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Schematic Question about BYOC vibrato

Started by woodstock_stomp_box, December 22, 2021, 07:22:12 AM

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woodstock_stomp_box

Hi all,
I'm reading through the schematic of the BYOC analog vibrato, and I'm having a little trouble determining where +Va is coming from. Is this the bias voltage set by the 25k trimpot? or should all +Va points tie into the emitter of the NPN transistor near the bottom left of the schematic? It seems like the former is unlikely, as the BBD/clock driver use +Va as their supply voltage while adjacent op-amps pull directly from the wiper pin of the trimpot.

Any help here would be greatly appreciated, I'm prototyping some mods on breadboard and just don't want to potentially damage the BBD.

Zerro

Yes it goes from that emitter of Q in left corner. It gives slightly lower value, but stabilized. So, modulator has little lower values going to driving process, not overhelming values at opamps in signal track. It is very frequently used hack - modulation circuit has lower top values not to overfuell modulation pins in opamps.
"Nudíte se? Kupte si našeho cvičeného ježka! Pobaví vás svými veselýmí kousky!"

woodstock_stomp_box

Quote from: Zerro on December 22, 2021, 09:20:03 AM
Yes it goes from that emitter of Q in left corner. It gives slightly lower value, but stabilized. So, modulator has little lower values going to driving process, not overhelming values at opamps in signal track. It is very frequently used hack - modulation circuit has lower top values not to overfuell modulation pins in opamps.

Thanks for the response! So why is a transistor used rather than a voltage divider here? Is that where stabilization comes into the picture/is this a way to isolate the slightly lower +Va bus from the rest of the power?

Also, when you say overwhelming the values at opamps in signal track, do you mean overwhelming the values coming out of the output pins? I guess I'm a bit confused as to how the power input for the modulator/clock driver section effects the signal coming from the output pins of the op amps in the signal path. Thanks again, for the guidance here

Zerro

#3
I would say it this way: if opamps in audiosignal have 9V power, then it is better, when driving values, that makes some modulation signals, are little lower, for example cca 8V. So, top of this driving curves or clock rates has some "reserve" not to clip and so distort process of modulation.
Transistor gives here stronger current to power rail, than divider from resistors. This is the simplest way how to get little lower voltage output without using opamp source with some divider at input.
"Nudíte se? Kupte si našeho cvičeného ježka! Pobaví vás svými veselýmí kousky!"

woodstock_stomp_box

Quote from: Zerro on December 22, 2021, 10:18:39 AM
I would say it this way: if opamps in audiosignal have 9V power, then it is better, when driving values, that makes some modulation signals, are little lower, for example cca 8V. So, top of this driving curves has some "reserve" not to clip and so distort process of modulation.
Transistor gives here stronger current to power rail, than divider from resistors. This is the simplest way how to get little lower voltage output without using opamp source with some divider at input.

I see, thanks for the explanation!

Zerro

BTW, that +Va output will be little noisy because of transistor transient characteristic. So, I would block this power output against ground with come caps - cca 10uF electrolytic, paralelly with 100nF ceramic.
"Nudíte se? Kupte si našeho cvičeného ježka! Pobaví vás svými veselýmí kousky!"