A time-varying flux induces a voltage
i → Ψ di
dt→ dΨ
dt→ v
If |k| = 1, we say that the inductors (windings) areperfectly coupled. In practice |k| < 1 but using amagnetic core, we can come close to ideal.
A. M. Niknejad University of California, Berkeley EECS 142 Lecture 8 p. 17/28 – p. 17/28
Ideal Transformer
V1
I1
V2
I2
+
−
+
−
V1 =1
NV2
I1 = NI2
The ideal transformer has the above current/voltagerelationship
In fact, the second equation is superfluous if wedemand energy conservation (e.g. the transformer is apassive device). It is understood that that an idealtransformer does not work at DC but at any ACfrequency, the above relations hold.
How do we build such a transformer from coupledinductors?
A. M. Niknejad University of California, Berkeley EECS 142 Lecture 8 p. 18/28 – p. 18/28
Coupled Inductor Equivalent Circuitideal transformer
N : 1Lx
Ly
We can show that the above circuit is totally equivalentto two coupled inductors. In particular, note that anideal transformer is embedded into the model. To seethis, note that
V1 = jωLxI1 + (I1 +I2
N)jωLy = jω(Lx + Ly)I1 +
jωLy
NI2
V2 =Vy
N= (I1 +
I2
N)jωLy
N= I1
jωLy
N+
jωLy
N2I2
A. M. Niknejad University of California, Berkeley EECS 142 Lecture 8 p. 19/28 – p. 19/28
Eq. Circuit Parameters
We can solve the following three equations to find theequivalence
L1 = Lx + Ly L2 =Ly
N2M =
Ly
N
Solving for the parameters, we have
Ly = nM = N2L2
n =M
L2
=k√
L1L2
L2
= k
√
L1
L2
Ly = N2L2 = k2L1
L2
L2 = k2L1
Lx = L1 Ly = (1 k2)L1
A. M. Niknejad University of California, Berkeley EECS 142 Lecture 8 p. 20/28 – p. 20/28
Leakage/Magnetization Inductance
ideal transformer
k2L1
(1 − k2)L1
N : 1
ideal transformer
k2L1
N : 1
We identify the parasitic inductors as the “leakageinductance” (because it goes away as k → 1) and themagnetization inductance.
We see that even if k = 1, two coupled inductors do notbehave like an ideal transformer unless L1 → ∞A large shunt inductor correctly predicts the DC/ACtransition (DC currents are shorted)
A. M. Niknejad University of California, Berkeley EECS 142 Lecture 8 p. 21/28 – p. 21/28
Transformer Constructionp+s+ p-
s-
The transformer on the left is a typical low frequencytransformer which uses a core to couple the flux. Byisolating the primary and the secondary, there isnegligible capacitive coupling.
The second transformer is a high frequency version,where the wires are twisted together to maximizecoupling. The core is inactive at high frequency but theboosted inductance is useful for rejecting commonmode currents in the circuit.
A. M. Niknejad University of California, Berkeley EECS 142 Lecture 8 p. 22/28 – p. 22/28
High Frequency Transformers
In high frequency applications, we cannot make L1
large. In fact, above around 100MHz, magnetic coresare lossy. Without a core, high inductance requiresmany windings, which ultimately results in excessivewinding capacitance and non-magnetic behavior.
Thus in a high frequency applications |k| < 1 and L isfinite. If possible, we can ensure that ωL ≫ Z0, whereZ0 is the typical impedances presented to thetransformer.
A. M. Niknejad University of California, Berkeley EECS 142 Lecture 8 p. 23/28 – p. 23/28
Tuned Transformers
k2L1
N : 1
Otherwise, we can use capacitors to tune away theeffects of finite L, resulting in more narrowbandmatches.
A. M. Niknejad University of California, Berkeley EECS 142 Lecture 8 p. 24/28 – p. 24/28
On Chip Transformer Layout
On-chip structures are planar. Two square spiralswound together form a 1:1 transformer. The “bifilar”layout and “symmetric” bifilar have coupling factors of0.7 0.8.
A. M. Niknejad University of California, Berkeley EECS 142 Lecture 8 p. 25/28 – p. 25/28
N:1 Transformers
An n : 1 transformer is designed by using fewer turns.
Alternatively, turns from the secondary can be put inparallel to lower the resistance.
A. M. Niknejad University of California, Berkeley EECS 142 Lecture 8 p. 26/28 – p. 26/28
Multi-Layer Transformers
Multiple metal layers can be utilized to form morecompact and higher coupling factor structures.
The primary is connected on the top two metal layers(in series), forming a high inductance. The secondary isa single layer inductor.
A. M. Niknejad University of California, Berkeley EECS 142 Lecture 8 p. 27/28 – p. 27/28
Baluns
+
V1− +
V3−
+
V2−
+
V1
−
+
V2
−
A balun is useful for converting single-ended single intobalanced (differential) signals. The inverse is also easilyaccomplished.
A symmetric single layer transformer balun layout isshown above. The common center can be connected toAC ground.
A. M. Niknejad University of California, Berkeley EECS 142 Lecture 8 p. 28/28 – p. 28/28