[Pulse wave with no DC]/4066

[Gene Zumchak]

Harry,

    I'm not sure what you mean by biasing the ground pin at -.7 volt.  The inputs are +/- some
number of volts.  Just put a pair of parallel diodes across the input.  As you say, when the
switch is on, its on resistance is low compared with the diodes.  When the switch is off, the
diodes conduct and prevent the transmission gate inputs from seeing more than +/- .7 volts.

Gene


Harry Bissell wrote:

> Sure, you can use 4051/4066 if you locate them at the summing junction of an
> op-amp (after the input resistors), and bias the Ground pin at -0.7V. If the
> switch is on, the voltage across the switch is set by the ratio of the on
> resistance and the input resistor, and if it is off the input resistor limits
> the current to the input pin (diodes conduct) who cares.
> The Prophet V did it this way.
>
> harrybissell at netscape.net
>
> jorgen.bergfors at idg.se wrote:
> Hi synthesists.
>
> Yesterday evening I tested a little circuit that cancels the DC component that
> you get when you modulate the pulse width on a VCO. Roman Sowa came up with
> the excellent idea to just add the CV to the output.
>
> I added an extra op-amp after the comparator that generates the pulse wave. A
> 62k resistor goes from the comparator output to the inverting input. A 22k
> feedback resistor from output to inveriing input
> A 22k resistor from non-inverting input to ground.
> A 22k resistor from PW CV to inverting input. A 1k resistor from output to
> output jack, as usual.
>
> Now the wave will be automatically shifted to cancel the DC. You also have the
> added bonus that the signal level vill be reduced to 10V p-p without passive
> voltage divider.
>
> A gotcha with this circuit is that you can't use 4066/4051 for waveform
> selection anymore. This is because the peaks now can go between -10 and + 10
> volts.
>
> Happy soldering
> /Jorgen
>
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