How exactly does Super / Hyper Fuzz octaving work?

Started by GreenBear, May 18, 2021, 06:18:57 AM

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GreenBear

Hi all,

So for some reason I can't get my head around this.

I understand the principle - FWR by phase splitting and recombining the positive half cycles - but I don't quite see how the transistor 'mirror' accomplishes this. My assumption is that they clip the negative half while amplifying the positive, but the collector voltage is sitting at around 3.5V so where does this asymmetry come from?

"Talking about music is like dancing about architecture"
-Steve Martin

antonis

#1


For positive signal (Green), right transistor amp is activated where left one is cut-off due to phase splitter Collector negative waveform..
For negative signal (Red), left transistor amp is activated (due to Collector signal phase reversal) where right one is cut-off..

There isn't any "asymmetry" 'cause at any instant one of BJT amps works in forward active mode (the other one is in cut-off region)

P.S.
You'll get the idea if you delete 10k/1k8/10μF and replace each group of 470R/100k/22k/BJT with a series diode.. :icon_wink:
(without amplification, of course..)

edit: Welcome..!!! :icon_biggrin:
"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..

GreenBear

Quote from: antonis on May 18, 2021, 07:06:18 AM

For positive signal (Green), right transistor amp is activated where left one is cut-off due to phase splitter Collector negative waveform..
For negative signal (Red), left transistor amp is activated (due to Collector signal phase reversal) where right one is cut-off..

There isn't any "asymmetry" 'cause at any instant one of BJT amps works in forward active mode (the other one is in cut-off region)

P.S.
You'll get the idea if you delete 10k/1k8/10μF and replace each group of 470R/100k/22k/BJT with a series diode.. :icon_wink:
(without amplification, of course..)

edit: Welcome..!!! :icon_biggrin:

Oh thank you! And thanks for the welcome  :icon_mrgreen:

That image helped a lot, I'd been staring at the schematic for days haha. So any signal that falls below Vb switches its corresponding transistor off, right? Explains why it's so important to match the base voltages...
"Talking about music is like dancing about architecture"
-Steve Martin

antonis

I'm not sure about your query but let's face it in a, more or less, brute approximation..

Consider phase splitter as an ideal Emitter follower in parallel with a CE amp of unity gain, each one driving one of long tail pair configuration BJT..

For a XV positive signal on phase splitter Base, we have XV (positive) signal on right BJT Base (actively biased) and -XV (due to CE amp phase reversal) on left BJT Base (cut-off) and vice-versa..

Actual signal amplitude (XV) doesn't mind (neither needs to be higher than VBE nor need transistor to be matced) 'cause each of long tail pair BJTs Collector is already biased at 3.5V and acts as "independent" CE amp..
(with ideal gain of about 180 or so, neglecting - heavy, due to shunt diode pair - loading due to next stage impedance..)


What I'm trying to say is ONLY one of full wave rectification BJTs is active at any instance..
(just like the 2 of 4 diodes into a bridge rectifier..)
You should agree with there is no need for diode forward voltage drop matching for (no precision) full wave rectifier, shouldn't you..?? :icon_wink:
"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..