The reason underdamped LRC circuits oscillate is because the energy keeps flowing between the inductor and capacitor. The energy is being constantly exchanged between the capacitor and inductor resulting in the oscillations - the fact that energy is being lost to heat explains the asymptote and why the amplitude of the oscillations keeps decreasing.

I'm having trouble understanding why this doesn't happen for over damped and critically damped circuits though. Sure, I can set up the differential equation and solve for the function. But that's not the part I'm having trouble with - I'm having trouble intuitively understanding what's different about over damped and critically damped circuits.

So can someone please give me an intuitive explanation of why critically damped and over damped circuits don't oscillate? Why isn't the energy being exchanged back and forth between the capacitor and inductor?

  • 3
    \$\begingroup\$ Imagine a child's swing- it goes back and forth with decreasing oscillations, even if you drag your feet a bit. Drag your feet enough and you come to a stop without going back and forth at all. \$\endgroup\$ Oct 26, 2014 at 21:12

5 Answers 5


Electrical circuits and micro-mechanical systems have the same differential equations if you write them out and it's much easier to picture mechanical systems. See here


A resistor is like friction from a damper.

A capacitor is like a spring.

An inductor is like mass of an object.


Zeta is the damping ratio and a function of your system, \$\zeta=\frac {B}{B_c}\$; Where \$B\$ is the actual damping in your system and \${B_c}\$ is the critical damping value that if existed your system will reach the target value without oscillations. By knowing this one number zeta we will know how the system will respond. You may need to read more on derivation here. But for now just know that Zeta <1 means an underdamped system Zeta = 1 is critically damped and Zeta >1 is overdamped.

Analogy 1

Imagine a weight and a spring like so:

enter image description here

In a perfect friction less world if I pull the mass to the right and then release it (like a step input to a circuit), the the weight will continually oscillate back and forth forever. A circuit with a step input and only a capacitor and inductor will also oscillate forever to a step input. This is like ζ = 0

Analogy 2

Now let's take the same system and add a resistor to the circuit and a friction damper to the mechanical system:

enter image description here

You can imagine if the friction is very low ζ = 0.1 and 0.2 the system will still oscillate for a while before everything comes to a stop. As we keep increasing R in the electrical domain or friction in the mechanical domain the oscillations will decrease until we get to ζ = 1. This is when the system is critically damped. It will return to it's original neutral state as quickly as possible without oscillations. I will pull the weight to the right let it go and because of the high friction it will slowly roll back to it's original position. If we keep increasing Resistance and increase ζ > 1 it will take the system longer and longer to get back to it's original state. The weight will roll back but even slower than in the ζ = 1 state.

All the Systems

Now let's plot all this behavior on a graph. We can see that the black graph is like our analogy 1, with only a capacitor and inductor. The green, blue, red and orange graphs shows what happens as we increase resistance of the circuit. Until we hit the purple graph which is critically damped. The dark blue show what happens as we keep increasing the resistance.

enter image description here


So in closing the reason that a critically and overdamped circuit doesn't oscillate is that the resistance (or friction) is too high to allow an energy exchange between energy storage elements like capacitors and inductors. All the energy gets dissipated in the resistance of the circuit before it can cause any oscillations - like a damper that doesn't let a spring bounce back and forth.

  • \$\begingroup\$ How about this ;) \$\endgroup\$
    – EM Fields
    Oct 26, 2014 at 23:08
  • \$\begingroup\$ The same principles apply. It's the same as analogy 2 but with a stiffer spring which will equal 2k and a more frictioned dashpot equal to 2c. This assumes bidirectional springs, when in reality springs are usually more linear in either compression or extension depending on construction. \$\endgroup\$
    – EasyOhm
    Oct 26, 2014 at 23:28
  • \$\begingroup\$ You missed the point, which was that oscillation is much more clearly visualized when it repeatedly crosses zero. :-) \$\endgroup\$
    – EM Fields
    Oct 26, 2014 at 23:50
  • \$\begingroup\$ Ah. The engineer in me is once again solving the technical side of the problem.... \$\endgroup\$
    – EasyOhm
    Oct 27, 2014 at 4:04

Imagine we are trying to design a "Servo-system". Its a system that takes an input angle and rotates a motor until its pointing at the input angle. The block diagram of this system could be represented like this

enter image description here

The proportional controller outputs tourque based of the difference between the current angle and the desired angle. We will test this system againest step inputs

You might be asking now "What does this has to do with my circuit". The answer would be "They are both second order systems" they both are charactarized by a second order time domain differential equation.

Now defining the variables

\$r\$ : The input reference angle

\$e\$ : The difference between the current motor angle and the required angle

\$T\$ : The proportional controller output that is the motor tourque

\$J\$ : The rotational intertia of the rotating body

\$B\$ : The amount of damping we provide to the system

Imagine we are not providing any kind of damping in our system, once the motor is rotating it will never stop rotating since there is no friction [assuming no frictions at all]

Here is a simulation for the case where this system has no damping value at all

enter image description here

Here is the expected time domain response for this system.

enter image description here

Eventhough the system reached the target value at \$t=T_r\$ and at this point the proportional controller unit output tourque is zero the motor is still rotating at a constant speed since there is no friction to decrease the motor rotational speed and rotational kinetic energy and once the motor is pointing to different angle [more than the desired angle] the proportional controller will drive the motor in the oposite direction and you will have this endless oscillation [NO FRICTION EXISTS]

\$B=0\$ is not a choice then, the system will never be stable if no friction is provided, so how much B should we provide to the system, and how would the system respond to different amounts of damping values.

It turns out that there is a term that charactarizes the system based on the damping value B which is the system damping ratio \$\zeta=\frac {B}{B_c}\$

Ratio between the system current damping value and the crictical damping value \$B_c\$

We can define \$B_c\$ as the value of the damping that is enough to stop the rotating body at the target value; that is once the motor is pointing to the required angle and the proportional controller output tourque is zero, the rotating body has no energy to rotate anymore.

enter image description here

So if \$\zeta ≥ 1\$ this means you are providing a damping value which is greater than or equal the crictical damping value [NO WAY YOUR SYSTEM WILL OSCILLATE]

The more you increase \$\zeta\$ the more it takes for your system to reach the target value

This is the simulation for our system when \$\zeta=1\$ where the system is circtically damped

enter image description here

And as we increase the damping ratio the time taken to reach the target value is incearsed

enter image description here

If \$\zeta < 1\$ this means that you are still providing damping to your system but this damping value is not high enough to prevent the rotating body from exceding the target value, this is the expected time response for an under damped system

enter image description here

At time \$t=T_r\$ you can see that the rotating body keeps rotating as the case where B=0; But in this case as the rotating body is rotating its energy decreases until it reached a point where its oscillating close to the target value

enter image description here

So its a matter of how much energy is stored and how much damping exists that is sufficent to eat up this amount of stored energy before exceding the target value. The more damping you provide the slower the system approaches the target value. The less damping you provide the more your system oscillates around the target value while its stored energy is eaten up by the damper untill it dies out

If you think this answer is confusing or provides false information or even didnt answer the part you are asking about please let me know in the comments i`ll definetly delete it.


It's trying to, but it's being radiated away as heat by resistance in the circuit.

  • \$\begingroup\$ Shouldn't that just result in there being oscillations but the amplitude of the oscillations decreases faster? \$\endgroup\$
    – dfg
    Oct 26, 2014 at 21:04
  • 1
    \$\begingroup\$ No, decaying oscillations still means underdamped (Q smaller but still > 1). Critical damping is defined as the level of damping where oscillations stop altogether. \$\endgroup\$ Oct 26, 2014 at 21:17

The dampening ratio is essentially the ratio of the time constants of the oscillations and the inverse exponential decay of energy. When the time constant of the oscillation is sufficiently larger than that of the decay, the energy that would oscillate is dissipated before the oscillation can "come around".

In the case of \$\zeta=\frac{1}{\sqrt{2}}\approx0.7\$, \$2\pi\$ time constants of decay are gone by the time the oscillation would make one full cycle. This is because when \$\zeta=\frac{1}{\sqrt{2}}\$, \$\sigma\$ and \$\omega\$ are exactly the same. However, \$\sigma = \frac{1}{s}\$, while \$\omega = \frac{rad}{s}\$. As a general guideline, 5 time constants dissipate 99% of the energy in the system. \$2\pi\$ is around \$6.28\$.


So can someone please give me an intuitive explanation of why critically damped and over damped circuits don't oscillate?

Imagine a child on a swing, which is an oscillating system. (Indeed, it actually has very similar math). An adult is pushing the kid higher and higher as the kid swings back and forth.

This adult would be the person applying energy to the oscillating system.

Now "dampening" would be the thickness of the air, or the fluid in which everyone is standing in. Imagine if the swing were in water... or worse... imagine if the swing and the child were in Honey or even Jello.

If everyone is in a vat of extremely thick liquid, eventually it becomes impossible to swing back-and-forth. Because the "thickness" or viscosity of liquid just prevents the swinging action all together.

Now we can imagine that there's a "critical" liquid, one where if you were just SLIGHTLY less viscous, swinging would occur. This "critical" hypothetical liquid would be the thinnest possible liquid to prevent swinging. Any thicker would be wasteful, any thinner would still have swinging occurring.

If you actually did all the math for a child and a swing, it would 100% match the dampening / oscillation math believe it or not. Air-resistance (or water-resistance, in the case of a "thicker" example) is your parameter for dampening.


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