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### Topic: Better understanding of filter/values for (best) audio output? (Read 6339 times)previous topic - next topic

#### oric_dan

#15
##### Jun 19, 2013, 08:07 pm
Corner frequency comes from Bode Plots, you can ignore the equations and just look at
the graphs. Also, to tell what your ckt is doing, you should plot out the Bode Plot for your
amp, ie measure Gain vs Frequency.
http://en.wikipedia.org/wiki/Bode_plot

The one formula you have is useful - ONLY because the low-pass and high-pass frequencies
are 100X apart, so you can use the formula for both cases [ie, each cap separately] with
little effect from the other cap.
Quote
corner frequency (f=1/(2*pi*R*C)

For low-pass, R = R8 || (R5+R6) = 5K, and C = 10nF.
For high-pass, R = R8+R5+R6 = 20K, and C = 1uF,

so  Flo = 8 hz,  and Fhi = 3185 hz.

Your gain in the "mid-band" = R6 / (R8+R5+R6) = 0.1, so you're losing most of your
signal right there.

AFAICT, the purpose of R7 & C15 is to prevent high-frequency oscillations in the LM386
at high-gains, as the corner frequency for that network is very high = 318 Khz.

#### xl97

#16
##### Jun 19, 2013, 09:29 pm
whoosh!..  hahaha..

can you 'dumb it up' a bit?

appreciate the response.. just wish I could use it better..

apply it to my project/problems..

what is the || (pipe) in your equation for?  I usually default that to be an OR operand?

Quote

For low-pass, R = R8 || (R5+R6) = 5K, and C = 10nF.
For high-pass, R = R8+R5+R6 = 20K, and C = 1uF,

the NEW values (not sure if you read the thread? or using first post/image only now?)

R8 = 1.4k (using a 2k for the moment..but using the calculator link.. it says R8=1.4k & C18 = 10nF  to give me around the 11kHz cut-off...

I also am not following this:

Quote

Your gain in the "mid-band" = R6 / (R8+R5+R6) = 0.1, so you're losing most of your
signal right there.

R6 / (R8+R5+R6)

2k / (2k+8k+2k) = 0.1666...

thought this 'WAS' correct now? going in the right direction?

(sorry Im just getting more confused)  haha..

#### oric_dan

#17
##### Jun 19, 2013, 09:46 pm
I used your original values. "||" = parallel combination, standard electronics terminology.

I did dumb it down, just did the calcs and no theory. You cannot understand why you
should use R = R8 || (R5+R6), etc,  unless you understand a fair amount of theory. The
equivalent R values are as I indicated, you cannot use just one Rx in the corner freq eqn.
All 3 Rs factor into both corner frequencies. Ask dc42.

When you swap out values and say "as I didnt notice any? difference", you won't unless
you do a complete Bode Plot, as I indicated.

If you provide a list of all of your final value, Rx = ..., Cx = ..., etc, I'll recalculate the
corner frequencies.

#### xl97

#18
##### Jun 19, 2013, 10:40 pm
(reading the Bode Plot link after the boss leaves!  lol)

ahh.. "parallel" (got it!)..

my current values are as follows:

(image if location reference only)

http://forum.arduino.cc/index.php?action=dlattach;topic=171578.0;attach=46278;image

R8 = 2.2k  (although my target value was 1.4k.... I also used a 1K value in place as well)
C18 = 10nF
C21 = 1uF
R5 = 8k
R6 = 2k

#### oric_dan

#19
##### Jun 19, 2013, 11:13 pm

R8 = 2.2k  (although my target value was 1.4k.... I also used a 1K value in place as well)
C18 = 10nF
C21 = 1uF
R5 = 8k
R6 = 2k

For low-pass, R = R8 || (R5+R6) = 1.8K, and C = C18 = 10nF.
For high-pass, R = R8+R5+R6 = 12.2K, and C = C21 = 1uF,
so  Fhi-pass =  13 hz,  and Flow-pass = 8846 hz.

Your gain in the "mid-band" = R6 / (R8+R5+R6) = 2K/12.2K = 0.16, so you're "still" losing most
of your signal right there.

I would think you'd do better with the previous low-pass freq, since it's purpose is
filter out the hi-freq transitions on the DAC output.

#### xl97

#20
##### Jun 19, 2013, 11:38 pm
(math aside)...all I can say is that is much louder/better than before using this:

R8 = 2.2k
C18 = 10nF
C21 = 1uF
R5 = 8k
R6 = 2k

than the previous:

R8 = 10k
C18 = 10nF
C21 = 1uF
R5 = 8k
R6 = 2k

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