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1 Chapter 6 Mixers
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Chapter 6 Mixers
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Sections to be covered• 6.1 General Considerations• 6.2 Passive Downconversion Mixers• 6.3 Active Downconversion Mixers
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Chapter Outline
General Considerations
Active Mixers
Port-to-Port Feedthrough Single-Balanced and
Double-Balanced Mixers Passive and Active Mixers
Conversion Gain
Conversion Gain
Passive Mixers
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Recall: Generic TX & RX
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General Considerations (I) Mixers perform frequency translation by multiplying two waveforms.
VLO turns the switch on and off, yielding
Example: mixer using an ideal switch
multiplication of the RF input by a square wave toggling between 0 and 1, even if VLO is a sinusoid.
or 0IF RF IFV V V
𝑣 𝑡 𝑣 𝑡 ⋅ 𝑆 𝑡
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General Considerations (II) Mixers perform frequency translation by multiplying two waveforms (and
possibly their harmonics).
The circuits mixes the RF input with all of the LO harmonics, producing “mixing spurs”. The LO port of this mixer is very nonlinear.
The RF port must remain sufficiently linear to satisfy the compression and intermodulation requirements.
Example: mixer using an ideal switch
𝑉 𝑓 𝑉 𝑓
RFV𝑣 𝑡 𝑣 𝑡 ⋅ 𝑆 𝑡
Performance Parameters: Port-to-Port Feedthrough
Owing to device capacitances, mixers suffer from unwanted coupling (feedthrough) from one port to another.
feedthrough from the LO port to the RF and IF ports.
gate-source capacitances
gate-drain capacitances
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Example of LO-RF Feedthrough in MixerConsider the mixer shown below, where VLO = V1 cos ωLOt + V0 and CGS denotes the gate-source overlap capacitance of M1. Neglecting the on-resistance of M1 and assuming abrupt switching, determine the dc offset at the output for RS = 0 and RS> 0. Assume RL >> RS.
The LO leakage to node X is expressed as
Basic component of VLO (square wave) can be expressed as
The dc component：
The output dc offset vanishes if RS = 0.
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Single-Balanced Mixers
The circuit provides differential outputs easing the design of subsequent stages.
The LO-RF feedthrough at ωLO (dc component) vanishes if the circuit is symmetric due to the differential output.
𝑆 𝜔 𝑡 1 cos 𝜔 𝑡 00 cos 𝜔 𝑡 0
12 1
2𝑛𝜋 cos 𝑛 𝜔 𝑡
𝑆 𝜔 𝑡 1 cos 𝜔 𝑡 01 cos 𝜔 𝑡 0
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𝑛𝜋, , ,...,
cos 𝑛 𝜔 𝑡
The simple mixer operate with a single-ended RF input and a single-ended LO. Discarding the RF signal for half of the LO
𝑣 𝑡 𝑣 𝑡 ⋅ 𝑆 𝜔 𝑡12 𝑣 𝑡
2𝜋 𝑣 𝑡 cos 𝜔 𝑡
23𝜋 𝑣 𝑡 cos 3 𝜔 𝑡 . . .
𝑣 𝑡 𝑣 𝑡 ⋅ 𝑆 𝜔 𝑡4𝜋 𝑣 𝑡 cos 𝜔 𝑡
43𝜋 𝑣 𝑡 cos 3 𝜔 𝑡
. . .
Single-balanced mixer: Two switches are driven
by differential LO phases;
transform the RF input to the two outputs.
Single-balanced mixer: Two switches are driven
by differential LO phases;
transform the RF input to the two outputs.
significant LO-IF feedthrough!!!!
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Single-Balanced Mixers
The circuit provides differential outputs easing the design of subsequent stages.
The LO-RF feedthrough at ωLO (dc component) vanishes if the circuit is symmetric due to the differential output.
𝑆 𝜔 𝑡 1 cos 𝜔 𝑡 00 cos 𝜔 𝑡 0
12 1
2𝑛𝜋 cos 𝑛 𝜔 𝑡
The simple mixer operate with a single-ended RF input and a single-ended LO. Discarding the RF signal for half of the LO
𝑣 𝑡 𝑣 𝑡 ⋅ 𝑆 𝜔 𝑡12 𝑣 𝑡
2𝜋 𝑣 𝑡 cos 𝜔 𝑡
23𝜋 𝑣 𝑡 cos 3 𝜔 𝑡 . . .
Single-balanced mixer: Two switches are driven
by differential LO phases;
transform the RF input to the two outputs.
Single-balanced mixer: Two switches are driven
by differential LO phases;
transform the RF input to the two outputs.
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Single-Balanced Mixers
The circuit provides differential outputs easing the design of subsequent stages.
The LO-RF feedthrough at ωLO (dc component) vanishes if the circuit is symmetric due to the differential output.
𝑆 𝜔 𝑡 1 cos 𝜔 𝑡 00 cos 𝜔 𝑡 0
12 1
2𝑛𝜋 cos 𝑛 𝜔 𝑡
𝑆 𝜔 𝑡 1 cos 𝜔 𝑡 01 cos 𝜔 𝑡 0
14
𝑛𝜋, , ,...,
cos 𝑛 𝜔 𝑡
The simple mixer operate with a single-ended RF input and a single-ended LO. Discarding the RF signal for half of the LO
𝑣 𝑡 𝑣 𝑡 ⋅ 𝑆 𝜔 𝑡12 𝑣 𝑡
2𝜋 𝑣 𝑡 cos 𝜔 𝑡
23𝜋 𝑣 𝑡 cos 3 𝜔 𝑡 . . .
𝑣 𝑡 𝑣 𝑡 ⋅ 𝑆 𝜔 𝑡4𝜋 𝑣 𝑡 cos 𝜔 𝑡
43𝜋 𝑣 𝑡 cos 3 𝜔 𝑡
. . .
Single-balanced mixer: Two switches are driven
by differential LO phases;
transform the RF input to the two outputs.
Single-balanced mixer: Two switches are driven
by differential LO phases;
transform the RF input to the two outputs.
significant LO-IF feedthrough!!!!
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LO-RF/LO-IF Feedthrough in Direct-Conversion RX
In the direct-conversion receiver: LO-RF feedthrough is entirely
determined by the symmetry of the mixer circuit and LO waveforms. (cancelled by balance structure)
The LO-IF feedthrough is not harmful because it is heavily suppressed by the baseband low-pass filter(s).
Power of RF signal is smaller than that of LO, which makes LO-RF feedthrough more harmful.
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RF-IF/RF-LO Feedthrough in Direct-Conversion RX
A large in-band interferer can couple to the LO and injection-pull it. Effect?
corrupting the LO spectrum. Solution?
interpose a buffer between the LO and the mixer
The RF-IF feedthrough corrupts the baseband signal by the beat component resulting from even-order distortion in the RF path. (cos2ωRFt related to IP2, but not discussed in this course)
RF frequency is equal to LO frequency.
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LO-RF/LO-IF Feedthrough in Heterodyne RX
The LO-RF feedthrough is relatively unimportant The LO leakage falls outside the band
Attenuated by the selectivity of the LNA, the front-end band-select filter, and the antenna.
Even there is residual LO leakage the dc offset appearing at the output of the RF mixer It can be removed by a high-pass filter (due to the IF output).
The LO-RF feedthrough is relatively unimportant The LO leakage falls outside the band
Attenuated by the selectivity of the LNA, the front-end band-select filter, and the antenna.
Even there is residual LO leakage the dc offset appearing at the output of the RF mixer It can be removed by a high-pass filter (due to the IF output).
Special case:LO frequency is far away from RF frequency while close to IF frequency.
The LO-IF feedthrough becomes serious if ωIF and ωLO are too close, We can not remove ωLO by filtering.
LO is generated locally, with a high power level; LO feedthrough may desensitize the IF mixers if its level is comparable with their
1-db compression point.
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Double-Balanced Mixers The single-balanced mixer
suffers from significant LO-IF feedthrough.
The single-balanced mixer suffers from significant LO-IF feedthrough.
𝑆 𝜔 𝑡 1 cos 𝜔 𝑡 01 cos 𝜔 𝑡 0
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𝑛𝜋, , ,...,
cos 𝑛 𝜔 𝑡
𝑣 𝑡 𝑣 𝑡 ⋅ 𝑆 𝜔 𝑡4𝜋 𝑣 𝑡 cos 𝜔 𝑡4
3𝜋 𝑣 𝑡 cos 3 𝜔 𝑡. . .
𝑣 𝑡 𝑣 𝑡 𝑣 𝑡2𝑣 𝑡 ⋅ 𝑆 𝜔 𝑡8𝜋 𝑣 𝑡 cos 𝜔 𝑡
83𝜋 𝑣 𝑡 cos 3 𝜔 𝑡
. . .
The circuit operates with both balanced LO waveforms and balanced RF inputs.
We connect two single-balanced mixers their output LO feedthroughs cancel， but their output signals do not.
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Double-Balanced Mixers The single-balanced mixer
suffers from significant LO-IF feedthrough.
The single-balanced mixer suffers from significant LO-IF feedthrough.
2
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1,3,5,...,
1 cos 0( )
1 cos 0
4( 1) cos
LOLO
LO
n
LOn
tS t
t
n tn
2( ) ( ) ( )4 ( ) cos
4 ( ) cos33...
out RF LO
RF LO
RF LO
v t v t S t
v t t
v t t
We connect two single-balanced mixers their output LO feedthroughs cancel， but their output signals do not.
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Double-Balanced Mixers The single-balanced mixer
suffers from significant LO-IF feedthrough.
The single-balanced mixer suffers from significant LO-IF feedthrough.
2
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1,3,5,...,
1 cos 0( )
1 cos 0
4( 1) cos
LOLO
LO
n
LOn
tS t
t
n tn
2( ) ( ) ( )4 ( ) cos
4 ( ) cos33...
out RF LO
RF LO
RF LO
v t v t S t
v t t
v t t
2( ) ( ) ( ) 2 ( ) ( )8 8( )cos ( ) cos3
3...
out out out RF LO
RF LO RF LO
v t v t v t v t S t
v t t v t t
The circuit operates with both balanced LO waveforms and balanced RF inputs.
We connect two single-balanced mixers their output LO feedthroughs cancel， but their output signals do not.
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Effect of Feedthrough
Direct-Conversion Heterodyne
LO-RF Harmful, DC component
Harmless
LO-IF Harmless Harmful, Desensitization
RF-LO Harmful, Large interferer, Injection-pull
Harmless
RF-IF Harmless Harmless
IF-LO Harmless Harmful,Oscillator pulling
IF-RF Harmless Harmless
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Passive and Active mixers
Mixers can be broadly classified into “passive” and “active” topologies;
As to a passive mixer, its transistors do not operate as amplifying devices.
Each type can be realized as a single-balanced or a double-balanced circuit.
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Passive Downconversion Mixers: Gain The input is multiplied by a square wave toggling between 0 and 1. The first harmonic has a peak amplitude of 2/ π and can be expressed as (2/ π)
cosωLOt. The convolution of an RF signal with these impulses creates the IF signal with a
gain of 1/ π (≈-10 dB).
The conversion gain is equal to 1/π for abrupt LO switching.
We call this topology a “return-to-zero” (RZ) mixer because the output falls to zero when the switch turns off.
We call this topology a “return-to-zero” (RZ) mixer because the output falls to zero when the switch turns off.
1
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1
1 cos 0( )
0 cos 0
1 2( 1) cos2
LOLO
LO
n
LOn
tS t
t
n tn
1( ) ( ) ( )
1 2 2( ) ( ) cos ( )cos3 ...2 3
IF RF LO
RF RF LO RF LO
v t v t S t
v t v t t v t t
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Example of Downconversion Gain of Single-Balanced Topology
Determine the conversion gain if the circuit is converted to a single-balanced topology.
The second output is similar to the first but shifted by 180 °.
The differential output contains twice the amplitude of each single-ended output.
The conversion gain is equal to 2/π (≈ -4 dB).
Solution:𝑆 𝜔 𝑡 1 cos 𝜔 𝑡 0
1 cos 𝜔 𝑡 0 14
𝑛𝜋, , ,...,
cos 𝑛 𝜔 𝑡
𝑣 𝑡 𝑣 𝑡 ⋅ 𝑆 𝜔 𝑡4𝜋 𝑣 𝑡 cos 𝜔 𝑡
43𝜋 𝑣 𝑡 cos 3 𝜔 𝑡 . . .
Providing differential outputs and twice the gain, this circuit is superior to the single-
ended topology.
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Example of Downconversion Gain of Double-Balanced TopologyDetermine the voltage conversion gain of a double-balanced version. (Decompose the differential output to return-to-zero waveforms.)
Vout1 - Vout2 contains an IF amplitude of (1/π)(4V0).
The peak differential input is equal to 2V0, the circuit provides a voltage conversion gain of 2/π, equal to that of the single-balanced counterpart.
Solution:
Solution:
28 8( ) ( ) ( ) 2 ( ) ( ) ( ) cos ( )cos3 ...
3out out out RF LO RF LO RF LOv t v t v t v t S t v t t v t t
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Are there any ways to improve the gain?
When the switch turns off, what can we do?
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Sampling Mixer: the Idea If the resistor is replaced with a capacitor, such an arrangement operates as a
sample-and-hold circuit and exhibits a higher gain because the output is held— rather than reset —when the switch turns off.
The output waveform can be decomposed into waveforms.
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If realized as a single-balanced topology, the circuit provides a gain twice this value, 1.186≈1.48dB about 5.5dB higher than its return-to-zero counterpart.
The total IF output is therefore equal to
Sampling Mixer: Conversion Gain
Same as the switch
Passive circuit (single-ended sampling mixer) has a voltage conversion gain greater than unity.
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Active Downconversion Mixers: Function and Typical Realization
Mixers can be realized so as to achieve conversion gain in one stage.
Called active mixers, such topologies perform three functions: convert the RF voltage to a current, Drive the RF current by the LO, convert the IF current to voltage.
transconductor
We call M2 and M3
the “switching pair.”
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Active Downconversion Mixers: Double-Balanced Topology
One advantage of double-balanced mixers over single-balanced mixers: rejection of amplitude noise in the LO waveform.
We call M2, M3, M5 and M6 the “switching quad.”
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Conversion GainWith abrupt LO switching, the circuit reduces to that shown in figure below (left).
We have for R1 = R2 = RD
The waveform exhibits a fundamental amplitude equal to 4/π, yielding an output given by
𝑉 𝑉 𝐼 𝑅 𝑉 𝐼 𝑅 𝐼 𝑅 𝐼 𝑅
VoltageGain
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Ideal LO Waveform The LO waveform must ideally be a square wave to ensure
abrupt switching maximum conversion gain
Otherwise, there will be energy “waste”.
The LO waveform must ideally be a square wave to ensure abrupt switching maximum conversion gain
Otherwise, there will be energy “waste”.
VLO and VLO Remain approximately equal for ΔT; All four transistors are on;
There is no differential component for 2 ΔT seconds each period. Energy “waste”.
VLO and VLO Remain approximately equal for ΔT; All four transistors are on;
There is no differential component for 2 ΔT seconds each period. Energy “waste”.
Mixers multiply the RF input by a square wave they can downconvert interferers located at the LO harmonics, which is a serious
issue in broadband receiver. For example, 3fLO is attenuated by about only 10dB.
At very high frequencies, the LO waveforms approximate to sinusoids. We choose a relatively large amplitude
To obtain a high slew rate To ensure a minimum overlap time, ΔT.
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Performance Parameters: Noise, Linearity and Gain
Noise and Linearity: The design of downconversion mixers entails a compromise between the noise figure and the IP3 (or P1dB).
In a receive chain, the input noise of the mixer following the LNA is divided by the LNA gain when referred to the RX input.
the IP3 of the mixer is scaled down by the LNA gain.
The noise figure of mixers is rarely less than 8dB, we typically allocate a gain of 10 to 15 dB to the LNA, seeking to maximize its linearity while not raising its NF.
Gain: mixer gain is critical in suppression of noise while retaining linearity.
“Voltage conversion gain” denote the ratio of the rms voltage of the IF signal to the rms voltage of the RF signal-----different frequencies.
Downconversion mixers must provide sufficient gain to adequately suppress the noise contributed by subsequent stages.
Low supply voltages make it difficult to achieve a gain of more than 10 dB while retaining linearity.
LNA gain
Mixer gain
Mixer Noise Figures: SSB Noise Figure
The mixer exhibits a flat frequency response at its input from the image band to the signal band.
The noise figure of a noiseless mixer is 3 dB. This quantity is called the “single-sideband” (SSB) noise.
For simplicity, let us consider a noiseless heterodyne mixer with unity gain.
Mixer Noise Figures: DSB Noise FigureNow, consider the direct-conversion mixer shown below.
In this case, only the noise in the signal band is translated to the baseband, thereby yielding equal input and output SNRs if the mixer is noiseless.
The noise figure is thus equal to 0 dB. This quantity is called the “double-sideband” (DSB) noise figure
Noise Behavior in Heterodyne Receiver (Ⅰ)
A student designs the heterodyne receiver shown below for two cases: (1) ωLO1 is far from ωRF ; (2) ωLO1 lies inside the band and so does the image. Study the noise behavior of the receiver in the two cases.
In the first case, the selectivity of the antenna, the BPF, and the LNA suppresses the thermal noise in the image band. Of course, the RF mixer still folds its own noise. The overall behavior is illustrated below, where SAdenotes the noise spectrum at the output of the LNA and Smix the noise in the inputnetwork of the mixer itself. Thus, the mixer downconverts three significant noise components to IF: the amplified noise of the antenna and the LNA around ωRF , its own noise around ωRF , and its image noise around ωim.
Solution:
Noise Behavior in Heterodyne Receiver (Ⅱ)
A student designs the heterodyne receiver shown below for two cases: (1) ωLO1 is far from ωRF ; (2) ωLO1 lies inside the band and so does the image. Study the noise behavior of the receiver in the two cases.
In the second case, the noise produced by the antenna, the BPF, and the LNA exhibits a flat spectrum fromthe image frequency to the signal frequency. As shown on the right, the RF mixer now downconverts four significant noise components to IF: the output noise of the LNA around ωRF and ωim, and the input noise of the mixer around ωRF and ωim. We therefore conclude that the noise figure of the second frequency plan is substantially higher than that of the first. In fact, if the noise contributed by the mixer is much less than that contributed by the LNA, the noise figure penalty reaches 3 dB. The low-IF receivers of Chapter 4, on the other hand, do not suffer from this drawback because they employ image rejection.
Solution:
NF of Direct-Conversion ReceiversIt is difficult to define a noise figure for receivers that translate the signal to a zero IF.
This is the most common NF definition for direct-conversion receivers. The SNR in the final combined output would serve as a more accurate measure
of the noise performance, but it depends on the modulation scheme.
Example of Noise Spectrum of a Simple Mixer (Ⅰ)
Consider the simple mixer shown below. Assuming RL >> RS and the LO has a 50% duty cycle, determine the output noise spectrum due to RS, i.e., assume RL is noiseless.
Since Vout is equal to the noise of RS for half of the LO cycle and equal to zero for the other half, we expect the output power density to be simply equal to half of that of the input, i.e., 2kTRS.
• To prove this conjecture, we view Vn,out(t) as the product of Vn,RS(t) and a square wave toggling between 0 and 1. The output spectrum is thus obtained by convolving the spectra of the two. (shown in next slide)
Solution:
Example of Noise Spectrum of a Simple Mixer (Ⅱ)
The output spectrum consists of (a) 2kTRS × 0.52, (b) 2kTRS shifted to the right and to the left by ± fLO and multiplied by (1/π)2, (c) 2kTRS shifted to the right and to the left by ± 3fLOand multiplied by [1/(3π)]2, etc. We therefore write
It follows that the two-sided output spectrum is equal to kTRS and hence the one-sided spectrum is given by
