ECE37 Advanced Analog Circuits INTRODUCTION TO DELTA-SIGMA ADCS Richard Schreier richard.schreier@analog.com NLCOTD: Level Translator VDD > VDD2, e.g. 3-V logic? -V logic VDD < VDD2, e.g. -V logic? 3-V logic Constraints: CMOS -V and 3-V devices no static current ECE37 2
Highlights (i.e. What you will learn today) st -order modulator (MOD) Structure and theory of operation 2 Inherent linearity of binary modulators 3 Inherent anti-aliasing of continuous-time modulators 4 2 nd -order modulator (MOD2) 5 Good FFT practice ECE37 3. Background (Stuff you already know) The SQNR * of an ideal n-bit ADC with a full-scale sine-wave input is (6.2n +.76) db 6 db = bit. The PSD at the output of a linear system is the product of the input s PSD and the squared magnitude of the system s frequency response i.e. X H(z) Y Syy () f = He ( j 2πf ) 2 S xx () f The power in any frequency band is the integral of the PSD over that band *. SQNR = Signal-to-Quantization-Noise Ratio ECE37 4
. What is Σ? Σ is NOT a fraternity Simplified Σ ADC structure: Analog In Loop Filter Coarse ADC Digital Out (to digital filter) DAC Key features: coarse quantization, filtering, feedback and oversampling Quantization is often quite coarse ( bit!), but the effective resolution can still be as high as 22 bits. ECE37 5 What is Oversampling? Oversampling is sampling faster than required by the Nyquist criterion For a lowpass signal containing energy in the frequency range (, f B ), the minimum sample rate required for perfect reconstruction is f s = 2f B. The oversampling ratio is OSR f s ( 2f B ) For a regular ADC, OSR 2 3 To make the anti-alias filter (AAF) feasible For a Σ ADC, OSR 3 To get adequate quantization noise suppression. Signals between and ~ are removed digitally. f B ECE37 6 f s
Oversampling Simplifies AAF OSR ~ : Desired Signal Undesired Signals OSR = 3: f s 2 f First alias band is very close Wide transition band f s 2 f Alias far away ECE37 7 How Does A Σ ADC Work? Coarse quantization lots of quantization error. So how can a Σ ADC achieve 22-bit resolution? A Σ ADC spectrally separates the quantization error from the signal through noise-shaping analog input u Σ v Decimation w digital ADC Filter bit @f output s n@2f B desired signal t undesired signals f B f s 2 f s 2 f B shaped noise Nyquist-rate PCM Data ECE37 8 t f B
digital input (interpolated) A Σ DAC System u v Reconstruction signal Σ Modulator analog Filter bit @f output s shaped noise analog output f B f s 2 f B f s 2 f B f s Mathematically similar to an ADC system Except that now the modulator is digital and drives a low-resolution DAC, and that the out-of-band noise is handled by an analog reconstruction filter. t w ECE37 9 Why Do It The Σ Way? ADC: Simplified Anti-Alias Filter Since the input is oversampled, only very high frequencies alias to the passband. A simple RC section often suffices. If a continuous-time loop filter is used, the anti-alias filter can often be eliminated altogether. DAC: Simplified Reconstruction Filter The nearby images present in Nyquist-rate reconstruction can be removed digitally. + Inherent Linearity Simple structures can yield very high SNR. + Robust Implementation Σ tolerates sizable component errors. ECE37
2. MOD: st -Order Σ Modulator [Ch. 2 of Schreier & Temes] U z - Σ z - V Y DAC Q ECE37 V Quantizer (-bit) v Feedback DAC v Since two points define a line, a binary DAC is inherently linear. y v MOD Analysis Exact analysis is intractable for all but the simplest inputs, so treat the quantizer as an additive noise source: U z - Y V(z) = Y(z) + E(z) z - Y(z) = ( U(z) z - V(z) ) / ( z - ) ( z - ) V(z) = U(z) z - V(z) + ( z - )E(z) Q V Y E V V(z) = U(z) + ( z )E(z) ECE37 2
The Noise Transfer Function (NTF) In general, V(z) = STF(z) U(z) + NTF(z) E(z) For MOD, NTF(z) = z The quantization noise has spectral shape! Poles & zeros: 4 3 NTF e j 2πf ( ) 2 2 ω 2 for ω «..2.3.4.5 Normalized Frequency (f /f s ) The total noise power increases, but the noise power at low frequencies is reduced ECE37 3 In-band Quant. Noise Power Assume that e is white with power σ2 e i.e. S ee ( ω) = σ 2 e π The in-band quantization noise power is IQNP = ω B H( e jω ) 2 S ee ( ω)dω π π 2 σ2 Since OSR ------ e, IQNP = ------------ ( OSR) ω B 3 3 For MOD, an octave increase in OSR increases SQNR by 9 db.5-bit/octave SQNR-OSR trade-off. σ e 2 ω B ----- ω π 2 dω ECE37 4
A Simulation of MOD dbfs/nbw 2 4 6 Full-scale test tone SQNR = 55 db @ OSR = 28 Shaped Noise 8 2 db/decade 3 2 Normalized Frequency NBW = 5.7x 6 ECE37 5 CT Implementation of MOD R i /R f sets the full-scale; C is arbitrary Also observe that an input at f s is rejected by the integrator inherent anti-aliasing Integrator R f C Latched Comparator R i u y clock D Q DFF CK QB v ECE37 6
MOD-CT Waveforms u = u =.6 v v y y 5 5 2 Time 5 5 2 Time With u=, v alternates between + and With u>, y drifts upwards; v contains consecutive +s to counteract this drift ECE37 7 MOD-CT STF = Recall z = e s z ---------------- s s-plane ω Zeros @ s = 2kπi σ Pole-zero cancellation @ s = ECE37 8
MOD-CT Frequency Responses NTF = z db 2 STF = z s 3 4 Quant. Noise Notch Inherent Anti-Aliasing 5 2 3 Frequency (Hz) ECE37 9 Summary Σ works by spectrally separating the quantization noise from the signal Requires oversampling. OSR f s ( 2f B ). Noise-shaping is achieved by the use of filtering and feedback A binary DAC is inherently linear, and thus a binary Σ modulator is too MOD has NTF (z) = z Arbitrary accuracy for DC inputs..5 bit/octave SQNR-OSR trade-off. MOD-CT has inherent anti-aliasing ECE37 2
3V V: 3V NLCOTD V V 3V 3V V 3V: 3V 3V 3V 3V ECE37 2 3. MOD2: 2 nd -Order Σ Modulator [Ch. 3 of Schreier & Temes] Replace the quantizer in MOD with another copy of MOD in a recursive fashion: U E Q E V z - z - z - z - V(z) = U(z) + ( z )E (z), E (z) = ( z )E(z) V(z) = U(z) + ( z ) 2 E(z) ECE37 22
U Simplified Block Diagrams z z z Q E V NTF( z) = ( z ) 2 STF( z) = z E U z z Q V - -2 NTF z ( ) = ( z ) 2 STF( z) = z 2 ECE37 23 NTF Comparison NTF ( e j2πf ) (db) 2 4 6 8 MOD MOD2 MOD2 has twice as much attenuation as MOD at all frequencies 3 2 Normalized Frequency ECE37 24
In-band Quant. Noise Power For MOD2, As before, IQNP = H( e jω ) 2 S ee ( ω)dω and S ee ( ω) = σ 2 e π So now IQNP He ( jω ) 2 ω 4 = ω B π 4 σ2 ------------ e ( OSR) 5 5 With binary quantization to ±, = 2 and thus σ2 e = 2 2 = 3. An octave increase in OSR increases MOD2 s SQNR by 5 db (2.5 bits) ECE37 25 Simulation Example Input at 75% of FullScale 5 5 2 Sample number ECE37 26
dbfs/nbw 2 4 6 8 2 Simulated MOD2 PSD Input at 5% of FullScale SQNR = 86 db @ OSR = 28 Simulated spectrum (smoothed) 4 db/decade Theoretical PSD (k = ) NBW = 5.7 6 4 3 2 Normalized Frequency ECE37 27 2 SQNR vs. Input Amplitude MOD & MOD2 @ OSR = 256 SQNR (db) 8 6 4 Predicted SQNR MOD2 Simulated SQNR MOD 2 8 6 4 2 Input Amplitude (dbfs) ECE37 28
2 SQNR vs. OSR SQNR (db) 8 6 4 MOD2 (Theoretical curve assumes -3 dbfs input) MOD (Theoretical curve assumes dbfs input) 2 Predictions for MOD2 are optimistic. Behavior of MOD is erratic. 4 8 6 32 64 28 256 52 24 ECE37 29 Audio Demo: MOD vs. MOD2 [dsdemo4] Sine Wave Slow Ramp Speech MOD MOD2 ECE37 3
MOD + MOD2 Summary Σ ADCs rely on filtering and feedback to achieve high SNR despite coarse quantization They also rely on digital signal processing. Σ ADCs need to be followed by a digital decimation filter and Σ DACs need to be preceded by a digital interpolation filter. Oversampling eases analog filtering requirements Anti-alias filter in an ADC; image filter in a DAC. Binary quantization yields inherent linearity MOD2 is better than MOD 5 db/octave vs. 9 db/octave SQNR-OSR trade-off. Quantization noise more white. Higher-order modulators are even better. ECE37 3 4. Good FFT Practice [Appendix A of Schreier & Temes] Use coherent sampling I.e. have an integer number of cycles in the record. Use windowing A Hann window works well. Use enough points Recommend N = 64 OSR. Scale (and smooth) the spectrum A full-scale sine wave should yield a -dbfs peak. State the noise bandwidth wn ( ) = ( cos( 2πn N) ) 2 For a Hann window, NBW =.5 N. N ECE37 32
Coherent vs. Incoherent Sampling db Incoherent Coherent 2 3..2.3.4.5 Normalized Frequency Coherent sampling: only one non-zero FFT bin Incoherent sampling: spectral leakage ECE37 33 db Windowing Σ data is usually not periodic Just because the input repeats does not mean that the output does too! A finite-length data record = an infinite record multiplied by a rectangular window: wn ( ) =, n < N Windowing is unavoidable. Multiplication in time is convolution in frequency Frequency response of a 32-point rectangular window: 2 3 4 5 6 Slow roll-off out-of-band Q. noise may appear in-band 7 8 9.25.25.375.5 ECE37 34
4 2 Example Spectral Disaster Rectangular window, N = 256 Out-of-band quantization noise obscures the notch! db 2 4 6.25.5 Normalized Frequency, f Actual Σ spectrum W( f) w 2 Windowed spectrum ECE37 35 Window Comparison (N = 6) (db) Wf () --------------- W ( ) 2 3 4 5 6 7 8 9 Rectangular Hann Hann 2.25.25.375.5 Normalized Frequency, f ECE37 36
Window Properties Window Rectangular Hann wn ( ), n =,,, N ( wn ( ) = otherwise) Number of non-zero FFT bins w 2 2 = wn ( ) 2 W ( ) = wn ( ) NBW = w 2 2 --------------- W ( ) 2 2πn cos---------- N ------------------------------- 2 Hann 2 2πn cos---------- 2 N ------------------------------- 2 3 5 N 3N/8 35N/28 N N/2 3N/8 /N.5/N 35/8N. MATLAB s hann function causes spectral leakage of tones located in FFT bins unless you add the optional argument periodic. ECE37 37 Window Length, N Need to have enough in-band noise bins to Make the number of signal bins a small fraction of the total number of in-band bins <2% signal bins >5 in-band bins N > 3 OSR 2 Make the SNR repeatable N = 3 OSR yields std. dev. ~.4 db. N = 64 OSR yields std. dev. ~. db. N = 256 OSR yields std. dev. ~.5 db. N = 64 OSR is recommended ECE37 38
FFT Scaling The FFT implemented in MATLAB is X M k + N ( ) = x M ( n + )e j n = If xn ( ) = Asin( 2πfn N), then Xk ( ) = -------------- 2πkn N AN --------, k = f or N f 2, otherwise Need to divide FFT by ( N 2) to get A.. f is an integer in (, N 2). I ve defined Xk ( ) X M ( k + ), xn ( ) x M ( n + ) since Matlab indexes from rather than. ECE37 39 x The Need For Smoothing The FFT can be interpreted as taking sample from the outputs of N complex FIR filters: h ( n) h ( n) h k ( n) y ( N) = X ( ) y ( N) = X( ) y k ( N) = X( k) h k ( n) e j 2πk ----------n = N, n < N, otherwise h N ( n) y N ( N) = X( N ) an FFT yields a high-variance spectral estimate ECE37 4
How To Do Smoothing Average multiple FFTs Implemented by MATLAB s psd() function 2 Take one big FFT and filter the spectrum Implemented by the Σ Toolbox s logsmooth() function logsmooth() averages an exponentially-increasing number of bins in order to reduce the density of points in the high-frequency regime and make a nice log-frequency plot ECE37 4 Raw and Smoothed Spectra 2 4 dbfs 6 8 Raw FFT logsmooth 2 2 Normalized Frequency ECE37 42
Simulation vs. Theory (MOD2) 2 2 4 Simulated Spectrum NTF = Theoretical Q. Noise? dbfs 6 8 2 4 Slight Discrepancy (~4 db) 6 4 3 2 Normalized Frequency ECE37 43 What Went Wrong? We normalized the spectrum so that a full-scale sine wave (which has a power of.5) comes out at db (whence the dbfs units) We need to do the same for the error signal. i.e. use S ee () f = 4 3. But this makes the discrepancy 3 db worse. 2 We tried to plot a power spectral density together with something that we want to interpret as a power spectrum Sine-wave components are located in individual FFT bins, but broadband signals like noise have their power spread over all FFT bins! The noise floor depends on the length of the FFT. ECE37 44
Spectrum of a Sine Wave + Noise ( dbfs ) Ŝ x () f 2 3 N = 2 6 N = 2 8 N = 2 N = 2 2 dbfs Sine Wave dbfs Noise SNR = db 3 db/ octave 4.25.5 Normalized Frequency, f ECE37 45 Observations The power of the sine wave is given by the height of its spectral peak The power of the noise is spread over all bins The greater the number of bins, the less power there is in any one bin. Doubling N reduces the power per bin by a factor of 2 (i.e. 3 db) But the total integrated noise power does not change. ECE37 46
So How Do We Handle Noise? Recall that an FFT is like a filter bank The longer the FFT, the narrower the bandwidth of each filter and thus the lower the power at each output We need to know the noise bandwidth (NBW) of the filters in order to convert the power in each bin (filter output) to a power density For a filter with frequency response Hf (), NBW = Hf () 2 df --------------------------- Hf ( ) 2 NBW f Hf () f ECE37 47 FFT Noise Bandwidth Rectangular Window h k ( n) = exp j 2πk ----------n, N H k () f = h k ( n) exp( j 2πfn) n = N k f = ----, H N k ( f ) = = N H k () f 2 = h k ( n) 2 = N n = N [Parseval] H k () f 2 df N NBW = ------------------------------- = ------ = H k ( f ) 2 N 2 ---- N ECE37 48
2 2 Better Spectral Plot Simulated Spectrum Theoretical Q. Noise dbfs/nbw 4 6 8 2 4 passband for OSR = 28 NBW =.5 / N = 2 5 6 4 3 2 Normalized Frequency 4 -- NTF() f 3 2 NBW N = 2 6 ECE37 49