Prelab: RLC Frequency Response
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In this week's lab, we'll be building an interesting circuit called a "buck converter," whose goal is to convert a high voltage into a low voltage as efficiently as possible.
To help prepare for the lab, in this exercise we will explore ways we can convert a high input voltage to a lower output voltage. These types of circuits, called "converters" allow us to connect together different systems which may require varying current and voltage capabilities.
Let’s consider an example, where we want to connect a 9Volt battery to power a 5Volt microcontroller (MCU). Throughout this semester, we have used resistive voltage dividers in similar applications, so let’s see what happens if we use a resistor divider here, like so:
(Reminder: power = voltage *
current (so, for a fixed resistor (using Ohm’s law), power = V^2/R)
R_{\rm MCU} (in Ohms) =
R_1 (in Ohms) =
Power dissipated through R_1 (in Watts) =
Total power (in Watts) =
Efficiency:
This means about half of the power coming from the battery is "lost" before reaching the MCU; what a waste!
While some low power applications may be able to tolerate this low efficiency, we typically want high efficiency, to fully utilize our source. In addition to the poor efficiency, we also know in our resistor divider circuit that if the power dissipated through the MCU changes (i.e. the R_{\rm MCU} changes) then the voltage across the MCU will also change. Lastly, as the voltage across the battery changes with state-of-charge the MCU voltage will also change.
So let’s investigate a different circuit topology which will allow us to more efficiently and accurately convert voltage.
One way to achieve this is with a circuit called a "buck converter" (which we'll build in lab this week). For now let’s consider a simplified version of a buck converter, where we have a square waveform as our source (we will talk about where this square waveform comes from later) connected to an inductor and capacitor at the output:
Our square waveform source has the following shape:
Let’s say we design L and C such that the frequency of V_{\rm SW} is well below the natural frequency of the LC circuit. In that case, similarly to the results we found analyzing PWM systems for RC and RL networks, the resulting output voltage (shown below) has the same average voltage as V_{\rm SW} but with some "ripple":
If we then modify our circuit such that the frequency of V_{\rm SW} is 10 times higher than the natural frequency of our LC circuit, then the average value of the output voltage will remain the same, but the amount of ripple will be dramatically reduces almost no ripple:
This means if we design our filter such that 1\over \sqrt{LC} is significantly lower than the frequency of our converter, then we can approximate the output voltage as the average of the square waveform.
The average value (or "DC component") can be found by taking the integral (or area under the curve) and dividing by the time period. This average is now our output voltage. In this case, the output voltage would be half the input voltage:
Now we want to find a way to vary our output voltage, so we can achieve the 5 V output we want for our MCU. In reality, our square waveform is being generated by two switches (or a switch and a diode) connected to the input voltage, as shown below. We won’t go into too much detail about how this circuit operates, if you want to learn more about this you should take 6.622 or 6.222. And either way, here is the circuit:
In a buck converter the switches (S_1) and (S_1') operate alternately, meaning when one is conducting the other is open. We call the percent of time S_1 is on during the period the "duty cycle", usually denoted with D. Above we assumed this switch was on for 50% of the time, but instead we can leave it on for a different portion of the period (anywhere from 0 to 1).
Now our V_{\rm SW} waveform will look like the following, where D can be anywhere between 0 and 1:
Now let’s take a look at the power coming from our source. Assuming we have a fixed 9 V battery, the current through the battery looks like this:
This is a pretty cool result! Of course, unfortunately, like any real circuit, buck converters are not actually lossless. In reality, we have a number of different losses to worry about (for example, the inductor has a parasitic resistance where we lose heat; and the switches are typically implemented with transistors that require power to turn on and off; etc). However, in the real world, it is often feasible to achieve >90% efficiency despite these losses with proper design and optimization, making this design much better for power applications compared to a resistor divider.
We can also view this filter from a frequency-domain perspective, reasoning about its behavior in sinusoidal steady-state as another way to reason about this input/output relationship. Let's go back to our earlier view of the circuit as a little RLC network being driven by a square wave source:
What is the frequency response of this circuit?
Enter your answer as a Python expression below. Use omega, j, R (or R_MCU), L, and C, as well as any other constants you might need.
In case it comes up, remember that exponentiation is **, not ^.
In lab, we'll reason about this circuit in both the time and frequency domain, and hopefully also see some of the benefits of using this design instead of a simpler design like a voltage divider.