I understand that this circuit could be used as a treble control circuit with high frequency gain occuring when R3 is set so a=b (we'll call this k=0), and high frequency attenuation occuring when R3 is 22kOhms across a and b (k=1). So therefore, depending on R3's setting, this circuit is either a high pass (k=0) or low pass (k=1) filter.

When comparing this circuit to high and low pass filters, I do not understand what is happening: C2 will always have a lower impedance for high frequencies and so surely adjusting R3 will only alter positive gain for high frequencies.

I also understand that the capacitors act as an open circuit for low frequencies and so surely all low frequnencies would be attenuated.

Can you help me understand this?

Link to circuitlab.com schematic: https://www.circuitlab.com/circuit/x66cq6/basic-frequency-control-circuit/

enter image description here

By the way, I realise I have discussed this schematic in previous questions.

Please note: I am asking something different to any of my previous questions. This is no duplicate question.

  • \$\begingroup\$ It's not a duplicate, but you are trying to understand this circuit without having clear what the simgle parts do. Did you understand how the feedback works for an amplifier? \$\endgroup\$ – clabacchio May 16 '12 at 9:44
  • \$\begingroup\$ @clabacchio Yes, I think so. I understand that DC feedback is needed to control any fluctuations in voltage between the virtual ground input nodes to the amplifier. It does this by outputing the signal with a DC offset opposite to that at the inputs. Although AC feedback can travel over a capacitor, for any DC signal it acts like an open circuit. \$\endgroup\$ – user1083734 May 16 '12 at 9:50
  • \$\begingroup\$ I am a beginner but am finding all of the answers I have recieved on electronics.stackexchange.com incredibly valuble. I realise the power of this resourse and certainly do not want to waste anybody's time \$\endgroup\$ – user1083734 May 16 '12 at 9:52
  • \$\begingroup\$ It's not about that: you have to understand that DC feedback and AC feedback are, in principle, feedbacks. From your first comment I see some confusion, and I don't blame you :) but you should take things a bit more step by step. I've also edited my answer to give a bit more general notions, check it. \$\endgroup\$ – clabacchio May 16 '12 at 10:00
  • \$\begingroup\$ I don't see how this can be a treble control. The cutoff frequencies on both sides vary between 6.5Hz and 66Hz. \$\endgroup\$ – Federico Russo May 16 '12 at 10:15

The AC feedback via C2 is negative feedback - it will decrease the gain.
Increasing the negative feedback with frequency will reduce the gain as frequency increases.
C2 impedance decreases as frequency increases so feedback via C2 increases so gain decreases if all feedback is applied to inverting input.

Final result will depend on where the feedback goes / how it is applied. Changing R3 split changes the overall "transfer function".

It's not obvious that this is a formal design. Can you give a reference to where you got it from?


Rather than thinking of the capactior as such, think of it as simply a resistor who's value decreases linearly with the increasing frequency of the current flowing through it.

An inductor can be considered in the same way except the frequency relationship is not inverse. (in this case rising 'reactance' with rising frequency)

Now imagine an number identical circuits to the one you draw, e.g. three of them. Each drawn with a 'frequency dependent resistor' who's value is infinate at zero Hz but is replaced in each redrawn circuit to one that is a lot smaller than the other gain components at some frequency, e.g. 10kHz.

You can now imagine what the gain does. 'Resistors' that reduce gain with increasing value will have the effect of a low pass filter. 'Resistors' that increase gain will have the effect of a high pass filter.

  • 1
    \$\begingroup\$ The variable impedance of a capacitor doesn't explain at all how this is a treble control! The capacitors are fixed, but the frequency characteristic varies with the potmeter position. \$\endgroup\$ – Federico Russo May 16 '12 at 15:19
  • \$\begingroup\$ @user1083734 - I can't believe you understand any of this. It simply makes no sense! \$\endgroup\$ – stevenvh May 16 '12 at 15:31
  • \$\begingroup\$ @stevenvh OK, maybe the third paragraph is a little confusing but what Jason Morgan said in the first paragraph made a lot of sense to me. It may just be a coincidence that I realised how the feedback impedance/input impedance ratio would vary over a frequency range for both settings of R3, when reading this answer. OK, not a great answer, but something about it made me realise enough about what is going on for me to consider my question answered. I am, of course, open to any more detailed explanations. \$\endgroup\$ – user1083734 May 16 '12 at 16:10
  • \$\begingroup\$ @user1083724 - That third paragraph is gibberish, I don't understand it either. The first paragraph just tells you the behavior of a capacitor, that it's impedance decreases with higher frequencies. Like Federico says the answer nowhere tells how you change the frequency response with the potentiometer, or how R1 and R2 play a role. \$\endgroup\$ – stevenvh May 17 '12 at 8:41

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