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)


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.


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.


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.


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.


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.


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


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.


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.


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.


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.


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.


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.


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


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.


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.


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?


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


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


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)


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.


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.


Tuned Transformers

k2L1

N : 1

Otherwise, we can use capacitors to tune away theeffects of finite L, resulting in more narrowbandmatches.


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.


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.


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.


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.


lecture 18: balanced mixers/pnoise and...

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