Cursed circuits #5: capacitance multiplier

Section A: Op-Amp Voltage Follower

An ideal op-amp does one thing and one thing only: it calculates the difference between the voltages on its two input pins (Vin- and Vin+), multiplies it by a humongous constant factor (AOL, typically 1,000,000 or more), and then outputs the resulting voltage in relation to the midpoint of the supply (Vmid). We can write this as:

In practical terms, it means that if Vin- is noticably less than Vin+, the output voltage swings toward the positive supply of the chip; conversely, if Vin- exceeds Vin+, the output swing dives toward the negative rail. Intermediate output voltages are possible only in a very narrow, microvolt-range linear region of Vin- ≈ Vin+.

The simplest op-amp circuit - and the only one we need today - is the voltage follower:

The circuit loops the output voltage onto one of its differential input pins. If the input signal on Vin+ creeps higher in relation to Vin-, this makes the amplified differential signal more positive, forcing the output voltage to rise until the Vin- ≈ Vin+ equilibrium is restored. In the same vein, if Vin- drops, the difference becomes more negative, sending the amplified signal toward a lower equilibrium point. In effect, the output voltage tracks the input signal with sub-millivolt accuracy.

Section B: Capacitor Charging Through a Resistor

To make sense of the next section, we need just two other tidbits of electronic theory. First, Ohm's law: the current flowing through a resistor is proportional to the electromotive force (voltage) applied to its terminals, divided by the component's value (resistance): I = V/R. Second, we need to know that a capacitor subjected to a voltage across its terminals will admit (nearly-arbitrary) charging current until the pushback force created by the accumulated charge matches the external voltage. Higher component value (higher capacitance) means that proportionately more electrons can be shuffled around in response to the same electromotive force; in other words, higher C means more current over time.

If the above paragraph sounds confusing, you should review the article on core concepts in electronic circuits before venturing forth.

A Capacitor, With a Twist

This brings us to our guest of honor: the capacitance multiplier. It's the kind of a circuit that usually doesn't make sense at first glance because we can't pattern-match it to anything we know:

To unravel the mystery, it suffices to break it down into two mostly-separate building blocks. Section A is just an op-amp configured as a voltage follower. No matter what else is going on in the circuit, it takes some voltage from section B and then mirrors that signal on its output leg.

Section B is a capacitor that's charging through a resistor; although the voltage across the capacitor's terminals (Vcap) will change over time, let's model this in a freeze-frame view. In this setting, the op-amp is mirroring the capacitor's current charge state; the voltage on the right terminal of R2 is almost exactly the same as Vcap.

Moving onto R1, we can conclude from Ohm's law that the current flowing through the resistor depends only on the component's value and the momentary voltage across its terminals. No matter what else is going on in the circuit, the upper terminal of R1 is at the same voltage as the input leg (Vsignal) and the lower terminal is always at Vcap. Therefore, the resistor current is:

For clarity, let's shorten the Vsignal - Vcap expression to v. This gives us:

Next, let's have a look at the other resistor, R2. The component's left terminal is at Vsignal; the right terminal must be at Vcap by the operation of the voltage follower. This means that the current flowing through the resistor is just:

In essence, we have two resistors in parallel between the input leg and Vcap. Naturally, the currents through the resistors are flowing in the same direction. If Vsignal > Vcap, the current flows in via the input terminal. In the inverse case, it feeds back into the signal supply. In both cases, the total input current is:

Next, let's calculate the ratio of Itotal to the current actually diverted to the capacitor via R1. We'll label this ratio n:

This tells us that the capacitor is charging n times more slowly than it would if the circuit consisted just of the passive R-C layout in section B. Another way to look at it is that the situation is indistinguishable from charging a proportionately larger capacitance - n · C - through a pair of parallel resistors R1 and R2. If R1 ≫ R2, this can be approximated as a model of n · C charging via R2.

And that's it: you can turn a cheap 1 µF capacitor into a fancy 1 mF by tossing in an op-amp and two resistors with carefully-chosen values. The circuit doesn't emulate increased energy storage - you can't use this for backup power - but in applications such as signal filtering or timekeeping, it's notionally a pretty neat trick. You can use components that are smaller, cost less, and have better specs.

In practice, it's less useful than it used to be. Capacitor technology has improved dramatically over the past 20-30 years. And if you truly need an ultra low-frequency filter or a long-interval timer, chances are, the same idea can be realized with greater flexibility and fidelity with a digital chip. Still - it's neat, right?

Postscript: Trading Places

As a voluntary homework assignment, consider the following variant of the earlier design:

The only change is that C and R1 are swapped. Your mission, should you accept it, is to figure out what this circuit does. If you're stumped, note that for a capacitor to charge, there must be a matching motion of charges on both terminals - i.e., some current must flow "through" the component. In the circuit, if the capacitor is initially discharged and the input voltage suddenly jumps to 5 V, the charging process is hindered by R1, so the voltage across the capacitor stays near 0 V; the electromotive force simply couples across the dielectric gap, putting Vin+ at 5 V - 0 V. It's only after a while that a sufficient number of electrons can make its way through R1, nudging Vcap toward 5 V and Vin+ toward 0 V.

Some of the earlier articles in this series: