This question already has an answer here:

back at it again with a problem that I am stuck on and need some clear explanation of.

1/2: I was wondering How those Bass controls work, for example the ones in the car you allow to change the bass from -10 to 0 to 10 and the bass would increase to barely hearing the drums. Is this due to a gain change or a Fc(Cut off frequency) change?

2/2 I been having a problem on a filter, and I am sure I am over thinking it, however in the picture below using an Input source of 1.92Vpk @ 1kHz. The Fc of the High pass filter is ~250Hz however the 1kHz is not at 0dB which means I am losing some sound on this common frequency in audio which isn't good. I was wondering to over come this problem would a higher order filter would work? Essentially making the roll off more sharper ?

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marked as duplicate by Dave Tweed Jun 22 '18 at 15:37

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  • \$\begingroup\$ 1. Frequency-dependent gain -- bass means "low f" so here a bass control will provide non-zero dB gain at low freq, either + or -, moving back towards flat gain at higher f. 2. Higher order filter is a way. 3. You want +/- dB gain at low frequency since you are talking about bass control, even if the flat gain part is just 0 dB. You also want steep cutoff for some reason at the high end, it seems. Answering your question is difficult because of the conflation of ideas in it, so I think as it is that it would require a too-long answer for me to consider. \$\endgroup\$ – jonk Jun 21 '18 at 20:54
  • \$\begingroup\$ What is the purpose of this filter? You clearly don’t need gain and driving 620 Ohms is already possible. Define all parameters , load , power and spectrum and anything else that matters (Application) \$\endgroup\$ – Sunnyskyguy EE75 Jun 21 '18 at 21:09
  • \$\begingroup\$ Its a pre amplifier, Just to control the input signal's bass and volume control. Its going into a power amplifier so this pre amp wont see the load. \$\endgroup\$ – Pllsz Jun 21 '18 at 21:12
  • \$\begingroup\$ Nothing you have posted is any kind of an amplifier. You have posted a couple of examples of simple single pole filters RC filters. No gain to be had there. \$\endgroup\$ – JRE Jun 21 '18 at 21:17
  • \$\begingroup\$ Ahh I see what you mean, sorry about that what I meant is I am going to use this low pass filter in a pre amplifier. This is getting out of topic. The problem is stated as how to overcome the Fc curve and what does it mean to control bass? is it a gain increase / decrease or is it a FC shift \$\endgroup\$ – Pllsz Jun 21 '18 at 21:20

You are not being very clear here. Do you want a preamp circuit or a equalizer circuit? What is your load?

  1. C1 generally appears in these circuits as a bypass capacitor, not some sort of bass control capacitor. You generally don't want it to attenuate frequencies you want to reproduce.
  2. First pick which frequencies you want and which ones you want to attenuate, and by how much. Then figure out the order of the filter you need. Assuming you just want some half-assed bass attenuation... you can change your filter's cutoff frequency down by increasing R1. My first instinct would be to try a 30k pot in series with that 620R resistor and see what the bode plot yields you. Be aware that you can't make the attenuation steeper with that circuit - for that you need a higher order filter.
  3. Yes. It should be trivial to reduce the gain in circuits involving op-amps. The easiest way is to use a voltage divider at the input.
  • \$\begingroup\$ Yeah sorry about that once again... Anyways I want a pre amp circuit with the functionality to control the bass frequencies (You can hear more bass and switch it to less bass). 1. What? Yeah its a typical first order High pass filter, I was planning to use a potentiometer on the resistor there from 620ohms to 8.2k so I can control the FC from 20Hz to 250Hz 2. 20Hz to 250Hz is the bass frequencies I want to control 3. Sorry I am lost here? Reduce the gain? I want the gains to be constant within my band pass? 20Hz to 250Hz. \$\endgroup\$ – Pllsz Jun 21 '18 at 21:24
  • \$\begingroup\$ You complained some butterworth circuit has too high of a gain. You can reduce it easily for all frequencies with a voltage divider if that's a problem \$\endgroup\$ – FrancoVS Jun 21 '18 at 21:42
  • \$\begingroup\$ I did, however it drastically changed the Fc. \$\endgroup\$ – Pllsz Jun 21 '18 at 21:43
  • \$\begingroup\$ @Pllsz If you are using passive filters, of course adding a load will change the response Fc. I'd assumed you were thinking of active filters, but now see that you've added passive schematics. So, I guess that's what you are doing? \$\endgroup\$ – jonk Jun 21 '18 at 21:46
  • \$\begingroup\$ No load! The butterworth I used was an active filter so the input and output were isolated from each other. I meant I made the NFB to a voltage follower. \$\endgroup\$ – Pllsz Jun 21 '18 at 21:49

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