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Subject: Re: ring mod leakage

From: "Paul Schreiber" <synth1@...
Date: 1999-07-11

>Now that I have VCOs (beautiful! amazing!) I have started patching all
>kinds of neat stuff, and have noticed a few things about my older modules.
>
>Should I expect to be able to trim the MOTM-110 ring modulator to absolute
>silence with one input zeroed? I feed a high-C sine wave into X IN with X
>LEVEL at 10, Y LEVEL at 0, X and Y both AC-coupled, Y IN unconnected. By
>turning the X trimpot I can render the original sine wave nearly
>inaudible, but a frequency-doubled version appears at the output and
>cannot be trimmed away. Exchange X and Y, and the doubled wave is much
>louder (clearly audible at "reasonable" volume levels). In addition the
>output clips if I turn Y LEVEL past 9, but I guess this is to be expected
>(driving from a 300).

Again, this subject has re-appeared again, so I will reiterate.

1) The MOTM-110 is specifically designed for LARGE signals. NOT guitars, NOT
the output
jacks of synths.

2) The best way to boost signals to be '110 happy are:

a) use the headphone output
b) use the SEND buss of a mixer
c) use a 'direct box'
d) gnash teeth while waiting for 1U triple pre-amp

3) ALL ring-mods have carrier leakage. The carrier "nulling" is input
amplitude ∗dependent∗.
If you adjust the 2 trims pots at one level (say 2V pk-pk) then if you input
8V pk-pk the null
is greatly reduced. So, Rule #1 when setting the trimpots: use the largest
signal you plan to see.
Like, VCO outputs.

Also, the null amount is basically ∗not∗ all that great: say 30-35 dB down.
You ∗will∗ be able to hear that.
But, in general a good null point, coupled with enveloping BOTH X & Y
inputs, will keep the
∗perceived∗ leakage to a minimum.

Of course, the UNBAlance control ∗adds∗ leakage!

4) Remember: the best RM sounds use a SINE or TRI wave for Y and the "audio"
into X. Shoving 2
presets into X & Y will get noisy fast! In fact, you may HEAR "extra noise"
coming out of the '110
which is the by-product of the RM.

5) The '110 has a ∗built-in∗ gain REDUCTION of 5. The reason: if you put in
2 5V pk signals, 5 ∗ 5 /5 = 5V out
which is what we want.

You put in 2 1Vpk-pk signals, like the OUT of a synth, and you get out
(.5)(.5)/5 = 50mv = not much there!!!

Careful level adjustments plus proper nulling using the trimmers will get
you some great effects!

POP QUIZ!!!

Why is Chris seeing a "doubled" waveform??

Paul S.