ECE145C: RF CMOS Communication Circuits and Systems Prof

ECE145C: RF CMOS Communication Circuits and Systems Prof. James Buckwalter

• email: [email protected] • Lecture: Girvetz Hall 1112 8-9:15 • Faq: Piazza access code: ece145c (Gauchospace?) • Please allow 24-48 hour turnaround • Computing Lab: E1 • TAs – Di Li • Office Hours: T/Th 12-1, TBD • OH Location: ESB-2205C

• refine fundamental understanding of RF circuits and systems to analyze modern wireless technology. • present a comprehensive understanding from devices to systems. • teach applications of RF CMOS as well as III-V • analyze RF transmitter/receiver architectures. Modern cellular and RF technologies are a mash-up of communication theory and devices. One needs to understand device limitations to understand communication system limits and vice versa. © James Buckwalter 3 Topics for our class

• Propagation, Noise and Distortion Budgeting • Basics of Modulation / Cellular Standards • Receiver Filtering, Mixing, and Architectures • Power Amplifiers (Linear and Nonlinear): Output power and Efficiency • High-efficiency transmitters • RF Architectures (Putting it all together)

© James Buckwalter 4 Lecture Topic Lecture Topic 1 (3/31) System Perspective: 2 (4/2) System Perspective: Link Budget Interference 3 (4/7) System Perspective: EVM 4 (4/9) System Perspective: Reciprocal Mixing 5 (4/14) Mixers (1) 6 (4/16) Mixers(2) 7 (4/21) Tunable Filters (I) 8 (4/23) Midterm 9 (4/28) Tunable Filters (II) 10 (4/30) RX Architectures: Mixer First 11 (5/5) RX Architectures: Direct 12 (5/7) Power Amplifiers: Classes Sampling 13 (5/12) Power Amplifiers: Classes 14 (5/14) Power Amplifiers: Spectral Regrowth 15 (5/19) Outphasing 16 (5/21) Outphasing Modulators Modulators…Transmitters 17 (5/26) Doherty Transmitters 18 (5/28) Doherty Transmitters 19 (6/1) Envelope Tracking 20 (6/3) Midterm Transmitters Reference Material

• Razavi…for now. • These notes. • Thomas Lee, Design of RF CMOS Integrated Circuits • H. Darabi and A. Mirzaei, Integration of Passive RF Front End Components in SoCs • Steve Cripps, RF Power Amplifiers for Wireless Communication Grading

• In-class midterms 60% • 4-5 homeworks, laboratory 25% • Final project 15%

• Limited collaboration among students on the homework problems is encouraged. Such collaboration may include verbal discussion of problems, and the use of scratch paper or writing boards to discuss concepts and approaches to solving specific problems. It is also OK for students to verbally compare the final answers obtained for a given problem as a method of checking their work. The following academic honesty rules should be considered in force at all times: 1) Never show any draft of a homework solution to another student in the class until after the homework due date and after that person has handed in his/her own solution set. 2) Never look at any draft of another person's homework solution until after the homework due date and after you have handed in your solution set. 3) Never use another person's simulation files or supply your simulation files to another person until after the due date and after you have both handed in your solution sets. • If any of the above academic honesty rules are violated by any student in the course, the student will receive a failing grade for the course and the incident will be reported to the dean of the student's college, in the case of an undergraduate student, or to the assistant dean of graduate studies, in the case of a graduate student, for further administrative action. RF Wireless Systems • Communicate information reliably as quickly as possible with as little power consumption as possible. • Today RF systems are about coexistence Transmitter Receiver

Coding: Try to encode error correction PA: Power Amplifier DAC: Digital to analog convertor LNA: Low Noise Amplifier Mixer: Translate signal to RF carrier ADC: Analog to digital convertor LO: Local Oscillator Coding: Look for errors and correct data PA: Power Amplifier © James Buckwalter Shannon capacity

C BWlog 1 SNR • Foundation of communication theory (Shannon, 1948). • Information can be transmitted reliably at rate C. • BW is bandwidth of the RF channel. This is typically fixed by regulations/standards. • SNR is the signal-to-noise. This is where circuits play the most important role. Typically, we want high SNR with low power consumption.

• We refer to RF bands and RF channels. • Bands refers to a collection of RF frequency spectrum that is set aside for licensed or unlicensed communication. • Channels are allocations of frequency spectrum within the band for users. – TDD/FDD multiple access of channel – E.g. Use one channel within band at a time.

• Channels can be 1.4, 3, 5, 10, 15, or 20 MHz. How many channels exist in each band? © James Buckwalter 12 Available Bandwidth

• RF spectrum is valuable. Verizon paid $5B for 22 MHz between 746-757 and 776-787 nationwide. • Spectral efficiency quantifies how many bits are packed into 1 Hz of bandwidth C SE = = log2 (1+ SNR) BW • In reality, the modulation “spreads” over more than one hertz and degrades this ideal value without signal processing.

• BPSK

– mI(t) = {-1, 1} • QPSK

– mI(t) = {-1, 1} – mQ(t) = {-1, 1} • QAM – E.g. 16-QAM

– mI(t) = {-3, -1, 1, 3} – mQ(t) = {-3, -1, 1, 3}

Modulation means that large power variations can occur in the output waveform!

22 P3 3  3  18 22 P2 1  3  10 22 P1 1  1  2

1 Pavg = (4×18 + 8×10 + 4×2) = 10 16 © James Buckwalter 16 Peak-to-Average Power Ratio

• PAR or PAPR: Ratio of the peak power for the waveform relative to the average power. • For 16-QAM P 18 PAPRdBpeak 2.5 Pavg 10 • PAPR increases for “filtered” signals. • Big PAPR is a problem for energy-efficient circuits.

• Unfiltered Modulation • PAPR increases (dramatically for filtered signals)

Modulation Number of PAPR Dynamic Range % data at % data Symbol Power highest above Levels power average power 16-QAM 3 1.8 2.5 dB 9 9.5 dB 25% 25% 32-QAM 5 1.7 2.3 dB 17 12.3 dB 25% 50% 64-QAM 9 2.3 3.7 dB 49 16.9 dB 6.3% 50% 256-QAM 32 2.7 4.2 dB 225 23.4 dB 4.6% 45%

• QPSK Illustration E = m2 t + m2 t s I ( ) Q ( ) • Energy-per-bit • Noise d = 2E s

Eb SNREbno No

E No E = s b M More on this next lecture © James Buckwalter 19 Signal-to-Noise, SNR • Ebno - Energy per bit directly effects bit error rate • 10-6 is generally acceptable for wireless • 10-12 is generally acceptable for wireline

¥ -x2 æ ö 1 2N E P = e o dx = Qç b ÷ = Q SNR e ò ç N ÷ ( ) 2p E è o ø b

1 ¥ -t2 Q(x) = ò e 2 dt 2p x © James Buckwalter 20 Spectral Efficiency and Distortion

• Wideband signals passing through a nonlinear circuit cause spectral energy to “spill” outside of bandwidth. • Signal processing can eliminate the energy leakage into neighboring bands but greatly increases the peak to

average ratio. © James Buckwalter 21 Distortion Contributions

In-band distortion: AM-AM (gain) compression, AM-PM compression BW

Out-of-band distortion BW

• Compare the power of the signal in-band against the power leaking into adjacent bands. P( f + BW ) ACLR = o P( fo ) © James Buckwalter 23 Tradeoffs • Modulation – Higher-order QAM, higher capacity. • Peak-to-average ratio (PAPR) – Higher PAPR, more distortion (in band and out of band) • SNR – Higher SNR, higher capacity • We want circuit solutions that offer the highest SNR, the lowest distortion, and the lowest power consumption possible. Summary of Digital Communication

• We want to transmit as many bits per second as possible. • I/Q are two orthogonal spaces to transmit information. • As more bits are transmitted per second, more SNR is required to achieve the same BER. • We guarantee circuit performance in terms of EVM. This is the ultimate measurement.

Gt Pt p(r) = 2 4pr • How much of the transmit power reaches the receiver? • Power density, p(r) • Received power Pt Pr = Gt 2 Ar 4pr • Gt is the transmit antenna gain, Ar is the receiver area

2  PPGrttr G  4 r

• Gr is the receive gain and λ is the wavelength of the RF signal. This gain basically depends on the size of the antenna (geometry) and type of antenna c • Remember l = f • 2 The path loss is æ 4prö L = çè l ÷ø

2 æ 4prö L = çè l ÷ø

• Gt and Gr are the antenna gains. If the antennas are isotropic radiators, Gt = Gr = 1. • Higher frequencies have inherently more loss • A popular way to express the Friis Transmission Equation is æ l ö P = P + G + G + 20log r t t r 10 çè 4pr÷ø

• What is the wavelength at – 800 MHz? 2.4 GHz? 60 GHz 37.5 cm, 12.5 cm, 5mm • What is the path loss over 1m at – 800 MHz? 2.4 GHz? 60 GHz -30 dB, -40 dB, -68 dB • Now, what is the path loss over 1km at – 800 MHz? 2.4 GHz? 60 GHz -90 dB, -100 dB, -128 dB

Gant =hant D æU ö D = 4p ç max ÷ è Prad ø Radiation Pattern in Azimuth/Elevation Umax = Bo 2p p

Prad = ò ò U (q,f)sinq dq df 0 0

• Short dipole: 1.76dB • Half-wave dipole: 2.15dB • Horn: 10-20dB • Parabolic: 40-50dB

• Antenna converts incident power to a voltage source with impedance, Zs. • This impedance comprises

– Radiation resistance Rr – Resistive losses Rl REF: Antenna Theory, Balanis – Reactance Xs

• Find E and H field ke jkr l /2 EjI x y z edzsin, , jkz cos     generated by current 4 r l /2 E H     • Find average Poynting 2 kl   kl  vector 2 cos cos  cos   1 Io 22    WEHav aˆ   a ˆ     a ˆ r  28 22r  sin  

© James Buckwalter 34 Example: Radiation Resistance of Half- wave Dipole (II) • Determine the radiated power 2 PW dsW rd d 2 sin   radavav   00 2 2P  0.609 Io rad P  2.435 Rrad   73  rad 2  8 Io 2 2 l Shorter antennas have lower Rrad  80   radiation resistance! • Caveat: If maximum current does not occur at input terminals, we need to consider the input resistance rather than the radiation resistance.

©James Buckwalter 35 Example: Radiation Resistance of Half- wave Dipole (III) • What about reactance of antenna? → Poynting vector does not capture transverse fields representing imaginary power

l Note that a corresponds to log1 dipole wire width. 2a X 120  m kl tan  2

• For half-wave dipole, tangent in denominator should be very large so reactance should be small!

©James Buckwalter 36 Here is the problem…

• We don’t have ideal voltage and current sources. • The source resistance and load resistance create the need to “match” the source and load impedances. • RF wavelengths are relatively short c  0.3mGHz @ 1 f • We need to treat RF signals as “WAVES” • We will spend much of today understanding how we study microwave circuits.

©James Buckwalter 37 Loss and Frequency

• RF energy can interact with the atmosphere and make the situation worse

• At low frequencies, atmospheric effects are low. • Atmosphere is decomposed into O2 and H20 components • Depending on conditions (raining or not) the attenuation is exacerbated at higher frequency.

• Multipath • Propagation consists of line of sight (LOS) and non-line of sight (NLOS) components.

• Slow fading -coherence time of the channel is large relative to the delay constraint of the channel. – The amplitude and phase change imposed by the channel can be considered roughly constant over the period of use. – Slow fading can be caused by events such as shadowing, where a large obstruction such as a hill or large building obscures the main signal path between the transmitter and the receiver. – The amplitude change caused by shadowing is often modeled using a log-normal distribution with a standard deviation according to the log-distance path loss model. • Fast fading - coherence time of the channel is small relative to the delay constraint of the channel. – The amplitude and phase change imposed by the channel varies considerably over the period of use.

• Signals rarely propagate from point-to-point in a straight line • When going through multiple reflections losses behave as 1/r4 . • Rayleigh Fading model. rexp r22 / 4 m  pr   2m • Fading is time varying. • Rule of thumb: Fading can add an additional 10 dB of additional loss.

d Dealing with Multipath

• In flat fading, the coherence bandwidth of the channel is larger than the bandwidth of the signal. – All frequency components of the signal will experience the same magnitude of fading. – Use single carrier • In frequency-selective fading, the coherence bandwidth of the channel is smaller than the bandwidth of the signal. – Different frequency components of the signal therefore experience decorrelated fading. – Use multiple carriers (OFDM)

• Most standards use OFDM Orthogonal Frequency Division Multiplexing

• Break bandwidth into N subcarriers • Separate data is transmitted on separate subcarriers • Each subcarrier is “orthogonal” © James Buckwalter 44 What does OFDM fix?

Single Carrier Multi-Carrier Fading Selective Fading Flat Fading on subcarrier Symbol period Short pulse (1/BW) Long Pulse (N/BW) Equalization Large number of taps (high ISI) Short number of taps (little ISI) Spectral efficiency Poor High Guard bands Moderate None

K jkt2/ T v tX  e k k 1

• Multi-carrier systems impact the circuit design requirements. • More linearity is required. • For 16-QAM

PAPRdB~ 10

• Free space path loss

• Atmospheric absorption

• Multipath fading

• A geosynchronous satellite is located 21,000km from earth. • The GPS band is at 1.57 GHz • How much path loss (in dB) exists between the satellite and earth?

• A geosynchronous satellite is located 21,000km from earth. • The GPS band is at 1.57 GHz • How much path loss exists in dB between the satellite and earth?

2 æ m ö 2 3´108 æ c ö ç ÷ L = = s = -182dB çè f 4pr ÷ø ç 4p ×1.57GHz ×21,000,000m÷ çè ÷ø

• If you have -182 dB of path loss, how much signal power is required to keep the receive above the minimum detectable power level?

•

Transmitter

How much power gets from the transmitter to the receiver?

Receiver Or given receiver performance, how far can I reliably communicate? At what EVM? With what modulation? A link budget helps put this together

• We need to define a specification for our circuits. • This can be done by figuring out how much power will travel over the RF communication link. • The link budget is a means to put a HUGE number of parameters together and figure out the specification for the transmitter and the receiver.

• RF Frequency: 2.4 GHz • Data Rate: 10Mb/s • Modulation: 16-QAM • BER: 10-3 • Link distance: 100m

• Specify a data rate: 107 b/s • Channel bandwidth: 2.5 MHz

• Now we can assess the noise floor of our receiver. • Background noise: kT – k is Boltzmann’s constant: 1.38x10-23 J/K – T is the background temperature • kT is -174 dBm per 1 Hz • Noise figure (NF) is a ratio that describes how much worse than the receiver is than this fundamental background noise limit in dB. More on this later.

SensitivityNFkTBWSNR 10log() • For example, NF is 10 dB

SensitivitydBdBmdBdBdBm10(  174)641288 

Background Digital Circuit Noise Modulation

• What is -88dBm? 9uV rms

• We know the minimum power at the receiver. • How much power must be transmitted? • Start with the desired distance of transmission and the RF frequency: 1km at 2.4 GHz. • Friis Path Loss: 100 dB 2 æ 4prö L = çè l ÷ø 2 æ 1km ö L = ç4p ÷ =100dB 0.125m è ø © James Buckwalter 59 Link Budget (vi)

• Other channel effects? – Multipath fading: 10 dB – Atmospheric absorption: 0dB • Total propagation loss is

Lprop = Lpath + Lmultipath L =100dB+10dB =110dB prop

• Now we can find the average transmit power

PTX ,avg = Sensitivity + Lprop P = -88dBm+110dB = 22dBm TX ,avg • The peak power can now be found.

PTX ,pk = PTX ,avg + PAPR P = 22dBm+ 2.5dB = 24.5dBm TX ,pk • 24.5 dBm peak power is challenging for CMOS

• Would it be better to use more bandwidth and a less dense modulation format? • QPSK needs 8 dB to achieve the same BER Sensitivity NF  kTBW10log( SNR ) SensitivitydBdBmdB10(  174)  dBdBm 67889 • Okay – we saved 1 dB but we have wasted twice as much bandwidth. The same performance could be sold to twice as many users. © James Buckwalter 62 Link Budget (ix)

• What if we have more power or a different bandwidth?

• Rework the problem with the circuit specifications to determine for instance the range of transmission (next slide)

© James Buckwalter 63 Link Budget (Summary) 1 element TX and RX The following equations Transmitter Max Pout, i.e. near OP1dB dBm 20 are expressed in dB Backoff, i.e. ~PAPR dB 5.5 Pout on-chip dBm 14.5 PPPAPRTX,, outTX avg

Antenna Element directivity dB 6.0 Antenna Efficiency, dB dB -3 Antenna gain dB 3 GDEantantant

Link Budget Effective Isotropic Radiated Power (EIRP) dBm 17.5 EIRP = Gant,TX + Pout Bandwidth MHz 5 kTB dBm -107 Receiver NF dBm 10 Receiver noise floor dBm -97 Required AWGN SNR dB 12 Receiver sensitivity dBm -85 Psens = NF +10log10 kTB + SNRmin Receiver antenna gain dBm 3 ( ) System gain dB 105.5 Gsystem = EIRP +GANT ,RX - Psens

Free space loss exponent, nPL 2 Free space path loss at 400 meters dB 92 æ l ö Lpath = 10× nPL log10 ç ÷ © James Buckwalter è 4pr ø 64

Equation Cheat Sheet

C BWlog 1 SNR

M Mfs = fb N = 2

PPPAPRTX,, outTX avg Sensitivity NF kTBWSNR10log()

 Lpath  20log10  MDS = NF + kT +10log(BW) 4 r æ l ö P = P + G + G + 20log r t t r 10 çè 4pr ÷ø © James Buckwalter 65