[music-dsp] hey guys! who has a Minimoog?
rbj at audioimagination.com
Thu Mar 27 21:56:18 EDT 2008
> ----- Original Message -----
> From: "James Chandler Jr" <jchandjr at bellsouth.net>
> To: "A discussion list for music-related DSP" <music-dsp at music.columbia.edu>
> Subject: Re: [music-dsp] hey guys! who has a Minimoog?
> Date: Thu, 27 Mar 2008 20:35:48 -0400
> In that 904a.jpg, Q22 and Q23 make a differential input stage.
i'm familiar (even if it is an old memory) with diff amps. what makes them work the best is if the emitters are coupled into a "constant current source". the R15, R16, and R18 resistors make a "pi" configuation that has an equivalant "Y" configuation with the two Y resistors connected to the emitters being about 500 ohms each and the single resistor at the bottom of the Y being about 5K. so the constant current source has an output impedance of about 5K (doesn't seem so good to me - you want that value to be as high as possible, but that might be as good as they can do with -6 volts to draw current).
anyway, what i had to figure out looking at the circuit was: "is the feedback gain change an equal number of dB for an equal twist of the knob?" that is what had direct effect in determining how the control signal would work my emulation. i knew the max feedback gain would be 12 dB, but until i looked at the circuit, i didn't know if the feedback gain would be reduced by equal increments of dB with equal twists of the knob. but if reverse audio taper (identified as "REV. AUDIO" on the schematic) is what i think it is, this works out well.
assuming reasonably high input impedance into the base of Q22, the voltage divider gain (from the output of Q20) is about
gain = R10/(R10+R11+R22)
if R22 is considered much larger than R10+R11 (or if these constant resistances are part of the design of the reverse audio taper), the gain is roughly proportional to 1/R22. for a regular audio taper, it might be
Rtaper = R1*(e^(alpha*x) - 1)
for 0 <= x <= 1 and x increasing with CW rotation,
and where R1 = (Rmax/(e^alpha - 1)).
and you would expect a resistor of about R1 in series with it, giving the linear gain of a non-inverting op-amp circuit true exponential gain. but if it's reverse audio taper, i might expect it to be the same, but with 1-x replacing x. so then, if it is the input resistor to a voltage divider, the gain would be
R1/(R1 + R1*(e^(alpha*(1-x)) - 1))
and the gain (in linear terms) increases exponentially with increasing x, or an equal dB for each equal twist of the knob.
that was the issue i was trying to settle. we want the feel of the resonance control to be somewhat similar to the feel of the corresponding control on the real moog. i assumed before that the feedback gain was controlled as dB, but i didn't know.
now it looks like that's the case, if my
R1 = R10+R11 = 2.8K and my
R1*(e^alpha - 1) = R22 = 50K
of course, the degree of the taper has to do with alpha and how many dB those pots do for their entire range, but i consider that a linear scaling issue. we are using a feedback gain range of -6 dB to +12 dB. the filter doesn't sound much deader than it already is at -6 dB feedback so, we're not including the gain values of less on our range of resonance settings.
anyway, Thanks James and thanks Paul.
r b-j rbj at audioimagination.com
"Imagination is more important than knowledge."
More information about the music-dsp