# [sdiy] 1-quadrant multiplier with 2164

Rutger Vlek rutgervlek at gmail.com
Tue Jan 18 11:42:56 CET 2022

```Hi all,

@David: thanks for the quick and detailed solution! I'm glad you took into
account the linearization as well, as I'll need that too. One thing in your
solution caught my eye, and that is that you're summing signals in front of
the 2164's input. I know you can sum signals on the input of the I-V
convertor, but did not know it could be done even earlier, before the 2164.
Are you sure it would not cause interactions between the summed signals
(i.e. can the input of the 2164 also be considered virtual ground)?

multipliers. Here's how I understand it:

- A 4-quadrant multiplier is capable of taking into account the magnitude
and sign (+ or -) of both (biplolar) inputs and flip the output according
to both signs (as in normal math multiplications). This would be the case
for a ring modulator

- A 2-quadrant multiplier can take into account the magnitude of both
inputs, but only the sign of 1 input (the second input should be unipolar
and cannot sign-flip the output). This would be the case for a normal audio
VCA.

- A 1-quadrant multiplier can take into account only the magnitude of both
inputs, and not deal with either of the signs to cause sign-flipping of the
output. In the common case this requires two unipolar signals, but I feel
my case of a bipolar wave of which the amplitude is decreased up to it's
lower bound also fits this... what do you think?

Rutger

Op di 18 jan. 2022 om 02:48 schreef David G Dixon <dixon at mail.ubc.ca>:

> Hi Rutger,
>
> It is really easy to do what you want to do.
>
> If I understand you correctly, for an input signal of 10Vpp centred on 0V,
> you basically want the output signal to grow from a baseline of -5V, so
> that when the VCA is off, the output is -5V DC, at 50% gain, the output is
> 5Vpp from -5V to 0V, and at 100% the output is 10Vpp from -5V to +5V.
>
> Let's assume that you have built the standard linearized VCA circuit and
> that you have good 5V and -5V reference voltages available.  Presuming that
> your input resistor into the amplifying 2164 VCA is 30k, and the feedback
> resistor on the I-V converting opamp is also 30k, then all you have to do
> is this:  Sum -5V into the 2164 VCA, in parallel with the input signal,
> through a 30k resistor, and sum +5V into the I-V converter also through a
> 30k resistor.  The +5V into the I-V converter will apply a constant bias of
> -5V to the output signal.  The -5V into the VCA will counteract this bias
> in proportion to the VCA's gain, such that, then the VCA is off, the bias
> is -5V, and when the VCA is at unity gain, the bias is 0V.
>
> Easy peasy, lemon squeezy.  Schematic attached.
>
> Cheers,
> Dave Dixon
> ------------------------------
> *From:* Synth-diy [mailto:synth-diy-bounces at synth-diy.org] *On Behalf Of *Rutger
> Vlek via Synth-diy
> *Sent:* Monday, January 17, 2022 1:04 PM
> *To:* SDIY List
> *Subject:* [sdiy] 1-quadrant multiplier with 2164
>
> [*CAUTION:* Non-UBC Email]
> Dear all,
>
> I think i need a bit of wisdom from the 2164 guru's around here. The case
> is as follows:
> I'm in need of a 1-quadrant multiplier (VCA) that works on a bipolar
> signal. The input would be a 10Vpp triangle wave centered around 0V. The
> output should be the same with the VCA fully open, but... when closing I
> would like to reduce the level with respect to the -5V power bound, so that
> a VCA half-open would result in a triangle of half the amplitude, sitting
> between -5V and 0V and a fully closed VCA would give -5V DC.
>
> Of course I could level-shift both input and output of a 2164 bases VCA,
> but I feel it could be simpler. Would it work to connect a -5V signal to
> the ground pin of a 2164? And would that also require a shifted control
> voltage?
>
> Regards,
>
> Rutger
>
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