[sdiy] 1-quadrant multiplier with 2164
cheater00social at gmail.com
Thu Jan 20 16:25:12 CET 2022
I wonder if it's possible to build a frequency shifter that shifts
higher harmonics more than lower harmonics.
On Wed, Jan 19, 2022 at 6:48 PM David G Dixon <dixon at mail.ubc.ca> wrote:
> I must confess that I've lost the thread of this argument just a little bit.
> However, what I like about my approach (which I have used many times in many
> different contexts) is that, in order to build a nice linear VCA from 2164,
> you really need to have a clean 5V source anyway. I keep a pile of LM336Z5
> for just this purpose, and use two opamps to buffer and invert this to get
> low-output-impedance +5V and -5V references on all my multipliers. If one
> uses precisely matched resistors on the inverter, then one can get those
> references within a mV of each other -- the actual voltage doesn't matter
> (and it is usually around 4.90V), but as long as they are equal and
> opposite, then they can be used for precise multiplication. This is one of
> the keys to the precision of my Freak Shift frequency shifter circuit.
> I don't really understand how adding a stable DC value to a signal increases
> the noise of that signal. I must confess that I also don't care at all
> about it. My method is the simplest. You don't have to pre-condition the
> incoming signals at all. The CV signal is unchanged, and the DC reference
> levels are simply summed to the incoming signal.
> If you want to change the actual levels, you can simply change the resistor
> values. I do this all the time. One of the keys to my one-VCA
> four-quadrant-multiplier circuit (of which there are two in the Freak Shift,
> made from a single 2164 chip) is to lift and diminish the CV such that the
> zero point of the multiplier is at +5V and full +/- unity-gain
> multiplication occurs between +2.5V and +7.5V. This gives lots of headroom
> -- it essentially makes it impossible for the CV in the multiplier to hit
> zero at the 2164 control pin (because the incoming CV signal will never be
> anywhere near 20Vpp), which would give a dead zone on the multiplication. I
> achieve this simply by bringing the CV in through 200k while using 100k on
> the reference voltages. Of course, the signal is now cut in half as well,
> so I simply double the feedback resistor on the I-V converter. As long as
> all of these 100k and 200k resistors are within 0.1% of each other (and the
> 100k and 200k resistors don't need to be in a precise ratio -- they only
> need to be precise within their own values), and all incoming signals are AC
> coupled through big back-to-back electrolytics, then the four-quadrant
> multiplication is very tight, which is important for frequency shifting.
> -----Original Message-----
> From: Synth-diy [mailto:synth-diy-bounces at synth-diy.org] On Behalf Of
> cheater cheater via Synth-diy
> Sent: Wednesday, January 19, 2022 4:23 AM
> To: Neil Johnson
> Cc: SDIY List
> Subject: Re: [sdiy] 1-quadrant multiplier with 2164
> [CAUTION: Non-UBC Email]
> I wonder if it matters that Dave's version will create theoretically more
> distortion on the positive swing of whatever vs the negative swing, whereas
> my version will apply distortion (non-linearity) more or less
> symmetrically... do the numbers show that it matters at all? I bet it would
> matter with some, let's say, crappy devices.
> On Tue, Jan 18, 2022 at 1:57 PM Neil Johnson via Synth-diy
> <synth-diy at synth-diy.org> wrote:
> > > This is certainly true but note also the importance of zero when
> multiplying. The zero signal stays zero no matter what you multiply by. In
> Rutger's case that zero is in fact -5V, so the origin of Neil's graph should
> be at -5V signal and zero control voltage. That is why the level-shifting
> solution is so effective and it is also why I believe Rutger is correct to
> call this a one quadrant multiplier.
> > Yes, this is just a bit of algebraic juggling.
> > If we take Dave's approach:
> > - convert the bipolar +/- 5V input to a unipolar 0 to -10V input
> > - add a -5V offset to the output _after_ the VCA (so no bearing on the
> > quadrantiness of the VCA itself)
> > With a unipolar CV and a unipolar signal ... a 1-quadrant VCA.
> > And don't forget that as-drawn the linearised VCA is inverting.
> > Cheers,
> > Neil
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