# Damping Ratio Implications for an Increasing Resistance In an RLC Circuit

I have managed to derive the 2nd order ODE for a simple RLC circuit (this circuit is part of a booster dc-dc converter). I derived an expression for the damping ratio of the circuit and fact checked this with derivations in literature. This is shown below:

Now from the expression above the damping ratio is inversely proportional to the load resistance given L and C are kept constant (which they are). However, this seems counter-intuitive to me. I would have thought for certain that as resistance is increased a higher damping ratio is expected. Likewise, from Ohm's law if resistance increases (given the initial and final transient voltage is kept constant) you would have a lower current which should result in a larger damping ratio.

Please could someone explain why physically increasing the resistance decreases the damping ratio?

Thank you.

• This is true for a parallel rlc
– user16222
Nov 16, 2021 at 18:52
• Sorry I am not sure I follow, could you please elaborate @JonRB Nov 16, 2021 at 19:08
• The equation that you pulled that expression from was talking about the effect of a resistor parallel to the coil, not a resistor in series. Nov 16, 2021 at 19:20
• Since damping ratio usually isn't the most important thing you're worried about in a boost converter, could you please edit your question with a schematic of the converter in question. Also, when you cite "the literature" always include a proper citation saying where it came from, and either a link if it's online or a short, relevant quote if it's not. Nov 16, 2021 at 19:21
• So a resistor in parallel means the resistance seen by the coil is actually decreased? Nov 16, 2021 at 19:22

There is at least two node connections for an RLC circuit

1. Series
2. Parallel

There are some weird combinations of series-parallel and parallel-series but sticking to the two types.

## Series

simulate this circuit – Schematic created using CircuitLab

This arrangement has a Quality factor of: $$\ Q = \frac{1}{R}\cdot\sqrt{\frac{L}{C}} \$$

and knowing that the damping factor is $$\ \zeta = \frac{1}{2Q} \$$ we therefore have a damping expression of:

$$\\zeta = \frac{R}{2}\cdot \sqrt{\frac{C}{L}} \$$

With increase resistance, the damping factor increases.

## Parallel

simulate this circuit

This arrangement has a Quality factor of: $$\ Q = R\cdot\sqrt{\frac{C}{L}} \$$

and thus $$\\zeta = \frac{1}{2R}\cdot \sqrt{\frac{L}{C}} \$$

With decrease resistance, the damping factor increases.

If you think about why this is the case, a low damping (ie high Q) implies that there is a lot of energy flowing between the two energy storage devices. For the series case, zero resistance would imply infinite current could flow, likewise infinite resistance would imply zero current would flow THUS: the higher the resistance the higher the damping.

Now consider the parallel case. The damping resistance is across the network. If this resistance was zero it would be shorting out the energy storage devices and thus no current would flow between them to resonate. Likewise if this resistance was infinite this current would cycle between the two energy storage devices. THUS: the lower the resistance, the higher the damping

• Right ok, that does make a lot of sense, whilst my circuit isn't with all parallel components (capacitor and resitor in parallel with the inductor in series), the same principles apply. Thank you very much Nov 16, 2021 at 19:53