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I’ve been using low pass RC filters in the design of the Ladybug Shield. Up until now, I haven’t had an intuitive feel for what is going on. Rather I relied on the terminology and topology. A low pass filter lets waveforms of low frequency through but blocks high frequency waveforms. It consists of a resistor and capacitor in series. Perhaps that is all I should know. Then I started taking the edX course from Berkeley – EE40LX Electronic Interfaces. They did a great job visualizing a low pass filter. So great it got me exploring visualizing a bit more.

# The Goal

The goal of this post is to solidify why a resistor and capacitor in series acts as a low pass filter

# Thanks to Those That Went Before

- Chris Gammell for covering these topics within his excellent Contextual Electronics courses. I am thoroughly enjoying the new courses going on now!
- edX for the EE40LX course.
- Paul and his excellent DVD downstream tutorials on circuit analysis.

# How a Low Pass Filter works

When I learn something in electronics, I like to absorb:

- The rule of thumb. I stated this above for the low pass filter. A low pass filter (RC in series) lets waveforms of low frequency through but blocks high frequency waveforms. It consists of a resistor and capacitor in series. If all I ever wanted to do was to sketch together a circuit, I’d stop here…but I don’t really understand why this is true.
- A take-it-slow sketchplanation of the math and physics.
- A simulation or prototype that lets me visualize, explore, think about the concept.

## Take it Slow Sketchplanation

This image from an edX video

- has a clear way of explaining why the low pass filter works. Since the current through a capacitor is proportional to the change in voltage over time (i = C(dV/dT) ), as the frequency (w in the image) of a waveform goes to 0 – which is the case for a DC circuit – the change in voltage goes to 0, so i = 0. If there is no current, there is an open circuit. When the frequency gets to be very high, the change in voltage/time gets higher, i keeps growing and eventually the current rushes to ground – a short circuit.
- assumes you have prior knowledge of the math/physics behind capacitors. Stuff like the formula for v(t) and i = C(dv/dt). The edX course covers this. I found the DVD downstream of these tutorials on circuit analysis to help me better understand. There are great books – like the Art of Electronics as well as “Google is your friend.”

So the key to understanding the RC filter from an intuitive math/physics perspective is getting a firm grasp that the current through a capacitor is proportional to the slope of the change in voltage/change in time.

### Low Frequency, Approaching a DC Circuit

As the change in voltage approaches 0 (basically DC), the current through the capacitor approaches 0, the capacitor’s symbol is in full force – an open circuit. Since there is no load in an open circuit and voltage always has potential, the voltage drop across the capacitor = V(t).

### High Frequency

The higher the waveform’s frequency, the stronger the current, to the point where the current flows through like a short circuit. In the case of a short circuit, there is no resistance and so no voltage drop across the capacitor. So high frequency waveforms get filtered out.

## Visualize Through Simulation

I just switched over to the Mac version of LTSpice IV. Once I bumbled through the differences in UI, I prefer the Mac UI to the Windows running under Parallels UI since it doesn’t look like a bitmap image and actions like zoom in/out use the Mac “way” instead of the Windows “way”.

Here is an image of the simulation I ran:

I’m asking the simulation to make two runs. The first run has a 10Hz Sine wave with an amplitude of 1V. The second run uses a 1KHz Sine wave for input. Here are the results. The blue lines are input, black output:

When the frequency is 10Hz, the Vout is close to Vin. When the frequency of Vin is increased to 1K, Vout is close to 0.

Ah yes. The simulation was very helpful in grasping the basic concepts behind a low pass filter. Because i = C(dv/dt), when there is little change in voltage, there is little current. The less current, the more Vout = Vin. The reverse of course is also true. The more change in voltage, the higher the current. The higher the current, the more Vout = short circuit = 0.

You probably already had a firm grasp. I am still gasping for air in the deep end of the electronics pool. These sort of step backs and HUHs? Help me better understand core electronic concepts.

Thanks for reading this far. Please find many things to smile about.