RLC Circuits in the Time Domain
On the importance of modeling systems with RLC circuits.....
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RC Circuits
RC circuits play a crucial role in electromagnetic compatibility and signal integrity modeling. Most CMOS digital circuits, coupling from heatsinks, ...... are just a few examples of real-world problems that are readily modeled using simple RC circuits. Also, Many important sources of common-mode emissions can be modeled as RC circuits and most...
Consider the simple RC circuit shown in the figure. Suppose we want to model the voltage across R2 and the capacitor before and after the switch closes. If we start by writing node and mesh equations and solving differential equations, this relatively simple circuit will quickly become very difficult to solve. However, there's a basic set of steps than we can follow to quickly solve for all of the voltages and currents in any RC circuit that has only one capacitor. Those steps are listed below.
1. We start by recognizing that any capacitor in a steady-state condition behaves like an open circuit. The current through a capacitor is C dV/dt. If the voltage isn't changing, no current can flow. So, prior to the switch closing, we can determine the voltage and currents everywhere in the circuit by simply assigning an infinite impedance to the capacitor. In this example, the switch is open, so no current flows anywhere and VC = 0. This is plotted in the graph on the right.
2. The voltage across a capacitor is incapable of changing instantaneously. It can only change as current begins to flow though it. At the moment immediately after the switch closes (t = 0+), the voltage across the capacitor is equal to the voltage across it at before the switch closed. In that moment, it can be modeled as an ideal voltage source. In this example, the initial voltage was zero, so it can be modeled as a short circuit. The voltages and currents everywhere in the circuit at that moment can be found by shorting the capacitor and doing a DC analysis of the rest of the circuit. Note that while the voltage across the capacitor can't change instantaneously, other circuit parameters can. If we were plotting the current through the capacitor, we would see it jump from 0 to 0.6 A at t=0.
3. Eventually, the circuit will reach a steady state again. At that point, the capacitor will behave like an open circuit again. So, we can determine the final steady-state voltages and currents by closing the switch and replacing the capacitor with an open circuit. In this example, the steady-state voltage across the capacitor is the same as the voltage across R2, which is 2.73 volts. A plot of the voltage before the switch closes and long after the switch closes is shown in the figure on the right.
4. All we are missing are the voltages and currents from the time immediately after the switch closes and the steady state. If the circuit has only one capacitor, its behavior during that time is governed by a first-order differential equation. Solutions to that equation are exponentials. All of the voltages and currents in the circuit will transition from their initial state to their final state exponentially, starting at a rapid pace and slowing as they approach the steady state. These transitions are governed by the equations,
While these expressions look complicated, when plotted, they simply show a smooth exponential transition from the initial value to the final value. The exponential starts with a steep slope and ends with zero slope. This transition is indicated in the figure on the right.
5. To complete the plot, we need to put a time scale on the transition. The time constant for an RC circuit is always τ = RC. In this equation, C is the value of the capacitor and R is the resistance viewed from the terminals of the capacitor. All of the voltages and currents in the circuit transition from their initial values to their final values with the same time constant.
In our example, C is 100 pF and R (when the switch is closed) is the parallel combination of 5 Ω and 50 Ω. The time constant is therefore, 455 ps. The 10% to 90% risetime is 2.2 time constants or 1.0 ns. The final plot is indicated on the right.
Now, let's use the same procedure to plot the voltage across the capacitor assuming the switch opens again after being closed for 1 ns. At that point, the voltage has risen to 90% of 2.73 volts, or 2.45 volts. That is the initial voltage across the capacitor. The voltage up to the time the switch opens is plotted in the figure on the right.
We can see that after the switch has been open for a long time, the capacitor will be fully discharged, and its voltage will be zero. This is state the circuit was in prior to closing the switch at t=0. The plot on the right includes the steady-state voltage reached long after the switch is opened.
The final step is to show the exponential waveform that connects the initial voltage to the final voltage. The 10% to 90% transition time is 2.2 time constants. However, the time constant with the switch open is very different than it was with the switch closed. The value of C is still 100 pF, but the value of R is 50 Ω (not the parallel combination of 5 Ω and 50 Ω). VC falls to 10% of its initial value (0.245 V) in 2.2 x 50 Ω x 100 pF = 11 ns.