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CGS48 VCO question

2009-06-21 by dhamaryder

Hello, I'm new to this forum. I started building the VCO and I just have a few more little parts to get and solder in. I'm not a super-experienced builder so there's one thing I'm a bit confused and insecure about. In the instructions he gives Modifications to the different versions of the PC board. I have version 0.4 of the PC board. The first mod in the REV. 0.4 is this:

["Cut the track between Pin 5 of the TL072 in the bottom right corner and the 47uF capacitor. Make the cut near the capacitor."]

I can't find anywhere that pin 5 of the TL072 is connected to the 47uF capacitor. Pin 5 seems to be connected to a pad that is supposed to be a jumper wire. And that's another thing, in the Modifications for version 0.2 it says to remove the jumper between pin 5 and the WS Trim. Should I still not put in the jumper or should I go ahead and do it since this is a later version.

["Jumper pin 5 to the Saw AC output terminal via a 100k. This will set PWM control to 50% at 0 volts."]

So, disregard the jumper that's marked on the PC board and just solder in a 100K resistor between the hole at pin5 and the hole at Saw AC output?

["It is possible to place the resistor in place of the link on the PCB though the track modification is still required."]

I'm not sure what "link" means here.

["Replace the 330k resistor near the SYNC pad with a 220k resistor. This should stop clipping."]

I couldn't find any 330k resistor in this area.

One more thing. He give a schematic for "Power rail decoupling for the VCO". I have no idea what decoupling means. I have a Blacet power supply with -15, 0, +15. Does that mean I don't have to do any modifications to the PC board?

Sorry if these are really dumb questions, any help would be greatly appreciated.Thank you for your patience!
steve

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