# [sdiy] 1-quadrant multiplier with 2164

Brian Willoughby brianw at audiobanshee.com
Tue Jan 18 02:27:42 CET 2022

```My first thought was that your bipolar input (±5V) would make this a 2-quadrant multiplier. I suppose that if you guarantee a level shift before and after your VCA, then it could be 1-quadrant, but the overall circuit would be 2-quadrant, right?

Brian

On Jan 17, 2022, at 13:50, Tom Wiltshire <tom at electricdruid.net> wrote:
> Honestly, it sounds like the level-shifting solution is easiest. Add +5V to the signal going in, process a 0-10V signal, then take 5V off again on the signal coming out. It's only a dual op-amp. It doesn't get a lot simpler than that.
>
> You've likely got an I-to-V stage after the VCA that would do the job for the output, and you might even already have an op-amp ahead of the VCA you could use for the input shift, in which case you don't even need to *add* any op-amps.
>
> HTH,
> Tom
>
> On 17 Jan 2022, at 21:03, Rutger Vlek via Synth-diy <synth-diy at synth-diy.org> wrote:
>> 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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