
Re: Equations for this bass boost circuit?
tempus fugit wrote:
> Hey all;
>
> I've designed the bass boost circuit below:
>
>
>
> .047 uF
>
> ||
> .----||--.
> | || |
> | |
> | ___ |
> o---___|-o---o
> | |
> | 80k |
> | |
> | |
> | |\| |
> .-----|-\ |
> | | >o--o----o
> In o---------------o|+/
> | |/|
> |
> |
> |
> |
> .-.
> | |
> | | 3.9k
> '-'
> |
> |
> ===
> GND
>
> I started out using a Baxandall filter circuit, then left out the
> attenuation side since I only want to boost, tried a couple of other things,
> and came up with this as the best sounding circuit for my application. It
> works fine, and I understand why it's boosting the bass, but I'm curious to
> know how to figure out what frequency the boost begins to roll off at (or
> reach the -3db point), and at what Q it rolls off at. I had thought of using
> the equations for the Baxandall filter, but the circuit is different enough
> (most notably lacking a series resistor) that I'm not sure how to go about
> it.
Since the opamp is connected in a non inverting
configuration, the amplifier has a minimum gain of 1, even
at frequencies so high that the feedback capacitor is
essentially a short circuit across the feedback resistor.
Once the frequency falls enough that the feedback divider
has a 1:1 division ratio, the gain rises to 2, and climbs in
proportion to the period (inverse to frequency, till the
capacitor impedance is about the same as the resistance of
the feedback resistor.
Below that the gain is fixed at (1 + 80k/3.9k)=21.5
So the frequency below which the gain starts to rise is
approximately where Xc=3.9k. From Xc=1/(2*pi*f*C), with Xc
= 3.9 k and C=0.047 uF, f=868 Hz. And the gain stops
increasing (actually increases a lot more slowly) below
about where Xc = 80k, or 42 Hz.
You could simulate this amplifier with LTspice and graph the
frequency response with its AC analysis.
http://www.linear.com/designtools/softw ... hercad.jsp