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 Forum index » DIY Hardware and Software » YuSynth
Simple VCA (DC version) is an inverting amplifier?
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Gypsum Fantastic



Joined: Mar 09, 2017
Posts: 7
Location: Sheffield, UK

PostPosted: Tue Jun 26, 2018 1:52 pm    Post subject: Simple VCA (DC version) is an inverting amplifier? Reply with quote  Mark this post and the followings unread

Hi Everyone

I've been building my Yusynth for 18 months now. I love it. It's amazing.

It's still quite a small system comparatively speaking: 2xVCO, 2xEG, dual slew, LFO2, Moog filter, Simple VCA (AC), Simple VCA (DC).

The problem I'm having: the VCAs both seem to be inverting amplifiers. Negative gain. I noticed this when calibrating them but didn't think much of it until I started trying to use them in a patch.

This is 100% fine with the AC version, but it's very awkward with the DC version.

For example, I was trying to modulate EG --> filter cutoff amount using an aux CV to control the DC VCA but the envelope was upside down: the attack phase takes the cutoff from high to low and then decay goes back high and so on.

I switched the patch lead to use the EG's inverting output but things still seemed backwards. I think I was getting a more pronounced effect at low VCA gain settings.

Is this how it's supposed to be? The simple VCA is an inverting amplifier? Is there an easy fix?

I'm feeling very confused.

cheers

Ju
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Grumble



Joined: Nov 23, 2015
Posts: 1294
Location: Netherlands
Audio files: 30

PostPosted: Tue Jun 26, 2018 10:19 pm    Post subject: Reply with quote  Mark this post and the followings unread

it would be very helpfull and you would get a better response to you question if you’d add schematics or at least provide a link to schematics when asking for help.
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Grumble



Joined: Nov 23, 2015
Posts: 1294
Location: Netherlands
Audio files: 30

PostPosted: Tue Jun 26, 2018 10:28 pm    Post subject: Reply with quote  Mark this post and the followings unread

that said, is it this vca?
if so, just switch the connections of R10 and R11 in a way that R10 connects to junction R8/Q1 and R11 to junction R9/Q2, effectively inverting the output of U1b.
edit: looking again at the schematics I think you have made a mistake somewhere because the output signal is NOT inverted.
U1a inverts the input voltage, so if the input voltage goes up, the output of U1a goes down, this will make the transistor Q1 less conductive, so the voltage of the collector of Q1 goes up wile the collector of Q2 does not change voltage, so the +input of U1b goes up, just like the input voltage.

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