I have these two circuits: enter image description here

With a 'normal' OPAMP (without capacitors), I do know how to calculate the amplification factor of the circuit (\$A=\dfrac{U_{uit}}{U_{in}}\$) . But how do I do it with those two capacitors? But also how do I calculate where the -3 dB point is of both circuits?

  • \$\begingroup\$ If you remove the opamp for a moment, can you calculate the -3dB point if it only was for C1/R1? Do you know the formula for A just for R1/C1? And if you look into the non-inverting input of the opamp, what input impedance do you expect? \$\endgroup\$ – jippie Dec 5 '13 at 19:09
  • \$\begingroup\$ well the impedance of the capacitor is 1/jwc (where w is omega) \$\endgroup\$ – SjonTeflon Dec 5 '13 at 19:13
  • \$\begingroup\$ And I think at the first circuit is is something like R1/((1/jwc)+R1) \$\endgroup\$ – SjonTeflon Dec 5 '13 at 19:14
  • \$\begingroup\$ My complex numbers is a bit rusty, but that seems correct to me. \$\endgroup\$ – jippie Dec 5 '13 at 19:15
  • \$\begingroup\$ Condensator in English is a capacitor BTW. \$\endgroup\$ – jippie Dec 5 '13 at 19:16

The opamp in both circuits is just a voltage follower with a gain of 1, so it is irrelevant for the purpose of calculating gain.

The left circuit is a simple R-C high pass filter, and the right circuit a simple R-C low pass filter. The gain of each of these is 1 well into the passband. Well into the stopband, the gain will decrease 6 dB/octave or 20 dB/decade of frequency.

The rolloff point of either type of filter is when the capacitor's impedance magnitude equals the resistance. The equation for the frequency is:

F = 1 / (2 π R C)

When R is in Ohms and C in Farads, then F is in Hertz. In your case you have 100 nF and 3.3 kΩ, so the rolloff frequency of each filter is about 480 Hz. At that frequency, the filter will attenuate by a factor of the square root of 2, or will have have a gain of -3 dB. The filter gain as a function of frequency varies smoothly, but after a octave or two in either direction it approaches 20 dB/decade down from the rolloff frequency on one side and unity gain on the other.

The left filter is a high pass, so for frequencies above 480 Hz it will approach unity gain as the frequency gets higher. After about 1 kHz the gain will be close enough to 1 for most purposes, certainly for any normal audio application. Well below 480 Hz, it will assymptotically approach attenuating by the ratio of 480 Hz to the actual frequency. For example, at 100 Hz it will attenuate close to 4.8 times, or the gain will be close to -14 dB.

The low pass filter on the right works the same way flipped in frequency around the 480 Hz rolloff value. At 100 Hz it's gain will be nearly 1, and at 3 kHz it will attenuate close to 3 kHz / 480 Hz = 6.25 times, for a gain of -16 dB.


But also how do I calculate where the -3 dB point is of both circuits?

In general, you would find the magnitude of the transfer function, set it equal to \$\dfrac{1}{\sqrt{2}}^*\$, and solve for the frequency.

For a simple 1st order filter, this is almost trivial. In the 2nd circuit, the voltage at the inverting input of the op-amp is:

$$V_+ = V_{in}\dfrac{1}{1 + j \omega RC} \tag{1}$$

Multiply by the conjugate to find the magnitude squared:

$$|\dfrac{V_+}{V_{in}}|^2 = \dfrac{1}{1 + j \omega RC}\dfrac{1}{1 - j \omega RC} = \dfrac{1}{1 + (\omega RC)^2} \tag{2}$$

Take the square root to find the magnitude:

$$ |\dfrac{V_+}{V_{in}}| = \dfrac{1}{\sqrt{1 + (\omega RC)^2}} \tag{3}$$

Now, it's easy to see that this equals \$\dfrac{1}{\sqrt{2}}\$ when \$\omega = \dfrac{1}{RC} \tag{4}\$

\$^*20\log(\frac{1}{\sqrt{2}}) \approx -3dB\$

  • 1
    \$\begingroup\$ Wow, way to make a simple thing almost impenetrable! But, +1 because it is correct and complete. \$\endgroup\$ – Olin Lathrop Dec 5 '13 at 21:09
  • 2
    \$\begingroup\$ @OlinLathrop Impenetrable? This is the most basic of math that any sophomore level engineering student deals with in circuit analysis classes. It's always good to know the math behind things. \$\endgroup\$ – krb686 Dec 5 '13 at 21:19
  • 1
    \$\begingroup\$ @OlinLathrop if ½√2 in this answer is correct, am I misreading yours then? "At that frequency, the filter will attenuate by a factor of 2" \$\endgroup\$ – jippie Dec 5 '13 at 22:48
  • \$\begingroup\$ My complex math is a bit rusty, I don't get how you get from (1) to (2). I get the \$|\dfrac{V_+}{V_{in}}|^2\$, but I don't get why the right may multiplied by 1/(1−jωRC). I'd expect that had to be 1/(1+jωRC). Is it simple to explain, help my fading memory a bit? Of course I see why you want the + rather than the -, but why is it allowed? \$\endgroup\$ – jippie Dec 5 '13 at 23:02
  • 1
    \$\begingroup\$ @jippie, the magnitude of a complex number or complex function \$Z\$ is \$|Z| = \sqrt{ZZ^*}\$ where \$Z^*\$ is the complex conjugate of \$Z\$. If \$Z = a + jb\$, then \$Z^* = a - jb\$. Then \$ZZ^* = (a + jb)(a - jb) = a^2 + b^2\$ which is real. \$\endgroup\$ – Alfred Centauri Dec 5 '13 at 23:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.