[sdiy] 1-quadrant multiplier with 2164

Mike Bryant mbryant at futurehorizons.com
Thu Jan 20 18:17:47 CET 2022


Yes :-)

But infinitely easier if you go digital 

-----Original Message-----
From: Synth-diy [mailto:synth-diy-bounces at synth-diy.org] On Behalf Of cheater cheater via Synth-diy
Sent: 20 January 2022 15:25
To: David G Dixon
Cc: synth-diy
Subject: Re: [sdiy] 1-quadrant multiplier with 2164

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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