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I need to design a low-pass RC filter that is connected to an oscilloscope at the output. The oscilloscope has a load impedance of 1MΩ||20pF. I have to choose values for R and C, taking this load impedance into account.

Originally I chose R = 1kΩ based off another post here (which I can't find now). Then, using the formula:$$\text{f}_c =\frac{\text{1}}{\text{2}\pi\text{RC}}$$ for my desired cutoff frequency of 10kHz, this gave me C = 15.9nF

Then I found this website which calculates the RC values for you if you enter the desired cutoff. It gave me R=160Ω and C=0.1uF

So here's the problem: using my values gives a real cutoff of 10,010Hz, and using the website's values gives 9947Hz. Clearly my values get closer to the chosen cutoff, but I haven't taken the load impedance into account because I don't know how it would affect the values. In a simulator, both circuits behave almost exactly the same, but I'm concerned that the simulator's oscilloscope doesn't have a 1MΩ impedance.

Can somebody explain how the load impedance would affect the choice of values, and which values I should choose? My set, the website's set, or a completely different set?

The circuits and their frequency responses are below, so you can see how they're basically the same.

enter image description here

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    \$\begingroup\$ The site you linked to makes its suggestion based on available parts. Yours is mathematically correct, but you will not be able to build it as calculated - you cannot buy a 15.9 nF capacitor. \$\endgroup\$
    – JRE
    Commented Mar 26, 2020 at 17:00
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    \$\begingroup\$ Given the relatively flat "cutoff" of a simple RC filter, it doesn't matter that much if the cutoff is exactly where you want it. \$\endgroup\$
    – JRE
    Commented Mar 26, 2020 at 17:02
  • \$\begingroup\$ @JRE thanks, that makes sense about the available parts and actually helps with another part of this. Are you able to help with the load impedance part of my question? I'm not sure whether the 1MΩ load actually affects the values or not if we're talking real-world building and not the simulator. The question I'm doing asks me to take the load impedance into account when choosing R and C, but I don't know how. \$\endgroup\$
    – MendelumS
    Commented Mar 26, 2020 at 17:16
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    \$\begingroup\$ The problem is the large R. Imagine you use 1M as the R in your filter. The R and the input impedance form a voltage divider to reduce the signal in general. With a large R you also use a small C. The C for the filter becomes so small that the capacitance of the scope input is as large as the filter C, which will change the cutoff. \$\endgroup\$
    – JRE
    Commented Mar 26, 2020 at 17:42
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    \$\begingroup\$ Exactly. I just don't know of a way or rule to figure the parts. \$\endgroup\$
    – JRE
    Commented Mar 26, 2020 at 19:20

1 Answer 1

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Well, we have the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

The transfer function is given by:

$$\underline{\mathcal{H}}\left(\text{j}\omega\right)=\frac{\frac{1}{\text{j}\omega\text{C}}\text{||}\text{R}_\text{L}\text{||}\frac{1}{\text{j}\omega\text{C}_\text{L}}}{\text{R}+\left(\frac{1}{\text{j}\omega\text{C}}\text{||}\text{R}_\text{L}\text{||}\frac{1}{\text{j}\omega\text{C}_\text{L}}\right)}\tag1$$

Now, we find that:

$$\left|\underline{\mathcal{H}}\left(\text{j}\omega\right)\right|=\frac{\text{R}_\text{L}}{\sqrt{\left(\text{R}+\text{R}_\text{L}\right)^2+\left(\omega\text{RR}_\text{L}\left(\text{C}+\text{C}_\text{L}\right)\right)^2}}\tag2$$

When \$\omega\to0\$ we get:

$$\lim_{\omega\to0}\left|\underline{\mathcal{H}}\left(\text{j}\omega\right)\right|=\frac{\text{R}_\text{L}}{\text{R}+\text{R}_\text{L}}\tag3$$

Now, we need to solve (in order to find the cutoff frequency):

$$\left|\underline{\mathcal{H}}\left(\text{j}\omega\right)\right|=\frac{1}{\sqrt{2}}\cdot\frac{\text{R}_\text{L}}{\text{R}+\text{R}_\text{L}}\space\Longleftrightarrow\space\omega=\frac{1}{\text{R}_\text{L}}\cdot\frac{\text{R}+\text{R}_\text{L}}{\text{CR}+\text{R}\text{C}_\text{L}}\tag4$$

Using your values we need to have:

$$2\pi\cdot10000=\frac{1}{10^6}\cdot\frac{\text{R}+10^6}{\text{R}\cdot\text{C}+\text{R}\cdot20\cdot10^{-12}}\tag5$$

Choosing a value of \$\text{R}=1\space\text{k}\Omega\$ we find for \$\text{C}\$:

$$\text{C}=\frac{5005 - 2 \pi}{100000000000\pi}\approx15.9114\space\text{nF}\tag6$$

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