Switched Capacitor Circuits I. Prof. Paul Hasler Georgia Institute of Technology


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1 Switched Capacitor Circuits I Prof. Paul Hasler Georgia Institute of Technology
2 Switched Capacitor Circuits Making a resistor using a capacitor and switches; therefore resistance is set by a digital clock and the capacitor. Filters built in this technology are set by external clocks, and ratio of capacitors (matching of 0.1% to 1%) The precision of the frequency response is realized by ratios of capacitors (1% to 0.1% better matching, larger caps; therefore more power/area), and a clock signal (which can be set precisely with a crystal reference)
3 Capacitor Circuits Capacitive Voltage Divider Q Capacitive Feedback Q = + C T Q C T C T Vfg Multiple Input Voltage Divider =   Q C2 Q C 3 C T = + + C T C T Q C T Vfg = (1 + )  Q C2
4 NonOverlapping Clocks We will always be using nonoverlapping clocks; therefore, we want a waveform like We effectively have four phases. t d (1) (2) (3) (4) cycle t
5 NonOverlapping Clocks We will always be using nonoverlapping clocks; therefore, we want a waveform like We effectively have four phases. t d (1) (2) (3) (4) cycle t Would want t d as small as possible for proper operation We will also assume that the input is held constant through the entire th cycle
6 NonOverlapping Clocks We will always be using nonoverlapping clocks; therefore, we want a waveform like We effectively have four phases. t d (1) (2) (3) (4) cycle Would want t d as small as possible for proper operation We will also assume that the input is held constant through the entire th cycle t Circuit to generate waveform Nstages of delay (sets t d ) Clock in
7 Basic Switched Capacitors
8 Basic Switched Capacitors (1), cycle I Q = ( [n1])
9 Basic Switched Capacitors (1), cycle I Q = ( [n1]) (3), cycle Q = ( ) I
10 Basic Switched Capacitors (1), cycle I Q = ( [n1]) (3), cycle If we assume the input changes slowly ( [n1] ~ ; therefore we are oversampling), we get Q = ( ) I I = Q f = f ( ) ; R = 1 / ( f) where f = clock frequency.
11 Basic Switched Capacitors ~ I R = 1 / ( f) where f = clock frequency.
12 Basic Switched Capacitors ~ I R = 1 / ( f) where f = clock frequency. For 0.1pF capacitor, and a 10kHz clock, we get a resistance of 1GOhm
13 Basic Switched Capacitors ~ I R = 1 / ( f) where f = clock frequency. For 0.1pF capacitor, and a 10kHz clock, we get a resistance of 1GOhm Rule of thumb: slow moving means we oversample the Nyquist frequency of the input signal by a factor of 20 or more.
14 Basic SwitchCap Integrator R 1
15 Basic SwitchCap Integrator R 1 We will step through all four phases, to get the proper result.
16 Basic SwitchCap Integrator (4), [n1] cycle [n1] Q =  [n1] Voltage = 0V [n1] (Voltage remains held) This case is important to understand our starting point charge is stored on a capacitor ; therefore we need to know the initial state
17 Basic SwitchCap Integrator (1), cycle: Q =  [n1] [n1] (Output unchanged) Charge up the capacitor with voltage
18 Basic SwitchCap Integrator (2), cycle Q 1 = Q =  [n1] [n1] (Output unchanged) We remove the capacitor from the input voltage. The voltage is stored across the capacitor
19 Basic SwitchCap Integrator (3), cycle: Q =  [n1] + = [n1]  ( / ) We connect the capacitor to the charge summing node The charge initially stored on the capacitor as well as the resulting charge from the second input ( ) contributes to the total charge
20 Basic SwitchCap Integrator (4), cycle Q 1 = 0 Q =  [n1] + Voltage = 0V = [n1]  ( / ) (Output unchanged) We disconnect the capacitor from the charge summing node, and return to our initial case = [n1]  ( / )
21 Basic SwitchCap Integrator = [n1]  ( / ) (z)  z 1 1 (z) = H(z) =  ( / ) 1
22 Basic SwitchCap Integrator = [n1]  ( / ) H(jω) =  ( / ) e jωt ~  ( / ) / jωt (z)  z 1 1 (z) = H(z) =  ( / ) 1 assumes ωt << 1; therefore we need to sample much higher (factor of 10 to 20) over frequencies of interest.
Switched Capacitor Circuits II. Dr. Paul Hasler Georgia Institute of Technology
Switched Capacitor Circuits II Dr. Paul Hasler Georgia Institute of Technology Basic SwitchCap Integrator = [n1]  ( / ) H(jω) =  ( / ) 1 1  e jωt ~  ( / ) / jωt (z)  z 1 1 (z) = H(z) =  ( / )
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