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If I am correct to calculate cutoff frequency in this LP filter I have to sum up the impedance of resistor and capacitor, right?

schematic

simulate this circuit – Schematic created using CircuitLab

So: $$ Z_1=R_1\\ Z_2 = \frac{Z_{R2}*Z_{C1}}{Z_{R2}+Z_{C1}} $$

My result is about 80 kHz but correct answer is 230 kHz. Who is wrong? :)

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  • \$\begingroup\$ I think you are overlooking that R1/R2 form a voltage divider even at low frequencies so your reference level should be adjusted accordingly. You want to calculate the -3dB from that reference level \$\endgroup\$
    – alexan_e
    Commented Jan 23, 2014 at 10:22

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$$ Z_{bot} = \frac{R2 \cdot \frac{1}{s \ C1}}{R2 + \frac{1}{s \ C1}} = \frac{R2}{1 + s \ C1 R2}$$

$$Z_{top} = R1$$

So

$$Gain = \frac{\frac{R2}{1 + s \ C1 R2}}{R1 + \frac{R2}{1 + s \ C1 R2}}= \frac{R2}{R1 + R2 + s C1 R1 R2} = \frac{R2}{R1+R2} \cdot \frac{1}{1 + s \ \frac{C1 R1 R2}{R1+R2}}$$

Filter thus as a pole when $$s \frac{C1 R1 R2}{R1+R2} = 1$$

Let $$s = 2 \pi f \Rightarrow f = \frac{1}{2 \pi} \cdot \frac{R1 + R2}{C1 R1 R2} = \frac{1}{2 \pi} \cdot \frac{2.2k \Omega + 2.8k \Omega}{560pF\cdot 2.2k \Omega \cdot 2.8k \Omega} \approx 230.69 kHz$$

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  • \$\begingroup\$ Hmm. Now I have to understand both answers! :D @Warren-Hill: Can't I just solve it for f with Gain=1/sqrt(2)? \$\endgroup\$
    – Ekci
    Commented Jan 23, 2014 at 10:44
  • \$\begingroup\$ Andy's answer uses the fact that you can replace any linear circuit with just a voltage source and an output impedance. So he first removes the capacitor and and finds that its an attenuator at all frequencies with output impedance equivalent to R1 and R2 in parallel. If your not sure why see Thevinin's theorem. Now he can use this resistance with the capacitor to calculate the cut off frequency \$\endgroup\$ Commented Jan 23, 2014 at 10:52
  • \$\begingroup\$ I understand it. I am just wondering why (in Your solution) I can't use |Gain(f)|=sqrt(1/2)... It was my way to solve it. Sorry if I can't explain something correctly - I am not english ;) \$\endgroup\$
    – Ekci
    Commented Jan 23, 2014 at 10:56
  • \$\begingroup\$ Gain is R2/(R1+R2) = 0.56 at DC so if you used Gain is 1/sqrt(2) there is no solution. You have to find where the gain changes and the corner frequency is where the real part has the same absolute value as the imaginary part. You could use Gain = (R2/(R1+R2))x(1/sqrt(2)) in this case however. \$\endgroup\$ Commented Jan 23, 2014 at 10:59
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    \$\begingroup\$ Strictly speaking it isn't I missed these || i.e. Absolute value. But a corner exists when the real and amaginary parts are equal. \$\endgroup\$ Commented Mar 26, 2018 at 6:49
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If you regard R2 as being in parallel with R1, you will obtain the correct frequency cut-off. This is just superposition. R1 and R2 produce a value of: -

\$\dfrac{1}{\frac{1}{R_1}+\frac{1}{R_2}}\$ = 1.232k ohms.

This forms a low pass filter with C1 of: -

\$\dfrac{1}{2\pi\times 1.232\times 10^3 \times 560\times 10^{-12}}\$ = 230.686 kHz

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