6
\$\begingroup\$

I have been struggling to see how C1 and R1 in this picture form a high-pass filter.

enter image description here

The voltage isn't taken across R1 like in a potential divider:

enter image description here

  1. Could someone shed some light on how this works please?
  2. How does this configuration form a band-pass filter: enter image description here
\$\endgroup\$
  • \$\begingroup\$ Hint: In both of these circuits we have negative feedback so the voltage at the amplifier inverting input is the same as the non-inverting input 0V. So what is the current in R1 and C1? That same current must be flowing in the feedback path R2 or R2 on parallel with C2. I'm happy to post a more detailed answer later, if necessary, but I'm sure we would all like to see you efforts first. \$\endgroup\$ – Warren Hill Nov 6 '17 at 20:24
  • \$\begingroup\$ @Warren Hill, so, please correct me if I am going in the wrong direction: Vin/(R1 + X1) = -Vout/(R2 || X2)? \$\endgroup\$ – Shibalicious Nov 6 '17 at 20:39
  • \$\begingroup\$ Yes that's the basic idea. Are you comfortable with complex numbers yet \$ j \cdot \omega \$ ? \$\endgroup\$ – Warren Hill Nov 6 '17 at 20:43
  • \$\begingroup\$ @Warren Hill I am, yes, I have done this all a long time ago, just brushing up on my rusty knowledge. Please continue :) \$\endgroup\$ – Shibalicious Nov 6 '17 at 20:44
4
\$\begingroup\$

As pointed out in my comments the inverting input is a virtual 0V so the gain of this circuit is :

$$Gain = -\dfrac{Z_f}{Z_i}$$

Where \$ Z_i \$ is our input impedance \$ R_1 \$ in series with \$ C_1 \$

$$Z_i = R_1 + \dfrac{1}{j \cdot \omega \cdot C_1} = \dfrac{1+ j \cdot \omega \cdot C_1 \cdot R_1}{j \cdot \omega \cdot C_1}$$

And \$ Z_f \$is the feedback impedance, taking your second example \$ Z_f \$ is \$R_2\$ in parallel with \$ C_2 \$.

$$Z_f = \dfrac{R_2 \cdot \dfrac{1}{j \cdot \omega \cdot C_2}}{R_2+\dfrac{1}{j \cdot \omega \cdot C_2}}= \dfrac{R_2}{1+ j \cdot \omega \cdot C_2 \cdot R_2}$$

$$Gain = - \dfrac{Zf}{Zi} = - \dfrac{\dfrac{R_2}{1+ j \cdot \omega \cdot C_2 \cdot R_2}}{\dfrac{1+ j \cdot \omega \cdot C_1 \cdot R_1}{j \cdot \omega \cdot C_1}} = - \dfrac{j \cdot \omega \cdot C_1 \cdot R_2}{\left( 1+ j \cdot \omega \cdot C_1 \cdot R_1 \right) \cdot \left( 1+ j \cdot \omega \cdot C_2 \cdot R_2 \right)}$$

This gives you zero gain at DC raising until the first pole where it levels out then falling at the second pole.

The poles being when \$ \omega \cdot C_1 \cdot R_1 = 1 \$

and when \$ \omega \cdot C_2 \cdot R_2 = 1 \$

This answer is a lot more mathematically rigorous than the one by @DaveTweed but that doesn't make his answer any less correct.

\$\endgroup\$
  • \$\begingroup\$ Great explanation. I was trying to think about it more intuitively, but I guess sometimes using maths is easier. Thank you. \$\endgroup\$ – Shibalicious Nov 6 '17 at 21:33
3
\$\begingroup\$

Consider the total impedance of R1 and C1 in series.

If the frequency is high enough so that the capacitive reactance is significantly less than the resistance, the total is dominated by the resistance — essentially constant. The gain of the circuit (feedback impedance over input impedance) is essentially flat.

On the other hand, if the frequency is low enough that the capacitor's reactance is greater than the resistance, it is the reactance that dominates. This impedance rises with decreasing frequency, which means that the gain of the circuit decreases — a high-pass effect.

For the second part, you can make essentially the same argument for C2 and R2, but in this case, since it's a parallel connection, the smaller of the two impedances dominates, and since it's in the feedback path, a decreasing impedance with increasing frequency results in decreasing gain — a low-pass effect.

\$\endgroup\$
  • \$\begingroup\$ I can see what you mean. But why do it this way rather than add an actual low-pass and high-pass filter? E.g: electronics-tutorials.ws/filter/fil33.gif \$\endgroup\$ – Shibalicious Nov 6 '17 at 20:41
  • 2
    \$\begingroup\$ Because the input node of the opamp is tied to "virtual ground" by the negative feedback (this is what makes the circuit gain equal to \$-\frac{Z_{fb}}{Z_{in}}\$). You can't put a simple voltage divider there. \$\endgroup\$ – Dave Tweed Nov 6 '17 at 21:18

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.