Lecture 16: I/Q Mixers; BJT...

EECS 142

Lecture 16: I/Q Mixers; BJT Mixers

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 16 p. 1/23 – p. 1/23

I/Q Hartley Mixer

RFcos(ωLOt)

sin(ωLOt)

90◦

90◦

IF

An I/Q mixer implemented as shown above is known asa Hartley Mixer.

We shall show that such a mixer can be designed toselect either the upper or lower sideband. For thisreason, it is sometimes called a single-sideband mixer.

We will also show that such a mixer can perform imagerejection.


Delay Operation

Consider the action of a 90◦ delay on an arbitrary signal.Clearley sin(x + 90◦) = cos(x). Even though this isobvious, consider the effect on the complexexponentials

sin(x −π

2) =

ejx−jπ/2 − e−jx+jπ/2

2j

=ejxe−jπ/2 − e−jxejπ/2

2j=

ejx(−j) − e−jx(j)

2j

= −ejx + e−jx

2= − cos(x)

Notice that positive frequencies get multiplied by −jand negative frequencies by +j. This is true for anywaveform when it is delayed by 90◦.


Complex Modulation

Consider multiplying a waveform f(t) by ejωt and takingthe Fourier transform

F{ejω0tf(t)

}=

∫∞

−∞

f(t)ejω0te−jωtdt

Grouping terms we have

=

∫∞

−∞

f(t)e−j(ω−ω0)tdt = F (ω − ω0)

It is clear that the action of multiplication by the complexexponential is a frequency shift.


Real Modulation

Now since cos(x) = (ejx + e−jx)/2, we see that theaction of time domain multiplication is to produce twofrequency shifts

F {cos(ω0t)f(t)} =1

2F (ω − ω0) +

1

2F (ω + ω0)

These are the sum and difference (beat) frequencycomponents.


Image Problem (Again)

RF+(ω)RF−(ω) RF+(ω − ω0)

RF+(ω + ω0)

IM+(ω)

IM−(ω)

IM+(ω − ω0)

IM−(ω + ω0)

LO−LO

IF−IF

ejωLOt

e−jωLOt

RF−(ω − ω0)

IM−(ω − ω0)

LO

LO

−LO

−LO

IM−(ω)

ejωLOt

e−jωLOt

RF−(ω) RF−(ω + ω0) RF+(ω)

IM+(ω) IM+(ω + ω0)

Complex Modulation (Positive Frequency)

Complex Modulation (Positive Frequency)

Real Modulation

We see that the image problem is due to tomultiplication by the sinusoid and not a complexexponential. If we could synthesize a complexexponential, we would not have the image problem.


Sine/Cosine Modulation

IF−IF LO−LO

IF−IF LO−LO

Cosine Modulation

Sine Modulation

IF−IF LO−LO

Delayed Sine Modulation

/j

/j/j

/j

Using the same approach, we can find the result ofmultipling by sin and cos as shown above. If we delaythe sin portion, we have a very desirable situation! Theimage is inverted with respect to the cos and can becancelled.


Image Rejection

The image rejection scheme just described is verysensitive to phase and gain match in the I/Q paths.Any mismatch will produce only finite image rejection.

The image rejection for a given gain/phase match isapproximately given by

IRR( dB) = 10 · log1

4

((δA

A

)

2 + (δθ)2)

For typical gain mismatch of 0.2 − 0.5 dB and phasemismtach of 1◦ − 4◦, the image rejection is about 30 dB -40 dB. We usually need about 60 − 70 dB of total imagerejection.


±45◦ Delay Element

RF

cos(ωLOt)

sin(ωLOt)

IF

The passive R/C andC/R lowpass andhighpass filters are anice way to implementthe delay. Note thattheir relative phasedifference is always90◦.

∠Hlp = ∠1

1 + jωRC= − arctan ωRC

∠Hhp = ∠jωRC

1 + jωRC=

π

2− arctan ωRC


Gain Match / Quadrature LO Gen

But to have equal gain, the circuit must operate at the1/RC frequency. This restricts the circuit to relativelynarrowband systems. Multi-stage polyphase circuitsremedy the situation but add insertion loss to the circuit.

The I/Q LO signal is usually generated directly ratherthan through an high-pass and low-pass network.

Two ways to generate the I/Q LO is through adivide-by-two circuit (requires 2 × LO) or a quadratureoscillator (requires two tanks).

A. M. Niknejad University of California, Berkeley EECS 142 Lecture 16 p. 10/23 – p

BJT with Large Sine Drive

vi

VA

IC

vi = V̂i cos ωt

IC = ISevBE

Vt

vBE = VA + V̂i cos ωt

Consider a bipolar device driven with a large sinesignal.

This occurs in many types of non-linear circuits, such asoscillators, frequency multipliers, mixers and class Camplifiers.


BJT Collector Current

The collector current can be factored into a DC biasterm and a periodic signal

IC = ISeVAVteV̂iVt

cos ωt

IC = ISeaeb cos ωt

Where the normalized bias is a = Va/Vt and thenormalized drive signal is b = V̂i/Vt.

Since IC is a periodic function, we can expand it into aFourier Series. Note that the Fourier coefficients ofeb cos ωt are modified Bessel functions In(b)

eb cos ωt = I0(b) + 2I1(b) cos ωt + 2I2(b) cos 2ωt + · · ·


BJT DC Current

Assume that the bias current of the amplifier isstabilized. Then

IC = ISeaI0(b)︸︷︷︸

IQ

(

1 +2I1(b)

I0(b)cos ωt +

2I2(b)

I0(b)cos 2ωt + · · ·

)

IC = ISeaeb cos ωt

=IQ

I0(b)eb cos ωt

1 2 3 4 5

1

2

3

4

5

6

IC

IQ


Collector Current Waveform

-0.4 -0.2 0 0.2 0.4

1

2

3

4

5

6

7

8

IC

IQ

b = .5

b = 4

b = 10

With increasing input drive, the current waveformbecomes “peaky”. The peak value can exceed the DCbias by a large factor.


Harmonic Current Amplitudes

2.5 5 7.5 10 12.5 15 17.5 20

0.2

0.4

0.6

0.8

1

In(b)

I0(b)

b

n = 1

n = 2

n = 3

n = 4

n = 5

n = 6

n = 7

The BJT output spectrum is rich in harmonics.


Small-Signal Region

0.05 0.1 0.15 0.2

0.002

0.004

0.006

0.008

0.01

In(b)

I0(b)

b

n = 1

n = 2

n = 3

If we zoom in on the curves to small b values, we enterthe small-signal regime, and the weakly non-linearbehavior is predicted by our power series analysis.


BJT with Stable BiasNeglecting base current (β ≫ 1), the voltage at the baseis given by

R1

R2 RE CE

+vi

V ′

A

V ′

A =R2

R1 + R2VCC

VE = V ′

A − VBE

IQ =VE

RE=

V ′

A − VBE

RE

We see that the bias is fixed since VBE does not varytoo much. Typically V ′

A is a few volts.

In this circuit CE is an emitter bypass capacitor used toshort RE at high frequency.


Differential Pair with Sine Drive

The large signal equation for IC1 is given by

IC1 + IC2 = IEE

VBE1 − VBE2 = Vi = Vt ln

(IC1

IC2

)

+Vi

2

Vi

2

+

IEE

IC1 =IEE

1 + e−vi/Vt

vi = V̂i cos ωt

IC1 =IEE

1 + e−b cos ωt


Diff Pair Waveforms

1 2 3 4 5 6

0.2

0.4

0.6

0.8

1

IC1

IEE

b = 1

b = 3b = 1 0

For large b, the waveform approaches a square wave

IC

IEE=

1

2+

2

π

(

cos ωt −1

3cos 3ωt +

1

5cos 5ωt + · · ·

)


Diff Pair Harmonic Currents

2.5 5 7.5 10 12.5 15 17.5 20

0.1

0.2

0.3

0.4

0.5

0.6

b

2

πn = 1

n = 3

n = 5

n = 7

Normalized Harmonic Currents

(negative)

(negative)

As expected, the ideal differential pair does not produceany even harmonics.


Mixer Analysis

As we have seen, a mixer has three ports, the LO, RF,and IF port.

Assume that a circuit is “pumped” with a periodic largesignal at the LO port with frequency ω0.

From the RF port, though, assume we apply a smallsignal at frequency ωs.

Since the RF input is small, the circuit response shouldbe linear (or weakly non-linear). But since the LO portchanges the operating point of the circuit periodically,we expect the overall response to the RF port to be alinear time-varying response

io(t) = g(t)vin


Mixer Assumptionsgm(t)

The transconductance varies periodically and can beexpanded in a Fourier series

g(t) = g0 + g1 cos ω0t + g2 cos 2ω0t + · · ·

Applying the input vin = V̂1 cos ωst

io(t) = (g0 + g1 cos ω0t + g2 cos 2ω0t + · · · ) × V̂1 cos ωst


Mixer Output Signal

Expanding the product, we have

io(t) = g0V̂1 cos ωst+g1

2V̂1 cos(ω0±ωs)t+

g2

2V̂1 cos(2ω0±ωs)t+· · ·

The first term is just the input amplified. The otherterms are all due to the mixing action of the lineartime-varying periodic circuit.

Let’s say the desired output is the IF at ω0 − ωs. Theconversion gain is therefore defined as

gconv =|IF output current|

|RF input signal voltage|=

g1

2


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