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Re: Name this pedal!?!?!?!

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10/4/1999 12:15 AM

Re: Name this pedal!?!?!?!

I haven't seen the schematic in at least a year, so I may be thinking of something else. There should be some kind of pickoff-and-buffer/rectify to measure the input signal. There should be some sort of RC low-pass filter in or at the end of that bit, the output of which should feed into the 3080's gain-control-current pin. One of the pots may be part of this circuit.

Point me to a schematic, please. Maybe I'll have something useful to say.

Then again...

10/4/1999 2:42 AM
Mark Hammer	I know Aron Nelson has it posted, and Jack Orman used to

10/4/1999 1:10 PM

With the understanding that I may well be blowing smoke:

Q2 buffers the 3080's output, both to the pedal output jack and to the _OUTPUT_ amplitude detector (Q3 & Q4). Both of these pull DOWN on the RC stack that controls the gain-current control (Q5).

I appears that Q3 may be (or control) the noise-gate function. A rising voltage at the RC junction would appear to increase the gain in the 3080 by making Q5 pass more current.

Dinking with the RC stack values and the gains of Q3 & Q4 would chnage the time contant(s).

Hope this is useful.

10/4/1999 4:40 PM

OK, had a chance to look at the schemo.

Q2 does a variety of things. It does buffer the output of the OTA, as the OTA wants to see a load of about 100K. The output of the compressor is taken at the emitter of Q2. Q2 also acts as a phase splitter, at collector and emitter.
the split signals drive Q3 and Q4 out of phase.

The diodes and 1M resistors to ground at the bases of Q3 and Q4 DC clamp the signal from the emitter/collector of Q2 and the signal is then current-mode rectified by the B-E junctions of the two transistors.

Their collectors pull down on the 150K resistor that feeds the 10uF capacitor to ground, giving full wave rectification since the signals at their bases are out of phase. The cap is normally charged to +9V, and holds the full wave rectified signal as a voltage *down* from +9V.

The 10uF cap is buffered by Q5, and its emitter drives the series combination of the 500K sustain pot and the 27K resistor into the bias pin of the OTA. The OTA bias pin looks like two diodes to ground.

The overall action is that with no signal, the 10uF cap charges up to +9V, and you get +9V-0.5V on the emitter of Q5. This puts max voltage across the sustain pot/27K, which runs the OTA to max gain. Signal is FWR'ed in Q2/Q3/Q4 and pulls down on the 10uF, which lowers the voltage on Q5's emitter, lowering the gain.

Near as I can tell, the attack time constant is whatever you get from the source impedances of Q2's emitter/collector into the B-E's of Q3/Q4 - it's set up to be as fast as possible, no series resistanct to slow down the attack at all. The 150K pullup resistor determines the release time constant.

The schematic at GEO" target="_blank">http://www.eden.com/dynacomp.gif">GEO is a bit easier to follow; it's linked by Aron's schemo collection.

10/4/1999 8:41 PM
Mark Hammer	Nice explanation, but I bet you haven't checked that URL lately, buddy.

10/5/1999 12:36 AM
R.G.	Oh, sure... check it out and see if I knew what I was doing... *< OK, what I really meant was http://www.geofex.com/dynacomp.gif" target="_blank">http://www.geofex.com/dynacomp.gif">http://www.geofex.com/dynacomp.gif

10/5/1999 1:49 PM

Nice explanation. Note the importance of the diodes at the bases of Q3 and Q4. If they were not there, the 0.01 caps from the phase splitter would charge with the first signal peaks and turn Q3 and Q4 off, and your peak detector would not work. Much like what happens with "blocking distortion" (grid clamping) in tube amps.

There's a little trick on the Dynacomp to get more dynamic range out of the CA3080.

The 15k/0.01u/1u network on the (+) input will send some highs to ground; as the signal ges to the (-) input the result is that the gain is lower for low frequencies (where you normally have more power in an audio signal). The attenuation at the input is frequency dependant. Then at the output you have a 150k/0.001 combo that will compensate this and try to make the response "flat" again. A "pre-emphasis/de-emphasis" thing.

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