# The Intuition for RLC circuits

I'm reading through Hayt's Engineering Circuit Analysis 9e and it does a fairly good job of giving the intuition for "where" the energy is going in a damped RLC circuit, but I'm hoping that someone here can help me get full closure.

Consider a parallel RLC circuit with no input and where the capacitor is initially at some non-zero i(0).

First of all, I'm hoping to clarify why the maximum voltage achieved by the system occurs in the underdamped case. Intuitively, is this because the resistor's value value is so large that most of the current will flow between the capacitor and the inductor (and thus energy is just exchanged between the two rather than being dissipated by the resistor).

If the above is true, then I am trying to square that with the fact that the slightly underdamped or critically damped case has the shortest settling time. My understanding is that settling time corresponds to the time required for all of the energy to be dissipated. If, in the slightly underdamped case, we are exchanging energy between the capacitor and the inductor in an oscillatory manner and the resistor is therefore not dissipating much energy, how does that system settle (well) before the overdamped case where the resistor (I think) dissipates a large amount of energy because resistance is low and much current therefore flows through the resistor? Or am I somehow confusing energy dissipation with the voltage here.

Generally, I'm looking for an intuitive explanation of "where the energy is going" and why it "goes away" quickest in the critically damped or slightly underdamped cases, rather than a mathematical demonstration.

• "underdamped case has the shortest settling time" citation needed. Jan 15 '20 at 15:09
• I should have been more rigorous. In general, the shortest settling time occurs for a slightly underdamped system (that is, there is a point of diminishing returns) - this comes directly from the Hayt book. But at any rate, I suppose my question could be more clearly put as - why does the critical damping case have a shorter settling time when it has a higher resistance in parallel (and then, according to my logic, which is clearly wrong, above, this would mean less current/energy going down it's path)? Jan 15 '20 at 15:17
• Sorry, just saw you finished your answer. Is it possible to directly adapt this analogy to the RLC circuit though? Describing where the energy is going specifically. Jan 15 '20 at 15:38
• Interactive RLC tool - increase R in value and the oscillations go on for a longer time. Decrease it and they settle down more quickly. This uses a step input into the inductor but there's no fundamental difference between this and the initial condition of current in C. Jan 15 '20 at 16:05
• @DKNguyen I re-read my comment and I can't see what you mean. Maybe you can shed light? Jan 16 '20 at 8:47

Resistors are the electrical equivalent of friction and produce losses to remove energy from the system. If you drop a frictionless pendulum to swing it, how long does it take to settle? Infinite time, because there are no losses so it just swings back and forth and oscillates around bottom position forever. The flip side is what happens if the pendulum has infinite friction? Well it just sits there at the high position forever taking infinite time to reach the bottom forever hanging on to its potential energy.

If you have non-infinite but high friction, your pendelum is going to take a VERY long time to reach the bottom position. It will never overshoot but it will take a very long time to get there. And if you have too little it will reach the bottom point faster but overshoot and then proceed to swing continuously about the bottom position for a very long time too. There is a middle ground that mutually minimizes the time it takes for energy to dissipate and the time it takes for the pendelum to travel one cycle to the bottom position.

Resistor = friction. Dissipated as heat (for the most part) in both systems.

Inductor = kinetic energy storage. In the pendulum this is the velocity of the weight combined with the mass.

Capacitor = potential energy storage. In the pendulum this is the gravitation potential energy due to the mass being at a height.

You might be wondering why the capacitor is the potential energy storage and why the inductor is the kinetic energy storage. This is because you can put charge in the capacitor and then just keep the energy there statically by disconnecting it (like sealing air in a pressurized tank or just holding the pendulum at a height). But in an inductor it requires current constantly flowing through it to store energy in its magnetic field otherwise the field collapses (just like a pendulum requires constant motion to store kinetic energy). You can't store energy in an inductor or kinetic energy statically. You can only store it dynamically.

So in both a pendulum energy is being exchanged between kinetic and potential. The pendulum exchanges speed for height back and forth. In the RLC circuit, the capacitor is exchanging energy with the inductor. With overly low resistance (parallel RLC), high resistance (series RLC), or high friction it takes a very long time for energy to completely transfer from one medium to the other, but it overshoots very little once the transfer is complete because so much is dissipated during the transfer with only one transfer occurring at the extreme but taking a very long time to do so. With very high resistance (parallel RLC), low resistance (series RLC), or low friction, a single transfer completes very quickly but so little energy is lost that much of it remains to be transferred back in the other direction so many, many transfers can occur.

• This is a fantastic answer! Jan 15 '20 at 22:23

The corrected statement is that the slightly underdamped case has the shortest settling time.

Now why is this the case when it's obvious that a highly underdamped system has a very long settling time?

The answer is dependent on the definition used for settling time : this is generally taken to mean the time to settle within a given percentage of the steady state value.

To settle precisely to the steady state value, a critically damped system gives the fastest response. An overdamped system will take longer to get there; an underdamped system will get there sooner but overshoot, and take longer overall to reach steady state. (Highly underdamped : it will pass through the steady state point many times)

But to settle within 1% (or 0.1% or 10%) of the steady state value, you can improve on the critically damped settling time by underdamping to such an extent that the first (and worst) overshoot is just within your permitted tolerance.

It still takes longer to reach the actual steady state value; but it is within tolerance faster than a critically damped system. So there is no real contradiction here.

• Well, also, in the idealized equations the time to reach exactly the steady state doesn't exist; at any time there's a tiny tiny bit of residual oscillation (underdamped) or offset (critically/overdamped). So we have to choose a point to "stop caring". Jan 17 '20 at 5:18

Generally, for an underdamped 2nd order system the 2% settling time is approximately $$\\frac{4}{\zeta\:\omega_n}\$$. But relating this ROT to, say, a RLC circuit is not intuitive since changing one component value will affect both $$\\zeta\$$ and $$\\omega_n\$$.

There are many ways to make a transmission line or RLC filter while the impedance $$\Zo=\sqrt { \dfrac{L}{C}} \$$

The optimum settling time is when reflections are nulled at the dominant spectrum such as that of a step pulse , when the load is matched to the characteristic value, Zo. ( related to Maximum power transfer theorem)

For higher than 2nd order systems Q<1 = 1/ζ Butterworth critical non-zero overshoot must use individual values of Q<1 and Bessel even lower or more damped, but in both cases fastest settling time occurs when R=Zo.

Here my criteria for Settling time was when the cursor showed the energy settling from 499.999 mW to 500.000 mW or 2 parts per million.

Below shows no damping with R removed and poles placed to give linear phase shift (Bessel) , smooth rolloff in Red, and fast settling time when matched R is attached.