[sdiy] What is a quadrature oscillator good for?
Don Tillman
don at till.com
Tue Sep 9 20:19:20 CEST 2003
> Date: 9 Sep 2003 16:48:49 +0200
> From: jhaible at debitel.net
>
> http://www.till.com/articles/QuadTrapVCO/index.html
>
> Great article, Don!
Thanks, JH!
> (And you obviously love to care about details, even down to
> the a little graphic like this one:
> http://www.till.com/articles/QuadTrapVCO/images/qthr.gif)
True. When I was testing the circuit I was really happy to see that
pattern on my scope, the dual traces showing the quadrature trapezoid
waves as the VCO runs through 0 Hz. This isn't something you see
every day. So I thought it would make a good logo, as it's a pattern
associated with the oscillator.
I made the graphic with a small Java program using the Java2D
library, which is really nice because it has anti-aliasing and a
healthy collection of drawing utilities. The slopes in that drawing
are all parabolas. (!!!)
> One question: You mentioned a jitter near 0Hz if the two halves
> of the 3280 don't zero-cross precisely. Can you get this adjusted
> so precisely that you don't mess up the tuning- and waveform-purity
> when the VCO "just touches" the 0Hz region ?
Yes; I've run the VCO around 0.05 Hz without problems.
The two OTAs are used to switch the currents into the two capacitors,
currents that are proportional the linear control voltage, inverted
and noninverted.
In the real world both OTA's will have an offset voltage. Any common
offset voltage is effectively the same as an offset of the input
control voltage. Any difference in offset voltage will show as a
difference in slope when switching between the OTA's. That would be
the "jitter".
Luckily it's very easy to null out the offset voltages -- just supply
a 0.0 V linear control voltage, disconnect the OTA outputs (short the
enable pin on the analog mux chip), and trim the OTA outputs to zero.
-- Don
--
Don Tillman
Palo Alto, California
don at till.com
http://www.till.com
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