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schematic

simulate this circuit – Schematic created using CircuitLab

What is the smallest value of C that will ensure that the 3dB frequency is not greater than 100Hz?

That is the question from the book but my question is not that exactly.

I am trying to find the cutoff frequency which is of course a step toward the solution and I have the transfer function:

$$\frac{sC_1R_2}{sC_1(R_1+R_2)+1}$$

now somehow the book goes straight from the transfer function to

$$\omega_0 = \frac{1}{C_1(R_1+R_2)}$$

I am having trouble understanding how we found \$\omega_0\$.

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  • \$\begingroup\$ Can you please point out the Vout "node" you used to calculate the transfer function (assuming your transfer function is Vout/V2)? Is it over R2? \$\endgroup\$ – Vicente Cunha Feb 12 '16 at 17:52
  • \$\begingroup\$ Yes sorry. Its been added \$\endgroup\$ – user125621 Feb 12 '16 at 17:55
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The rolloff frequency of a R-C filter is:

  F = 1 / (2πRC)

When R is in Ohms, C in Farads, then F is in Hz.

Therefore to get the capacitance:

  C = 1 / (2πFC) = 1 / [2π(100 Hz)(10 kΩ)] = 160 nF

Yes, it's really that easy.

R-C rolloff filter calculations come up a lot in electronics. I keep 1/2π permanently in a register in my calculator. I can then divide that by two of the frequency, capacitance, or resistance to get the third.

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From the transfer function, I ended at a rather interesting answer:

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

$$abs(H) = \frac{\omega C_1 R_1}{\sqrt{\left(\omega C_1 (R_1 + R_2)\right)^2 + 1}}$$

Finding cutoff frequency (step by step, so we don't trip):

$$\frac{\omega_0 C_1 R_1}{\sqrt{\left(\omega_0 C_1 (R_1 + R_2)\right)^2 + 1}} = \frac{1}{\sqrt{2}}$$

$$\frac{(\omega_0 C_1 R_1)^2}{\left(\omega_0 C_1 (R_1 + R_2)\right)^2 + 1} = \frac{1}{2}$$

$$(\omega_0 C_1 R_1)^2 = \frac{\left(\omega_0 C_1 (R_1 + R_2)\right)^2 + 1}{2}$$

$$(\omega_0 C_1 R_1)^2 - \frac{\left(\omega_0 C_1 (R_1 + R_2)\right)^2}{2} = \frac{1}{2}$$

$$(\omega_0 C_1)^2 \left(R_1^2-\frac{\left(R_1 + R_2\right)^2}{2}\right) = \frac{1}{2}$$

$$\omega_0 C_1 = \frac{1}{\sqrt{2\left(R_1^2-\frac{\left(R_1 + R_2\right)^2}{2}\right)}}$$

$$\omega_0 C_1 = \frac{1}{\sqrt{2\left(R_1^2-\frac{R_1^2 + 2R_1R_2 +R_2^2}{2}\right)}}$$

$$\omega_0 = \frac{1}{C_1\sqrt{R_1^2-2R_1R_2-R_2^2}}$$

I did not reach your book's very-well known formula, yet seemingly incorrect for this application. I could only bet the one who wrote the solution just botched it. Or I did. Eitherway, there's my two cents.

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    \$\begingroup\$ Your solution is almost on the right track; the filter has some built-in attenuation (in this case -20dB), so the right hand side should be modified to be \$\frac{R_1}{(R_1+R_2)\sqrt{2}}\$. \$\endgroup\$ – helloworld922 Feb 12 '16 at 19:56
  • \$\begingroup\$ Using a CAS program with this slight modification I do indeed get the book's solution, which also matches a Bode plot solution for \$V_o\$. \$\endgroup\$ – helloworld922 Feb 12 '16 at 19:59
  • \$\begingroup\$ @helloworld922 Can you possibly elaborate on how you came up with the modified part? \$\endgroup\$ – user125621 Feb 13 '16 at 4:57
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For this type of a circuit the cutoff frequency is not at Vin/Vin = 0.707. Because the R1 and R2 form a voltage divider, and "mid-band gain" (maximum of Vout) is equal to R2/(R1 + R2) = 0.9[V/V], therefore the cutoff frequency is at 0.9*0.707 = 0.636(the frequency at where the magnitude of Vout is 3dB less than the maximum of Vout).

And this is why right hand side of the equation should looks like this: $$\frac{R2}{R1+R2}*\frac{1}{\sqrt{2}}=\frac{R2}{(R1+R2)\sqrt{2}}$$

$$\frac{\omega_0 C_1 R_1}{\sqrt{\left(\omega_0 C_1 (R_1 + R_2)\right)^2 + 1}} = \frac{R2}{(R1+R2)\sqrt{2}}$$.

But there is a simpler method for finding the cutoff frequency. 1 - Finding the pole directly from transfer function

$$H(s)=\frac{sC_1R_2}{sC_1(R_1+R_2)+1}$$

And for this type of a circuit we can do it by inspection.

2 - We can find a time constant of the circuit. In this case the time constant is \$\ t=C1*(R1+R2)\$ and \$\ \Large Fc =\frac{1}{2*\pi * t} \$

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