[sdiy] 1-quadrant multiplier with 2164
David G Dixon
dixon at mail.ubc.ca
Wed Jan 19 18:47:55 CET 2022
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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