Need Help with Analogman KOT CMOS Adaption

Started by Mr. Lime, August 27, 2015, 05:40:45 AM

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Mr. Lime

Hi guys,

I would like to built a Bluesbreaker / KOT pedal around a CMOS CD4049UB chip and need some help concerning the filtering.
My idea is to simply put the modified CMOS KOT circuit in place of the TS circuit of the ROG UBE Screamer as I like the feature blending some clean signal.

I think it would look pretty similar to the UBE Screamer using two CMOS gain stages for the op-amp clipping stage.
My main problem is the filtering around the clipping stage, which values should I take for the resistor and cap in series infront of the CMOS stage?
Maybe some clever fellows could give me an advice..

Thanks for help

Mr. Lime

Maybe someone can explain to me how to calculate the corner frequencies formed by C1, C2, R1 and R2?
This would help me a lot..

Can I reduce C4, R10, R11 and C7 to only one cap with 0.1u and one resistor with 10k?
Thanks for help

GGBB

Quote from: Mr. Lime on August 29, 2015, 08:31:00 AM
Maybe someone can explain to me how to calculate the corner frequencies formed by C1, C2, R1 and R2?
This would help me a lot..

Can I reduce C4, R10, R11 and C7 to only one cap with 0.1u and one resistor with 10k?

Yes - BluesBreaker clones typically use 10k 0.1µ. It's the total resistance and capacitance of the series that matters for the calculation - so 2*4.7k and 0.22µ/2 is pretty close to 10k and 0.1µ.
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Mr. Lime

Am I right that C1 + C2 + R2 equals ~ 964 Hz and R1 + C2 ~ 589 Hz?

Does this make any sense?
I don't think so..

If I would take the average frequency I would choose about 720 Hz which is the same cut off freuquency as a TS.
Am I totally wrong?
Thanks for help

GGBB

Those frequencies are right I think, but you need to keep in mind gain as well. The R2->C2->C1 series has different gain than the R1->C1 series, so you end up with two shelves - one on top of the other. I don't think that the average frequency would be an accurate reproduction of what is actually going on, but it could be close enough for rock-n-roll.
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Mr. Lime

Good point, but is there really a noticeable difference in gain at 33k and 27k?
This should cause a gain of 3.3 at 964 Hz and 4.0 at 589 Hz.

The catch is that I have to keep it simple for the CMOS circuit, that's why I could imagine that a gain of ~ 3.8 at 720 Hz will do the trick too..

But if there's a way to get all of this into the CMOS circuit I would prefer it, any ideas?

On the other hand I really don't want to build another TS.. :D
Thanks for help

SISKO

C1+C2 + R2 gives me 241Hz.. (R2*(C1+C2)*2*PI)^-1
--Is there any body out there??--

GGBB

Quote from: SISKO on August 29, 2015, 07:23:55 PM
C1+C2 + R2 gives me 241Hz.. (R2*(C1+C2)*2*PI)^-1

C1 and C2 are in series, so the overall total capacitance is 1 / ( 1 / C1 + 1 / C2 ) not C1 + C2.
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SISKO

Right.

I calculated the impedance of that net. It gave me

Z(s) = {(R1R2C1C2)s^2 + (R1C2 + R1C1)s + 1}/ {(R1C1C2 + R2C1C2)s^2 + C2s}

Zeros at 589 and 482 Hz
Poles at 0 and 9.5KHz
--Is there any body out there??--

GGBB

Quote from: Mr. Lime on August 29, 2015, 07:21:57 PM
Good point, but is there really a noticeable difference in gain at 33k and 27k?
This should cause a gain of 3.3 at 964 Hz and 4.0 at 589 Hz.

The catch is that I have to keep it simple for the CMOS circuit, that's why I could imagine that a gain of ~ 3.8 at 720 Hz will do the trick too..

But if there's a way to get all of this into the CMOS circuit I would prefer it, any ideas?

On the other hand I really don't want to build another TS.. :D

No CMOS ideas unfortunately.
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Mr. Lime

Thanks for calculating the right values!  :)

Here's my idea:


Splitting the signal into two paths should give the option of emphasizing different frequencys and gain.
How does this look like to your guys?

Thanks for help

Mr. Lime

Am I right that the gain stage with the soft clipping diodes has gain of 22? (R13 / (R10+R11) = 220 / 10.

Sorry, still didn't get that frequencys at which gain?

QuoteZeros at 589 and 482 Hz
Poles at 0 and 9.5KHz

What does this mean?

Isn't the gain simply calculated by gain pot (100k + 1) / 27k = 4.1 and (100k + 1) / 33k = 3.4?

Is there really so little gain going on in the bluesbreaker circuit?
Thanks for help

Mr. Lime

So are the frequency values for the red circled part 482 Hz and 9500 Hz?

The gain of the path with 482 Hz should equal 3.4 and the path with 9.5 kHz equals 4.1.

I think there's a mistake with the gain values..  ???
Thanks for help