How High Pass Filters Work using a Resistor and Capacitor (RC)

I explain how high-pass RC filters work using the falstad circuit simulator and a circuit which is similar to the one used with the low pass filters in a previous video.
The simulations include the circuit diagram showing current flow, capacitor operation, time-domain voltage plots at various frequencies, and a frequency response plot.
Below is the text of the code I used in Falstad:
$ 1 0.000005 14.235633750745258 55 5 50 5e-11
v 176 256 176 80 0 2 85 5 0 0 0.5
r 176 80 336 80 0 187
c 336 80 336 256 0 0.000009999999999999999 -4.2244607677073445 0.001
w 176 256 336 256 0
o 0 16 0 x81016 20 0.2 0 14 2.5 0 1 3 0.05 0 2 3 0.05 0 1 3 0.05 0 2 3 0.05 0 2 3 0.05 0 1 3 0.05 0 2 3 0.05 0 1 3 0.05 0 2 3 0.05 0 1 0 2.5 0 1 3 0.05 0 2 0 2.5 0 2 3 0.05 0
38 2 F1 0 0.000001 0.000101 -1 Capacitance

[@Fundalf_the_greyt]

Great explanation, thanks! I didn't realize that high pass and low pass is basically the same circuit, just measuring voltage on a different component

[@JohnnyMoniz]

Quick question: First of all, fantastic video! I’m curious if a capacitor alone would work as a high pass filter in an audio experiment. My understanding is that capacitors naturally want to filter out low frequency. If I wanted my guitar effect pedal to be quieter at 200hz and below, why is it that I can’t simply put a capacitor of a specific value in series right at the output? (Or can I?) or what is it about the relationship with the resistor that makes both parts mandatory?

[@matthewjohnson2554]

Thank you. And good question. I would encourage you to simulate this to see for yourself. Go to falstad.com/circuit, circuits tab, passive filters, and High Pass RC filters. If you run the simulation as is, you can see the output voltage passes well at the higher frequencies, but not as well at the lower frequencies. Now see what happens if you remove the resistor. Just click on it and hit delete, then draw a wire where the resistors was (draw tab, add wire). You will see if you remove the resistor, the output voltage will be directly connected to the ground, so it will always just be 0, not very exciting. If you remove the wire connecting to ground, then you will see there will be no current flowing through the capacitor, and the output voltage will be the exact same as the input voltage, no matter what the frequency is. Also, not very exciting. So it turns out you need the resistor there to both allow current to flow while also preventing your output voltage from being connected directly to the ground. You can also change the value of the resistor and see how this changes things. At high resistance, say 1k, v_out just matches the input voltage no matter the frequency, like when there is no connection to ground. At low resistance (1 ohm) it acts more like the short circuit where there is barely any output voltage, although it's a little higher at the higher frequencies.

[@ThevonMusic]

Nice tutorial overall, but there's two things I can't figure out. I understand the order of the resistor and capacitor changed gives us a lowpass filter. I can also understand how depending on the load time of the capacitor certain frequencies are cutoff more than others. But... What I don't understand is the flow of the signal. If I look at this square wave generator, I see the pulse going from the + side one half cycle and then from the minus side one half cycle. In my mind It's like the + part it goes through the resistor first, which then passes our output point and then fills up the capacitor. On the minus side the other way around, first capacitor. I know this doesn't make sense, because the result would be one half cycle lowpass one half cycle highpass if that makes sense. I just don't really understand the flow of the signal and also when the capacitor discharges. Does it discharge to our output signal? Is this also causing our delay/phase shift? Second thing I don't understand is the resistor switch to double peak (10V).

[@matthewjohnson2554]

Hi Thevon, I see you are putting some serious though into this, and your question is challenging. I think what may be tripping you up is the idea that the order of resistor and capacitor (e.g., RC vs CR) gives you a high or low pass filter. It turns out the order doesn't matter. What matters is whether you are taking your output voltage across the capacitor (which would make it low pass) or across the resistor (which would make it high pass).

[@matthewjohnson2554]

As for the voltage across the resistor changing from 0 to -10 V, I attempt to explain this briefly around the the @3:58 mark, but basically this 10 V swing is the immediate response to the 10 V swing in the voltage input. With sufficient time at +5 V input, the capacitor acts like an open circuit, there is no current across the resistor, and thus no voltage across it, 0 V. When the voltage source changes from +5 to -5, now the left side of the resistor is at -5, and the right side of the resistor is at 5 V. The right side is at 5 V because that's what the capacitor had charged up to. So the difference between that -5 and 5 V gives you the initial -10 V across the resistor, which dissipates back down to 0 once the capacitor is recharged in the reverse polarity this time.

[@ThevonMusic]

@@matthewjohnson2554 Thanks. I was thinking way too much in DC anyway. I asked my teacher as well and basically it's all about reactance to a certain frequency that changes the voltage drop over R and C, which makes sense when switching them gives a lowpass or highpass filter. I understand the 10V as well with your explanation, thanks for taking the time to reply!

[@matthewjohnson2554]

@@ThevonMusic Sometimes a filter can be referred to as RC and CR, which would be low and high pass, respectively, if the output voltage is taken over the second element. I imagine that may have a been source of your initial misunderstanding.

[@user-eg1zw3ky4s]

hello sir , first thank you very much for you're amazing explanation but i have a question. shouldn't the capacitor by connected directedly to the ac source ? and the resistor in parallel with the output?

[@mejiqal]

the resisitor is just constant suck in electrons and output them as heat.....

[@kkgt6591]

Great video, audio is a bit clanky and needs some fixing to be clear.

[@matthewjohnson2554]

Thanks, I recorded this one in a bedroom instead of a studio, so good ear.

[@aungkyawmoe8023]

isn't this low pass filter?