# RLC Circuits and Resonance

You've made it to the final topic of the beginner's course, well done!

Let's bring this course to closure by discussing the very interesting RLC circuits which occur **EVERYWHERE **in practical designs on circuit boards and the concept of resonance which will serve you incredibly well out in the real world!

## RLC Circuits

Recall that an **RLC Circuit** simply consists of a **power source**, **resistor**, **inductor **and **capacitor **as the name suggests. In this case we will be looking at circuits which have an AC voltage source with an example circuit shown below:

Let's now plot a spectrum with the following settings:

So there's some interesting activity from around 20kHz to 30kHz, let's update the range to zoom in:

So what we see is **minimum attenuation** at around **22.85kHz** of **-6.319dB** which then decreases to the left and right in frequency. What is happening here? Why do we see a curve like this? The answer lies in the interplay between the **capacitance **and **inductance**, i.e. the **reactive **components.

Recall that for **low frequencies** the **capacitor **acts as an **open circuit** and at **high frequencies** a **short circuit** and for an inductor acts in reverse. This means that at low frequencies the capacitor is the **higher impedance** and at high frequencies the inductor is **higher**. There is a frequency however, where the **reactance **of the **capacitor equals **the **impedance **of the inductor and this is known as resonance, as at this point the circuit behaves as just a simple resistive circuit with resistance R in the above circuit.

This frequency is where we observe the **maximum amplitude** shown in the curve above.

How can we determine this frequency? By setting Xc=Xl:

## Series RLC Circuits

Shown below is a series RLC circuit.

Let's display a frequency spectrum with the voltage taken across the capacitor with the following settings:

It looks remarkably similar to the frequency response of a low pass filter for an RC or RL circuit! Zooming in on the chart it appears the -3dB point occurs around 4Hz. Can we determine this by hand?

Let's begin by representing the circuit in the **Laplace Domain**:

We can then derive the Transfer Function by using the Voltage Divider Rule, recall it works for any number of components connected in series :)

Let's now convert back into the **Phasor Domain** to solve for poles in terms of frequency (the reason for this is that if we solved the denominator in terms of **s **we would be assuming the **solution is real valued** which may not be the case for **complex numbers**):

From here we can equate the denominator to zero and solve for the poles to find:

So we expect to see the -3dB point around 4Hz and then a **-20dB/decade rolloff** followed by another pole at 79.573kHz followed by a **-40dB/decade rolloff**, let's zoom in around these regions to confirm.

Yes, we have **-3dB** at **3.969Hz**!

If we compare the amplitude at **1MHz **vs **10MHz **we can check the **rolloff**. We have approx **-130dB** at **1MHz **and **-170dB** at **10MHz **which is **40dB lower** :)

## Parallel RLC Circuits

Let's now consider what is known as a parallel RLC circuit but with a slight twist to make it realistic!!!

The twist is the addition of **Rs**, which represents the **internal resistance** of the voltage source to **model a practical voltage source**. Let's then take a look at a frequency spectrum for this circuit probing at the node connecting Rs, R2, C1 and L1:

We can see that for **low frequencies** we have a **very low gain **which slowly increases with frequency until around **100kHz **which then starts to drop off again.

At **low frequencies **the impedance of the **inductor** will be very low and so the voltage source see's an effective impedance of the 1 ohm in series with almost a short circuit meaning the vast majority of the voltage is dropped across the 1 ohm resistor which explains the low gain observed.

At **high **frequencies the **capacitor **acts as a short. Around **100kHz **we see resonance and so we effectively have 1ohm in series with 1kohm and so the majority of the voltage drops across the 1k resistor which reflects the maximum gain across the frequency response.

Let's perform a similar exercise to the series RLC circuit to determine the **Transfer Function** and then calculate any **poles **and **zeroes **to characterize this type of circuit. As always, let's begin by redrawing it in the **Laplace Domain**:

From here we can determine the equivalent parallel impedance of R2, C1 and L1 and then apply the voltage divider rule to determine the transfer function:

Let's now use the Voltage Divider Rule:

Let's now convert back into the **Phasor Domain** to solve in terms of frequency:

We can see from inspection that we have a **Zero **when **ω=0**. This means we will have a **rise **of **20dB/decade** from **0Hz**.

From here we can equate the denominator to zero and solve for the poles to find:

We can also equate the **resonant frequency** by reusing our formula: