Understanding the EHX Y-Triggered Filter

Started by Mark Hammer, April 21, 2016, 09:37:36 PM

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Mark Hammer

Nah.  Scruffie accidentally put a "ground" triangle symbol where a circle was intended.  The 470k is the terminating resistance, and the output switch selects between the 470k/0.1uf junction or the input.

robthequiet

Ah. Got it. Still trying to parse the salad around the left side of OTA. Thx!

Mark Hammer

Quote from: duck_arse on April 22, 2016, 11:43:22 AM
that's funny, I've just been playing with up/blank/down. I'd hazard a guesss, under duress, that C5 and C9 might be timing sweeps, being tants and all.
Well, on mine C9 is the only tantalum on the board. C5 is a regular electro.

BUT, you win the prize, my clever friend.  Upping the value of either of them slows the sweep time.  As shown in the schematic, the down sweep time was noticeably faster than the up sweep time.  Which probably accounts for why there was a 5uf electro tacked on the copper side in parallel with the 10uf.  I removed it just to hear what happened, and the sweep time shortened, confirming both your guess, and the reason why the extra 5uf was there.

I think we are making progress!  :icon_biggrin:

So are these "one-way" envelopes? I.E., there is an attack but no decay?

duck_arse

the "original" scribbled copy of this that I have gleaned from the interwires shows no connection between the collectors of Q2 and Q4. I would have said that 470k looks like a de-pop for clinton bypass, but I really don't understand the drawing of the input jack, on the original or the redraw.
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robthequiet

@duck_arse, that was my question. It makes more sense if C21 and R2 form the envelope input path, and R27, R26 and R28 form the audio input --  but the output is a bit weird, though it seems to come out of pin 8 of the OTA with C15 as the output cap after R25, R24 being feedback (possibly) and R30 being the load/impedance setter... not too many spice models of CA3094 about, apparently. 

StephenGiles

Quote from: robthequiet on April 25, 2016, 05:35:21 PM
@duck_arse, that was my question. It makes more sense if C21 and R2 form the envelope input path, and R27, R26 and R28 form the audio input --  but the output is a bit weird, though it seems to come out of pin 8 of the OTA with C15 as the output cap after R25, R24 being feedback (possibly) and R30 being the load/impedance setter... not too many spice models of CA3094 about, apparently. 

I used to have an EH design note on the CA 3094, but my paper copy has gone awol. This explained the various configurations required when using either pin 6 or pin 8 as its output. I'll keep looking.
"I want my meat burned, like St Joan. Bring me pickles and vicious mustards to pierce the tongue like Cardigan's Lancers.".

robthequiet

@StephenGiles -- Thank you -- I have located a datasheet or two and even found a story someone posted online having built one from discrete components. The EH notes would certainly be invaluable if they yield a clue. Of course, I am being lazy in not just hacking a model or two myself, although building a simulation out of the box may be more effort than it's worth if we could just build the actual, albeit having q's regarding the schematic.

snap

#27
this might be of help: depending on which side of the buffer (pin6 or 8 ) you take off the outputsignal, the inputs (pin2 and 3) each are inverting OR non-inverting:
http://www.mudpods.com/CA3094.gif


http://www.synthdiy.com/files/2008/ca3094.pdf

anotherjim

That's what I thought -  or rather it had to be an open collector buffer output - akin to the LM13700 with its open emitter buffer. Hence the output pin 8 is connected to supply, but via R23 as collector resistor. There is no emitter resistor fitted from pin 6, but there is ac negative feedback via C13 to pin 1.


jatalahd

Let's get back to this one.

After thinking more about the sweep part of the circuit, I don't need to change much of my writings in earlier posts. The sweeps are not symmetrical as I first assumed, and the sweep time around Q3 is not that easy to determine because R12 makes Q3 a constant current source which loads up C9. Therefore, the sweep on the left side of R11 is a linear sawtooth sweep and the sweep on the right side is more exponential and it does not go down all the way to 0V because of the voltage divider R19 and R13.

Then the OTA filter. I made a simulation using the following simplified circuit:


For the XOTA I simply used a VCCS subcircuit model:

.SUBCKT OTAS        1   2   3   gm=0.009
RIN     1 2     10000k
Ggm   0 3 1 2 {gm}
.ENDS

and added the transistor buffer from discrete components with reference to the schematic snap posted. There is still some problems in the Y-triggered filter schematic, but I presume that the simplification of the input circuit is close enough. Don't mind the actual gain values I got, because R4 affects to those and I was not sure how to connect that properly due to the problems in the original schematic.

Here are the simulation results as a function of the OTA transconductance parameter:


It seems that it tries to mimic the original wah pedal frequency response with a combination of a high-pass filter (Chp) and a low-pass filter (Clp). The maximum transconductance value (gm=0.009) I calculated from the approximate formula gm = 20*Ibias, and started decreasing it from there. There are two options how it changes the "resonance" frequency. It is either using the fact that the output impedance of the OTA changes with Ibias and this forms a low-pass with the Clp capacitor or then Clp is used as a Miller capacitance in the feedback loop and its value changes with gain (which is controlled by Ibias). I am hoping it would use the former option, but the most obvious is to use the latter mechanism as the original wah pedal uses.

The transistor buffer with the ac feedback through Clp has a transfer function of a first order low-pass filter when it is connected to a voltage source with internal resistance Rs. The cut-off frequency is linearly related (but not directly) to the time constant of Rs and Clp. This would make it possible to alter the cut-off by modifying the source resistance, but I have not been able to verify how much the output resistance of the VCCS block changes with gm. This is why I would opt for the Miller theorem in this case.

Would be nice to know if anyone else has made progress in analysing the complete Y-triggered filter circuit.

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I have failed to understand.

Mark Hammer

Your diligence in this is deeply appreciated.  I am confident something useful will come out of this.

And thank you for confirming by your analysis of the circuit, something I've long held about the perceptual aspects of upward and downward sweep: they need to BE different in order to be PERCEIVED as similar but inverted.

rankot

#31
Am I the only one wondering why C16/C17/R34 etc are connected both to the ground and V/2?  :icon_question:

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Scruffie

Because that's how it was drawn in the factory schematic, nobody ever said that factory schematic was right, I just copied it verbatim so it would be readable.

rankot

Anyone having original pedal, so can check this?
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rankot

I have finally had some time for this, so I tried to trace images of PCB that Mark Hammer posted here: https://www.diystompboxes.com/smfforum/index.php?topic=120148.msg1124242#msg1124242. It seems that the schematic above is far from correct, and this is what I made. I'd be grateful to anyone who can confirm this schematic as a good one.

I don't have a CA3094, so I used it's schematic from the datasheet, hoping to re-create a discreet one.



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rankot

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Scruffie

First of all, there's no need to make a discrete OTA (and you'd need matched transistors etc.) just use a CA3080 or 1/2 a 13600/13700 with a discrete darlington output buffer.

Second... not to discourage you but having owned one (no I don't any more so can't take photos) I really wouldn't put that much effort in to copying it, it's not particularly wonderful. Different, yes, but the envelope triggering mechanism is a pain in practice.

Mark Hammer

Came across this YT demo from David Rolo,  Provides a nice compendium of the sounds possible, using both guitar and bass.  Gives you a good idea of the relative consistency of sweep.  It's clearly a very different sort of swept filter than most envelope-controlled units.  Would be worth understanding a little more; especially with an eye towards more control over the time constants of the envelope.  Glad the topic has been resurrected.


rankot

Quote from: Scruffie on November 14, 2019, 01:49:35 PM
First of all, there's no need to make a discrete OTA (and you'd need matched transistors etc.) just use a CA3080 or 1/2 a 13600/13700 with a discrete darlington output buffer.

Second... not to discourage you but having owned one (no I don't any more so can't take photos) I really wouldn't put that much effort in to copying it, it's not particularly wonderful. Different, yes, but the envelope triggering mechanism is a pain in practice.

Thanks! I though that it use CA3094 pins 1 and 8, which are not available on CA3080, so I wanted to try it with discrete OTA, since CA3094 is unobtainable.
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Kipper4

That's quite a nice filter there. Thanks for the vid Mark.
I noticed the envelope detector didn't trigger every time.
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