TFE4188 - Lecture 5

Switched-Capacitor Circuits

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Goal

Understand why we would use switched capacitor circuits

Introduction to discrete-time, and switched capacitor principles and the circuits we need

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Why

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Active-RC

H(s) = \frac{\left[ \frac{C_1}{C_B}s^2 + \frac{G_2}{C_B}s + (\frac{G_1G_3}{C_A C_B})\right]}{\left[ s^2 + \frac{G_5}{C_B}s + \frac{G_3 G_4}{C_A C_B}\right]}

\omega_{p\vert z} \propto \frac{G}{C} = \frac{1}{RC}

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Gm-C

H(s) = \frac{\left[ s^2\frac{C_X}{C_X + C_B} + s\frac{G_{m5}}{C_X + C_B} + \frac{G_{m2}G_{m4}}{C_A(C_X + C_B)}\right]} {\left[s^2 + s\frac{G_{m3}}{C_X + C_B} + \frac{G_{m1}G_{m2}}{C_A(C_X + C_B)} \right]}

\omega_{p\vert z} \propto \frac{G_m}{C}

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Switched capacitor

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Q_{\phi2\$} = C_1 V_{GND} = 0

Q_{\phi1\$} = C_1 V_{I}

Z_{I} = (V_{I} - V_{GND})/I_{I}

I_{I} = \frac{\Delta Q}{\Delta t} = \left(Q_{\phi1\$} - Q_{\phi2\$}\right) f_{\phi}

Z_{I} = \frac{V_{I} - V_{GND}}{\left(Q_{\phi1\$} - Q_{\phi2\$}\right) f_{\phi}}

Z_{I} = \frac{V_{I}}{\left(C_1 V_{I} - 0 \right) f_{\phi}} = \frac{1}{C_1 f_\phi}

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Z_{I} = \frac{V_{I} - V_{O}}{\left(Q_{\phi1\$} - Q_{\phi2\$}\right) f_{\phi}}

Q_{\phi1\$} = C_1 (V_I - V_O)

Q_{\phi2\$} = 0

Z_{I} = \frac{V_{I} - V_{O}}{C_1 \left(V_I - V_O\right) f_{\phi}} = \frac{1}{C_1 f_\phi}

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Z_{I} = \frac{ V_{I} - V_{O} }{ \left(Q_{\phi1\$} - Q_{\phi2\$}\right) f_{\phi}}

Q_{\phi1\$} = C_1 V_I

Q_{\phi2\$} = C_1 V_O

Z_{I} = \frac{V_{I} - V_{O}}{\left(C_1 V_I - C_1 V_O\right) f_{\phi}} = \frac{1}{C_1 f_\phi}

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A pipelined 5-Msample/s 9-bit analog-to-digital converter [@Lewis87]

\omega_{p\vert z} \propto f_{clk}\frac{C_1}{C_2}

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Discrete-Time Signals

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Define \(x_c (t)\) as a continuous time, continuous value signal

Define \(\ell(t) = \begin{cases} 1 & \text{if } t \geq 0 \\ 0 & \text{if } t < 0 \end{cases}\)

Define \(x_{sn}(t) = \frac{x_c(nT)}{\tau}[\ell(t-nT) - \ell(t - nT - \tau)]\)

Define \(x_s(t) = \sum_{n=-\infty}^{\infty}{x_{sn}(t)}\)

Think of a sampled version of an analog signal as an infinite sum of pulse trains where the area under the pulse train is equal to the analog signal.

Why do this?

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If \(x_s(t) = \sum_{n=-\infty}^{\infty}{x_{sn}(t)}\)

Then \(X_{sn}(s) = \frac{1}{\tau}\frac{1 - e^{-s\tau}}{s} x_c(nT)e^{-snT}\)

And \(X_s(s) = \frac{1}{\tau}\frac{1 - e^{-s\tau}}{s} \sum_{n=-\infty}^{\infty}x_c(nT)e^{-snT}\)

Thus \(\lim_{\tau \to 0} \rightarrow X_s(s) = \sum_{n=-\infty}^{\infty}x_c(nT)e^{-snT}\)

The spectrum of a sampled signal is an infinite sum of frequency shifted spectra

or equivalently

When you sample a signal, then there will be copies of the input spectrum at every \(nf_s\)

However, if you do an FFT of a sampled signal, then all those infinite spectra will fold down between \(0 \to f_{s1}/2\) or \(- f_{s1}/2 \to f_{s1}/2\) for a complex FFT

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dt.py - interactive

#- Create a time vector
N = 2**13
t = np.arange(N)

#- Create the "continuous time" signal with multiple 
#- "sinusoidal signals and some noise
#- f1 is deliberately halfway between FFT bins, so the
#- record is not coherent and the window has a job to do
f1 = 233.5/N
fd = 1/N*119
x_s = np.sin(2*np.pi*f1*t) + 1/1024*np.random.randn(N) + \
    0.5*np.sin(2*np.pi*(f1-fd)*t) + 0.5*np.sin(2*np.pi*(f1+fd)*t)

#- Create the sampling vector, and the sampled signal
t_s_unit = [1,1,0,0,0,0,0,0]
t_s = np.tile(t_s_unit,int(N/len(t_s_unit)))
x_sn = x_s*t_s

#- Convert to frequency domain with a hanning window to avoid FFT bin
#- energy spread
Hann = True
if(Hann):
    w = np.hanning(N+1)
else:
    w = np.ones(N+1)
X_s = np.fft.fftshift(np.fft.fft(np.multiply(w[0:N],x_s)))
X_sn = np.fft.fftshift(np.fft.fft(np.multiply(w[0:N],x_sn)))

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X_s(s) = \sum_{n=-\infty}^{\infty}x_c(nT)e^{-snT}

X_s(z) = \sum_{n=-\infty}^{\infty}x_c[n]z^{-n}

For discrete time signal processing we use Z-transform

If you're unfamiliar with the Z-transform, read the book or search https://en.wikipedia.org/wiki/Z-transform

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Pole-Zero plots

If you're not comfortable with pole/zero plots, have a look at

What does the Laplace Transform really tell us

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Z-domain

Spectra repeat every \(2\pi\)

Bi-linear transform

s = \frac{2}{T}\frac{z -1}{z + 1}

_{Warning: First-order approximation https://en.wikipedia.org/wiki/Bilinear_transform}

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First order filter

y[n+1] = bx[n] + ay[n] \Rightarrow Y z = b X + a Y

\(y[n] = b x[n-1] + ay[n-1] \Rightarrow Y = b X z^{-1} + a Y z^{-1}\)

H(z) = \frac{b}{z-a}

Infinite-impulse response (IIR)

h[n] = \begin{cases} k & \text{if } n < 1 \\ a^{n-1}b + a^n k & \text{if } n \geq 1 \end{cases}

_{Head's up: Fig 13.12 in AIC is wrong}

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Second order filter

y[n] = b x[n-1] + 2a\, y[n-1] - (a^2 + b^2)\, y[n-2]

H(z) = \frac{b z}{z^2 - 2a z + (a^2+b^2)}

\[z_p = a + jb\]

iir.py - interactive

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Finite-impulse response(FIR)

H(z) = \frac{1}{3}\sum_{i=0}^2 z^{-i} = \frac{1}{3}\left(1 + z^{-1} + z^{-2}\right)

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Switched-Capacitor

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Q_{1\phi_1\$} = C_1 V_1

Q_{2\phi_1\$} = 0

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Q_{1\phi_2\$} = 0

Q_{2\phi_2\$} = Q_{1\phi_1\$} = C_1 V_1 = C_2 V_2

\frac{V_2}{V_1} = \frac{C_1}{C_2}

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Switched capacitor gain circuit

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V_o[n+1] = \frac{C_1}{C_2}V_i[n]

V_o z = \frac{C_1}{C_2} V_i

\frac{V_o}{V_i} = H(z) = \frac{C_1}{C_2}z^{-1}

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Switched capacitor integrator

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V_o[n] = V_o[n-1] + \frac{C_1}{C_2}V_i[n-1]

V_o - z^{-1}V_o = \frac{C_1}{C_2}z^{-1}V_i

H(z) = \frac{C_1}{C_2}\frac{z^{-1}}{1-z^{-1} } = \frac{C_1}{C_2}\frac{1}{z-1}

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Both phases add noise, \(V_n^2 > \frac{2 k T}{C}\)

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Mean \(\overline{x(t)} = \lim_{T\to\infty} \frac{1}{T}\int^{+T/2}_{-T/2}{ x(t) dt}\)

Mean Square \(\overline{x^2(t)} = \lim_{T\to\infty} \frac{1}{T}\int^{+T/2}_{-T/2}{ x^2(t) dt}\)

Variance \(\sigma^2 = \overline{x^2(t)} - \overline{x(t)}^2\)

where \(\sigma\) is the standard deviation. If mean is removed, or is zero, then \(\sigma^2 = \overline{x^2(t)}\)

Assume two random processes, \(x_1(t)\) and \(x_2(t)\) with mean of zero (or removed). \(x_{tot}(t) = x_1(t) + x_2(t)\) \(x_{tot}^2(t) = x_1^2(t) + x_2^2(t) + 2x_1(t)x_2(t)\)

Variance (assuming mean of zero) \(\sigma^2_{tot} = \lim_{T\to\infty} \frac{1}{T}\int^{+T/2}_{-T/2}{ x_{tot}^2(t) dt}\) \(\sigma^2_{tot} = \sigma_1^2 + \sigma_2^2 + \lim_{T\to\infty} \frac{1}{T}\int^{+T/2}_{-T/2}{ 2x_1(t)x_2(t) dt}\)

Assuming uncorrelated processes (covariance is zero), then \(\sigma^2_{tot} = \sigma_1^2 + \sigma_2^2\)

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Sub-circuits for SC-circuits

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OTA

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Switches

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wulffern/sun_sar9b_sky130nm

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Non-overlapping clocks

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Example

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Summary

A switched capacitor is a resistor R = 1/(f C) made of ratio-accurate parts: SC filters get their time constants from C ratios and a clock
Sampling moves the world to discrete time: the z-domain, aliasing and the sample-rate theorem come with it
The parasitic-insensitive integrator and the SC gain stage are the two workhorse circuits; correlated double sampling throws in offset removal
Every sample costs kT/C of noise - capacitor sizes come from the noise budget, not the layout
Switches need non-overlapping clocks, and bootstrapping when the signal swings
Behind every SC circuit stands an OTA and its bias; the settling budget is half a clock period, verified in the transient

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Would you like to know more?

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Thanks!

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