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I have a low pass filter like this:

schematic

simulate this circuit – Schematic created using CircuitLab

\$V_{\text{out}}\$ is measured right after \$R_1\$, which I suppose means that it is measured over the parallel part.

\$R_2\$ is the load of the filter. When this circuit is measured with an oscilloscope it seems like it is not dependent on the frequency at all. I would like to investigate why.

I tried to calculate the transfer function for the filter, but I am not sure that it is right.

$$ H(j\omega) = \frac{1}{R_1\left(j\omega C+\frac{1}{R_2}\right)+1}$$

I'm using \$R_1 = 33\text{k}\Omega\$, \$R_2 = 1\text{k}\Omega\$, and \$C = 220\text{pF}\$.

If I plot the frequency response in Matlab with this, I just get a straight line going from the origin through (1,.5), (2,1) where (Hz, H(w)) and so on.

Is this correct?

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  • \$\begingroup\$ Yes, a schematic would be helpful. Also, you said you measured this circuit. What values of R1, R2, and C1 did you use? And what frequencies did you use for Vin? \$\endgroup\$ – Eric Nov 25 '14 at 17:01
  • \$\begingroup\$ Welcome to EE community! I allowed myself to edit your question and draw a schematic for you. Please check if this is what you mean. \$\endgroup\$ – Kamil Nov 25 '14 at 17:09
  • \$\begingroup\$ Thanks! The circuit looks right! I used a sinus frequency of between 1kHz up to 100kHz as input. My components has the values R1=33kOhm, R2=1kOhm, C1=220pF \$\endgroup\$ – theva Nov 25 '14 at 17:11
  • \$\begingroup\$ I added "box" after I read what is filter and what is load. You can edit it and enter component values if you want. \$\endgroup\$ – Kamil Nov 25 '14 at 17:16
  • \$\begingroup\$ Try changing your load to 1 Meg ohm. (The 1 k is loading the entire circuit down. Without the capacitor R1 and R2 make a voltage divider that is cutting your input down by a factor of 33!) \$\endgroup\$ – George Herold Nov 25 '14 at 17:36
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You can interpret this circuit as a voltage divider using $$R_2 \parallel \frac{1}{j\omega C} = \frac{R_2}{j\omega R_2 C + 1} $$ and \$R_1\$. The transfer function is therefore

$$H(j\omega) = \frac{R_2 \parallel \frac{1}{j\omega C}}{R_2 \parallel \frac{1}{j\omega C} + R_1} = \frac{\frac{R_2}{j\omega R_2 C + 1}}{\frac{R_2}{j\omega R_2 C + 1} + R_1} = \frac{R_2}{R_2 + R_1(j\omega R_2 C + 1)}$$

If you divide numerator and denominator by \$R_2\$ this is the same expression you calculated, but I think it's easier to understand the filter using my result. As \$\omega \to 0\$ $$H(j\omega) = H(0) = \frac{R_2}{R_2 + R_1}$$ which is what you would expect for a simple voltage divider using \$R_1\$ and \$R_2\$. As \$\omega \to \infty\$ the denominator dominates and \$|H(j\omega)| \to 0\$. This is a low pass filter so the output should depend on the frequency (provided you sweep to a high enough frequency).

Here is your circuit in CircuitLab setup so that you can simulate it within CircuitLab:

schematic

simulate this circuit – Schematic created using CircuitLab

And here is the frequency sweep on the circuit as reported by CircuitLab (click to make it larger):

enter image description here

You can use this to verify your Matlab code. If you post your Matlab code we might also be able to help you find a problem with it.

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  • \$\begingroup\$ I did actually tried to run my code again, with a higher frequency span this time. It now looks like the graphs you posted so I think its correct! :) What does the || notation mean in your calculations? \$\endgroup\$ – theva Nov 26 '14 at 7:43
  • \$\begingroup\$ Great! The || symbol denotes the parallel combination of two impedances. \$\endgroup\$ – Null Nov 26 '14 at 13:35
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Your equation is correct but written a bit unusually.

You can re-arrange to get:

$$H(\omega) = \frac{1}{C_1 R_1} \times \frac{1}{j\omega+\frac{R_1+R_2}{C_1R_1 R_2}}$$

so your pole frequency (or cutoff of your lowpass filter) is $$f=\frac{\frac{R_1+R_2}{C_1 R_1 R_2}}{2\pi}$$

For your values that works out to be \$745\$kHz which is well above your test signal's highest frequency (\$100\$kHz). So you won't see any rolloff.

Here's what the mag/phase versus frequency looks like:

enter image description here

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  • \$\begingroup\$ akellyirl - please recalculate the given cut-off frequency. It should simply be wc=1/RpC (Rp=R1||R2) and fc=wc/2Pi \$\endgroup\$ – LvW Nov 25 '14 at 18:23
  • \$\begingroup\$ That's the same thing..... \$\endgroup\$ – akellyirl Nov 26 '14 at 9:26
  • \$\begingroup\$ I don`t think so: f=2*pi....? \$\endgroup\$ – LvW Nov 26 '14 at 13:15
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You can rearrange the circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

But using Thévenin equivalence, you have this:

schematic

simulate this circuit

Which is a standard RC circuit with $$ V'_{in} = {R_2\over{R_1+R_2}}V_{in} $$ and $$ R_p = R_1||R_2 = {R_1R_2\over{R_1+R_2}}$$

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