PCIRF_5_MIXER_4

7/29/2019 PCIRF_5_MIXER_4

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EECS 142

Lecture 20: BJT/FET Mixers/Mixer Noise

Prof. Ali M. Niknejad

University of California, Berkeley

Copyright c 2008 by Ali M. Niknejad

A. M. Nikne ad Universit of California Berkele EECS 142 Lecture 18 . 1/35


2/35

A BJT Mixer

C

L1

C1

L2

C2

L3

C

C3

LO

RF

IF

The transformer is used to sum the LO and RF signals at the

input. The winding inductance is used to form resonant tanks atthe LO and RF frequencies.

The output tank is tuned to the IF frequency.

Large capacitors are used to form AC grounds.



3/35

AC Eq. Circuit

L1

C1

L2

C2

L3C3LO

RF

IF

The AC equivalent circuit is shown above.



4/35

BJT Mixer Analysis

When we apply the LO alone, the collector current of the mixeris given by

IC = IQ

1 + 2I1(b)

I0(b)cos t + 2I

2(b)I0(b)

cos2t +

We can therefore define a time-varying gm(t) by

gm(t) =IC(t)

Vt=

qIC(t)

kT

The output current when the RF is also applied is thereforegiven by iC(t) = gm(t)vs

iC =qIQ

kT 1 +2I1(b)

I0(b)

cos t +2I2(b)

I0(b)

cos2t + Vs cos stA. M. Nikne ad Universit of California Berkele EECS 142 Lecture 18 . 4/35


5/35

BJT Mixer Analysis (cont)

The output at the IF is therefore given by

iC|IF = VsqIQkTgmQ

I1(b)

I0(b)cos(0 s

IF)t

The conversion gain is given by

gconv = gmQI1(b)

I0(b)

2 4 6 8 10

0.2

0.4

0.6

0.8

1

gconv

gmQ

b =Vi

V



6/35

LO Signal Drive

For now, lets ignore the small-signal input and determine theimpedance seen by the LO drive. If we examine the collectorcurrent

IC = IQ

1 +

2I1(b)

I0(b)cos t +

2I2(b)

I0(b)cos2t +

The base current is simply IC/, and so the input impedanceseen by the LO is given by

Zi|0

=Vo

iB,0=

Vo

iC,0=

Vo

IQ 2I1(b)I0(b)=

bVt

IQ 2I1(b)I0(b)

=b

2

gmQ

I0(b)

I1(b)=

Gm



7/35

RF Signal Drive

The impedance seen by the RF singal source is also the basecurrent at the s components. Typically, we have a high-Qcircuit at the input that resonates at RF.

iB(t) =iC(t)

= 1

qIQkT

Vs cos st + 2I

1(b)I0(b)

cos(0 s)t +

The input impedance is thus the same as an amplifier

Rin =Vs

|component in iB at s|=

kT

qIQ=

gmQ



8/35

Mixer Analysis: General Approach

If we go back to our original equations, our major assumptionwas that the mixer is a linear time-varying function relative tothe RF input. Lets see how that comes about

IC = ISevBE/Vt

where

vBE = vin + vo + VA

or

IC = ISeVA/Vt eb cos0t e

VsVt

cosst

If we assume that the RF signal is weak, then we canapproximate ex 1 + x



9/35

General Approach (cont)

Now the output current can be expanded into

IC = IQ1 + 2I1(b)I0(b) cos t + 2I2(b)I0(b) cos2t + 1 +

VsVt

cos stIn other words, the output can be written as

= BIAS + LO + Conversion Products

In general we would filter the output of the mixer and so the LOterms can be minimized. Likewise, the RF terms are undesiredand filtered from the output.



10/35

Distortion in Mixers

Using the same formulation, we can now insert a signal with twotones

vin = Vs1 cos s1t + Vs2 cos s2

IC = ISeVA/Vt eb cos0t e

Vs1Vt

coss1t+Vs2Vt

coss2t

The final term can be expanded into a Taylor series

IC = ISeVA/Vt eb cos0t

1 + Vs1 cos s1t + Vs2 cos s2t + ( )

2 + ( )3 +

The square and cubic terms produce IM products as before,but now these products are frequency translated to the IFfrequency



11/35

Another BJT Mixer

L3

C

C3LO

RF

IF

C0

CsLs

+

v0x

The signal from the LO driver is capacitively coupled to the BJTmixer



12/35

LO Capacitive Divider

CsLs

+

v0x

C0

+

v0x

C0

C

s

Assume that LO > RF, or a high side injectionNote beyond resonance, the input impedance of the tankappears capacitive. Thus Cs is the effective capacitance of thetank. The equivalent circuit for the LO drive is therefore a

capacitive divider

vo =Co

Co + Csvox



13/35

Harmonic Mixer

Second

Harmonic

Mixer

LO

RF I F

We can use a harmonic of the LO to build a mixer.

Example, let LO = 500MHz, RF = 900MHz, and IF = 100MHz.

Note that IF = 2LO RF = 1000 900 = 100



14/35

Harmonic Mixer Analysis

The nth harmonic conversion tranconductance is given by

gconv,n

=|IF current out|

|input signal voltage|=

gn

2

For a BJT, we have

gconv,n = gmQ In(b)I0(b)

The advantage of a harmonic mixer is the use of a lower

frequency LO and the separation between LO and RF.The disadvantage is the lower conversion gain and higher noise.



15/35

FET Large Signal Drive

vo

VQ

ID

Consider the output current of a FET driven by a large LO signal

ID =Cox

2

W

L(VGS VT)

2(1 + VDS)

where VGS = VA + vLO = VA + Vo cos 0t. Here we implicitlyassume that Vo is small enough such that it does not take thedevice into cutoff.



16/35

FET Large Signal Drive (cont)

0.5 1 1.5 2 2.5 3

That means that VA + V0 cos 0t > VT, or VA V0 > VT, orequivalently V0 < VA VT. Under such a case we expand thecurrent

ID

(VA VT)

2 + V20 cos2 0t + 2(VA VT)V0 cos 0t



17/35

FET Current Components

The cos2 term can be further expanded into a DC and secondharmonic term.

Identifying the quiescent operating point

IQ =Cox

2

W

L(VA VT)

2(1 + VDS)

ID = IDQ + CoxW

L

1

4V20

bias point shift+ (VA VT)Vo cos 0t

LO modulation+

V204

cos(20t)

LO 2nd harmonic

(1 + VDS)



18/35

FET Time-Varying Transconductance

The transconductance of a FET is given by (assuming stronginversion operation)

g(t) = IDVGS

= Cox WL

(VGS VT)(1 + VDS)

VGS(t) = VA + V0 cos 0t

g(t) = CoxW

L(VA VT + V0 cos 0t)(1 + VDS)

g(t) = gmQ1 + V0VA VT cos 0t (1 + VDS)

This is an almost ideal mixer in that there is no harmoniccomponents in the transconductance.



19/35

MOS Mixer

vs

VQ

vo

IF FilterWe see that we can build a mixer

by simply injecting an LO + RFsignal at the gate of the FET (ig-nore output resistance)

i0 = g(t)vs = gmQ

1 +

V0VA VT

cos 0t

Vs cos st

i0|IF =gmQ

2

V0VA VT

cos(0 s)tVs

gc =

i0|IF

Vs =

gmQ

2

V0

VA VT



20/35

MOS Mixer Summary

But gmQ = CoxWL (VA VT)

gc =

Cox

2

W

L V0

which means that gc is independent of bias VA. The gain iscontrolled by the LO amplitude V0 and by the device aspect

ratio.Keep in mind, though, that the transistor must remain in forwardactive region in the entire cycle for the above assumptions tohold.

In practice, a real FET is not square law and the above analysisshould be verified with extensive simulation. Sub-thresholdconduction and output conductance complicate the picture.



21/35

Dual Gate Mixer

+

vs

M1

M2

VA

+

vo

VA

io

G1

G2D2

S1

shared junction

(no contact)

The dual gate mixer, or more commonly a cascode amplifier,can be turned into a mixer by applying the LO at the gate of M2and the RF signal at the gate of M1. Using two transistors inplace of one transistor results in area savings since the signalsdo not need to be combined with a transformer or capacitively .


D l G Mi O i


22/35

Dual Gate Mixer Operation

0.5 1 1.5 2 2.5 3

trio

de

saturation

LO

Without the LO signal, this is simply a cascode amplifier. Butthe LO signal is large enough to push M1 into triode during partof the operating cycle.

The transconductance of M1 is therefore modulated periodically

gm|sat = CoxW

L(VGS VT)

gm|triode = CoxWL

VDS



23/35

Dual Gate Waveforms

Active Region

Triode Region

gm,max

gm(t)

vLO,1

vLO,2

VGS2 is roughly constant since M1 acts like a current source.

VD1 = vLO VGS2 = VA2 + V0 cos 0t VGS2

g(t) =

Cox

WL (VGS1 VT) VD1 > VGS VT

Cox(VA2 VGS2 |V0 cos 0t|) VD1 < VGS VT



24/35

Realistic Waveforms

A more sophisticated analysis would take sub-thresholdoperation into account and the resulting g(t) curve would besmoother. A Fourier decomposition of the waveform would yield

the conversion gain coefficient as the first harmonic amplitude.


Mi A l i Ti D i


25/35

Mixer Analysis: Time Domain

A generic mixer operates with a periodic transfer functionh(t + T) = h(t), where T = 1/0, or T is the LO period. We canthus expand h(t) into a Fourier series

y(t) = h(t)x(t) =

cnej0ntx(t)

For a sinusoidal input, x(t) = A(t)cos 1t, we have

y(t) =

cn2

A(t)ej(1+0n)t + ej(1+0n)t

Since h(t) is a real function, the coefficients ck = ck are even.That means that we can pair positive and negative frequency

components.


Ti D i A l i ( t)


26/35

Time Domain Analysis (cont)

Take c1 and c1 as an example

= c1

ej(1+0)t + ej(1+0)t

2A(t)+c

1

ej(10)t + ej(10)t

2A(t)+

= c1A(t) cos(1 + 0)t + c1A(t)cos(1 0)t +

Summing together all the components, we have

y(t) =

cn cos(1 + n0)t

Unlike a perfect multiplier, we get an infinite number offrequency translations up and down by harmonics of 0.


Frequency Domain Analysis


27/35

Frequency Domain Analysis

Since multiplication in time, y(t) = h(t)x(t), is convolution in thefrequency domain, we have

Y(f) = H(f) X(f)

The transfer function H(f) =

cn(f nf0) has a discrete

spectrum. So the output is given by

Y(f) =

cn( nf0)X(f )d

=

cn

( nf0)X(f )d


Frequency Domain (cont)


28/35

Frequency Domain (cont)

noi

sefoldingfrom3LO

noisefoldingfrom

2LO

noise

fold

ingf

rom-2LO

noise

foldin

gfrom-3LO

fo fsfiffofs fif 2fo 3fo2fo3fo

By the frequency sifting property of the (f ) function, wehave

Y(f) =

cnX(f nf0)Thus, the input spectrum is shifted by all harmonics of the LOup and down in frequency.


Noise/Image Problem


29/35

Noise/Image Problem

Previously we examined the image problem. Any signalenergy a distance of IF from the LO gets downconverted in aperfect multiplier. But now we see that for a general mixer, any

signal energy with an IF of any harmonic of the LO will bedownconverted !

These other images are easy to reject because they are distantfrom the desired signal and a image reject filter will be able to

attenuate them significantly.

The noise power, though, in all image bands will fold onto the IFfrequency. Note that the noise is generated by the mixer sourceresistance itself and has a white spectrum. Even though thenoise of the antenna is filtered, new noise is generated by thefilter itself!


Current Commutating Mixers


30/35

Current Commutating Mixers

+LO LO

VCC

Q2 Q3

Q1+RF

IF

RIF

+LO LO

+RF

IF


Current Commutating Mixer Model


31/35

Current Commutating Mixer Model

+LO LO

VCC

Q1+RF

IF

RIF

If we model the circuit with ideal elements, we see that thecurrent IC1 is either switched to the output or to supply at the

rate of the LO signal.When the LO signal is positive, we have a cascode dumping itscurrent into the supply. When the LO signal is negative, though,we have a cascode amplifier driving the output.


Conversion Gain


32/35

Conversion Gain

t

TLO

+1

0

s(t)

We can now see that the output current is given by a periodictime varying transconductance

io = gm(t)vs = gmQs(t)vs

where s(t) is a square pulse waveform (ideally) switching

between 1 and 0 at the rate of the LO signal. A Fourierdecomposition yields

io = gmQvs0.5 +2

cos 0t

2

1

3cos30t +


Conversion Gain (cont)


33/35

Conversion Gain (cont)

So the RF signal vs is amplified (feed-thru) by the DC term andmixed by all the harmonics

ioVs

= gmQ2

12

cos st + 2

cos(0 s)t 23

cos(30 s)t +

The primary conversion gain is gc =1 gmQ.

Since the role of Q1 (or M1) is to simply create an RF current, itcan be degenerated to improve the linearity of the mixer.Inductance degeneration can be employed to also achieve animpedance match.

MOS version acts in a similar way but the conversion gain islower (lower gm) and it requires a larger LO drive.


Differential Output


34/35

Differential Output

+LO LO

+RF

IF

This block is commonly known as the Gilbert Cell

If we take the output signal differentially, then the output currentis given by

io = gm(t)vs = gmQs2(t)vs


Differential Output Gain


35/35

Differential Output Gain

t

TLO

+1

1

s2(t)

The pulse waveform s2(t) now switches between 1, and thushas a zero DC value

s2(t) =4

cos 0t

1

3

4

cos30t +

The lack of the DC term means that there is ideally no RFfeedthrough to the IF port. The conversion gain is doubled since

we take a differential output gc/gmQ =2


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