Explain these paralleled parts please

[lars-musik]

Dear friends,
I am just building a Rockett Blue Note and I wonder about this construction with the paralleled resistors and caps. Can't they just be combined to the values I added in red? What am I missing?

Thanks!


Lars

[Kipper4]

So it's 2x filter
1x hpf
1x lpf

Leaving the band in the middle?

[anotherjim]

Argh... complex.
It's a high pass, but with 2 turn-over knees.
At some high freq, it is effectively 560R for the gain calc when the impedance of both caps is negligible.
As Hz falls, the .1uF path becomes higher impedance first, leaving close to the 1k2 alone setting the gain, and it behaves like a normal single inverting input leg. That's the first response knee.
When Hz falls further, the .22uF impedance start to rise and gain falls. The second response knee.
In-between those points, there is err... some in-between gain applied.

[lars-musik]

Thanks, Jim!

I understand that!!!

[antonis]

It's 2 X high pass filters (like in ProcoRat..), placing 2 poles and attenuating freqs below:

f_c1 (right) = 1600 Hz -> 6db/oct (20db/dec)
f_c2 (left)  = 603 Hz -> 12db/oct (40db/dec)

This kind of filtering is used for severe attenuation of bass/mid harmonics before clipping action (low/mid end less clip) making frequency selective distortion...

edit: Agrrrrr, Jim..

[lars-musik]

Thanks for the values, antonis!
Did you use an online calculator and if so, which one!

[EBK]

f is just 1/(2×pi×R×C).  A spreadsheet is often handy for this stuff so you can try out all sorts of standard values.

As far as the slope, to shallower one is standard single pole filter slope, and the steeper one is standard two pole filter slope.

[Digital Larry]

f is just 1/(2×pi×R×C).  A spreadsheet is often handy for this stuff so you can try out all sorts of standard values.

As far as the slope, to shallower one is standard single pole filter slope, and the steeper one is standard two pole filter slope.

I made a spreadsheet showing impedance of caps at various frequencies.  For a single pole filter you then match that value to the driving or load impedance to see where the -3 dB point is.  I realize that it takes some practice to figure out what the impedance is, but say you have an op-amp output (LOW impedance, assume 0) driving a 1 k resistor in series then to a cap to ground.  The point between the R and C is a low pass filter whose corner frequency is that at which the impedance of the cap matches 1k ohms.

[antonis]

Thanks for the values, antonis!
Did you use an online calculator and if so, which one!

My humble mobile phone calculator..   

It isn't complicated if you set a fix "starting" point on your mind, e.g. 100nF/100k -> 16Hz

Any higher value of resistor or capacitor lowers filter's cut-off frequency by same analogy, e.g. double value - half frequency
(as long as you have in mind that R x C are multiplied so their values changes fight each other..)

Despite its rude simplicity, it's very effective for a rough estimation, taking in mind components value tolerance (especially for caps..) 

[blackieNYC]

So will this roll off at the first knee @ 6db per oct, and then at the lower knee at 12 per?

[anotherjim]

So will this roll off at the first knee @ 6db per oct, and then at the lower knee at 12 per?

Very good question Blackie.
I feel it can't be steeper than a single RC can be, just that the knee moves depending on the frequency. Might be easier to think of the effect on overall gain for different spot frequencies. I mean, think of the RC impedance setting Rin of the amp in a frequency dependant manner, not as filter roll-off.

[Vitrolin]

I think a RC filter has -3dB per octave
so 3 and 6dB per oct

[anotherjim]

I think that for roll-off to be steeper, the RC sections would be in series. I'm trying to visualise how series HP could work in this situation. Plain 2 pole HP filter is simple, In > cap > R to ground > cap > R to ground > Out. I don't know equivalent way to do it in the Rin path.
Except the old brain hurts & needs some beer now.

[blackieNYC]

Check this out- 2nd order filter calculator, But for low pass only.
http://sim.okawa-denshi.jp/en/CRCRtool.php
Though it's not the same filter, HP v LP, it is the same values. Looks steeper than 6 per oct. not sure.

[antonis]

I think a RC filter has -3dB per octave
so 3 and 6dB per oct

-3db is on cut-off point..
(which actually isn't sharp so it's placed in the middle of knee curve..)

One octave higher for LPF (or lower for HPF) has -6db attenuation..
(V_out/V_in = 1/2, <=> 20 x log(V_out/V_in) =-6)


<Except the old brain hurts & needs some beer now>
I should strongly recommend it, Jim..!! 
(put it on my account, of course..)

The upper arrangement isn't any kind of series RC filters, so there isn't any series 2nd pole (neither loading)..
It acts as 2 parallel independent HP filters with 2 different cut-off points - so the lower the frequency the most the attenuation..

e.g. 800Hz (1600/2) are -6db attenuated by right RC but left almost "untouched" by the left one..
      300Hz (600/2) are -6db attenuated by left RC but also more than -12db (300 is more than 2 octaves down from 1600) attenuated by        the right one..

Or maybe I need some of your beer, too..

[anotherjim]

Ok heck. SPice this somebody!

[antonis]

Same arrangement for 60Hz & 1k5Hz..

[EBK]

I threw some calculations into Excel (didn't have a circuit sim handy), so mistakes are highly likely, but this is what I got for the gain, considering those RC values, while ignoring the cap and diodes in the feedback loop and the filtering done on the input (i.e., this is only the isolated effect of the high pass filter).

The dots are a third of an octave apart.

[Groovenut]

Ok guys here are my findings. I feel you have to consider the interaction of the Fat control at min and the gain at max. So here's the freq plots for the 2 separate component chains vs the paralleled value single chain (560R>.33u).



The 10Ms are there to keep Circuitmaker from throwing an error and I guess-timated on the output load as it wasn't on the schem.

IMO I don't see a difference you would hear. The values in the 2 parallel chains are too close to each other to have a significant step difference. At the largest difference point, 766Hz, the difference between the two is only 0.7dB

[anotherjim]

Yeh, I would have thought you want at least x5 difference in the networks to see much of a kink in the curve.