OCD Buffer....double buffer?

Started by SprinkleSpraycan, March 19, 2025, 07:08:41 PM

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SprinkleSpraycan

Even though i dont love the OCD drive i do really enjoy the buffer. I am curious about this schematics buffer bypass option, why does it go buffer to second buffer? The signal path when the pedal is on is the first buffer>FX>out. I just cant make sense of the bypass doubling up on it. Knowledge wanted.





GGBB

Probably to keep absolute phase intact. Both buffers are phase inverting.
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SprinkleSpraycan

Quote from: GGBB on March 19, 2025, 09:48:52 PMProbably to keep absolute phase intact. Both buffers are phase inverting.

That makes sense. I suppose if I put this in a box for a mid chain buffer I could have a phase toggle. Perhaps that's not worth it though.

antonis

I shouldn't rate "buffers" CS amps (despite their volatge gain) of 10k output impedance..
"I'm getting older while being taught all the time" Solon the Athenian..
"I don't mind  being taught all the time but I do mind a lot getting old" Antonis the Thessalonian..

SprinkleSpraycan

Quote from: antonis on March 20, 2025, 06:21:39 AMI shouldn't rate "buffers" CS amps (despite their volatge gain) of 10k output impedance..

I sensed the snark, but I'll need it explained.

I looked at this thread -
https://www.diystompboxes.com/smfforum/index.php?topic=116842.0

Seems like everything is on the level.

PRR

Quote from: SprinkleSpraycan on March 20, 2025, 01:51:20 PMI sensed the snark, but

I think post #1083409 is off-track.

I can't read your image. Fuzzy. Missing lines.

Where do you see "buffer"? Reading around the missing lines I think I see two common Source stages. There's no official definition of "buffer" but it usually offers high input Z and low output Z. "Low" is relative of course, but 10K is not particularly low, when a Source follower could give 1k Ohms, an Emitter follower could give 30 Ohms, a TL072 under 10 Ohms output.

While the diagram is unclear (even manipulated), I think Q5 Q6 are flavor stages which happen to have high Input impedance.
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antonis

#6
Quote from: SprinkleSpraycan on March 20, 2025, 01:51:20 PMI sensed the snark, but I'll need it explained.

No snark at all.. Just trying to set correct terms for agreement.. :icon_wink:

Quote from: SprinkleSpraycan on March 20, 2025, 01:51:20 PMI looked at this thread -
https://www.diystompboxes.com/smfforum/index.php?topic=116842.0
Seems like everything is on the level.

I presume you didn't pay attention on R.G.'s 3rd paragraph.. :icon_wink:

The major difference (except for phase reversal) between a buffer and a unity gain CS amp is output impedance..!!
For the former is 1/gm where for the later is Drain resistor value..
Put it another way, the former is considered current amplifier where the later voltage amplifier

edit: Paul is faster..!!
"I'm getting older while being taught all the time" Solon the Athenian..
"I don't mind  being taught all the time but I do mind a lot getting old" Antonis the Thessalonian..

SprinkleSpraycan

Quote from: PRR on March 20, 2025, 02:16:18 PMWhile the diagram is unclear (even manipulated), I think Q5 Q6 are flavor stages which happen to have high Input impedance.


So Q5/6 are part of a buffer bypass. Q5 feeds to the actual drive circuit when the pedal is on and Q6 is bypassed.

And so sorry for a crappy picture, hopefully this is better.


Unfortunately I understand the concept of Z but not the nuts and bolts (math) of it all. I may have the opportunity to take some electronics courses soon.

SprinkleSpraycan

Quote from: antonis on March 20, 2025, 02:18:26 PMPut it another way, the former is considered current amplifier where the later voltage amplifier


Helpful, thank you. Ill go and absorb some more of this.

See above comment, Z and more so, math, is a huge challenge for me. Hopefully i can get into a classroom soon and raise my hand a lot.  ???

PRR

#9
OK, with those values and a very high-Z load, the voltage gain is unity to better than 5%. The interstage impedance of about 505k is very high-Z compared to the ~~10k Zout of the first stage. However I note they use a 50k load on the 2nd stage which may cause a mild loss. It is inverting. The P-P output is less than half the 9V supply, tho this may not matter even on hot pickups.



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Rob Strand

#10
Quote from: SprinkleSpraycan on March 20, 2025, 05:41:59 PMHelpful, thank you. Ill go and absorb some more of this.

See above comment, Z and more so, math, is a huge challenge for me. Hopefully i can get into a classroom soon and raise my hand a lot.  ???

When I first looked at this circuit and saw you calling it a buffer I thought pretty much what antonis has said.  However if you think about it it's not that clear cut.

While I agree the use of a CE/CS amplifier as a buffer is a bit of a stretch.    Kind of a this is a buffer "and use that term loosely" type of thing.

The truth is no buffer is ideal.  So you can pick holes in circuits which people would accept as buffers.  Like emitter followers and source followers.

The key point is while a CE/CS amplifier fundamentally has an output current, the collector/drain resistor is part of the amplifier.  So when you draw a box around the whole amplifier it has an output impedance equal to collector/drain resistor.  (Perhaps look-up Thevenin and Norton equivalent circuits).  Outside the box you cannot know if inside the box is fundamentally current or voltage based - you cannot tell. 

In this case the drain resistor is 10k.  For a 1M load is 10k output impedance is pretty low.  So works as a voltage buffer.

The thing about JFET buffers (source followers) is the gain tends to be a bit less than unity.  With the CS amplifier you can tweak the source resistors so the gain ends up closer to unity.

As far as why the designer chose this circuit.  I see two possible reasons:
- The effect part of the circuit is inverting so by adding an inverting buffer at the input it corrects the phase of the effect.  Then once that inversion is in place we need a second inverting stage to make the bypass phase correct.
- to prevent the loss seen in a JFET (source follower) buffer.

As far as motivation goes - I'd go with the first scenario.

The thing about the using the CS stage as a buffer is you loose a lot of headroom.  Maybe not what you want for a pedal bypass.  Not long ago I posted a BJT unit gain inverter stage which had low noise and didn't loose so much headroom.

https://www.diystompboxes.com/smfforum/index.php?topic=131557.0


If you want to be pedantic you could argue a buffer shouldn't invert.   To park that idea the two cascaded stages like in the ADHD are non inverting overall.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

SprinkleSpraycan

I guess it shouldn't surprise me that the dude from fulltone has spent so much time praising his proprietary buffer.

Still, what drew me too it was it [to my ear] does clean up signal chain noise and gives some high end *sparkle*. It's not transparent or whatever word I'm supposed to use.

I'm gonna dig into this and see what i can learn. Thanks for it.

Quote from: Rob Strand on March 20, 2025, 08:41:47 PMThe thing about the using the CS stage as a buffer is you loose a lot of headroom.  Maybe not what you want for a pedal bypass.  Not long ago I posted a BJT unit gain inverter stage which had low noise and didn't loose so much headroom.

https://www.diystompboxes.com/smfforum/index.php?topic=131557.0