Switched Capacitor Circuitry
Digital Circuits
What would happen to the voltage across capacitor C_{2} if the following steps were followed, over and over again:

 Connect capacitor C_{1} to battery, allow to fully charge
 Disconnect capacitor C_{1} from battery
 Connect capacitor C_{1} to capacitor C_{2}, allow for charges to equalize
 Disconnect capacitor C_{1} from capacitor C_{2}
 Repeat
Describe what happens to V_{out} (the voltage across capacitor C_{4}) as time goes on, assuming the relay is continuously toggled by the oscillator circuit at a high frequency. Assume that the input voltage (V_{in}) is constant over time:

This type of circuit is often referred to as a flying capacitor circuit, with C_{3} being the “flying” capacitor. Explain why this is, and what possible benefit might be realized by using a flying capacitor circuit to sample a voltage.
Suppose an engineer decided to use a flying capacitor circuit to sample voltage across a shunt resistor, to measure AC current from an electrical generator:

The frequency of the alternator’s output is 50 Hz. How does this affect the design of the flying capacitor circuit, so we ensure a fairly accurate reproduction of the AC signal at the output of the flying capacitor circuit? Generalize your answer to cover all conditions where the input signal varies over time.
In this circuit, a capacitor is alternately connected to a voltage source, then a load, by means of two MOSFET transistors that are never conducting at the same time:

Note: the φ_{1} and φ_{2} pulse signals are collectively referred to as a nonoverlapping, twophase clock.
Consider the average amount of current through the load resistor, as a function of clock frequency. Assume that the “on” resistance of each MOSFET is negligible, so that the time required for the capacitor to charge is also negligible. As the clock frequency is increased, does the load resistor receive more or less average current over a span of several clock cycles? Here is another way to think about it: as the clock frequency increases, does the load resistor dissipate more or less power?
Now suppose we have a simple tworesistor circuit, where a potentiometer (connected as a variable resistor) throttles electrical current to a load:

It should be obvious in this circuit that the load current decreases as variable resistance R increases. What might not be so obvious is that the aforementioned switched capacitor circuit emulates the variable resistor R in the second circuit, so that there is a mathematical equivalence between f and C in the first circuit, and R in the second circuit, so far as average current is concerned. To put this in simpler terms, the switched capacitor network behaves sort of like a variable resistor.
Calculus is required to prove this mathematical equivalence, but only a qualitative understanding of the two circuits is necessary to choose the correct equivalency from the following equations. Which one properly describes the equivalence of the switched capacitor network in the first circuit to the variable resistor in the second circuit?

Be sure to explain the reasoning behind your choice of equations.
In this circuit, a capacitor is alternately connected in series between a voltage source and a load, then shorted, by means of two MOSFET transistors that are never conducting at the same time:

Note: the φ_{1} and φ_{2} pulse signals are collectively referred to as a nonoverlapping, twophase clock.
Consider the average amount of current through the load resistor, as a function of clock frequency. Assume that the “on” resistance of each MOSFET is negligible, so that the time required for the capacitor to charge is also negligible. As the clock frequency is increased, does the load resistor receive more or less average current over a span of several clock cycles? Here is another way to think about it: as the clock frequency increases, does the load resistor dissipate more or less power?
Now suppose we have a simple tworesistor circuit, where a potentiometer (connected as a variable resistor) throttles electrical current to a load:

It should be obvious in this circuit that the load current decreases as variable resistance R increases. What might not be so obvious is that the aforementioned switched capacitor circuit emulates the variable resistor R in the second circuit, so that there is a mathematical equivalence between f and C in the first circuit, and R in the second circuit, so far as average current is concerned. To put this in simpler terms, the switched capacitor network behaves sort of like a variable resistor.
Calculus is required to prove this mathematical equivalence, but only a qualitative understanding of the two circuits is necessary to choose the correct equivalency from the following equations. Which one properly describes the equivalence of the switched capacitor network in the first circuit to the variable resistor in the second circuit?

Be sure to explain the reasoning behind your choice of equations.
Identify the polarity of voltage across the load resistor in the following switched capacitor circuit (called a transresistor circuit). Note: φ_{1} and φ_{2} are twophase, nonoverlapping clock signals, and the switches are just generic representations of transistors.

Identify the polarity of voltage across the load resistor in the following switched capacitor circuit. Note: the only difference between this circuit and the last is the switching sequence.

What difference would it make to the output signal of this operational amplifier circuit if the switching sequence of the switched capacitor network were changed? What difference would it make if the switching frequency were changed?

Many switched capacitor circuits require nonoverlapping, twophase clock signals. Shown here is a schematic diagram of a gate circuit that converts a single clock signal into two complementary, nonoverlapping clock signals:

Assume that the only gates possessing propagation delays are the doubleNOT gates (inside the dotted boxes), whose sole purpose it is to provide a short time delay in the feedback signals. Draw a timing diagram showing the twophase clock signals (φ_{1} and φ_{2}) in relation to the input clock waveform, and be prepared to explain how and why this circuit works:

The fact that switched capacitor networks can behave equivalently to resistors is exploited in a variety of integrated circuits. At first, this may seem strange, as switched capacitor networks generally require at least two switching transistors and a twophase, nonoverlapping clock (in addition to the capacitor itself) in order to function, which seems like a lot of peripheral circuitry compared to a single resistor. What possible advantage is there to using switched capacitor networks in integrated circuits instead of resistors? Support your answer with research, if possible.
Published under the terms and conditions of the Creative Commons Attribution License