How to Simulate Series Capacitors in CircuitLab

First, you should be aware that the treatment of circuit theory in The Art of Electronics is very brief. An actual circuit analysis textbook would cover time constants in much more detail and provide more examples.

Your circuit appears to be Figure 1.31 from section 1.13. You left off the switch and the initial conditions, and mislabeled the voltage source. The battery is \$V_i\$, not \$V_1\$. Here's a corrected version:

^{simulate this circuit – Schematic created using CircuitLab}

It is implied (but not stated) that the capacitor is discharged at t = 0, so that V starts out at zero volts. With the switch open, there is no DC path from V to ground, so we have to make an assumption like this.

Once the switch is closed, current can start to flow. The mathematical way of approaching this is to write an equation using Kirchhoff's Current Law (KCL). Because of the capacitor, this will be a first-order differential equation:

$$current\ out\ of\ V_i = current\ into\ C$$ $$\frac{V_i - V}{R} = C\frac{dV}{dt}$$

(A differential equation is an equation involving a rate of change. Here, \$\frac{dV}{dt}\$ is the rate of change of V with respect to time.) You can then solve this to get an equation of the form:

$$V = V_i + Ae^{-t/RC}$$

where e is the base of the natural logarithm (~2.718) and A is an unknown constant. You can solve for the constant using your initial condition of \$V_{t=0} = 0\$.

An alternate way of looking at it is to say that a capacitor acts like an open circuit at DC, and like a short circuit when the voltages in the circuit are changing rapidly. At the instant we close the switch, we have a rapid change -- \$V_i\$ is suddenly applied all at once. The capacitor acts like a short to ground, so the current is \$V_i/R\$. After a long time, the voltage has stabilized and we effectively have a DC circuit. The capacitor acts like an open circuit, so \$V = V_i\$ and no current flows.

The transition between these two states in an exponential decay. This means that the equation for V will have a term like \$e^{-t/\tau}\$, where \$\tau\$ (tau), called the "time constant", determines the rate of decay. At t = 0, this exponential term is equal to 1. As t -> infinity, the exponential term decays to 0. We can use this to get an equation for V:

$$V = (final\ condition) - (difference\ between\ final\ and\ initial\ conditions) * (exponential\ term)$$

At t = 0, when the exponential term is equal to 1, this gives us:

$$V = (final\ condition) - (difference\ between\ final\ and\ initial\ conditions) = (initial\ condition)$$

At t = infinity, when the exponential term is equal to 0, it gives us:

$$V = (final\ condition) - 0 = (final\ condition)$$

In this circuit, our initial condition is \$V = 0\$. Our final condition is \$V = V_i\$. The difference between them is \$V_i - 0 = V_i\$. We would normally have to solve the differential equation to get the time constant \$\tau\$, but what the book is telling us is that for an R-C circuit, \$\tau = RC\$. Now we can write the final equation:

$$V = V_i - V_ie^{-t/RC}$$

When the initial condition is zero (as it is here), we can write the equation as:

$$V = (final\ condition) * (1 - (exponential\ term))$$

which gives:

$$V = V_i(1 - e^{-t/RC})$$

And that's exactly what's in the book.

First, break out the ESR element as a separate element in your model. Now the Q variable is proportional to the voltage across the ideal capacitor part of the model, so it's easy to probe:

Now if you use an X-Y plot to plot the voltage across the whole (nonideal) capacitor vs the voltage across the ideal capacitor (again, as an analog of the charge variable), you'll get what you expect (with a bit of a starting transient as the first cycle gets started):

Some tips:

• You need to simulate several cycles of the sinusoid to get the system to steady state.

• You need to choose a frequency near the pole frequency \$1/(2\pi{}RC)\$. Too low and you won't be able to detect the hysteresis in the graph. Too high and you the hysteresis will get so opened up it will just look like the oval in your figure 1b.