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<DIV dir=ltr align=left><SPAN class=366473601-18012022>Hi Rutger,</SPAN></DIV>
<DIV dir=ltr align=left><SPAN class=366473601-18012022></SPAN> </DIV>
<DIV dir=ltr align=left><SPAN class=366473601-18012022>It is really easy to do
what you want to do.</SPAN></DIV>
<DIV dir=ltr align=left><SPAN class=366473601-18012022></SPAN> </DIV>
<DIV dir=ltr align=left><SPAN class=366473601-18012022>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.</SPAN></DIV>
<DIV dir=ltr align=left><SPAN class=366473601-18012022></SPAN> </DIV>
<DIV dir=ltr align=left><SPAN class=366473601-18012022>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.</SPAN></DIV>
<DIV dir=ltr align=left><SPAN class=366473601-18012022></SPAN> </DIV>
<DIV dir=ltr align=left><SPAN class=366473601-18012022>Easy peasy,
lemon squeezy. Schematic attached.</SPAN></DIV>
<DIV dir=ltr align=left><SPAN class=366473601-18012022></SPAN> </DIV>
<DIV dir=ltr align=left><SPAN class=366473601-18012022>Cheers,</SPAN></DIV>
<DIV dir=ltr align=left><SPAN class=366473601-18012022>Dave
Dixon </SPAN><BR></DIV>
<DIV dir=ltr lang=en-us class=OutlookMessageHeader align=left>
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<FONT size=2 face=Tahoma><B>From:</B> Synth-diy
[mailto:synth-diy-bounces@synth-diy.org] <B>On Behalf Of </B>Rutger Vlek via
Synth-diy<BR><B>Sent:</B> Monday, January 17, 2022 1:04 PM<BR><B>To:</B> SDIY
List<BR><B>Subject:</B> [sdiy] 1-quadrant multiplier with
2164<BR></FONT><BR></DIV>
<DIV></DIV>
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<DIV dir=auto>Dear all,
<DIV dir=auto><BR></DIV>
<DIV dir=auto>I think i need a bit of wisdom from the 2164 guru's around here.
The case is as follows:
<DIV dir=auto>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.</DIV>
<DIV dir=auto><BR></DIV>
<DIV dir=auto>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?</DIV>
<DIV dir=auto><BR></DIV>
<DIV dir=auto>Regards,</DIV>
<DIV dir=auto><BR></DIV>
<DIV dir=auto>Rutger</DIV></DIV></DIV></BODY></HTML>